3.24 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=157 \[ \frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (B+i A) \cot (c+d x)}{d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {(2 B+3 i A) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+4 a^3 x (B+i A)-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d} \]

[Out]

4*a^3*(I*A+B)*x+4*a^3*(I*A+B)*cot(d*x+c)/d+1/12*a^3*(15*A-14*I*B)*cot(d*x+c)^2/d+4*a^3*(A-I*B)*ln(sin(d*x+c))/
d-1/4*a*A*cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2/d-1/6*(3*I*A+2*B)*cot(d*x+c)^3*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.42, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3593, 3591, 3529, 3531, 3475} \[ \frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (B+i A) \cot (c+d x)}{d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {(2 B+3 i A) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+4 a^3 x (B+i A)-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

4*a^3*(I*A + B)*x + (4*a^3*(I*A + B)*Cot[c + d*x])/d + (a^3*(15*A - (14*I)*B)*Cot[c + d*x]^2)/(12*d) + (4*a^3*
(A - I*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2)/(4*d) - (((3*I)*A + 2*B)*Cot[c
+ d*x]^3*(a^3 + I*a^3*Tan[c + d*x]))/(6*d)

Rule 3475

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

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 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 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]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 (2 a (3 i A+2 B)-2 a (A-2 i B) \tan (c+d x)) \, dx\\ &=-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \left (-2 a^2 (15 A-14 i B)-2 a^2 (9 i A+10 B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot ^2(c+d x) \left (-48 a^3 (i A+B)+48 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot (c+d x) \left (48 a^3 (A-i B)+48 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=4 a^3 (i A+B) x+\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\left (4 a^3 (A-i B)\right ) \int \cot (c+d x) \, dx\\ &=4 a^3 (i A+B) x+\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}\\ \end {align*}

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Mathematica [B]  time = 9.12, size = 1007, normalized size = 6.41 \[ a^3 \left (\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (A \cos \left (\frac {3 c}{2}\right )-i B \cos \left (\frac {3 c}{2}\right )-i A \sin \left (\frac {3 c}{2}\right )-B \sin \left (\frac {3 c}{2}\right )\right ) \left (-4 i \tan ^{-1}(\tan (4 c+d x)) \cos \left (\frac {3 c}{2}\right )-4 \tan ^{-1}(\tan (4 c+d x)) \sin \left (\frac {3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (A \cos \left (\frac {3 c}{2}\right )-i B \cos \left (\frac {3 c}{2}\right )-i A \sin \left (\frac {3 c}{2}\right )-B \sin \left (\frac {3 c}{2}\right )\right ) \left (2 \cos \left (\frac {3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-2 i \log \left (\sin ^2(c+d x)\right ) \sin \left (\frac {3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {x (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (16 i A \cos ^3(c)+16 B \cos ^3(c)-4 A \cot (c) \cos ^3(c)+4 i B \cot (c) \cos ^3(c)+24 A \sin (c) \cos ^2(c)-24 i B \sin (c) \cos ^2(c)-16 i A \sin ^2(c) \cos (c)-16 B \sin ^2(c) \cos (c)-4 A \sin ^3(c)+4 i B \sin ^3(c)+(A-i B) \cot (c) (4 \cos (3 c)-4 i \sin (3 c))\right ) \sin ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(i A+B) (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) (4 d x \cos (3 c)-4 i d x \sin (3 c)) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {1}{6} \cos (3 c)-\frac {1}{6} i \sin (3 c)\right ) (-15 i A \sin (d x)-13 B \sin (d x)) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (-6 i A \cos (c)-2 B \cos (c)+15 A \sin (c)-9 i B \sin (c)) \left (\frac {1}{12} \cos (3 c)-\frac {1}{12} i \sin (3 c)\right ) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {1}{6} \cos (3 c)-\frac {1}{6} i \sin (3 c)\right ) (3 i A \sin (d x)+B \sin (d x)) \sin (c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (\frac {1}{4} i A \sin (3 c)-\frac {1}{4} A \cos (3 c)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

a^3*(((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(-1/4*(A*Cos[3*c]) + (I/4)*A*Sin[3*c]))/(d*(Cos[d*x] + I*Sin[d
*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Csc[c/2]*Sec[c/2]*(Cos[
3*c]/6 - (I/6)*Sin[3*c])*((3*I)*A*Sin[d*x] + B*Sin[d*x])*Sin[c + d*x])/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c +
d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Csc[c/2]*Sec[c/2]*((-6*I)*A*Cos[c] - 2*B
*Cos[c] + 15*A*Sin[c] - (9*I)*B*Sin[c])*(Cos[3*c]/12 - (I/12)*Sin[3*c])*Sin[c + d*x]^2)/(d*(Cos[d*x] + I*Sin[d
*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Csc[c/2]*Sec[c/2]*(Cos[
3*c]/6 - (I/6)*Sin[3*c])*((-15*I)*A*Sin[d*x] - 13*B*Sin[d*x])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*
Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(A*Cos[(3*c)/2] - I*B*Cos[(3*c)/2
] - I*A*Sin[(3*c)/2] - B*Sin[(3*c)/2])*((-4*I)*ArcTan[Tan[4*c + d*x]]*Cos[(3*c)/2] - 4*ArcTan[Tan[4*c + d*x]]*
Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c +
d*x])^3*(B + A*Cot[c + d*x])*(A*Cos[(3*c)/2] - I*B*Cos[(3*c)/2] - I*A*Sin[(3*c)/2] - B*Sin[(3*c)/2])*(2*Cos[(3
*c)/2]*Log[Sin[c + d*x]^2] - (2*I)*Log[Sin[c + d*x]^2]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x]
)^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*((16*I)*A*Cos[c]^3 + 16*
B*Cos[c]^3 - 4*A*Cos[c]^3*Cot[c] + (4*I)*B*Cos[c]^3*Cot[c] + 24*A*Cos[c]^2*Sin[c] - (24*I)*B*Cos[c]^2*Sin[c] -
(16*I)*A*Cos[c]*Sin[c]^2 - 16*B*Cos[c]*Sin[c]^2 - 4*A*Sin[c]^3 + (4*I)*B*Sin[c]^3 + (A - I*B)*Cot[c]*(4*Cos[3
*c] - (4*I)*Sin[3*c]))*Sin[c + d*x]^4)/((Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I*A +
B)*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(4*d*x*Cos[3*c] - (4*I)*d*x*Sin[3*c])*Sin[c + d*x]^4)/(d*(Cos[d*
x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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fricas [A]  time = 0.89, size = 228, normalized size = 1.45 \[ -\frac {2 \, {\left (12 \, {\left (3 \, A - 2 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (23 \, A - 19 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (27 \, A - 23 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (15 \, A - 13 i \, B\right )} a^{3} - 6 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(12*(3*A - 2*I*B)*a^3*e^(6*I*d*x + 6*I*c) - 3*(23*A - 19*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 2*(27*A - 23*I*B)
*a^3*e^(2*I*d*x + 2*I*c) - (15*A - 13*I*B)*a^3 - 6*((A - I*B)*a^3*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^3*e^(6*I
*d*x + 6*I*c) + 6*(A - I*B)*a^3*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^3)*log
(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^
(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 4.28, size = 322, normalized size = 2.05 \[ -\frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 456 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 408 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1536 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 768 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {1600 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1600 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 456 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 408 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^3*tan(1/2*d*x + 1/2*c)^3 -
108*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 456*I*A*a^3*tan(1/2*d*x + 1/2*c) + 408*
B*a^3*tan(1/2*d*x + 1/2*c) + 1536*(A*a^3 - I*B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) - 768*(A*a^3 - I*B*a^3)*log(
tan(1/2*d*x + 1/2*c)) + (1600*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 1600*I*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 456*I*A*a^3
*tan(1/2*d*x + 1/2*c)^3 - 408*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 108*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*I*B*a^3*tan
(1/2*d*x + 1/2*c)^2 + 24*I*A*a^3*tan(1/2*d*x + 1/2*c) + 8*B*a^3*tan(1/2*d*x + 1/2*c) + 3*A*a^3)/tan(1/2*d*x +
1/2*c)^4)/d

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maple [A]  time = 0.47, size = 189, normalized size = 1.20 \[ \frac {4 i A \,a^{3} c}{d}-\frac {4 i a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {4 i A \cot \left (d x +c \right ) a^{3}}{d}+4 i A x \,a^{3}+\frac {2 A \,a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {4 a^{3} A \ln \left (\sin \left (d x +c \right )\right )}{d}+4 a^{3} B x +\frac {4 B \cot \left (d x +c \right ) a^{3}}{d}+\frac {4 a^{3} B c}{d}-\frac {3 i a^{3} B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {i A \,a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{d}-\frac {A \,a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{3} B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

4*I/d*A*a^3*c-4*I/d*a^3*B*ln(sin(d*x+c))+4*I/d*A*cot(d*x+c)*a^3+4*I*A*x*a^3+2/d*A*a^3*cot(d*x+c)^2+4*a^3*A*ln(
sin(d*x+c))/d+4*a^3*B*x+4/d*B*cot(d*x+c)*a^3+4/d*a^3*B*c-3/2*I/d*a^3*B*cot(d*x+c)^2-I/d*A*a^3*cot(d*x+c)^3-1/4
/d*A*a^3*cot(d*x+c)^4-1/3/d*a^3*B*cot(d*x+c)^3

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maxima [A]  time = 0.91, size = 134, normalized size = 0.85 \[ \frac {12 \, {\left (d x + c\right )} {\left (4 i \, A + 4 \, B\right )} a^{3} - 24 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 48 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) - \frac {{\left (-48 i \, A - 48 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} - 6 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + {\left (12 i \, A + 4 \, B\right )} a^{3} \tan \left (d x + c\right ) + 3 \, A a^{3}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*(4*I*A + 4*B)*a^3 - 24*(A - I*B)*a^3*log(tan(d*x + c)^2 + 1) + 48*(A - I*B)*a^3*log(tan(d*x
+ c)) - ((-48*I*A - 48*B)*a^3*tan(d*x + c)^3 - 6*(4*A - 3*I*B)*a^3*tan(d*x + c)^2 + (12*I*A + 4*B)*a^3*tan(d*
x + c) + 3*A*a^3)/tan(d*x + c)^4)/d

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mupad [B]  time = 6.57, size = 114, normalized size = 0.73 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,A\,a^3-\frac {B\,a^3\,3{}\mathrm {i}}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,B\,a^3+A\,a^3\,4{}\mathrm {i}\right )-\frac {A\,a^3}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{3}+A\,a^3\,1{}\mathrm {i}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {8\,a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(tan(c + d*x)^2*(2*A*a^3 - (B*a^3*3i)/2) + tan(c + d*x)^3*(A*a^3*4i + 4*B*a^3) - (A*a^3)/4 - tan(c + d*x)*(A*a
^3*1i + (B*a^3)/3))/(d*tan(c + d*x)^4) + (8*a^3*atan(2*tan(c + d*x) + 1i)*(A*1i + B))/d

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sympy [A]  time = 6.49, size = 235, normalized size = 1.50 \[ \frac {4 a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 30 A a^{3} + 26 i B a^{3} + \left (108 A a^{3} e^{2 i c} - 92 i B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (- 138 A a^{3} e^{4 i c} + 114 i B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (72 A a^{3} e^{6 i c} - 48 i B a^{3} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

4*a**3*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-30*A*a**3 + 26*I*B*a**3 + (108*A*a**3*exp(2*I*c) - 92*I
*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (-138*A*a**3*exp(4*I*c) + 114*I*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (72*A*a**
3*exp(6*I*c) - 48*I*B*a**3*exp(6*I*c))*exp(6*I*d*x))/(-3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp(6*I*d
*x) - 18*d*exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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