\(\int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\) [59]

Optimal result

Integrand size = 34, antiderivative size = 185 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {(A+15 i B) x}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {i A-15 B}{16 a^4 d (1+i \tan (c+d x))}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3} \]

[Out]

1/16*(A+15*I*B)*x/a^4-B*ln(cos(d*x+c))/a^4/d+1/16*(-I*A+15*B)/a^4/d/(1+I*tan(d*x+c))-1/16*(I*A-7*B)*tan(d*x+c)
^2/a^4/d/(1+I*tan(d*x+c))^2+1/8*(I*A-B)*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^4+1/12*(A+3*I*B)*tan(d*x+c)^3/a/d/(a
+I*a*tan(d*x+c))^3

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3676, 3670, 3556, 12, 3607, 8} \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {(-7 B+i A) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {-15 B+i A}{16 a^4 d (1+i \tan (c+d x))}+\frac {x (A+15 i B)}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}+\frac {(-B+i A) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

((A + (15*I)*B)*x)/(16*a^4) - (B*Log[Cos[c + d*x]])/(a^4*d) - (I*A - 15*B)/(16*a^4*d*(1 + I*Tan[c + d*x])) - (
(I*A - 7*B)*Tan[c + d*x]^2)/(16*a^4*d*(1 + I*Tan[c + d*x])^2) + ((I*A - B)*Tan[c + d*x]^4)/(8*d*(a + I*a*Tan[c
+ d*x])^4) + ((A + (3*I)*B)*Tan[c + d*x]^3)/(12*a*d*(a + I*a*Tan[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3556

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

Rule 3607

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/(2*a*f*m)), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3670

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 3676

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac {\int \frac {\tan ^3(c+d x) (4 a (i A-B)+8 i a B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^2(c+d x) \left (-12 a^2 (A+3 i B)-48 a^2 B \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = -\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x) \left (-24 a^3 (i A-7 B)-192 i a^3 B \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = -\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {i \int \frac {24 a^4 (A+15 i B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{192 a^7}+\frac {B \int \tan (c+d x) \, dx}{a^4} \\ & = -\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {(i A-15 B) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {i A-15 B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {(A+15 i B) \int 1 \, dx}{16 a^4} \\ & = \frac {(A+15 i B) x}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {i A-15 B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (18 i A-48 B+8 (-4 i A+21 B) \cos (2 (c+d x))+2 \cos (4 (c+d x)) (7 i A-60 B+(-3 i A+93 B) \log (i-\tan (c+d x))+3 (i A+B) \log (i+\tan (c+d x)))+16 A \sin (2 (c+d x))+144 i B \sin (2 (c+d x))-11 A \sin (4 (c+d x))-117 i B \sin (4 (c+d x))+6 A \log (i-\tan (c+d x)) \sin (4 (c+d x))+186 i B \log (i-\tan (c+d x)) \sin (4 (c+d x))-6 A \log (i+\tan (c+d x)) \sin (4 (c+d x))+6 i B \log (i+\tan (c+d x)) \sin (4 (c+d x)))}{192 a^4 d (-i+\tan (c+d x))^4} \]

[In]

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

[Out]

(Sec[c + d*x]^4*((18*I)*A - 48*B + 8*((-4*I)*A + 21*B)*Cos[2*(c + d*x)] + 2*Cos[4*(c + d*x)]*((7*I)*A - 60*B +
((-3*I)*A + 93*B)*Log[I - Tan[c + d*x]] + 3*(I*A + B)*Log[I + Tan[c + d*x]]) + 16*A*Sin[2*(c + d*x)] + (144*I
)*B*Sin[2*(c + d*x)] - 11*A*Sin[4*(c + d*x)] - (117*I)*B*Sin[4*(c + d*x)] + 6*A*Log[I - Tan[c + d*x]]*Sin[4*(c
+ d*x)] + (186*I)*B*Log[I - Tan[c + d*x]]*Sin[4*(c + d*x)] - 6*A*Log[I + Tan[c + d*x]]*Sin[4*(c + d*x)] + (6*
I)*B*Log[I + Tan[c + d*x]]*Sin[4*(c + d*x)]))/(192*a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {31 i x B}{16 a^{4}}+\frac {x A}{16 a^{4}}+\frac {13 \,{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 d \,a^{4}}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{8 d \,a^{4}}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} B}{4 d \,a^{4}}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 d \,a^{4}}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{16 d \,a^{4}}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{24 d \,a^{4}}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )} B}{128 d \,a^{4}}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )} A}{128 d \,a^{4}}+\frac {2 i B c}{d \,a^{4}}-\frac {B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,a^{4}}\)	\(197\)
derivativedivides	\(-\frac {17 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {49 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {15 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {31 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {3 i B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {7 A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}\)	\(219\)
default	\(-\frac {17 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {49 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {15 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {31 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {3 i B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {7 A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}\)	\(219\)

[In]

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

[Out]

31/16*I*x/a^4*B+1/16*x/a^4*A+13/16/d/a^4*exp(-2*I*(d*x+c))*B-1/8*I/d/a^4*exp(-2*I*(d*x+c))*A-1/4/d/a^4*exp(-4*
I*(d*x+c))*B+3/32*I/d/a^4*exp(-4*I*(d*x+c))*A+1/16/d/a^4*exp(-6*I*(d*x+c))*B-1/24*I/d/a^4*exp(-6*I*(d*x+c))*A-
1/128/d/a^4*exp(-8*I*(d*x+c))*B+1/128*I/d/a^4*exp(-8*I*(d*x+c))*A+2*I*B/d/a^4*c-B/d/a^4*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 \, {\left (A + 31 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 384 \, B e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 24 \, {\left (2 i \, A - 13 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 12 \, {\left (-3 i \, A + 8 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 \, {\left (2 i \, A - 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

[In]

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

[Out]

1/384*(24*(A + 31*I*B)*d*x*e^(8*I*d*x + 8*I*c) - 384*B*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24*(
2*I*A - 13*B)*e^(6*I*d*x + 6*I*c) - 12*(-3*I*A + 8*B)*e^(4*I*d*x + 4*I*c) - 8*(2*I*A - 3*B)*e^(2*I*d*x + 2*I*c
) + 3*I*A - 3*B)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

Sympy [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.94 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=- \frac {B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} + \begin {cases} \frac {\left (\left (24576 i A a^{12} d^{3} e^{12 i c} - 24576 B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (- 131072 i A a^{12} d^{3} e^{14 i c} + 196608 B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (294912 i A a^{12} d^{3} e^{16 i c} - 786432 B a^{12} d^{3} e^{16 i c}\right ) e^{- 4 i d x} + \left (- 393216 i A a^{12} d^{3} e^{18 i c} + 2555904 B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {A + 31 i B}{16 a^{4}} + \frac {\left (A e^{8 i c} - 4 A e^{6 i c} + 6 A e^{4 i c} - 4 A e^{2 i c} + A + 31 i B e^{8 i c} - 26 i B e^{6 i c} + 16 i B e^{4 i c} - 6 i B e^{2 i c} + i B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A + 31 i B\right )}{16 a^{4}} \]

[In]

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

[Out]

-B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d) + Piecewise((((24576*I*A*a**12*d**3*exp(12*I*c) - 24576*B*a**12*d*
*3*exp(12*I*c))*exp(-8*I*d*x) + (-131072*I*A*a**12*d**3*exp(14*I*c) + 196608*B*a**12*d**3*exp(14*I*c))*exp(-6*
I*d*x) + (294912*I*A*a**12*d**3*exp(16*I*c) - 786432*B*a**12*d**3*exp(16*I*c))*exp(-4*I*d*x) + (-393216*I*A*a*
*12*d**3*exp(18*I*c) + 2555904*B*a**12*d**3*exp(18*I*c))*exp(-2*I*d*x))*exp(-20*I*c)/(3145728*a**16*d**4), Ne(
a**16*d**4*exp(20*I*c), 0)), (x*(-(A + 31*I*B)/(16*a**4) + (A*exp(8*I*c) - 4*A*exp(6*I*c) + 6*A*exp(4*I*c) - 4
*A*exp(2*I*c) + A + 31*I*B*exp(8*I*c) - 26*I*B*exp(6*I*c) + 16*I*B*exp(4*I*c) - 6*I*B*exp(2*I*c) + I*B)*exp(-8
*I*c)/(16*a**4)), True)) + x*(A + 31*I*B)/(16*a**4)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 1.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.83 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, {\left (-i \, A + 31 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {25 i \, A \tan \left (d x + c\right )^{4} - 775 \, B \tan \left (d x + c\right )^{4} - 260 \, A \tan \left (d x + c\right )^{3} + 1924 i \, B \tan \left (d x + c\right )^{3} + 522 i \, A \tan \left (d x + c\right )^{2} + 1866 \, B \tan \left (d x + c\right )^{2} + 388 \, A \tan \left (d x + c\right ) - 772 i \, B \tan \left (d x + c\right ) - 103 i \, A - 103 \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[In]

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

[Out]

-1/384*(12*(-I*A - B)*log(tan(d*x + c) + I)/a^4 - 12*(-I*A + 31*B)*log(tan(d*x + c) - I)/a^4 - (25*I*A*tan(d*x
+ c)^4 - 775*B*tan(d*x + c)^4 - 260*A*tan(d*x + c)^3 + 1924*I*B*tan(d*x + c)^3 + 522*I*A*tan(d*x + c)^2 + 186
6*B*tan(d*x + c)^2 + 388*A*tan(d*x + c) - 772*I*B*tan(d*x + c) - 103*I*A - 103*B)/(a^4*(tan(d*x + c) - I)^4))/
d

Mupad [B] (verification not implemented)

Time = 8.09 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {29\,B}{4\,a^4}+\frac {A\,7{}\mathrm {i}}{4\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {15\,A}{16\,a^4}+\frac {B\,49{}\mathrm {i}}{16\,a^4}\right )-\frac {A\,1{}\mathrm {i}}{3\,a^4}+\frac {7\,B}{4\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {61\,A}{48\,a^4}+\frac {B\,97{}\mathrm {i}}{16\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{32\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,31{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d} \]

[In]

int((tan(c + d*x)^4*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(tan(c + d*x)^2*((A*7i)/(4*a^4) - (29*B)/(4*a^4)) - tan(c + d*x)^3*((15*A)/(16*a^4) + (B*49i)/(16*a^4)) - (A*1
i)/(3*a^4) + (7*B)/(4*a^4) + tan(c + d*x)*((61*A)/(48*a^4) + (B*97i)/(16*a^4)))/(d*(tan(c + d*x)*4i - 6*tan(c
+ d*x)^2 - tan(c + d*x)^3*4i + tan(c + d*x)^4 + 1)) + (log(tan(c + d*x) + 1i)*(A*1i + B))/(32*a^4*d) - (log(ta
n(c + d*x) - 1i)*(A + B*31i)*1i)/(32*a^4*d)