3.44 \(\int \cot ^7(c+d x) (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=162 \[ \frac{67 a^4 \cot ^4(c+d x)}{60 d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}-\frac{4 a^4 \cot ^2(c+d x)}{d}-\frac{8 i a^4 \cot (c+d x)}{d}-\frac{8 a^4 \log (\sin (c+d x))}{d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-8 i a^4 x \]

[Out]

(-8*I)*a^4*x - ((8*I)*a^4*Cot[c + d*x])/d - (4*a^4*Cot[c + d*x]^2)/d + (((8*I)/3)*a^4*Cot[c + d*x]^3)/d + (67*
a^4*Cot[c + d*x]^4)/(60*d) - (8*a^4*Log[Sin[c + d*x]])/d - (Cot[c + d*x]^6*(a^2 + I*a^2*Tan[c + d*x])^2)/(6*d)
- (((7*I)/15)*Cot[c + d*x]^5*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.334155, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3553, 3593, 3591, 3529, 3531, 3475} \[ \frac{67 a^4 \cot ^4(c+d x)}{60 d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}-\frac{4 a^4 \cot ^2(c+d x)}{d}-\frac{8 i a^4 \cot (c+d x)}{d}-\frac{8 a^4 \log (\sin (c+d x))}{d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-8 i a^4 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^7*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-8*I)*a^4*x - ((8*I)*a^4*Cot[c + d*x])/d - (4*a^4*Cot[c + d*x]^2)/d + (((8*I)/3)*a^4*Cot[c + d*x]^3)/d + (67*
a^4*Cot[c + d*x]^4)/(60*d) - (8*a^4*Log[Sin[c + d*x]])/d - (Cot[c + d*x]^6*(a^2 + I*a^2*Tan[c + d*x])^2)/(6*d)
- (((7*I)/15)*Cot[c + d*x]^5*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3593

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3591

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[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]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3475

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

Rubi steps

\begin{align*} \int \cot ^7(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{1}{6} \int \cot ^6(c+d x) (a+i a \tan (c+d x))^2 \left (-14 i a^2+10 a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac{1}{30} \int \cot ^5(c+d x) (a+i a \tan (c+d x)) \left (134 a^3+106 i a^3 \tan (c+d x)\right ) \, dx\\ &=\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac{1}{30} \int \cot ^4(c+d x) \left (240 i a^4-240 a^4 \tan (c+d x)\right ) \, dx\\ &=\frac{8 i a^4 \cot ^3(c+d x)}{3 d}+\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac{1}{30} \int \cot ^3(c+d x) \left (-240 a^4-240 i a^4 \tan (c+d x)\right ) \, dx\\ &=-\frac{4 a^4 \cot ^2(c+d x)}{d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}+\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac{1}{30} \int \cot ^2(c+d x) \left (-240 i a^4+240 a^4 \tan (c+d x)\right ) \, dx\\ &=-\frac{8 i a^4 \cot (c+d x)}{d}-\frac{4 a^4 \cot ^2(c+d x)}{d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}+\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac{1}{30} \int \cot (c+d x) \left (240 a^4+240 i a^4 \tan (c+d x)\right ) \, dx\\ &=-8 i a^4 x-\frac{8 i a^4 \cot (c+d x)}{d}-\frac{4 a^4 \cot ^2(c+d x)}{d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}+\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\left (8 a^4\right ) \int \cot (c+d x) \, dx\\ &=-8 i a^4 x-\frac{8 i a^4 \cot (c+d x)}{d}-\frac{4 a^4 \cot ^2(c+d x)}{d}+\frac{8 i a^4 \cot ^3(c+d x)}{3 d}+\frac{67 a^4 \cot ^4(c+d x)}{60 d}-\frac{8 a^4 \log (\sin (c+d x))}{d}-\frac{\cot ^6(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac{7 i \cot ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}\\ \end{align*}

Mathematica [B]  time = 1.97959, size = 363, normalized size = 2.24 \[ \frac{a^4 \csc (c) \csc ^6(c+d x) \left (-600 i d x \sin (c)-450 i d x \sin (c+2 d x)-345 \sin (c+2 d x)+450 i d x \sin (3 c+2 d x)+345 \sin (3 c+2 d x)+180 i d x \sin (3 c+4 d x)+120 \sin (3 c+4 d x)-180 i d x \sin (5 c+4 d x)-120 \sin (5 c+4 d x)-30 i d x \sin (5 c+6 d x)+30 i d x \sin (7 c+6 d x)-780 i \cos (c+2 d x)-510 i \cos (3 c+2 d x)+366 i \cos (3 c+4 d x)+150 i \cos (5 c+4 d x)-86 i \cos (5 c+6 d x)-300 \sin (c) \log \left (\sin ^2(c+d x)\right )-225 \sin (c+2 d x) \log \left (\sin ^2(c+d x)\right )+225 \sin (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+90 \sin (3 c+4 d x) \log \left (\sin ^2(c+d x)\right )-90 \sin (5 c+4 d x) \log \left (\sin ^2(c+d x)\right )-15 \sin (5 c+6 d x) \log \left (\sin ^2(c+d x)\right )+15 \sin (7 c+6 d x) \log \left (\sin ^2(c+d x)\right )-490 \sin (c)+860 i \cos (c)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^7*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*Csc[c]*Csc[c + d*x]^6*((860*I)*Cos[c] - (780*I)*Cos[c + 2*d*x] - (510*I)*Cos[3*c + 2*d*x] + (366*I)*Cos[3
*c + 4*d*x] + (150*I)*Cos[5*c + 4*d*x] - (86*I)*Cos[5*c + 6*d*x] - 490*Sin[c] - (600*I)*d*x*Sin[c] - 300*Log[S
in[c + d*x]^2]*Sin[c] - 345*Sin[c + 2*d*x] - (450*I)*d*x*Sin[c + 2*d*x] - 225*Log[Sin[c + d*x]^2]*Sin[c + 2*d*
x] + 345*Sin[3*c + 2*d*x] + (450*I)*d*x*Sin[3*c + 2*d*x] + 225*Log[Sin[c + d*x]^2]*Sin[3*c + 2*d*x] + 120*Sin[
3*c + 4*d*x] + (180*I)*d*x*Sin[3*c + 4*d*x] + 90*Log[Sin[c + d*x]^2]*Sin[3*c + 4*d*x] - 120*Sin[5*c + 4*d*x] -
(180*I)*d*x*Sin[5*c + 4*d*x] - 90*Log[Sin[c + d*x]^2]*Sin[5*c + 4*d*x] - (30*I)*d*x*Sin[5*c + 6*d*x] - 15*Log
[Sin[c + d*x]^2]*Sin[5*c + 6*d*x] + (30*I)*d*x*Sin[7*c + 6*d*x] + 15*Log[Sin[c + d*x]^2]*Sin[7*c + 6*d*x]))/(2
40*d)

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Maple [A]  time = 0.057, size = 131, normalized size = 0.8 \begin{align*} -4\,{\frac{{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}-8\,{\frac{{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-8\,i{a}^{4}x-{\frac{{\frac{4\,i}{5}}{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{d}}-{\frac{8\,i{a}^{4}c}{d}}-{\frac{8\,i\cot \left ( dx+c \right ){a}^{4}}{d}}+{\frac{7\,{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{\frac{8\,i}{3}}{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{6}}{6\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x)

[Out]

-4*a^4*cot(d*x+c)^2/d-8*a^4*ln(sin(d*x+c))/d-8*I*a^4*x-4/5*I/d*a^4*cot(d*x+c)^5-8*I/d*a^4*c-8*I*a^4*cot(d*x+c)
/d+7/4*a^4*cot(d*x+c)^4/d+8/3*I*a^4*cot(d*x+c)^3/d-1/6/d*a^4*cot(d*x+c)^6

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Maxima [A]  time = 1.66716, size = 166, normalized size = 1.02 \begin{align*} -\frac{480 i \,{\left (d x + c\right )} a^{4} - 240 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 480 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - \frac{-480 i \, a^{4} \tan \left (d x + c\right )^{5} - 240 \, a^{4} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} \tan \left (d x + c\right )^{3} + 105 \, a^{4} \tan \left (d x + c\right )^{2} - 48 i \, a^{4} \tan \left (d x + c\right ) - 10 \, a^{4}}{\tan \left (d x + c\right )^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/60*(480*I*(d*x + c)*a^4 - 240*a^4*log(tan(d*x + c)^2 + 1) + 480*a^4*log(tan(d*x + c)) - (-480*I*a^4*tan(d*x
+ c)^5 - 240*a^4*tan(d*x + c)^4 + 160*I*a^4*tan(d*x + c)^3 + 105*a^4*tan(d*x + c)^2 - 48*I*a^4*tan(d*x + c) -
10*a^4)/tan(d*x + c)^6)/d

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Fricas [A]  time = 2.2347, size = 738, normalized size = 4.56 \begin{align*} \frac{4 \,{\left (270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} - 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 86 \, a^{4} - 30 \,{\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} - 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{15 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} - 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} - 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

4/15*(270*a^4*e^(10*I*d*x + 10*I*c) - 855*a^4*e^(8*I*d*x + 8*I*c) + 1350*a^4*e^(6*I*d*x + 6*I*c) - 1125*a^4*e^
(4*I*d*x + 4*I*c) + 486*a^4*e^(2*I*d*x + 2*I*c) - 86*a^4 - 30*(a^4*e^(12*I*d*x + 12*I*c) - 6*a^4*e^(10*I*d*x +
10*I*c) + 15*a^4*e^(8*I*d*x + 8*I*c) - 20*a^4*e^(6*I*d*x + 6*I*c) + 15*a^4*e^(4*I*d*x + 4*I*c) - 6*a^4*e^(2*I
*d*x + 2*I*c) + a^4)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d
*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 7.85581, size = 253, normalized size = 1.56 \begin{align*} - \frac{8 a^{4} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{72 a^{4} e^{- 2 i c} e^{10 i d x}}{d} - \frac{228 a^{4} e^{- 4 i c} e^{8 i d x}}{d} + \frac{360 a^{4} e^{- 6 i c} e^{6 i d x}}{d} - \frac{300 a^{4} e^{- 8 i c} e^{4 i d x}}{d} + \frac{648 a^{4} e^{- 10 i c} e^{2 i d x}}{5 d} - \frac{344 a^{4} e^{- 12 i c}}{15 d}}{e^{12 i d x} - 6 e^{- 2 i c} e^{10 i d x} + 15 e^{- 4 i c} e^{8 i d x} - 20 e^{- 6 i c} e^{6 i d x} + 15 e^{- 8 i c} e^{4 i d x} - 6 e^{- 10 i c} e^{2 i d x} + e^{- 12 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**7*(a+I*a*tan(d*x+c))**4,x)

[Out]

-8*a**4*log(exp(2*I*d*x) - exp(-2*I*c))/d + (72*a**4*exp(-2*I*c)*exp(10*I*d*x)/d - 228*a**4*exp(-4*I*c)*exp(8*
I*d*x)/d + 360*a**4*exp(-6*I*c)*exp(6*I*d*x)/d - 300*a**4*exp(-8*I*c)*exp(4*I*d*x)/d + 648*a**4*exp(-10*I*c)*e
xp(2*I*d*x)/(5*d) - 344*a**4*exp(-12*I*c)/(15*d))/(exp(12*I*d*x) - 6*exp(-2*I*c)*exp(10*I*d*x) + 15*exp(-4*I*c
)*exp(8*I*d*x) - 20*exp(-6*I*c)*exp(6*I*d*x) + 15*exp(-8*I*c)*exp(4*I*d*x) - 6*exp(-10*I*c)*exp(2*I*d*x) + exp
(-12*I*c))

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Giac [A]  time = 1.8455, size = 332, normalized size = 2.05 \begin{align*} -\frac{5 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 48 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 240 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 880 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2835 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 30720 \, a^{4} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 15360 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 10080 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{37632 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 10080 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2835 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 880 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 240 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/1920*(5*a^4*tan(1/2*d*x + 1/2*c)^6 - 48*I*a^4*tan(1/2*d*x + 1/2*c)^5 - 240*a^4*tan(1/2*d*x + 1/2*c)^4 + 880
*I*a^4*tan(1/2*d*x + 1/2*c)^3 + 2835*a^4*tan(1/2*d*x + 1/2*c)^2 - 30720*a^4*log(tan(1/2*d*x + 1/2*c) + I) + 15
360*a^4*log(abs(tan(1/2*d*x + 1/2*c))) - 10080*I*a^4*tan(1/2*d*x + 1/2*c) - (37632*a^4*tan(1/2*d*x + 1/2*c)^6
- 10080*I*a^4*tan(1/2*d*x + 1/2*c)^5 - 2835*a^4*tan(1/2*d*x + 1/2*c)^4 + 880*I*a^4*tan(1/2*d*x + 1/2*c)^3 + 24
0*a^4*tan(1/2*d*x + 1/2*c)^2 - 48*I*a^4*tan(1/2*d*x + 1/2*c) - 5*a^4)/tan(1/2*d*x + 1/2*c)^6)/d