3.16 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=139 \[ -\frac {a^2 (4 B+5 i A) \cot ^3(c+d x)}{12 d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}+\frac {2 a^2 (B+i A) \cot (c+d x)}{d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}+2 a^2 x (B+i A)-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[Out]

2*a^2*(I*A+B)*x+2*a^2*(I*A+B)*cot(d*x+c)/d+a^2*(A-I*B)*cot(d*x+c)^2/d-1/12*a^2*(5*I*A+4*B)*cot(d*x+c)^3/d+2*a^
2*(A-I*B)*ln(sin(d*x+c))/d-1/4*A*cot(d*x+c)^4*(a^2+I*a^2*tan(d*x+c))/d

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Rubi [A]  time = 0.29, antiderivative size = 139, 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} \[ -\frac {a^2 (4 B+5 i A) \cot ^3(c+d x)}{12 d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}+\frac {2 a^2 (B+i A) \cot (c+d x)}{d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}+2 a^2 x (B+i A)-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

2*a^2*(I*A + B)*x + (2*a^2*(I*A + B)*Cot[c + d*x])/d + (a^2*(A - I*B)*Cot[c + d*x]^2)/d - (a^2*((5*I)*A + 4*B)
*Cot[c + d*x]^3)/(12*d) + (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])
)/(4*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))^2 (A+B \tan (c+d x)) \, dx &=-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) (a (5 i A+4 B)-a (3 A-4 i B) \tan (c+d x)) \, dx\\ &=-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^3(c+d x) \left (-8 a^2 (A-i B)-8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^2(c+d x) \left (-8 a^2 (i A+B)+8 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot (c+d x) \left (8 a^2 (A-i B)+8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=2 a^2 (i A+B) x+\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\left (2 a^2 (A-i B)\right ) \int \cot (c+d x) \, dx\\ &=2 a^2 (i A+B) x+\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}\\ \end {align*}

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Mathematica [B]  time = 9.15, size = 902, normalized size = 6.49 \[ a^2 \left (\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (A \cos (c)-i B \cos (c)-i A \sin (c)-B \sin (c)) \left (-2 i \tan ^{-1}(\tan (3 c+d x)) \cos (c)-2 \tan ^{-1}(\tan (3 c+d x)) \sin (c)\right ) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (A \cos (c)-i B \cos (c)-i A \sin (c)-B \sin (c)) \left (\cos (c) \log \left (\sin ^2(c+d x)\right )-i \log \left (\sin ^2(c+d x)\right ) \sin (c)\right ) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {x (\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \left (6 i A \cos ^2(c)+6 B \cos ^2(c)-2 A \cot (c) \cos ^2(c)+2 i B \cot (c) \cos ^2(c)+6 A \sin (c) \cos (c)-6 i B \sin (c) \cos (c)-2 i A \sin ^2(c)-2 B \sin ^2(c)+(A-i B) \cot (c) (2 \cos (2 c)-2 i \sin (2 c))\right ) \sin ^3(c+d x)}{(\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(i A+B) (\cot (c+d x)+i)^2 (B+A \cot (c+d x)) (2 d x \cos (2 c)-2 i d x \sin (2 c)) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) \left (\frac {1}{3} \cos (2 c)-\frac {1}{3} i \sin (2 c)\right ) (-8 i A \sin (d x)-7 B \sin (d x)) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) (-4 i A \cos (c)-2 B \cos (c)+9 A \sin (c)-6 i B \sin (c)) \left (\frac {1}{6} \cos (2 c)-\frac {1}{6} i \sin (2 c)\right ) \sin (c+d x)}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c) \left (\frac {1}{3} \cos (2 c)-\frac {1}{3} i \sin (2 c)\right ) (2 i A \sin (d x)+B \sin (d x))}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^2 (B+A \cot (c+d x)) \csc (c+d x) \left (\frac {1}{4} i A \sin (2 c)-\frac {1}{4} A \cos (2 c)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

a^2*(((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c + d*x]*(-1/4*(A*Cos[2*c]) + (I/4)*A*Sin[2*c]))/(d*(Cos[d
*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*(Co
s[2*c]/3 - (I/3)*Sin[2*c])*((2*I)*A*Sin[d*x] + B*Sin[d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*S
in[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*((-4*I)*A*Cos[c] - 2*B*Cos[c] + 9*A*Sin[c] -
(6*I)*B*Sin[c])*(Cos[2*c]/6 - (I/6)*Sin[2*c])*Sin[c + d*x])/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*
Sin[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Csc[c]*(Cos[2*c]/3 - (I/3)*Sin[2*c])*((-8*I)*A*Sin
[d*x] - 7*B*Sin[d*x])*Sin[c + d*x]^2)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I +
Cot[c + d*x])^2*(B + A*Cot[c + d*x])*(A*Cos[c] - I*B*Cos[c] - I*A*Sin[c] - B*Sin[c])*((-2*I)*ArcTan[Tan[3*c +
d*x]]*Cos[c] - 2*ArcTan[Tan[3*c + d*x]]*Sin[c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] +
B*Sin[c + d*x])) + ((I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*(A*Cos[c] - I*B*Cos[c] - I*A*Sin[c] - B*Sin[c])
*(Cos[c]*Log[Sin[c + d*x]^2] - I*Log[Sin[c + d*x]^2]*Sin[c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*C
os[c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*((6*I)*A*Cos[c]^2 + 6*B*Cos[c]^2
- 2*A*Cos[c]^2*Cot[c] + (2*I)*B*Cos[c]^2*Cot[c] + 6*A*Cos[c]*Sin[c] - (6*I)*B*Cos[c]*Sin[c] - (2*I)*A*Sin[c]^
2 - 2*B*Sin[c]^2 + (A - I*B)*Cot[c]*(2*Cos[2*c] - (2*I)*Sin[2*c]))*Sin[c + d*x]^3)/((Cos[d*x] + I*Sin[d*x])^2*
(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I*A + B)*(I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*(2*d*x*Cos[2*c] - (2
*I)*d*x*Sin[2*c])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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fricas [A]  time = 0.46, size = 227, normalized size = 1.63 \[ -\frac {2 \, {\left (3 \, {\left (7 \, A - 5 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (12 \, A - 11 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (29 \, A - 25 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (8 \, A - 7 i \, B\right )} a^{2} - 3 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(3*(7*A - 5*I*B)*a^2*e^(6*I*d*x + 6*I*c) - 3*(12*A - 11*I*B)*a^2*e^(4*I*d*x + 4*I*c) + (29*A - 25*I*B)*a^
2*e^(2*I*d*x + 2*I*c) - (8*A - 7*I*B)*a^2 - 3*((A - I*B)*a^2*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^2*e^(6*I*d*x
+ 6*I*c) + 6*(A - I*B)*a^2*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^2)*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 = 3.03, size = 322, normalized size = 2.32 \[ -\frac {3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 240 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 768 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 384 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {800 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 800 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 216 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*I*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^2*tan(1/2*d*x + 1/2*c)^3 -
60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 240*I*A*a^2*tan(1/2*d*x + 1/2*c) + 216*B
*a^2*tan(1/2*d*x + 1/2*c) + 768*(A*a^2 - I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) - 384*(A*a^2 - I*B*a^2)*log(ta
n(1/2*d*x + 1/2*c)) + (800*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 800*I*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 240*I*A*a^2*tan
(1/2*d*x + 1/2*c)^3 - 216*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2*tan(1/2*
d*x + 1/2*c)^2 + 16*I*A*a^2*tan(1/2*d*x + 1/2*c) + 8*B*a^2*tan(1/2*d*x + 1/2*c) + 3*A*a^2)/tan(1/2*d*x + 1/2*c
)^4)/d

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maple [A]  time = 0.48, size = 188, normalized size = 1.35 \[ \frac {a^{2} A \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a^{2} A \ln \left (\sin \left (d x +c \right )\right )}{d}+2 a^{2} B x +\frac {2 B \cot \left (d x +c \right ) a^{2}}{d}+\frac {2 B \,a^{2} c}{d}+\frac {2 i A \,a^{2} c}{d}-\frac {2 i A \,a^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 i B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {i B \,a^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{d}+2 i A \,a^{2} x +\frac {2 i A \cot \left (d x +c \right ) a^{2}}{d}-\frac {a^{2} A \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.76, size = 132, normalized size = 0.95 \[ \frac {12 \, {\left (d x + c\right )} {\left (2 i \, A + 2 \, B\right )} a^{2} - 12 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 24 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac {{\left (-24 i \, A - 24 \, B\right )} a^{2} \tan \left (d x + c\right )^{3} - 12 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + {\left (8 i \, A + 4 \, B\right )} a^{2} \tan \left (d x + c\right ) + 3 \, A a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*(2*I*A + 2*B)*a^2 - 12*(A - I*B)*a^2*log(tan(d*x + c)^2 + 1) + 24*(A - I*B)*a^2*log(tan(d*x
+ c)) - ((-24*I*A - 24*B)*a^2*tan(d*x + c)^3 - 12*(A - I*B)*a^2*tan(d*x + c)^2 + (8*I*A + 4*B)*a^2*tan(d*x +
c) + 3*A*a^2)/tan(d*x + c)^4)/d

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mupad [B]  time = 6.44, size = 113, normalized size = 0.81 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A\,a^2-B\,a^2\,1{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,B\,a^2+A\,a^2\,2{}\mathrm {i}\right )-\frac {A\,a^2}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{3}+\frac {A\,a^2\,2{}\mathrm {i}}{3}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {4\,a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

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)^2,x)

[Out]

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

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sympy [A]  time = 1.39, size = 235, normalized size = 1.69 \[ \frac {2 a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 16 A a^{2} + 14 i B a^{2} + \left (58 A a^{2} e^{2 i c} - 50 i B a^{2} e^{2 i c}\right ) e^{2 i d x} + \left (- 72 A a^{2} e^{4 i c} + 66 i B a^{2} e^{4 i c}\right ) e^{4 i d x} + \left (42 A a^{2} e^{6 i c} - 30 i B a^{2} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*a**2*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-16*A*a**2 + 14*I*B*a**2 + (58*A*a**2*exp(2*I*c) - 50*I*
B*a**2*exp(2*I*c))*exp(2*I*d*x) + (-72*A*a**2*exp(4*I*c) + 66*I*B*a**2*exp(4*I*c))*exp(4*I*d*x) + (42*A*a**2*e
xp(6*I*c) - 30*I*B*a**2*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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