3.31 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=156 \[ -\frac{a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac{a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac{(2 B+5 i A) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-8 a^4 x (B+i A)-\frac{3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d} \]

[Out]

-8*a^4*(I*A + B)*x - (a^4*(A - (4*I)*B)*Log[Cos[c + d*x]])/d - (a^4*(7*A - (4*I)*B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3)/(2*d) - (((5*I)*A + 2*B)*Cot[c + d*x]*(a^2 + I*a^2*Tan[c + d*x])^2)
/(2*d) - (3*A*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.442379, antiderivative size = 156, 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} \[ -\frac{a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac{a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac{(2 B+5 i A) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-8 a^4 x (B+i A)-\frac{3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-8*a^4*(I*A + B)*x - (a^4*(A - (4*I)*B)*Log[Cos[c + d*x]])/d - (a^4*(7*A - (4*I)*B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3)/(2*d) - (((5*I)*A + 2*B)*Cot[c + d*x]*(a^2 + I*a^2*Tan[c + d*x])^2)
/(2*d) - (3*A*(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 ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 (a (5 i A+2 B)+a (A+2 i B) \tan (c+d x)) \, dx\\ &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac{(5 i A+2 B) \cot (c+d x) \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 (7 A-4 i B)+6 i a^2 A \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac{(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 A \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 (7 A-4 i B)-2 a^3 (i A+4 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac{(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) \left (-2 a^4 (7 A-4 i B)-16 a^4 (i A+B) \tan (c+d x)\right ) \, dx+\left (a^4 (A-4 i B)\right ) \int \tan (c+d x) \, dx\\ &=-8 a^4 (i A+B) x-\frac{a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac{(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\left (a^4 (7 A-4 i B)\right ) \int \cot (c+d x) \, dx\\ &=-8 a^4 (i A+B) x-\frac{a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac{a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac{(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [B]  time = 10.3433, size = 1116, normalized size = 7.15 \[ a^4 \left (\frac{x (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (-\frac{71}{2} i A \cos ^4(c)-22 B \cos ^4(c)+7 A \cot (c) \cos ^4(c)-4 i B \cot (c) \cos ^4(c)-\frac{145}{2} A \sin (c) \cos ^3(c)+50 i B \sin (c) \cos ^3(c)+75 i A \sin ^2(c) \cos ^2(c)+60 B \sin ^2(c) \cos ^2(c)+\frac{1}{2} i A \cos ^2(c)+2 B \cos ^2(c)+40 A \sin ^3(c) \cos (c)-40 i B \sin ^3(c) \cos (c)+\frac{3}{2} A \sin (c) \cos (c)-6 i B \sin (c) \cos (c)-\frac{19}{2} i A \sin ^4(c)-14 B \sin ^4(c)-\frac{3}{2} i A \sin ^2(c)-6 B \sin ^2(c)+(4 \cos (2 c) A+3 A-4 i B \cos (2 c)) \csc (c) \sec (c) (i \sin (4 c)-\cos (4 c))-\frac{1}{2} A \sin ^4(c) \tan (c)+2 i B \sin ^4(c) \tan (c)-\frac{1}{2} A \sin ^2(c) \tan (c)+2 i 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)) (7 A \cos (2 c)-4 i B \cos (2 c)-7 i A \sin (2 c)-4 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)) (A \cos (2 c)-4 i B \cos (2 c)-i A \sin (2 c)-4 B \sin (2 c)) \left (\frac{1}{2} i \log \left (\cos ^2(c+d x)\right ) \sin (2 c)-\frac{1}{2} \cos (2 c) \log \left (\cos ^2(c+d x)\right )\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)) (7 A \cos (2 c)-4 i B \cos (2 c)-7 i A \sin (2 c)-4 B \sin (2 c)) \left (\frac{1}{2} i \log \left (\sin ^2(c+d x)\right ) \sin (2 c)-\frac{1}{2} \cos (2 c) \log \left (\sin ^2(c+d x)\right )\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 \cos (4 c)-8 d x \sin (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{B (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \sec (c) (\cos (4 c)-i \sin (4 c)) \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{(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) (\cos (4 c)-i \sin (4 c)) (4 i A \sin (d x)+B \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} i A \sin (4 c)-\frac{1}{2} A \cos (4 c)\right ) \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]^3*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

a^4*(((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*(-(A*Cos[4*c])/2 + (I/2)*A*Sin[4*c])*Sin[c + d*x]^3)/(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])*Csc[c]*(Co
s[4*c] - I*Sin[4*c])*((4*I)*A*Sin[d*x] + B*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])*(7*A*Cos[2*c] - (4*I)*B*Cos[2*c] - (7*I)*A
*Sin[2*c] - 4*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 + Cot[c + d*x])^4*(B + A*Cot[c + d*x])
*(A*Cos[2*c] - (4*I)*B*Cos[2*c] - I*A*Sin[2*c] - 4*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])*(7*A*Cos[2*c] - (4*I)*B*Cos[2*c] - (7*I)*A*Sin[2*c] - 4*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*I)*d*x*C
os[4*c] - 8*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*((I/2)*A*Cos[c]^2 + 2*B*Cos[c]^2 - ((71*I)/2)*A*Cos[
c]^4 - 22*B*Cos[c]^4 + 7*A*Cos[c]^4*Cot[c] - (4*I)*B*Cos[c]^4*Cot[c] + (3*A*Cos[c]*Sin[c])/2 - (6*I)*B*Cos[c]*
Sin[c] - (145*A*Cos[c]^3*Sin[c])/2 + (50*I)*B*Cos[c]^3*Sin[c] - ((3*I)/2)*A*Sin[c]^2 - 6*B*Sin[c]^2 + (75*I)*A
*Cos[c]^2*Sin[c]^2 + 60*B*Cos[c]^2*Sin[c]^2 + 40*A*Cos[c]*Sin[c]^3 - (40*I)*B*Cos[c]*Sin[c]^3 - ((19*I)/2)*A*S
in[c]^4 - 14*B*Sin[c]^4 + (3*A + 4*A*Cos[2*c] - (4*I)*B*Cos[2*c])*Csc[c]*Sec[c]*(-Cos[4*c] + I*Sin[4*c]) - (A*
Sin[c]^2*Tan[c])/2 + (2*I)*B*Sin[c]^2*Tan[c] - (A*Sin[c]^4*Tan[c])/2 + (2*I)*B*Sin[c]^4*Tan[c]))/((Cos[d*x] +
I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (B*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Sec[c]*(Cos[4*
c] - I*Sin[4*c])*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])))

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Maple [A]  time = 0.074, size = 166, normalized size = 1.1 \begin{align*} -{\frac{A{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-8\,B{a}^{4}x+{\frac{B{a}^{4}\tan \left ( dx+c \right ) }{d}}-8\,{\frac{B{a}^{4}c}{d}}+{\frac{4\,iB{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{4\,iA\cot \left ( dx+c \right ){a}^{4}}{d}}+{\frac{4\,iB{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-7\,{\frac{A{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-8\,iAx{a}^{4}-{\frac{8\,iA{a}^{4}c}{d}}-{\frac{A{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{\cot \left ( dx+c \right ) B{a}^{4}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 2.10328, size = 149, normalized size = 0.96 \begin{align*} -\frac{16 \,{\left (d x + c\right )}{\left (i \, A + B\right )} a^{4} - 2 \,{\left (4 \, A - 4 i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (7 \, A - 4 i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) - 2 \, B a^{4} \tan \left (d x + c\right ) - \frac{2 \,{\left (-4 i \, A - B\right )} a^{4} \tan \left (d x + c\right ) - A a^{4}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(16*(d*x + c)*(I*A + B)*a^4 - 2*(4*A - 4*I*B)*a^4*log(tan(d*x + c)^2 + 1) + 2*(7*A - 4*I*B)*a^4*log(tan(d
*x + c)) - 2*B*a^4*tan(d*x + c) - (2*(-4*I*A - B)*a^4*tan(d*x + c) - A*a^4)/tan(d*x + c)^2)/d

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Fricas [A]  time = 1.54735, size = 678, normalized size = 4.35 \begin{align*} \frac{10 \, A a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \,{\left (A - 2 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \,{\left (2 \, A - i \, B\right )} a^{4} -{\left ({\left (A - 4 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (A - 4 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (A - 4 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - 4 i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) -{\left ({\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (7 \, A - 4 i \, 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)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(10*A*a^4*e^(4*I*d*x + 4*I*c) + 2*(A - 2*I*B)*a^4*e^(2*I*d*x + 2*I*c) - 4*(2*A - I*B)*a^4 - ((A - 4*I*B)*a^4*e
^(6*I*d*x + 6*I*c) - (A - 4*I*B)*a^4*e^(4*I*d*x + 4*I*c) - (A - 4*I*B)*a^4*e^(2*I*d*x + 2*I*c) + (A - 4*I*B)*a
^4)*log(e^(2*I*d*x + 2*I*c) + 1) - ((7*A - 4*I*B)*a^4*e^(6*I*d*x + 6*I*c) - (7*A - 4*I*B)*a^4*e^(4*I*d*x + 4*I
*c) - (7*A - 4*I*B)*a^4*e^(2*I*d*x + 2*I*c) + (7*A - 4*I*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 = 16.9671, size = 228, normalized size = 1.46 \begin{align*} \frac{\frac{10 A a^{4} e^{- 2 i c} e^{4 i d x}}{d} + \frac{\left (2 A a^{4} - 4 i B a^{4}\right ) e^{- 4 i c} e^{2 i d x}}{d} - \frac{\left (8 A a^{4} - 4 i 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 A a^{4} d - 8 i B a^{4} d\right ) + 7 A^{2} a^{8} - 32 i A B a^{8} - 16 B^{2} a^{8}, \left ( i \mapsto i \log{\left (\frac{i d e^{- 2 i c}}{3 A a^{4}} + e^{2 i d x} + \frac{\left (4 A - 4 i B\right ) e^{- 2 i c}}{3 A} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

(10*A*a**4*exp(-2*I*c)*exp(4*I*d*x)/d + (2*A*a**4 - 4*I*B*a**4)*exp(-4*I*c)*exp(2*I*d*x)/d - (8*A*a**4 - 4*I*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*A*a**4*d - 8*I*B*a**4*d) + 7*A**2*a**8 - 32*I*A*B*a**8 - 16*B**2*a**8, Lambda(_i, _i*l
og(_i*d*exp(-2*I*c)/(3*A*a**4) + exp(2*I*d*x) + (4*A - 4*I*B)*exp(-2*I*c)/(3*A))))

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Giac [B]  time = 1.73924, size = 433, normalized size = 2.78 \begin{align*} -\frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 \,{\left (8 \, A a^{4} - 8 i \, B a^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 8 \,{\left (A a^{4} - 4 i \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 8 \,{\left (A a^{4} - 4 i \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 8 \,{\left (7 \, A a^{4} - 4 i \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{8 \,{\left (A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{4} + 4 i \, B a^{4}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{84 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, 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 )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^4*tan(1/2*d*x + 1/2*c)^2 - 16*I*A*a^4*tan(1/2*d*x + 1/2*c) - 4*B*a^4*tan(1/2*d*x + 1/2*c) - 16*(8*A*
a^4 - 8*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) + 8*(A*a^4 - 4*I*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 8*
(A*a^4 - 4*I*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 8*(7*A*a^4 - 4*I*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c))
) - 8*(A*a^4*tan(1/2*d*x + 1/2*c)^2 - 4*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 - 2*B*a^4*tan(1/2*d*x + 1/2*c) - A*a^4
+ 4*I*B*a^4)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (84*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 48*I*B*a^4*tan(1/2*d*x + 1/2*c)
^2 - 16*I*A*a^4*tan(1/2*d*x + 1/2*c) - 4*B*a^4*tan(1/2*d*x + 1/2*c) - A*a^4)/tan(1/2*d*x + 1/2*c)^2)/d