3.51 \(\int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=191 \[ -\frac{(-3 B+i A) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(7 A+25 i B) \tan (c+d x)}{8 a^3 d}-\frac{(-3 B+i A) \log (\cos (c+d x))}{a^3 d}-\frac{x (7 A+25 i B)}{8 a^3}+\frac{(-B+i A) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \]

[Out]

-((7*A + (25*I)*B)*x)/(8*a^3) - ((I*A - 3*B)*Log[Cos[c + d*x]])/(a^3*d) + ((7*A + (25*I)*B)*Tan[c + d*x])/(8*a
^3*d) + ((I*A - B)*Tan[c + d*x]^4)/(6*d*(a + I*a*Tan[c + d*x])^3) + ((5*A + (11*I)*B)*Tan[c + d*x]^3)/(24*a*d*
(a + I*a*Tan[c + d*x])^2) - ((I*A - 3*B)*Tan[c + d*x]^2)/(2*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.473484, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3595, 3525, 3475} \[ -\frac{(-3 B+i A) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(7 A+25 i B) \tan (c+d x)}{8 a^3 d}-\frac{(-3 B+i A) \log (\cos (c+d x))}{a^3 d}-\frac{x (7 A+25 i B)}{8 a^3}+\frac{(-B+i A) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A + (25*I)*B)*x)/(8*a^3) - ((I*A - 3*B)*Log[Cos[c + d*x]])/(a^3*d) + ((7*A + (25*I)*B)*Tan[c + d*x])/(8*a
^3*d) + ((I*A - B)*Tan[c + d*x]^4)/(6*d*(a + I*a*Tan[c + d*x])^3) + ((5*A + (11*I)*B)*Tan[c + d*x]^3)/(24*a*d*
(a + I*a*Tan[c + d*x])^2) - ((I*A - 3*B)*Tan[c + d*x]^2)/(2*d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 3595

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

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && 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]

Rubi steps

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

Mathematica [B]  time = 7.01278, size = 1251, normalized size = 6.55 \[ \frac{\sec ^3(c+d x) (-B \cos (3 c-d x)+B \cos (3 c+d x)-i B \sin (3 c-d x)+i B \sin (3 c+d x)) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^3}{2 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) (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) \left (-\frac{1}{2} A \cos ^3(c)-\frac{3}{2} i B \cos ^3(c)-2 i A \sin (c) \cos ^2(c)+6 B \sin (c) \cos ^2(c)+3 A \sin ^2(c) \cos (c)+9 i B \sin ^2(c) \cos (c)+\frac{1}{2} A \cos (c)+\frac{3}{2} i B \cos (c)+2 i A \sin ^3(c)-6 B \sin ^3(c)+i A \sin (c)-3 B \sin (c)-\frac{1}{2} A \sin ^3(c) \tan (c)-\frac{3}{2} i B \sin ^3(c) \tan (c)-\frac{1}{2} A \sin (c) \tan (c)-\frac{3}{2} i B \sin (c) \tan (c)+i (A+3 i B) (\cos (3 c)+i \sin (3 c)) \tan (c)\right ) (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{(11 A+23 i B) \cos (2 d x) \sec ^2(c+d x) \left (\frac{1}{16} i \cos (c)-\frac{\sin (c)}{16}\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{(7 B-5 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{\sec ^2(c+d x) \left (-i A \cos \left (\frac{3 c}{2}\right )+3 B \cos \left (\frac{3 c}{2}\right )+A \sin \left (\frac{3 c}{2}\right )+3 i B \sin \left (\frac{3 c}{2}\right )\right ) \left (\cos \left (\frac{3 c}{2}\right ) \log (\cos (c+d x))+i \sin \left (\frac{3 c}{2}\right ) \log (\cos (c+d x))\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{(A+i B) \cos (6 d x) \sec ^2(c+d x) \left (\frac{1}{48} i \cos (3 c)+\frac{1}{48} \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{(7 A+25 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{(11 A+23 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{(5 A+7 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} \cos (3 c)-\frac{1}{48} i \sin (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[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((11*A + (23*I)*B)*Cos[2*d*x]*Sec[c + d*x]^2*((I/16)*Cos[c] - Sin[c]/16)*(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) + (((-5*I)*A + 7*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*S
in[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (Sec[c + d*x]^2*((-I)*A*Cos[(3*c)/2] + 3*B*Cos[(3*c)/2] + A*Sin[(3*c)
/2] + (3*I)*B*Sin[(3*c)/2])*(Cos[(3*c)/2]*Log[Cos[c + d*x]] + I*Log[Cos[c + 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) + ((A + I*B)*
Cos[6*d*x]*Sec[c + d*x]^2*((I/48)*Cos[3*c] + Sin[3*c]/48)*(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) + ((7*A + (25*I)*B)*Sec[c + d*x]^2*(-(d*x*Cos[3*c])
/8 - (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) + ((11*A + (23*I)*B)*Sec[c + d*x]^2*(Cos[c]/16 + (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*Tan[c + d*x])^3) + ((5*A
+ (7*I)*B)*Sec[c + d*x]^2*(-Cos[c]/32 + (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*(Cos[3*c]/48 -
(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) + (Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^3*(-(B*Cos[3*c - d*x]) + B*Cos[3*c + d*
x] - I*B*Sin[3*c - d*x] + I*B*Sin[3*c + d*x])*(A + B*Tan[c + d*x]))/(2*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin
[c/2])*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3) + (x*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])
^3*((A*Cos[c])/2 + ((3*I)/2)*B*Cos[c] - (A*Cos[c]^3)/2 - ((3*I)/2)*B*Cos[c]^3 + I*A*Sin[c] - 3*B*Sin[c] - (2*I
)*A*Cos[c]^2*Sin[c] + 6*B*Cos[c]^2*Sin[c] + 3*A*Cos[c]*Sin[c]^2 + (9*I)*B*Cos[c]*Sin[c]^2 + (2*I)*A*Sin[c]^3 -
6*B*Sin[c]^3 - (A*Sin[c]*Tan[c])/2 - ((3*I)/2)*B*Sin[c]*Tan[c] - (A*Sin[c]^3*Tan[c])/2 - ((3*I)/2)*B*Sin[c]^3
*Tan[c] + I*(A + (3*I)*B)*(Cos[3*c] + I*Sin[3*c])*Tan[c])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d
*x])*(a + I*a*Tan[c + d*x])^3)

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Maple [A]  time = 0.036, size = 219, normalized size = 1.2 \begin{align*}{\frac{iB\tan \left ( dx+c \right ) }{{a}^{3}d}}-{\frac{A}{6\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{6}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{49\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{16\,{a}^{3}d}}+{\frac{{\frac{15\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{{a}^{3}d}}+{\frac{17\,A}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{31\,i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{7\,i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{9\,B}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{16\,{a}^{3}d}}+{\frac{{\frac{i}{16}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/d/a^3*B*tan(d*x+c)-1/6/d/a^3/(tan(d*x+c)-I)^3*A-1/6*I/d/a^3/(tan(d*x+c)-I)^3*B-49/16/d/a^3*ln(tan(d*x+c)-I)*
B+15/16*I/d/a^3*ln(tan(d*x+c)-I)*A+17/8/d/a^3/(tan(d*x+c)-I)*A+31/8*I/d/a^3/(tan(d*x+c)-I)*B+7/8*I/d/a^3/(tan(
d*x+c)-I)^2*A-9/8/d/a^3/(tan(d*x+c)-I)^2*B+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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.46397, size = 506, normalized size = 2.65 \begin{align*} -\frac{12 \,{\left (15 \, A + 49 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (12 \,{\left (15 \, A + 49 i \, B\right )} d x - 66 i \, A + 330 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} -{\left (51 i \, A - 117 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (-13 i \, A + 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} -{\left ({\left (-96 i \, A + 288 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (-96 i \, A + 288 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(12*(15*A + 49*I*B)*d*x*e^(8*I*d*x + 8*I*c) + (12*(15*A + 49*I*B)*d*x - 66*I*A + 330*B)*e^(6*I*d*x + 6*I
*c) - (51*I*A - 117*B)*e^(4*I*d*x + 4*I*c) - (-13*I*A + 19*B)*e^(2*I*d*x + 2*I*c) - ((-96*I*A + 288*B)*e^(8*I*
d*x + 8*I*c) + (-96*I*A + 288*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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Sympy [A]  time = 35.9561, size = 292, normalized size = 1.53 \begin{align*} - \frac{2 B e^{- 2 i c}}{a^{3} d \left (e^{2 i d x} + e^{- 2 i c}\right )} - \frac{\left (\begin{cases} 15 A x e^{6 i c} - \frac{11 i A e^{4 i c} e^{- 2 i d x}}{2 d} + \frac{5 i A e^{2 i c} e^{- 4 i d x}}{4 d} - \frac{i A e^{- 6 i d x}}{6 d} + 49 i B x e^{6 i c} + \frac{23 B e^{4 i c} e^{- 2 i d x}}{2 d} - \frac{7 B e^{2 i c} e^{- 4 i d x}}{4 d} + \frac{B e^{- 6 i d x}}{6 d} & \text{for}\: d \neq 0 \\x \left (15 A e^{6 i c} - 11 A e^{4 i c} + 5 A e^{2 i c} - A + 49 i B e^{6 i c} - 23 i B e^{4 i c} + 7 i B e^{2 i c} - i B\right ) & \text{otherwise} \end{cases}\right ) e^{- 6 i c}}{8 a^{3}} + \frac{\left (- i A + 3 B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*B*exp(-2*I*c)/(a**3*d*(exp(2*I*d*x) + exp(-2*I*c))) - Piecewise((15*A*x*exp(6*I*c) - 11*I*A*exp(4*I*c)*exp(
-2*I*d*x)/(2*d) + 5*I*A*exp(2*I*c)*exp(-4*I*d*x)/(4*d) - I*A*exp(-6*I*d*x)/(6*d) + 49*I*B*x*exp(6*I*c) + 23*B*
exp(4*I*c)*exp(-2*I*d*x)/(2*d) - 7*B*exp(2*I*c)*exp(-4*I*d*x)/(4*d) + B*exp(-6*I*d*x)/(6*d), Ne(d, 0)), (x*(15
*A*exp(6*I*c) - 11*A*exp(4*I*c) + 5*A*exp(2*I*c) - A + 49*I*B*exp(6*I*c) - 23*I*B*exp(4*I*c) + 7*I*B*exp(2*I*c
) - I*B), True))*exp(-6*I*c)/(8*a**3) + (-I*A + 3*B)*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)

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Giac [A]  time = 2.37201, size = 194, normalized size = 1.02 \begin{align*} \frac{\frac{6 \,{\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac{6 \,{\left (-15 i \, A + 49 \, B\right )} \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{96 i \, B \tan \left (d x + c\right )}{a^{3}} - \frac{165 i \, A \tan \left (d x + c\right )^{3} - 539 \, B \tan \left (d x + c\right )^{3} + 291 \, A \tan \left (d x + c\right )^{2} + 1245 i \, B \tan \left (d x + c\right )^{2} - 171 i \, A \tan \left (d x + c\right ) + 981 \, B \tan \left (d x + c\right ) - 29 \, A - 259 i \, B}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(6*(I*A + B)*log(tan(d*x + c) + I)/a^3 - 6*(-15*I*A + 49*B)*log(I*tan(d*x + c) + 1)/a^3 + 96*I*B*tan(d*x
+ c)/a^3 - (165*I*A*tan(d*x + c)^3 - 539*B*tan(d*x + c)^3 + 291*A*tan(d*x + c)^2 + 1245*I*B*tan(d*x + c)^2 - 1
71*I*A*tan(d*x + c) + 981*B*tan(d*x + c) - 29*A - 259*I*B)/(a^3*(tan(d*x + c) - I)^3))/d