3.57 \(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=183 \[ -\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(-B+3 i A) \log (\sin (c+d x))}{a^3 d}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (25 A+7 i B)}{8 a^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

[Out]

-1/8*(25*A+7*I*B)*x/a^3-1/8*(25*A+7*I*B)*cot(d*x+c)/a^3/d-(3*I*A-B)*ln(sin(d*x+c))/a^3/d+1/6*(A+I*B)*cot(d*x+c
)/d/(a+I*a*tan(d*x+c))^3+1/24*(11*A+5*I*B)*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^2+1/2*(3*A+I*B)*cot(d*x+c)/d/(a^3
+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.53, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(-B+3 i A) \log (\sin (c+d x))}{a^3 d}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (25 A+7 i B)}{8 a^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-((25*A + (7*I)*B)*x)/(8*a^3) - ((25*A + (7*I)*B)*Cot[c + d*x])/(8*a^3*d) - (((3*I)*A - B)*Log[Sin[c + d*x]])/
(a^3*d) + ((A + I*B)*Cot[c + d*x])/(6*d*(a + I*a*Tan[c + d*x])^3) + ((11*A + (5*I)*B)*Cot[c + d*x])/(24*a*d*(a
+ I*a*Tan[c + d*x])^2) + ((3*A + I*B)*Cot[c + d*x])/(2*d*(a^3 + I*a^3*Tan[c + d*x]))

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 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^2(c+d x) (a (7 A+i B)-4 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^2(c+d x) \left (3 a^2 (13 A+3 i B)-3 a^2 (11 i A-5 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (6 a^3 (25 A+7 i B)-48 a^3 (3 i A-B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-48 a^3 (3 i A-B)-6 a^3 (25 A+7 i B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=-\frac {(25 A+7 i B) x}{8 a^3}-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(3 i A-B) \int \cot (c+d x) \, dx}{a^3}\\ &=-\frac {(25 A+7 i B) x}{8 a^3}-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(3 i A-B) \log (\sin (c+d x))}{a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 7.39, size = 1282, normalized size = 7.01 \[ \frac {\csc \left (\frac {c}{2}\right ) \csc (c+d x) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (\frac {1}{2} i A \cos (3 c-d x)-\frac {1}{2} i A \cos (3 c+d x)-\frac {1}{2} A \sin (3 c-d x)+\frac {1}{2} A \sin (3 c+d x)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{2 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(5 B-7 i A) \cos (4 d x) \sec ^2(c+d x) \left (\frac {\cos (c)}{32}-\frac {1}{32} i \sin (c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(11 B-23 i A) \cos (2 d x) \sec ^2(c+d x) \left (\frac {\cos (c)}{16}+\frac {1}{16} i \sin (c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {\sec ^2(c+d x) \left (-3 i A \cos \left (\frac {3 c}{2}\right )+B \cos \left (\frac {3 c}{2}\right )+3 A \sin \left (\frac {3 c}{2}\right )+i B \sin \left (\frac {3 c}{2}\right )\right ) \left (\tan ^{-1}(\tan (d x)) \sin \left (\frac {3 c}{2}\right )-i \tan ^{-1}(\tan (d x)) \cos \left (\frac {3 c}{2}\right )\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {\sec ^2(c+d x) \left (-3 i A \cos \left (\frac {3 c}{2}\right )+B \cos \left (\frac {3 c}{2}\right )+3 A \sin \left (\frac {3 c}{2}\right )+i B \sin \left (\frac {3 c}{2}\right )\right ) \left (\frac {1}{2} \cos \left (\frac {3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )+\frac {1}{2} i \sin \left (\frac {3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {x \sec ^2(c+d x) (-6 A \cos (c)-2 i B \cos (c)+3 i A \cot (c) \cos (c)-B \cot (c) \cos (c)-3 i A \sin (c)+B \sin (c)+(B-3 i A) \cot (c) (\cos (3 c)+i \sin (3 c))) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(B-i A) \cos (6 d x) \sec ^2(c+d x) \left (\frac {1}{48} \cos (3 c)-\frac {1}{48} i \sin (3 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(25 A+7 i B) \sec ^2(c+d x) \left (-\frac {1}{8} d x \cos (3 c)-\frac {1}{8} i d x \sin (3 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(23 A+11 i B) \sec ^2(c+d x) \left (-\frac {\cos (c)}{16}-\frac {1}{16} i \sin (c)\right ) \sin (2 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(7 A+5 i B) \sec ^2(c+d x) \left (\frac {1}{32} i \sin (c)-\frac {\cos (c)}{32}\right ) \sin (4 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3}+\frac {(A+i B) \sec ^2(c+d x) \left (\frac {1}{48} i \sin (3 c)-\frac {1}{48} \cos (3 c)\right ) \sin (6 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-7*I)*A + 5*B)*Cos[4*d*x]*Sec[c + d*x]^2*(Cos[c]/32 - (I/32)*Sin[c])*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c
+ d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (((-23*I)*A + 11*B)*Cos[2*d*x]*Sec[
c + d*x]^2*(Cos[c]/16 + (I/16)*Sin[c])*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*
Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (Sec[c + d*x]^2*((-3*I)*A*Cos[(3*c)/2] + B*Cos[(3*c)/2] + 3*A*Sin[(3
*c)/2] + I*B*Sin[(3*c)/2])*((-I)*ArcTan[Tan[d*x]]*Cos[(3*c)/2] + ArcTan[Tan[d*x]]*Sin[(3*c)/2])*(Cos[d*x] + I*
Sin[d*x])^3*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (Sec[c + d*
x]^2*((-3*I)*A*Cos[(3*c)/2] + B*Cos[(3*c)/2] + 3*A*Sin[(3*c)/2] + I*B*Sin[(3*c)/2])*((Cos[(3*c)/2]*Log[Sin[c +
d*x]^2])/2 + (I/2)*Log[Sin[c + d*x]^2]*Sin[(3*c)/2])*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c + d*x]))/(d*(A*Co
s[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (x*Sec[c + d*x]^2*(-6*A*Cos[c] - (2*I)*B*Cos[c] + (3*
I)*A*Cos[c]*Cot[c] - B*Cos[c]*Cot[c] - (3*I)*A*Sin[c] + B*Sin[c] + ((-3*I)*A + B)*Cot[c]*(Cos[3*c] + I*Sin[3*c
]))*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^
3) + (((-I)*A + B)*Cos[6*d*x]*Sec[c + d*x]^2*(Cos[3*c]/48 - (I/48)*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^3*(A + B*
Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + ((25*A + (7*I)*B)*Sec[c + d*x]
^2*(-1/8*(d*x*Cos[3*c]) - (I/8)*d*x*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*
x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + ((23*A + (11*I)*B)*Sec[c + d*x]^2*(-1/16*Cos[c] - (I/16)*Sin[
c])*(Cos[d*x] + I*Sin[d*x])^3*Sin[2*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*T
an[c + d*x])^3) + ((7*A + (5*I)*B)*Sec[c + d*x]^2*(-1/32*Cos[c] + (I/32)*Sin[c])*(Cos[d*x] + I*Sin[d*x])^3*Sin
[4*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + ((A + I*B)*Sec[
c + d*x]^2*(-1/48*Cos[3*c] + (I/48)*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^3*Sin[6*d*x]*(A + B*Tan[c + d*x]))/(d*(A
*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (Csc[c/2]*Csc[c + d*x]*Sec[c/2]*Sec[c + d*x]^2*(Co
s[d*x] + I*Sin[d*x])^3*((I/2)*A*Cos[3*c - d*x] - (I/2)*A*Cos[3*c + d*x] - (A*Sin[3*c - d*x])/2 + (A*Sin[3*c +
d*x])/2)*(A + B*Tan[c + d*x]))/(2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3)

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fricas [A]  time = 0.51, size = 175, normalized size = 0.96 \[ -\frac {12 \, {\left (49 \, A + 15 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (12 \, {\left (49 \, A + 15 i \, B\right )} d x - 330 i \, A + 66 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (117 i \, A - 51 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (19 i \, A - 13 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left ({\left (-288 i \, A + 96 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (288 i \, A - 96 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 2 i \, A + 2 \, B}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(12*(49*A + 15*I*B)*d*x*e^(8*I*d*x + 8*I*c) - (12*(49*A + 15*I*B)*d*x - 330*I*A + 66*B)*e^(6*I*d*x + 6*I
*c) - (117*I*A - 51*B)*e^(4*I*d*x + 4*I*c) - (19*I*A - 13*B)*e^(2*I*d*x + 2*I*c) - ((-288*I*A + 96*B)*e^(8*I*d
*x + 8*I*c) + (288*I*A - 96*B)*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 2*I*A + 2*B)/(a^3*d*e^(8*I*
d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))

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giac [A]  time = 2.28, size = 186, normalized size = 1.02 \[ -\frac {\frac {6 \, {\left (-49 i \, A + 15 \, B\right )} \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {6 \, {\left (i \, A + B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {96 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac {96 \, {\left (-3 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) + A\right )}}{a^{3} \tan \left (d x + c\right )} + \frac {539 \, A \tan \left (d x + c\right )^{3} + 165 i \, B \tan \left (d x + c\right )^{3} - 1821 i \, A \tan \left (d x + c\right )^{2} + 579 \, B \tan \left (d x + c\right )^{2} - 2085 \, A \tan \left (d x + c\right ) - 699 i \, B \tan \left (d x + c\right ) + 819 i \, A - 301 \, B}{a^{3} {\left (i \, \tan \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*(-49*I*A + 15*B)*log(I*tan(d*x + c) + 1)/a^3 + 6*(I*A + B)*log(I*tan(d*x + c) - 1)/a^3 + 96*(3*I*A -
B)*log(tan(d*x + c))/a^3 + 96*(-3*I*A*tan(d*x + c) + B*tan(d*x + c) + A)/(a^3*tan(d*x + c)) + (539*A*tan(d*x +
c)^3 + 165*I*B*tan(d*x + c)^3 - 1821*I*A*tan(d*x + c)^2 + 579*B*tan(d*x + c)^2 - 2085*A*tan(d*x + c) - 699*I*
B*tan(d*x + c) + 819*I*A - 301*B)/(a^3*(I*tan(d*x + c) + 1)^3))/d

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maple [A]  time = 0.69, size = 252, normalized size = 1.38 \[ -\frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}-\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}-\frac {3 i A \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {A}{a^{3} d \tan \left (d x +c \right )}+\frac {5 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {17 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {49 i \ln \left (\tan \left (d x +c \right )-i\right ) A}{16 d \,a^{3}}-\frac {15 \ln \left (\tan \left (d x +c \right )-i\right ) B}{16 d \,a^{3}}+\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

-1/16/d/a^3*B*ln(tan(d*x+c)+I)-1/16*I/d/a^3*A*ln(tan(d*x+c)+I)-3*I/a^3/d*A*ln(tan(d*x+c))+1/a^3/d*B*ln(tan(d*x
+c))-1/a^3/d*A/tan(d*x+c)+5/8*I/a^3/d/(tan(d*x+c)-I)^2*A-3/8/d/a^3/(tan(d*x+c)-I)^2*B-17/8/d/a^3/(tan(d*x+c)-I
)*A-7/8*I/d/a^3/(tan(d*x+c)-I)*B+49/16*I/a^3/d*ln(tan(d*x+c)-I)*A-15/16/d/a^3*ln(tan(d*x+c)-I)*B+1/6/d/a^3/(ta
n(d*x+c)-I)^3*A+1/6*I/d/a^3/(tan(d*x+c)-I)^3*B

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.88, size = 197, normalized size = 1.08 \[ -\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {25\,A}{8\,a^3}+\frac {B\,7{}\mathrm {i}}{8\,a^3}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {17\,B}{8\,a^3}+\frac {A\,63{}\mathrm {i}}{8\,a^3}\right )+\frac {A\,1{}\mathrm {i}}{a^3}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {71\,A}{12\,a^3}+\frac {B\,17{}\mathrm {i}}{12\,a^3}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,3{}\mathrm {i}\right )}{a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-15\,B+A\,49{}\mathrm {i}\right )}{16\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)^2*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(log(tan(c + d*x) - 1i)*(A*49i - 15*B))/(16*a^3*d) - (log(tan(c + d*x))*(A*3i - B))/(a^3*d) - (log(tan(c + d*x
) + 1i)*(A*1i + B))/(16*a^3*d) - (tan(c + d*x)^3*((25*A)/(8*a^3) + (B*7i)/(8*a^3)) - tan(c + d*x)^2*((A*63i)/(
8*a^3) - (17*B)/(8*a^3)) + (A*1i)/a^3 - tan(c + d*x)*((71*A)/(12*a^3) + (B*17i)/(12*a^3)))/(d*(tan(c + d*x)*1i
- 3*tan(c + d*x)^2 - tan(c + d*x)^3*3i + tan(c + d*x)^4))

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sympy [A]  time = 1.22, size = 342, normalized size = 1.87 \[ - \frac {2 i A}{a^{3} d e^{2 i c} e^{2 i d x} - a^{3} d} + \begin {cases} - \frac {\left (\left (512 i A a^{6} d^{2} e^{6 i c} - 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (5376 i A a^{6} d^{2} e^{8 i c} - 3840 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (35328 i A a^{6} d^{2} e^{10 i c} - 16896 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- 49 A - 15 i B}{8 a^{3}} + \frac {i \left (49 i A e^{6 i c} + 23 i A e^{4 i c} + 7 i A e^{2 i c} + i A - 15 B e^{6 i c} - 11 B e^{4 i c} - 5 B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (49 A + 15 i B\right )}{8 a^{3}} - \frac {i \left (3 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

-2*I*A/(a**3*d*exp(2*I*c)*exp(2*I*d*x) - a**3*d) + Piecewise((-((512*I*A*a**6*d**2*exp(6*I*c) - 512*B*a**6*d**
2*exp(6*I*c))*exp(-6*I*d*x) + (5376*I*A*a**6*d**2*exp(8*I*c) - 3840*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (3
5328*I*A*a**6*d**2*exp(10*I*c) - 16896*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3),
Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*(-(-49*A - 15*I*B)/(8*a**3) + I*(49*I*A*exp(6*I*c) + 23*I*A*exp(4*I*c
) + 7*I*A*exp(2*I*c) + I*A - 15*B*exp(6*I*c) - 11*B*exp(4*I*c) - 5*B*exp(2*I*c) - B)*exp(-6*I*c)/(8*a**3)), Tr
ue)) - x*(49*A + 15*I*B)/(8*a**3) - I*(3*A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)

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