3.1476 \(\int \frac{\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=424 \[ -\frac{4 b^4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac{b^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac{2 b^4 \left (-9 a^2 b^2+10 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{7/2}}-\frac{2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{3 b^5 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b^5 \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.589752, antiderivative size = 424, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 2648, 2664, 2754, 12, 2660, 618, 204} \[ -\frac{4 b^4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac{b^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{7/2}}-\frac{2 b^4 \left (-9 a^2 b^2+10 a^4+3 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{7/2}}-\frac{2 b^5 \left (2 a^2-b^2\right ) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{3 b^5 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b^5 \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[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 2897

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 3770

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

Rule 3767

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 8

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

Rule 2648

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 2664

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] && Integer
Q[2*n]

Rule 2754

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 12

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

Rule 2660

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 618

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 204

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

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\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 \tanh ^{-1}(\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{\operatorname{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 ) \operatorname{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 \tanh ^{-1}(\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 ) \operatorname{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 ) \tan ^{-1}\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 \tanh ^{-1}(\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 ) \operatorname{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 ) \tan ^{-1}\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 \tanh ^{-1}(\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 ) \operatorname{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 ) \operatorname{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 ) \tan ^{-1}\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 ) \tan ^{-1}\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 \tanh ^{-1}(\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 ) \operatorname{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 ) \tan ^{-1}\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 ) \tan ^{-1}\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 ) \tan ^{-1}\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 \tanh ^{-1}(\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]  time = 6.34985, size = 379, normalized size = 0.89 \[ 4 \left (-\frac{b^5 \cos (c+d x)}{8 a^2 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^2}+\frac{4 b^7 \cos (c+d x)-11 a^2 b^5 \cos (c+d x)}{8 a^3 d (a-b)^3 (a+b)^3 (a+b \sin (c+d x))}-\frac{3 b^4 \left (-7 a^2 b^2+10 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{4 a^4 d \left (a^2-b^2\right )^{7/2}}-\frac{3 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 a^4 d}+\frac{3 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 a^4 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{8 a^3 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{8 a^3 d}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{4 d (a+b)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{4 d (a-b)^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right ) \]

Antiderivative was successfully verified.

[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))

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Maple [B]  time = 0.194, size = 829, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)-13/d*b^6/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2
*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^3+6/d*b^8/(a-b)^3/(a+b)^3/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*ta
n(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3-12/d*b^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/
2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-19/d*b^7/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/
a^2*tan(1/2*d*x+1/2*c)^2+10/d*b^9/(a-b)^3/(a+b)^3/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(
1/2*d*x+1/2*c)^2-35/d*b^6/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/
2*c)+14/d*b^8/(a-b)^3/(a+b)^3/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)-12/d*
b^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+5/d*b^7/(a-b)^3/(a+b)^3/a^2/(tan(1/2*d
*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2-30/d*b^4/(a-b)^3/(a+b)^3/(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))+21/d*b^6/(a-b)^3/(a+b)^3/a^2/(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))-6/d*b^8/(a-b)^3/(a+b)^3/a^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/d/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^4*b*ln(tan(1/2*d*x+1/2*c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 20.927, size = 4748, normalized size = 11.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.27455, size = 855, normalized size = 2.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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