3.2.90 \(\int \frac {\cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [190]

Optimal. Leaf size=238 \[ \frac {2 \left (a^4-5 a^2 b^2+4 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^5 \sqrt {a^2-b^2} d}-\frac {b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))} \]

[Out]

-b*(3*a^2-4*b^2)*arctanh(cos(d*x+c))/a^5/d+1/3*(7*a^2-12*b^2)*cot(d*x+c)/a^4/d-(a^2-2*b^2)*cot(d*x+c)*csc(d*x+
c)/a^3/b/d+1/3*(3*a^2-4*b^2)*cot(d*x+c)*csc(d*x+c)/a^2/b/d/(a+b*sin(d*x+c))-1/3*cot(d*x+c)*csc(d*x+c)^2/a/d/(a
+b*sin(d*x+c))+2*(a^4-5*a^2*b^2+4*b^4)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^5/d/(a^2-b^2)^(1/2)

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Rubi [A]
time = 0.47, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2803, 3134, 3080, 3855, 2739, 632, 210} \begin {gather*} \frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {2 \left (a^4-5 a^2 b^2+4 b^4\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*(a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*Sqrt[a^2 - b^2]*d) - (b*(3
*a^2 - 4*b^2)*ArcTanh[Cos[c + d*x]])/(a^5*d) + ((7*a^2 - 12*b^2)*Cot[c + d*x])/(3*a^4*d) - ((a^2 - 2*b^2)*Cot[
c + d*x]*Csc[c + d*x])/(a^3*b*d) + ((3*a^2 - 4*b^2)*Cot[c + d*x]*Csc[c + d*x])/(3*a^2*b*d*(a + b*Sin[c + d*x])
) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*a*d*(a + b*Sin[c + d*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 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 2803

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(3*a*f*Sin[e + f*x]^3)), x] + (-Dist[1/(3*a^2*b*(m + 1)), Int[((a + b*Sin[e + f*x
])^(m + 1)/Sin[e + f*x]^3)*Simp[6*a^2 - b^2*(m - 1)*(m - 2) + a*b*(m + 1)*Sin[e + f*x] - (3*a^2 - b^2*m*(m - 2
))*Sin[e + f*x]^2, x], x], x] - Simp[(3*a^2 + b^2*(m - 2))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(3*a^2*b
*f*(m + 1)*Sin[e + f*x]^2)), x]) /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3080

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

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 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*} \int \frac {\cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc ^3(c+d x) \left (6 \left (a^2-2 b^2\right )-a b \sin (c+d x)-\left (3 a^2-8 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 a^2 b}\\ &=-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc ^2(c+d x) \left (-2 b \left (7 a^2-12 b^2\right )+4 a b^2 \sin (c+d x)+6 b \left (a^2-2 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 a^3 b}\\ &=\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (6 b^2 \left (3 a^2-4 b^2\right )+6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 a^4 b}\\ &=\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac {\left (b \left (3 a^2-4 b^2\right )\right ) \int \csc (c+d x) \, dx}{a^5}+\frac {\left (a^4-5 a^2 b^2+4 b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^5}\\ &=-\frac {b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac {\left (2 \left (a^4-5 a^2 b^2+4 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^5 d}\\ &=-\frac {b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}-\frac {\left (4 \left (a^4-5 a^2 b^2+4 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^5 d}\\ &=\frac {2 \left (a^4-5 a^2 b^2+4 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^5 \sqrt {a^2-b^2} d}-\frac {b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac {\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 6.17, size = 403, normalized size = 1.69 \begin {gather*} \frac {2 \left (a^4-5 a^2 b^2+4 b^4\right ) \tan ^{-1}\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 )}{a^5 \sqrt {a^2-b^2} d}+\frac {\left (4 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-9 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{6 a^4 d}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}+\frac {\left (-3 a^2 b+4 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (3 a^2 b-4 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-4 a^2 \sin \left (\frac {1}{2} (c+d x)\right )+9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac {a^2 b \cos (c+d x)-b^3 \cos (c+d x)}{a^4 d (a+b \sin (c+d x))}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*(a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^
2]])/(a^5*Sqrt[a^2 - b^2]*d) + ((4*a^2*Cos[(c + d*x)/2] - 9*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^4*d)
+ (b*Csc[(c + d*x)/2]^2)/(4*a^3*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a^2*d) + ((-3*a^2*b + 4*b^3)*Lo
g[Cos[(c + d*x)/2]])/(a^5*d) + ((3*a^2*b - 4*b^3)*Log[Sin[(c + d*x)/2]])/(a^5*d) - (b*Sec[(c + d*x)/2]^2)/(4*a
^3*d) + (Sec[(c + d*x)/2]*(-4*a^2*Sin[(c + d*x)/2] + 9*b^2*Sin[(c + d*x)/2]))/(6*a^4*d) + (a^2*b*Cos[c + d*x]
- b^3*Cos[c + d*x])/(a^4*d*(a + b*Sin[c + d*x])) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*a^2*d)

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Maple [A]
time = 0.49, size = 287, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{3}-2 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+12 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}+\frac {\frac {2 \left (b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b \left (a^{2}-b^{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (a^{4}-5 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{5}}-\frac {1}{24 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-5 a^{2}+12 b^{2}}{8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (3 a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5}}}{d}\)	\(287\)
default	\(\frac {\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{3}-2 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+12 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}+\frac {\frac {2 \left (b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b \left (a^{2}-b^{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 \left (a^{4}-5 a^{2} b^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{a^{5}}-\frac {1}{24 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-5 a^{2}+12 b^{2}}{8 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (3 a^{2}-4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5}}}{d}\)	\(287\)
risch	\(\frac {12 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}+\frac {50 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}-\frac {22 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{3}-28 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+24 i b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-\frac {14 i a^{2} b}{3}-4 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+20 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-24 i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-14 i a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+8 i b^{3}+2 i a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-8 i b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-14 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+14 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{4}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{5}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}+\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{5}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}-\frac {4 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right ) b^{2}}{d \,a^{5}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}+\frac {4 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right ) b^{2}}{d \,a^{5}}\)	\(571\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/8/a^4*(1/3*tan(1/2*d*x+1/2*c)^3*a^2-2*a*b*tan(1/2*d*x+1/2*c)^2-5*a^2*tan(1/2*d*x+1/2*c)+12*b^2*tan(1/2*
d*x+1/2*c))+2/a^5*((b^2*(a^2-b^2)*tan(1/2*d*x+1/2*c)+a*b*(a^2-b^2))/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/
2*c)+a)+(a^4-5*a^2*b^2+4*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)))-1/24/a
^2/tan(1/2*d*x+1/2*c)^3-1/8*(-5*a^2+12*b^2)/a^4/tan(1/2*d*x+1/2*c)+1/4/a^3*b/tan(1/2*d*x+1/2*c)^2+1/a^5*b*(3*a
^2-4*b^2)*ln(tan(1/2*d*x+1/2*c)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c))^2,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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (227) = 454\).
time = 0.46, size = 1149, normalized size = 4.83 \begin {gather*} \left [-\frac {4 \, {\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left ({\left (a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{2} b - 4 \, b^{3} - 2 \, {\left (a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 4 \, a b^{2} - {\left (a^{3} - 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b^{2} - 4 \, b^{4} - 2 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 4 \, a b^{3} - {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b^{2} - 4 \, b^{4} - 2 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 4 \, a b^{3} - {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left ({\left (7 \, a^{3} b - 12 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{5} b d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b d \cos \left (d x + c\right )^{2} + a^{5} b d - {\left (a^{6} d \cos \left (d x + c\right )^{2} - a^{6} d\right )} \sin \left (d x + c\right )\right )}}, -\frac {4 \, {\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left ({\left (a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + a^{2} b - 4 \, b^{3} - 2 \, {\left (a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} - 4 \, a b^{2} - {\left (a^{3} - 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b^{2} - 4 \, b^{4} - 2 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 4 \, a b^{3} - {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b^{2} - 4 \, b^{4} - 2 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 4 \, a b^{3} - {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left ({\left (7 \, a^{3} b - 12 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{5} b d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b d \cos \left (d x + c\right )^{2} + a^{5} b d - {\left (a^{6} d \cos \left (d x + c\right )^{2} - a^{6} d\right )} \sin \left (d x + c\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(4*(2*a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + 3*((a^2*b - 4*b^3)*cos(d*x + c)^4 + a^2*b - 4*b^3 - 2*(a^2*b - 4
*b^3)*cos(d*x + c)^2 + (a^3 - 4*a*b^2 - (a^3 - 4*a*b^2)*cos(d*x + c)^2)*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*(a^4 - 2*a^2*b^2)*cos(d*x + c) +
3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 + (3*a^3*b -
4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((3*a^2*b^2 - 4*b
^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 + (3*a^3*b - 4*a*b^3 - (3*a^3*b
- 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*((7*a^3*b - 12*a*b^3)*cos(d*x + c)^3
- 3*(3*a^3*b - 4*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^5*b*d*cos(d*x + c)^4 - 2*a^5*b*d*cos(d*x + c)^2 + a^5*
b*d - (a^6*d*cos(d*x + c)^2 - a^6*d)*sin(d*x + c)), -1/6*(4*(2*a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + 6*((a^2*b - 4
*b^3)*cos(d*x + c)^4 + a^2*b - 4*b^3 - 2*(a^2*b - 4*b^3)*cos(d*x + c)^2 + (a^3 - 4*a*b^2 - (a^3 - 4*a*b^2)*cos
(d*x + c)^2)*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))) - 6*(a
^4 - 2*a^2*b^2)*cos(d*x + c) + 3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^
4)*cos(d*x + c)^2 + (3*a^3*b - 4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c
) + 1/2) - 3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 +
(3*a^3*b - 4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*((7*a^
3*b - 12*a*b^3)*cos(d*x + c)^3 - 3*(3*a^3*b - 4*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^5*b*d*cos(d*x + c)^4 - 2
*a^5*b*d*cos(d*x + c)^2 + a^5*b*d - (a^6*d*cos(d*x + c)^2 - a^6*d)*sin(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**4/(a + b*sin(c + d*x))**2, x)

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Giac [A]
time = 17.41, size = 356, normalized size = 1.50 \begin {gather*} \frac {\frac {24 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{5}} + \frac {48 \, {\left (a^{4} - 5 \, a^{2} b^{2} + 4 \, b^{4}\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 )}}{\sqrt {a^{2} - b^{2}} a^{5}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}} + \frac {48 \, {\left (a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} b - a b^{3}\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 )} a^{5}} - \frac {132 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 176 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(24*(3*a^2*b - 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 + 48*(a^4 - 5*a^2*b^2 + 4*b^4)*(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)))/(sqrt(a^2 - b^2)*a^5) + (a^4*t
an(1/2*d*x + 1/2*c)^3 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 15*a^4*tan(1/2*d*x + 1/2*c) + 36*a^2*b^2*tan(1/2*d*x
+ 1/2*c))/a^6 + 48*(a^2*b^2*tan(1/2*d*x + 1/2*c) - b^4*tan(1/2*d*x + 1/2*c) + a^3*b - a*b^3)/((a*tan(1/2*d*x +
1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^5) - (132*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 176*b^3*tan(1/2*d*x + 1/2
*c)^3 - 15*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/
(a^5*tan(1/2*d*x + 1/2*c)^3))/d

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Mupad [B]
time = 6.83, size = 973, normalized size = 4.09 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (28\,a^2\,b-40\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (8\,a\,b^2-\frac {14\,a^3}{3}\right )-\frac {a^3}{3}+\frac {4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (5\,a^4+4\,a^2\,b^2-16\,b^4\right )}{a}}{d\,\left (8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,b\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {16\,a^2+32\,b^2}{64\,a^4}+\frac {3}{8\,a^2}-\frac {2\,b^2}{a^4}\right )}{d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^3\,d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a^2-4\,b^2\right )}{a^5\,d}+\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )\,\left (\frac {2\,\left (a^9-8\,a^7\,b^2+8\,a^5\,b^4\right )}{a^8}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^7\,b-20\,a^5\,b^3+16\,a^3\,b^5\right )}{a^7}+\frac {\left (2\,a^2\,b-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^{10}-4\,a^8\,b^2\right )}{a^7}\right )\,\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )}{a^5}\right )\,1{}\mathrm {i}}{a^5}+\frac {\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )\,\left (\frac {2\,\left (a^9-8\,a^7\,b^2+8\,a^5\,b^4\right )}{a^8}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^7\,b-20\,a^5\,b^3+16\,a^3\,b^5\right )}{a^7}-\frac {\left (2\,a^2\,b-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^{10}-4\,a^8\,b^2\right )}{a^7}\right )\,\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )}{a^5}\right )\,1{}\mathrm {i}}{a^5}}{\frac {4\,\left (3\,a^6\,b-19\,a^4\,b^3+32\,a^2\,b^5-16\,b^7\right )}{a^8}+\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^6-14\,a^4\,b^2+28\,a^2\,b^4-16\,b^6\right )}{a^7}-\frac {\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )\,\left (\frac {2\,\left (a^9-8\,a^7\,b^2+8\,a^5\,b^4\right )}{a^8}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^7\,b-20\,a^5\,b^3+16\,a^3\,b^5\right )}{a^7}+\frac {\left (2\,a^2\,b-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^{10}-4\,a^8\,b^2\right )}{a^7}\right )\,\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )}{a^5}\right )}{a^5}+\frac {\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )\,\left (\frac {2\,\left (a^9-8\,a^7\,b^2+8\,a^5\,b^4\right )}{a^8}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,a^7\,b-20\,a^5\,b^3+16\,a^3\,b^5\right )}{a^7}-\frac {\left (2\,a^2\,b-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^{10}-4\,a^8\,b^2\right )}{a^7}\right )\,\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )}{a^5}\right )}{a^5}}\right )\,\sqrt {b^2-a^2}\,\left (a^2-4\,b^2\right )\,2{}\mathrm {i}}{a^5\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4/(a + b*sin(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^3*(28*a^2*b - 40*b^3) - tan(c/2 + (d*x)/2)^2*(8*a*b^2 - (14*a^3)/3) - a^3/3 + (4*a^2*b*tan
(c/2 + (d*x)/2))/3 + (tan(c/2 + (d*x)/2)^4*(5*a^4 - 16*b^4 + 4*a^2*b^2))/a)/(d*(8*a^5*tan(c/2 + (d*x)/2)^3 + 8
*a^5*tan(c/2 + (d*x)/2)^5 + 16*a^4*b*tan(c/2 + (d*x)/2)^4)) + tan(c/2 + (d*x)/2)^3/(24*a^2*d) - (tan(c/2 + (d*
x)/2)*((16*a^2 + 32*b^2)/(64*a^4) + 3/(8*a^2) - (2*b^2)/a^4))/d - (b*tan(c/2 + (d*x)/2)^2)/(4*a^3*d) + (b*log(
tan(c/2 + (d*x)/2))*(3*a^2 - 4*b^2))/(a^5*d) + (atan((((b^2 - a^2)^(1/2)*(a^2 - 4*b^2)*((2*(a^9 + 8*a^5*b^4 -
8*a^7*b^2))/a^8 + (2*tan(c/2 + (d*x)/2)*(5*a^7*b + 16*a^3*b^5 - 20*a^5*b^3))/a^7 + ((2*a^2*b - (2*tan(c/2 + (d
*x)/2)*(3*a^10 - 4*a^8*b^2))/a^7)*(b^2 - a^2)^(1/2)*(a^2 - 4*b^2))/a^5)*1i)/a^5 + ((b^2 - a^2)^(1/2)*(a^2 - 4*
b^2)*((2*(a^9 + 8*a^5*b^4 - 8*a^7*b^2))/a^8 + (2*tan(c/2 + (d*x)/2)*(5*a^7*b + 16*a^3*b^5 - 20*a^5*b^3))/a^7 -
((2*a^2*b - (2*tan(c/2 + (d*x)/2)*(3*a^10 - 4*a^8*b^2))/a^7)*(b^2 - a^2)^(1/2)*(a^2 - 4*b^2))/a^5)*1i)/a^5)/(
(4*(3*a^6*b - 16*b^7 + 32*a^2*b^5 - 19*a^4*b^3))/a^8 + (4*tan(c/2 + (d*x)/2)*(2*a^6 - 16*b^6 + 28*a^2*b^4 - 14
*a^4*b^2))/a^7 - ((b^2 - a^2)^(1/2)*(a^2 - 4*b^2)*((2*(a^9 + 8*a^5*b^4 - 8*a^7*b^2))/a^8 + (2*tan(c/2 + (d*x)/
2)*(5*a^7*b + 16*a^3*b^5 - 20*a^5*b^3))/a^7 + ((2*a^2*b - (2*tan(c/2 + (d*x)/2)*(3*a^10 - 4*a^8*b^2))/a^7)*(b^
2 - a^2)^(1/2)*(a^2 - 4*b^2))/a^5))/a^5 + ((b^2 - a^2)^(1/2)*(a^2 - 4*b^2)*((2*(a^9 + 8*a^5*b^4 - 8*a^7*b^2))/
a^8 + (2*tan(c/2 + (d*x)/2)*(5*a^7*b + 16*a^3*b^5 - 20*a^5*b^3))/a^7 - ((2*a^2*b - (2*tan(c/2 + (d*x)/2)*(3*a^
10 - 4*a^8*b^2))/a^7)*(b^2 - a^2)^(1/2)*(a^2 - 4*b^2))/a^5))/a^5))*(b^2 - a^2)^(1/2)*(a^2 - 4*b^2)*2i)/(a^5*d)

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