\(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1476]

Optimal result

Integrand size = 29, antiderivative size = 424 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {b^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \]

[Out]

-4*b^4*(2*a^2-b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(7/2)/d-b^4*(2*a^2+b^2)*arct
an((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(7/2)/d-2*b^4*(10*a^4-9*a^2*b^2+3*b^4)*arctan((b+a*
tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(7/2)/d+3*b*arctanh(cos(d*x+c))/a^4/d-cot(d*x+c)/a^3/d+1/2*
cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/2*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))-1/2*b^5*cos(d*x+c)/a^2/(a^2-b^2)^2
/d/(a+b*sin(d*x+c))^2-3/2*b^5*cos(d*x+c)/a/(a^2-b^2)^3/d/(a+b*sin(d*x+c))-2*b^5*(2*a^2-b^2)*cos(d*x+c)/a^3/(a^
2-b^2)^3/d/(a+b*sin(d*x+c))

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2976, 3855, 3852, 8, 2727, 2743, 2833, 12, 2739, 632, 210} \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac {b^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac {3 b^5 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{7/2}}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(-4*b^4*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(7/2)*d) - (b^4*(2*a^
2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(7/2)*d) - (2*b^4*(10*a^4 - 9*a^2*
b^2 + 3*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(7/2)*d) + (3*b*ArcTanh[Cos[c
+ d*x]])/(a^4*d) - Cot[c + d*x]/(a^3*d) + Cos[c + d*x]/(2*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a
- b)^3*d*(1 + Sin[c + d*x])) - (b^5*Cos[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) - (3*b^5*Cos
[c + d*x])/(2*a*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (2*b^5*(2*a^2 - b^2)*Cos[c + d*x])/(a^3*(a^2 - b^2)^3*
d*(a + b*Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2976

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3 b \csc (c+d x)}{a^4}+\frac {\csc ^2(c+d x)}{a^3}-\frac {1}{2 (a+b)^3 (-1+\sin (c+d x))}+\frac {1}{2 (a-b)^3 (1+\sin (c+d x))}-\frac {b^4}{a^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^3}-\frac {2 b^4 \left (2 a^2-b^2\right )}{a^3 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {-10 a^4 b^4+9 a^2 b^6-3 b^8}{a^4 \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}\right ) \, dx \\ & = \frac {\int \csc ^2(c+d x) \, dx}{a^3}+\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{2 (a-b)^3}-\frac {(3 b) \int \csc (c+d x) \, dx}{a^4}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^3}-\frac {b^4 \int \frac {1}{(a+b \sin (c+d x))^3} \, dx}{a^2 \left (a^2-b^2\right )}-\frac {\left (2 b^4 \left (2 a^2-b^2\right )\right ) \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{a^3 \left (a^2-b^2\right )^2}-\frac {\left (b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )^3} \\ & = \frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {b^4 \int \frac {-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 a^2 \left (a^2-b^2\right )^2}-\frac {\left (2 b^4 \left (2 a^2-b^2\right )\right ) \int \frac {a}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^3}-\frac {\text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}-\frac {\left (2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right )^3 d} \\ & = \frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {b^4 \int \frac {2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^3}-\frac {\left (2 b^4 \left (2 a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^3}+\frac {\left (4 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right )^3 d} \\ & = -\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (b^4 \left (2 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^3}-\frac {\left (4 b^4 \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d} \\ & = -\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (8 b^4 \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d}-\frac {\left (b^4 \left (2 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d} \\ & = -\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 b^4 \left (2 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d} \\ & = -\frac {4 b^4 \left (2 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {b^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 b^4 \left (10 a^4-9 a^2 b^2+3 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{7/2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {b^5 \cos (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 b^5 \cos (c+d x)}{2 a \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.43 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=4 \left (-\frac {3 b^4 \left (10 a^4-7 a^2 b^2+2 b^4\right ) \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{4 a^4 \left (a^2-b^2\right )^{7/2} d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}+\frac {3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^4 d}-\frac {3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^4 d}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{4 (a+b)^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{4 (a-b)^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {b^5 \cos (c+d x)}{8 a^2 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))^2}+\frac {-11 a^2 b^5 \cos (c+d x)+4 b^7 \cos (c+d x)}{8 a^3 (a-b)^3 (a+b)^3 d (a+b \sin (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}\right ) \]

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

4*((-3*b^4*(10*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sq
rt[a^2 - b^2]])/(4*a^4*(a^2 - b^2)^(7/2)*d) - Cot[(c + d*x)/2]/(8*a^3*d) + (3*b*Log[Cos[(c + d*x)/2]])/(4*a^4*
d) - (3*b*Log[Sin[(c + d*x)/2]])/(4*a^4*d) + Sin[(c + d*x)/2]/(4*(a + b)^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)
/2])) + Sin[(c + d*x)/2]/(4*(a - b)^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (b^5*Cos[c + d*x])/(8*a^2*(a
- b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])^2) + (-11*a^2*b^5*Cos[c + d*x] + 4*b^7*Cos[c + d*x])/(8*a^3*(a - b)^3*
(a + b)^3*d*(a + b*Sin[c + d*x])) + Tan[(c + d*x)/2]/(8*a^3*d))

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {2 b^{4} \left (\frac {\left (\frac {13}{2} a^{3} b^{2}-3 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (12 a^{4}+19 a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 b^{2} a \left (5 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (12 a^{2}-5 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (10 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{4}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(311\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {2 b^{4} \left (\frac {\left (\frac {13}{2} a^{3} b^{2}-3 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (12 a^{4}+19 a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 b^{2} a \left (5 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (12 a^{2}-5 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (10 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{4}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(311\)
risch	\(\text {Expression too large to display}\)	\(1106\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/2*tan(1/2*d*x+1/2*c)/a^3-1/2/a^3/tan(1/2*d*x+1/2*c)-3/a^4*b*ln(tan(1/2*d*x+1/2*c))-2*b^4/(a-b)^3/(a+b)^
3/a^4*(((13/2*a^3*b^2-3*a*b^4)*tan(1/2*d*x+1/2*c)^3+1/2*b*(12*a^4+19*a^2*b^2-10*b^4)*tan(1/2*d*x+1/2*c)^2+7/2*
b^2*a*(5*a^2-2*b^2)*tan(1/2*d*x+1/2*c)+1/2*a^2*b*(12*a^2-5*b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c
)+a)^2+3/2*(10*a^4-7*a^2*b^2+2*b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2)))-
1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/(a+b)^3/(tan(1/2*d*x+1/2*c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1028 vs. \(2 (399) = 798\).

Time = 1.58 (sec) , antiderivative size = 2140, normalized size of antiderivative = 5.05 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*a^11 - 12*a^9*b^2 + 12*a^7*b^4 - 4*a^5*b^6 + 2*(4*a^9*b^2 - 4*a^7*b^4 + 17*a^5*b^6 - 23*a^3*b^8 + 6*a
*b^10)*cos(d*x + c)^4 - 2*(4*a^11 - 10*a^9*b^2 + 14*a^7*b^4 + 7*a^5*b^6 - 21*a^3*b^8 + 6*a*b^10)*cos(d*x + c)^
2 - 3*(2*(10*a^5*b^5 - 7*a^3*b^7 + 2*a*b^9)*cos(d*x + c)^3 - 2*(10*a^5*b^5 - 7*a^3*b^7 + 2*a*b^9)*cos(d*x + c)
+ ((10*a^4*b^6 - 7*a^2*b^8 + 2*b^10)*cos(d*x + c)^3 - (10*a^6*b^4 + 3*a^4*b^6 - 5*a^2*b^8 + 2*b^10)*cos(d*x +
c))*sin(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*
cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 -
b^2)) - 6*(2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c)^3 - 2*(a^9*b^2 - 4*a^7*b^4 +
6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c) + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*cos(d*x
+ c)^3 - (a^10*b - 3*a^8*b^3 + 2*a^6*b^5 + 2*a^4*b^7 - 3*a^2*b^9 + b^11)*cos(d*x + c))*sin(d*x + c))*log(1/2*c
os(d*x + c) + 1/2) + 6*(2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c)^3 - 2*(a^9*b^2 -
4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c) + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^
11)*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 2*a^6*b^5 + 2*a^4*b^7 - 3*a^2*b^9 + b^11)*cos(d*x + c))*sin(d*x + c
))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^10*b - 6*a^8*b^3 + 6*a^6*b^5 - 2*a^4*b^7 + (8*a^10*b - 14*a^8*b^3 + 2
8*a^6*b^5 - 31*a^4*b^7 + 9*a^2*b^9)*cos(d*x + c)^2)*sin(d*x + c))/(2*(a^13*b - 4*a^11*b^3 + 6*a^9*b^5 - 4*a^7*
b^7 + a^5*b^9)*d*cos(d*x + c)^3 - 2*(a^13*b - 4*a^11*b^3 + 6*a^9*b^5 - 4*a^7*b^7 + a^5*b^9)*d*cos(d*x + c) + (
(a^12*b^2 - 4*a^10*b^4 + 6*a^8*b^6 - 4*a^6*b^8 + a^4*b^10)*d*cos(d*x + c)^3 - (a^14 - 3*a^12*b^2 + 2*a^10*b^4
+ 2*a^8*b^6 - 3*a^6*b^8 + a^4*b^10)*d*cos(d*x + c))*sin(d*x + c)), -1/2*(2*a^11 - 6*a^9*b^2 + 6*a^7*b^4 - 2*a^
5*b^6 + (4*a^9*b^2 - 4*a^7*b^4 + 17*a^5*b^6 - 23*a^3*b^8 + 6*a*b^10)*cos(d*x + c)^4 - (4*a^11 - 10*a^9*b^2 + 1
4*a^7*b^4 + 7*a^5*b^6 - 21*a^3*b^8 + 6*a*b^10)*cos(d*x + c)^2 - 3*(2*(10*a^5*b^5 - 7*a^3*b^7 + 2*a*b^9)*cos(d*
x + c)^3 - 2*(10*a^5*b^5 - 7*a^3*b^7 + 2*a*b^9)*cos(d*x + c) + ((10*a^4*b^6 - 7*a^2*b^8 + 2*b^10)*cos(d*x + c)
^3 - (10*a^6*b^4 + 3*a^4*b^6 - 5*a^2*b^8 + 2*b^10)*cos(d*x + c))*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(
d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3*(2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*co
s(d*x + c)^3 - 2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c) + ((a^8*b^3 - 4*a^6*b^5 +
6*a^4*b^7 - 4*a^2*b^9 + b^11)*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 2*a^6*b^5 + 2*a^4*b^7 - 3*a^2*b^9 + b^11
)*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 3*(2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8
+ a*b^10)*cos(d*x + c)^3 - 2*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c) + ((a^8*b^3 -
4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 2*a^6*b^5 + 2*a^4*b^7 - 3*a^
2*b^9 + b^11)*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - (2*a^10*b - 6*a^8*b^3 + 6*a^6*b^5 - 2
*a^4*b^7 + (8*a^10*b - 14*a^8*b^3 + 28*a^6*b^5 - 31*a^4*b^7 + 9*a^2*b^9)*cos(d*x + c)^2)*sin(d*x + c))/(2*(a^1
3*b - 4*a^11*b^3 + 6*a^9*b^5 - 4*a^7*b^7 + a^5*b^9)*d*cos(d*x + c)^3 - 2*(a^13*b - 4*a^11*b^3 + 6*a^9*b^5 - 4*
a^7*b^7 + a^5*b^9)*d*cos(d*x + c) + ((a^12*b^2 - 4*a^10*b^4 + 6*a^8*b^6 - 4*a^6*b^8 + a^4*b^10)*d*cos(d*x + c)
^3 - (a^14 - 3*a^12*b^2 + 2*a^10*b^4 + 2*a^8*b^6 - 3*a^6*b^8 + a^4*b^10)*d*cos(d*x + c))*sin(d*x + c))]

Sympy [F]

\[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**2*sec(c + d*x)**2/(a + b*sin(c + d*x))**3, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 633, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (10 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 2 \, b^{8}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}}{{\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}} + \frac {2 \, {\left (13 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{4} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 19 \, a^{2} b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, b^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b^{5} - 5 \, a^{2} b^{7}\right )}}{{\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} + \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(6*(10*a^4*b^4 - 7*a^2*b^6 + 2*b^8)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/
2*c) + b)/sqrt(a^2 - b^2)))/((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*sqrt(a^2 - b^2)) - (2*a^6*b*tan(1/2*d*x
+ 1/2*c)^3 - 6*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 - 2*b^7*tan(1/2*d*x + 1/2*c)^
3 - 5*a^7*tan(1/2*d*x + 1/2*c)^2 - 9*a^5*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^3*b^4*tan(1/2*d*x + 1/2*c)^2 + a*b^6
*tan(1/2*d*x + 1/2*c)^2 + 10*a^6*b*tan(1/2*d*x + 1/2*c) + 10*a^4*b^3*tan(1/2*d*x + 1/2*c) - 6*a^2*b^5*tan(1/2*
d*x + 1/2*c) + 2*b^7*tan(1/2*d*x + 1/2*c) + a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)/((a^10 - 3*a^8*b^2 + 3*a^6*b^
4 - a^4*b^6)*(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))) + 2*(13*a^3*b^6*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^8
*tan(1/2*d*x + 1/2*c)^3 + 12*a^4*b^5*tan(1/2*d*x + 1/2*c)^2 + 19*a^2*b^7*tan(1/2*d*x + 1/2*c)^2 - 10*b^9*tan(1
/2*d*x + 1/2*c)^2 + 35*a^3*b^6*tan(1/2*d*x + 1/2*c) - 14*a*b^8*tan(1/2*d*x + 1/2*c) + 12*a^4*b^5 - 5*a^2*b^7)/
((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + 6*b*l
og(abs(tan(1/2*d*x + 1/2*c)))/a^4 - tan(1/2*d*x + 1/2*c)/a^3)/d

Mupad [B] (verification not implemented)

Time = 15.58 (sec) , antiderivative size = 3122, normalized size of antiderivative = 7.36 \[ \int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + b*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - ((tan(c/2 + (d*x)/2)^6*(5*a^8 - 12*b^8 + 25*a^2*b^6 + 3*a^4*b^4 + 9*a^6*b^2))/(
a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - a^2 + (tan(c/2 + (d*x)/2)^4*(9*a^8 - 20*b^8 + 55*a^2*b^6 + 23*a^4*b^4 - 7
*a^6*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^2*(81*a^2*b^6 - 32*b^8 - 3*a^8 + 7*a^4*b^
4 + 37*a^6*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)*(7*a*b^7 - 8*a^7*b - 18*a^3*b^5 +
4*a^5*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (4*tan(c/2 + (d*x)/2)^3*(2*a^8*b - 5*b^9 + 12*a^2*b^7 + 4*a
^4*b^5 + 2*a^6*b^3))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(4*a^8*b - 10*b^9 + 17*
a^2*b^7 + 18*a^4*b^5 + 16*a^6*b^3))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d*(2*a^5*tan(c/2 + (d*x)/2)^7 -
tan(c/2 + (d*x)/2)^3*(2*a^5 + 8*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(2*a^5 + 8*a^3*b^2) - 2*a^5*tan(c/2 + (d*x)/2)
- 8*a^4*b*tan(c/2 + (d*x)/2)^2 + 8*a^4*b*tan(c/2 + (d*x)/2)^6)) - (3*b*log(tan(c/2 + (d*x)/2)))/(a^4*d) - (b^
4*atan(((b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^26*b + 24*a^6*
b^21 - 228*a^8*b^19 + 978*a^10*b^17 - 2454*a^12*b^15 + 3936*a^14*b^13 - 4170*a^16*b^11 + 2928*a^18*b^9 - 1338*
a^20*b^7 + 384*a^22*b^5 - 66*a^24*b^3) + 12*a^7*b^20 - 111*a^9*b^18 + 462*a^11*b^16 - 1119*a^13*b^14 + 1716*a^
15*b^12 - 1707*a^17*b^10 + 1086*a^19*b^8 - 417*a^21*b^6 + 84*a^23*b^4 - 6*a^25*b^2 - (3*b^4*(-(a + b)^7*(a - b
)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^30 + 8*a^10*b^20 - 78*a^12*b^18 + 342*a^14*b^
16 - 888*a^16*b^14 + 1512*a^18*b^12 - 1764*a^20*b^10 + 1428*a^22*b^8 - 792*a^24*b^6 + 288*a^26*b^4 - 62*a^28*b
^2) - 2*a^29*b + 2*a^11*b^19 - 18*a^13*b^17 + 72*a^15*b^15 - 168*a^17*b^13 + 252*a^19*b^11 - 252*a^21*b^9 + 16
8*a^23*b^7 - 72*a^25*b^5 + 18*a^27*b^3))/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*b^10 + 35*a^10*b^8 - 35*a^1
2*b^6 + 21*a^14*b^4 - 7*a^16*b^2)))*3i)/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*b^10 + 35*a^10*b^8 - 35*a^12
*b^6 + 21*a^14*b^4 - 7*a^16*b^2)) + (b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 +
(d*x)/2)*(6*a^26*b + 24*a^6*b^21 - 228*a^8*b^19 + 978*a^10*b^17 - 2454*a^12*b^15 + 3936*a^14*b^13 - 4170*a^16*
b^11 + 2928*a^18*b^9 - 1338*a^20*b^7 + 384*a^22*b^5 - 66*a^24*b^3) + 12*a^7*b^20 - 111*a^9*b^18 + 462*a^11*b^1
6 - 1119*a^13*b^14 + 1716*a^15*b^12 - 1707*a^17*b^10 + 1086*a^19*b^8 - 417*a^21*b^6 + 84*a^23*b^4 - 6*a^25*b^2
+ (3*b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^30 + 8*a^10*b^20
- 78*a^12*b^18 + 342*a^14*b^16 - 888*a^16*b^14 + 1512*a^18*b^12 - 1764*a^20*b^10 + 1428*a^22*b^8 - 792*a^24*b^
6 + 288*a^26*b^4 - 62*a^28*b^2) - 2*a^29*b + 2*a^11*b^19 - 18*a^13*b^17 + 72*a^15*b^15 - 168*a^17*b^13 + 252*a
^19*b^11 - 252*a^21*b^9 + 168*a^23*b^7 - 72*a^25*b^5 + 18*a^27*b^3))/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8
*b^10 + 35*a^10*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2)))*3i)/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*
b^10 + 35*a^10*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2)))/(2*tan(c/2 + (d*x)/2)*(18*a^4*b^20 - 189*a^6*b^
18 + 765*a^8*b^16 - 1575*a^10*b^14 + 1575*a^12*b^12 - 468*a^14*b^10 - 306*a^16*b^8 + 180*a^18*b^6) + 36*a^3*b^
21 - 342*a^5*b^19 + 1476*a^7*b^17 - 3690*a^9*b^15 + 5760*a^11*b^13 - 5706*a^13*b^11 + 3492*a^15*b^9 - 1206*a^1
7*b^7 + 180*a^19*b^5 - (3*b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6
*a^26*b + 24*a^6*b^21 - 228*a^8*b^19 + 978*a^10*b^17 - 2454*a^12*b^15 + 3936*a^14*b^13 - 4170*a^16*b^11 + 2928
*a^18*b^9 - 1338*a^20*b^7 + 384*a^22*b^5 - 66*a^24*b^3) + 12*a^7*b^20 - 111*a^9*b^18 + 462*a^11*b^16 - 1119*a^
13*b^14 + 1716*a^15*b^12 - 1707*a^17*b^10 + 1086*a^19*b^8 - 417*a^21*b^6 + 84*a^23*b^4 - 6*a^25*b^2 - (3*b^4*(
-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^30 + 8*a^10*b^20 - 78*a^12*b
^18 + 342*a^14*b^16 - 888*a^16*b^14 + 1512*a^18*b^12 - 1764*a^20*b^10 + 1428*a^22*b^8 - 792*a^24*b^6 + 288*a^2
6*b^4 - 62*a^28*b^2) - 2*a^29*b + 2*a^11*b^19 - 18*a^13*b^17 + 72*a^15*b^15 - 168*a^17*b^13 + 252*a^19*b^11 -
252*a^21*b^9 + 168*a^23*b^7 - 72*a^25*b^5 + 18*a^27*b^3))/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*b^10 + 35*
a^10*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2))))/(2*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*b^10 + 35*a^10
*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2)) + (3*b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*
b^2)*(tan(c/2 + (d*x)/2)*(6*a^26*b + 24*a^6*b^21 - 228*a^8*b^19 + 978*a^10*b^17 - 2454*a^12*b^15 + 3936*a^14*b
^13 - 4170*a^16*b^11 + 2928*a^18*b^9 - 1338*a^20*b^7 + 384*a^22*b^5 - 66*a^24*b^3) + 12*a^7*b^20 - 111*a^9*b^1
8 + 462*a^11*b^16 - 1119*a^13*b^14 + 1716*a^15*b^12 - 1707*a^17*b^10 + 1086*a^19*b^8 - 417*a^21*b^6 + 84*a^23*
b^4 - 6*a^25*b^2 + (3*b^4*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + 2*b^4 - 7*a^2*b^2)*(tan(c/2 + (d*x)/2)*(6*a^3
0 + 8*a^10*b^20 - 78*a^12*b^18 + 342*a^14*b^16 - 888*a^16*b^14 + 1512*a^18*b^12 - 1764*a^20*b^10 + 1428*a^22*b
^8 - 792*a^24*b^6 + 288*a^26*b^4 - 62*a^28*b^2) - 2*a^29*b + 2*a^11*b^19 - 18*a^13*b^17 + 72*a^15*b^15 - 168*a
^17*b^13 + 252*a^19*b^11 - 252*a^21*b^9 + 168*a^23*b^7 - 72*a^25*b^5 + 18*a^27*b^3))/(2*(a^18 - a^4*b^14 + 7*a
^6*b^12 - 21*a^8*b^10 + 35*a^10*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2))))/(2*(a^18 - a^4*b^14 + 7*a^6*b
^12 - 21*a^8*b^10 + 35*a^10*b^8 - 35*a^12*b^6 + 21*a^14*b^4 - 7*a^16*b^2))))*(-(a + b)^7*(a - b)^7)^(1/2)*(10*
a^4 + 2*b^4 - 7*a^2*b^2)*3i)/(d*(a^18 - a^4*b^14 + 7*a^6*b^12 - 21*a^8*b^10 + 35*a^10*b^8 - 35*a^12*b^6 + 21*a
^14*b^4 - 7*a^16*b^2))