3.30 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac{(-B+2 i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{a^4 (B+4 i A) \log (\sin (c+d x))}{d}+\frac{a^4 (7 B+4 i A) \log (\cos (c+d x))}{d}-8 a^4 x (A-i B)-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d} \]

[Out]

-8*a^4*(A - I*B)*x + (a^4*((4*I)*A + 7*B)*Log[Cos[c + d*x]])/d + (a^4*((4*I)*A + B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]*(a + I*a*Tan[c + d*x])^3)/d + (((2*I)*A - B)*(a^2 + I*a^2*Tan[c + d*x])^2)/(2*d) - (3*B*(a^4 +
I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.432015, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3593, 3594, 3589, 3475, 3531} \[ \frac{(-B+2 i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{a^4 (B+4 i A) \log (\sin (c+d x))}{d}+\frac{a^4 (7 B+4 i A) \log (\cos (c+d x))}{d}-8 a^4 x (A-i B)-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-8*a^4*(A - I*B)*x + (a^4*((4*I)*A + 7*B)*Log[Cos[c + d*x]])/d + (a^4*((4*I)*A + B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]*(a + I*a*Tan[c + d*x])^3)/d + (((2*I)*A - B)*(a^2 + I*a^2*Tan[c + d*x])^2)/(2*d) - (3*B*(a^4 +
I*a^4*Tan[c + d*x]))/d

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 3594

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

Rule 3589

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

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]

Rubi steps

\begin{align*} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\int \cot (c+d x) (a+i a \tan (c+d x))^3 (a (4 i A+B)+a (2 A+i B) \tan (c+d x)) \, dx\\ &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\frac{(2 i A-B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{1}{2} \int \cot (c+d x) (a+i a \tan (c+d x))^2 \left (2 a^2 (4 i A+B)+6 i a^2 B \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\frac{(2 i A-B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^3 (4 i A+B)-2 a^3 (4 A-7 i B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\frac{(2 i A-B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) \left (2 a^4 (4 i A+B)-16 a^4 (A-i B) \tan (c+d x)\right ) \, dx-\left (a^4 (4 i A+7 B)\right ) \int \tan (c+d x) \, dx\\ &=-8 a^4 (A-i B) x+\frac{a^4 (4 i A+7 B) \log (\cos (c+d x))}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\frac{(2 i A-B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\left (a^4 (4 i A+B)\right ) \int \cot (c+d x) \, dx\\ &=-8 a^4 (A-i B) x+\frac{a^4 (4 i A+7 B) \log (\cos (c+d x))}{d}+\frac{a^4 (4 i A+B) \log (\sin (c+d x))}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d}+\frac{(2 i A-B) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [B]  time = 10.5206, size = 1122, normalized size = 7.79 \[ a^4 \left (\frac{x (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (-22 A \cos ^4(c)+\frac{17}{2} i B \cos ^4(c)-4 i A \cot (c) \cos ^4(c)-B \cot (c) \cos ^4(c)+50 i A \sin (c) \cos ^3(c)+\frac{55}{2} B \sin (c) \cos ^3(c)+60 A \sin ^2(c) \cos ^2(c)-45 i B \sin ^2(c) \cos ^2(c)+2 A \cos ^2(c)-\frac{7}{2} i B \cos ^2(c)-40 i A \sin ^3(c) \cos (c)-40 B \sin ^3(c) \cos (c)-6 i A \sin (c) \cos (c)-\frac{21}{2} B \sin (c) \cos (c)-14 A \sin ^4(c)+\frac{37}{2} i B \sin ^4(c)-6 A \sin ^2(c)+\frac{21}{2} i B \sin ^2(c)+(4 \cos (2 c) B-3 B+4 i A \cos (2 c)) \csc (c) \sec (c) (\cos (4 c)-i \sin (4 c))+2 i A \sin ^4(c) \tan (c)+\frac{7}{2} B \sin ^4(c) \tan (c)+2 i A \sin ^2(c) \tan (c)+\frac{7}{2} B \sin ^2(c) \tan (c)\right ) \sin ^5(c+d x)}{(\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+B \cos (2 c)+4 A \sin (2 c)-i B \sin (2 c)) \left (-i \tan ^{-1}(\tan (5 c+d x)) \cos (2 c)-\tan ^{-1}(\tan (5 c+d x)) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+7 B \cos (2 c)+4 A \sin (2 c)-7 i B \sin (2 c)) \left (\frac{1}{2} \cos (2 c) \log \left (\cos ^2(c+d x)\right )-\frac{1}{2} i \log \left (\cos ^2(c+d x)\right ) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+B \cos (2 c)+4 A \sin (2 c)-i B \sin (2 c)) \left (\frac{1}{2} \cos (2 c) \log \left (\sin ^2(c+d x)\right )-\frac{1}{2} i \log \left (\sin ^2(c+d x)\right ) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(A-i B) (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (8 i d x \sin (4 c)-8 d x \cos (4 c)) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \sec (c) (\cos (4 c)-i \sin (4 c)) (A \sin (d x)-4 i B \sin (d x)) \tan (c+d x) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{A (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) (\cos (4 c)-i \sin (4 c)) \sin (d x) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (\frac{1}{2} B \cos (4 c)-\frac{1}{2} i B \sin (4 c)\right ) \tan ^2(c+d x) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

a^4*((A*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Csc[c]*(Cos[4*c] - I*Sin[4*c])*Sin[d*x]*Sin[c + d*x]^4)/(d*(
Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I
)*A*Cos[2*c] + B*Cos[2*c] + 4*A*Sin[2*c] - I*B*Sin[2*c])*((-I)*ArcTan[Tan[5*c + d*x]]*Cos[2*c] - ArcTan[Tan[5*
c + d*x]]*Sin[2*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Co
t[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I)*A*Cos[2*c] + 7*B*Cos[2*c] + 4*A*Sin[2*c] - (7*I)*B*Sin[2*c])*((Cos[2
*c]*Log[Cos[c + d*x]^2])/2 - (I/2)*Log[Cos[c + d*x]^2]*Sin[2*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*
(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I)*A*Cos[2*c] + B*Cos[2*c]
+ 4*A*Sin[2*c] - I*B*Sin[2*c])*((Cos[2*c]*Log[Sin[c + d*x]^2])/2 - (I/2)*Log[Sin[c + d*x]^2]*Sin[2*c])*Sin[c
+ d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((A - I*B)*(I + Cot[c + d*x])^4*(B
+ A*Cot[c + d*x])*(-8*d*x*Cos[4*c] + (8*I)*d*x*Sin[4*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[
c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Sin[c + d*x]^5*(2*A*Cos[c]^2 - ((7*
I)/2)*B*Cos[c]^2 - 22*A*Cos[c]^4 + ((17*I)/2)*B*Cos[c]^4 - (4*I)*A*Cos[c]^4*Cot[c] - B*Cos[c]^4*Cot[c] - (6*I)
*A*Cos[c]*Sin[c] - (21*B*Cos[c]*Sin[c])/2 + (50*I)*A*Cos[c]^3*Sin[c] + (55*B*Cos[c]^3*Sin[c])/2 - 6*A*Sin[c]^2
+ ((21*I)/2)*B*Sin[c]^2 + 60*A*Cos[c]^2*Sin[c]^2 - (45*I)*B*Cos[c]^2*Sin[c]^2 - (40*I)*A*Cos[c]*Sin[c]^3 - 40
*B*Cos[c]*Sin[c]^3 - 14*A*Sin[c]^4 + ((37*I)/2)*B*Sin[c]^4 + (-3*B + (4*I)*A*Cos[2*c] + 4*B*Cos[2*c])*Csc[c]*S
ec[c]*(Cos[4*c] - I*Sin[4*c]) + (2*I)*A*Sin[c]^2*Tan[c] + (7*B*Sin[c]^2*Tan[c])/2 + (2*I)*A*Sin[c]^4*Tan[c] +
(7*B*Sin[c]^4*Tan[c])/2))/((Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^
4*(B + A*Cot[c + d*x])*Sec[c]*(Cos[4*c] - I*Sin[4*c])*(A*Sin[d*x] - (4*I)*B*Sin[d*x])*Sin[c + d*x]^4*Tan[c + d
*x])/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*
x])*((B*Cos[4*c])/2 - (I/2)*B*Sin[4*c])*Sin[c + d*x]^3*Tan[c + d*x]^2)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c +
d*x] + B*Sin[c + d*x])))

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Maple [A]  time = 0.066, size = 165, normalized size = 1.2 \begin{align*} -8\,A{a}^{4}x+{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}-8\,{\frac{A{a}^{4}c}{d}}+{\frac{B{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+7\,{\frac{B{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,iA{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,iA{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,iB{a}^{4}c}{d}}+8\,iBx{a}^{4}-{\frac{4\,iB\tan \left ( dx+c \right ){a}^{4}}{d}}-{\frac{A\cot \left ( dx+c \right ){a}^{4}}{d}}+{\frac{B{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

-8*A*a^4*x+1/d*a^4*A*tan(d*x+c)-8/d*A*a^4*c+1/2/d*a^4*B*tan(d*x+c)^2+7*a^4*B*ln(cos(d*x+c))/d+4*I/d*A*a^4*ln(c
os(d*x+c))+4*I/d*A*a^4*ln(sin(d*x+c))+8*I/d*B*a^4*c+8*I*B*x*a^4-4*I/d*B*tan(d*x+c)*a^4-1/d*A*cot(d*x+c)*a^4+1/
d*B*a^4*ln(sin(d*x+c))

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Maxima [A]  time = 2.13928, size = 142, normalized size = 0.99 \begin{align*} \frac{B a^{4} \tan \left (d x + c\right )^{2} - 2 \,{\left (d x + c\right )}{\left (8 \, A - 8 i \, B\right )} a^{4} - 8 \,{\left (i \, A + B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (4 i \, A + B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) +{\left (2 \, A - 8 i \, B\right )} a^{4} \tan \left (d x + c\right ) - \frac{2 \, A a^{4}}{\tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(B*a^4*tan(d*x + c)^2 - 2*(d*x + c)*(8*A - 8*I*B)*a^4 - 8*(I*A + B)*a^4*log(tan(d*x + c)^2 + 1) + 2*(4*I*A
+ B)*a^4*log(tan(d*x + c)) + (2*A - 8*I*B)*a^4*tan(d*x + c) - 2*A*a^4/tan(d*x + c))/d

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Fricas [B]  time = 1.54956, size = 686, normalized size = 4.76 \begin{align*} \frac{10 \, B a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, A - 2 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-4 i \, A - 8 \, B\right )} a^{4} +{\left ({\left (4 i \, A + 7 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (4 i \, A + 7 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, A - 7 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-4 i \, A - 7 \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left ({\left (4 i \, A + B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (4 i \, A + B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, A - B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-4 i \, A - B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (6 i \, d x + 6 i \, c\right )} + d e^{\left (4 i \, d x + 4 i \, c\right )} - d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(10*B*a^4*e^(4*I*d*x + 4*I*c) + (-4*I*A - 2*B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*I*A - 8*B)*a^4 + ((4*I*A + 7*B)*a
^4*e^(6*I*d*x + 6*I*c) + (4*I*A + 7*B)*a^4*e^(4*I*d*x + 4*I*c) + (-4*I*A - 7*B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*
I*A - 7*B)*a^4)*log(e^(2*I*d*x + 2*I*c) + 1) + ((4*I*A + B)*a^4*e^(6*I*d*x + 6*I*c) + (4*I*A + B)*a^4*e^(4*I*d
*x + 4*I*c) + (-4*I*A - B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*I*A - B)*a^4)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I
*d*x + 6*I*c) + d*e^(4*I*d*x + 4*I*c) - d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [A]  time = 8.58648, size = 230, normalized size = 1.6 \begin{align*} \frac{\frac{10 B a^{4} e^{- 2 i c} e^{4 i d x}}{d} - \frac{\left (4 i A a^{4} + 2 B a^{4}\right ) e^{- 4 i c} e^{2 i d x}}{d} - \frac{\left (4 i A a^{4} + 8 B a^{4}\right ) e^{- 6 i c}}{d}}{e^{6 i d x} + e^{- 2 i c} e^{4 i d x} - e^{- 4 i c} e^{2 i d x} - e^{- 6 i c}} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (- 8 i A a^{4} d - 8 B a^{4} d\right ) - 16 A^{2} a^{8} + 32 i A B a^{8} + 7 B^{2} a^{8}, \left ( i \mapsto i \log{\left (\frac{i d e^{- 2 i c}}{3 B a^{4}} + e^{2 i d x} - \frac{\left (4 i A + 4 B\right ) e^{- 2 i c}}{3 B} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

(10*B*a**4*exp(-2*I*c)*exp(4*I*d*x)/d - (4*I*A*a**4 + 2*B*a**4)*exp(-4*I*c)*exp(2*I*d*x)/d - (4*I*A*a**4 + 8*B
*a**4)*exp(-6*I*c)/d)/(exp(6*I*d*x) + exp(-2*I*c)*exp(4*I*d*x) - exp(-4*I*c)*exp(2*I*d*x) - exp(-6*I*c)) + Roo
tSum(_z**2*d**2 + _z*(-8*I*A*a**4*d - 8*B*a**4*d) - 16*A**2*a**8 + 32*I*A*B*a**8 + 7*B**2*a**8, Lambda(_i, _i*
log(_i*d*exp(-2*I*c)/(3*B*a**4) + exp(2*I*d*x) - (4*I*A + 4*B)*exp(-2*I*c)/(3*B))))

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Giac [B]  time = 1.68531, size = 458, normalized size = 3.18 \begin{align*} \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 32 \,{\left (i \, A a^{4} + B a^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 2 \,{\left (4 i \, A a^{4} + 7 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \,{\left (-4 i \, A a^{4} - 7 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 2 \,{\left (-4 i \, A a^{4} - B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{8 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{12 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 4 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 46 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 16 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 i \, A a^{4} + 21 \, B a^{4}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(A*a^4*tan(1/2*d*x + 1/2*c) - 32*(I*A*a^4 + B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) + 2*(4*I*A*a^4 + 7*B*a^4)
*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*(-4*I*A*a^4 - 7*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(-4*I*A*
a^4 - B*a^4)*log(abs(tan(1/2*d*x + 1/2*c))) - (8*I*A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c) +
A*a^4)/tan(1/2*d*x + 1/2*c) - (12*I*A*a^4*tan(1/2*d*x + 1/2*c)^4 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^4 + 4*A*a^4*
tan(1/2*d*x + 1/2*c)^3 - 16*I*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 24*I*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 46*B*a^4*tan(
1/2*d*x + 1/2*c)^2 - 4*A*a^4*tan(1/2*d*x + 1/2*c) + 16*I*B*a^4*tan(1/2*d*x + 1/2*c) + 12*I*A*a^4 + 21*B*a^4)/(
tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d