\(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [34]

Optimal result

Integrand size = 24, antiderivative size = 160 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {8 a^4 \log (\cos (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d} \]

[Out]

8*I*a^4*x+8*a^4*ln(cos(d*x+c))/d-8*I*a^4*tan(d*x+c)/d+4*a^4*tan(d*x+c)^2/d+8/3*I*a^4*tan(d*x+c)^3/d-67/60*a^4*
tan(d*x+c)^4/d-1/6*tan(d*x+c)^4*(a^2+I*a^2*tan(d*x+c))^2/d-7/15*tan(d*x+c)^4*(a^4+I*a^4*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3637, 3675, 3673, 3609, 3606, 3556} \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}+\frac {4 a^4 \tan ^2(c+d x)}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {8 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d} \]

[In]

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

[Out]

(8*I)*a^4*x + (8*a^4*Log[Cos[c + d*x]])/d - ((8*I)*a^4*Tan[c + d*x])/d + (4*a^4*Tan[c + d*x]^2)/d + (((8*I)/3)
*a^4*Tan[c + d*x]^3)/d - (67*a^4*Tan[c + d*x]^4)/(60*d) - (Tan[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])^2)/(6*d)
- (7*Tan[c + d*x]^4*(a^4 + I*a^4*Tan[c + d*x]))/(15*d)

Rule 3556

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

Rule 3606

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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}+\frac {1}{6} a \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 (10 a+14 i a \tan (c+d x)) \, dx \\ & = -\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {1}{30} a \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \left (106 a^2+134 i a^2 \tan (c+d x)\right ) \, dx \\ & = -\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {1}{30} a \int \tan ^3(c+d x) \left (240 a^3+240 i a^3 \tan (c+d x)\right ) \, dx \\ & = \frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {1}{30} a \int \tan ^2(c+d x) \left (-240 i a^3+240 a^3 \tan (c+d x)\right ) \, dx \\ & = \frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {1}{30} a \int \tan (c+d x) \left (-240 a^3-240 i a^3 \tan (c+d x)\right ) \, dx \\ & = 8 i a^4 x-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\left (8 a^4\right ) \int \tan (c+d x) \, dx \\ & = 8 i a^4 x+\frac {8 a^4 \log (\cos (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 a^4 \log (i+\tan (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {7 a^4 \tan ^4(c+d x)}{4 d}-\frac {4 i a^4 \tan ^5(c+d x)}{5 d}+\frac {a^4 \tan ^6(c+d x)}{6 d} \]

[In]

Integrate[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-8*a^4*Log[I + Tan[c + d*x]])/d - ((8*I)*a^4*Tan[c + d*x])/d + (4*a^4*Tan[c + d*x]^2)/d + (((8*I)/3)*a^4*Tan[
c + d*x]^3)/d - (7*a^4*Tan[c + d*x]^4)/(4*d) - (((4*I)/5)*a^4*Tan[c + d*x]^5)/d + (a^4*Tan[c + d*x]^6)/(6*d)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(\frac {a^{4} \left (-8 i \tan \left (d x +c \right )+\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {4 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {7 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {8 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+4 \left (\tan ^{2}\left (d x +c \right )\right )-4 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(93\)
default	\(\frac {a^{4} \left (-8 i \tan \left (d x +c \right )+\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {4 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {7 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {8 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+4 \left (\tan ^{2}\left (d x +c \right )\right )-4 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(93\)
risch	\(-\frac {16 i a^{4} c}{d}+\frac {4 a^{4} \left (270 \,{\mathrm e}^{10 i \left (d x +c \right )}+855 \,{\mathrm e}^{8 i \left (d x +c \right )}+1350 \,{\mathrm e}^{6 i \left (d x +c \right )}+1125 \,{\mathrm e}^{4 i \left (d x +c \right )}+486 \,{\mathrm e}^{2 i \left (d x +c \right )}+86\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {8 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(110\)
parallelrisch	\(\frac {-48 i a^{4} \left (\tan ^{5}\left (d x +c \right )\right )+10 \left (\tan ^{6}\left (d x +c \right )\right ) a^{4}+160 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )-105 \left (\tan ^{4}\left (d x +c \right )\right ) a^{4}+480 i a^{4} x d -480 i a^{4} \tan \left (d x +c \right )+240 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )-240 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{60 d}\)	\(110\)
norman	\(\frac {4 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {7 a^{4} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{4} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}+8 i a^{4} x -\frac {8 i a^{4} \tan \left (d x +c \right )}{d}+\frac {8 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {4 i a^{4} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(125\)
parts	\(\frac {a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{4} \left (\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}-\frac {4 i a^{4} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {4 i a^{4} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {6 a^{4} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(206\)

[In]

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*a^4*(-8*I*tan(d*x+c)+1/6*tan(d*x+c)^6-4/5*I*tan(d*x+c)^5-7/4*tan(d*x+c)^4+8/3*I*tan(d*x+c)^3+4*tan(d*x+c)^
2-4*ln(1+tan(d*x+c)^2)+8*I*arctan(tan(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.59 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 \, {\left (270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 86 \, a^{4} + 30 \, {\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

4/15*(270*a^4*e^(10*I*d*x + 10*I*c) + 855*a^4*e^(8*I*d*x + 8*I*c) + 1350*a^4*e^(6*I*d*x + 6*I*c) + 1125*a^4*e^
(4*I*d*x + 4*I*c) + 486*a^4*e^(2*I*d*x + 2*I*c) + 86*a^4 + 30*(a^4*e^(12*I*d*x + 12*I*c) + 6*a^4*e^(10*I*d*x +
10*I*c) + 15*a^4*e^(8*I*d*x + 8*I*c) + 20*a^4*e^(6*I*d*x + 6*I*c) + 15*a^4*e^(4*I*d*x + 4*I*c) + 6*a^4*e^(2*I
*d*x + 2*I*c) + a^4)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d
*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.54 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {1080 a^{4} e^{10 i c} e^{10 i d x} + 3420 a^{4} e^{8 i c} e^{8 i d x} + 5400 a^{4} e^{6 i c} e^{6 i d x} + 4500 a^{4} e^{4 i c} e^{4 i d x} + 1944 a^{4} e^{2 i c} e^{2 i d x} + 344 a^{4}}{15 d e^{12 i c} e^{12 i d x} + 90 d e^{10 i c} e^{10 i d x} + 225 d e^{8 i c} e^{8 i d x} + 300 d e^{6 i c} e^{6 i d x} + 225 d e^{4 i c} e^{4 i d x} + 90 d e^{2 i c} e^{2 i d x} + 15 d} \]

[In]

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

[Out]

8*a**4*log(exp(2*I*d*x) + exp(-2*I*c))/d + (1080*a**4*exp(10*I*c)*exp(10*I*d*x) + 3420*a**4*exp(8*I*c)*exp(8*I
*d*x) + 5400*a**4*exp(6*I*c)*exp(6*I*d*x) + 4500*a**4*exp(4*I*c)*exp(4*I*d*x) + 1944*a**4*exp(2*I*c)*exp(2*I*d
*x) + 344*a**4)/(15*d*exp(12*I*c)*exp(12*I*d*x) + 90*d*exp(10*I*c)*exp(10*I*d*x) + 225*d*exp(8*I*c)*exp(8*I*d*
x) + 300*d*exp(6*I*c)*exp(6*I*d*x) + 225*d*exp(4*I*c)*exp(4*I*d*x) + 90*d*exp(2*I*c)*exp(2*I*d*x) + 15*d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} \tan \left (d x + c\right )^{6} - 48 i \, a^{4} \tan \left (d x + c\right )^{5} - 105 \, a^{4} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} \tan \left (d x + c\right )^{3} + 240 \, a^{4} \tan \left (d x + c\right )^{2} + 480 i \, {\left (d x + c\right )} a^{4} - 240 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 480 i \, a^{4} \tan \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

1/60*(10*a^4*tan(d*x + c)^6 - 48*I*a^4*tan(d*x + c)^5 - 105*a^4*tan(d*x + c)^4 + 160*I*a^4*tan(d*x + c)^3 + 24
0*a^4*tan(d*x + c)^2 + 480*I*(d*x + c)*a^4 - 240*a^4*log(tan(d*x + c)^2 + 1) - 480*I*a^4*tan(d*x + c))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (142) = 284\).

Time = 0.89 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.04 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 \, {\left (30 \, a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 600 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 86 \, a^{4}\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

4/15*(30*a^4*e^(12*I*d*x + 12*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 180*a^4*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 450*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 600*a^4*e^(6*I*d*x + 6*I*c)*log(e^
(2*I*d*x + 2*I*c) + 1) + 450*a^4*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 180*a^4*e^(2*I*d*x + 2*I*c
)*log(e^(2*I*d*x + 2*I*c) + 1) + 270*a^4*e^(10*I*d*x + 10*I*c) + 855*a^4*e^(8*I*d*x + 8*I*c) + 1350*a^4*e^(6*I
*d*x + 6*I*c) + 1125*a^4*e^(4*I*d*x + 4*I*c) + 486*a^4*e^(2*I*d*x + 2*I*c) + 30*a^4*log(e^(2*I*d*x + 2*I*c) +
1) + 86*a^4)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x
+ 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

Mupad [B] (verification not implemented)

Time = 4.88 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-4\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,8{}\mathrm {i}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}}{3}+\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}}{5}}{d} \]

[In]

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

-(8*a^4*log(tan(c + d*x) + 1i) + a^4*tan(c + d*x)*8i - 4*a^4*tan(c + d*x)^2 - (a^4*tan(c + d*x)^3*8i)/3 + (7*a
^4*tan(c + d*x)^4)/4 + (a^4*tan(c + d*x)^5*4i)/5 - (a^4*tan(c + d*x)^6)/6)/d