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

Optimal. Leaf size=160 \[ -\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}+\frac {4 a^4 \tan ^2(c+d x)}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {8 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d} \]

[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]  time = 0.27, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3556, 3594, 3592, 3528, 3525, 3475} \[ -\frac {67 a^4 \tan ^4(c+d x)}{60 d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}+\frac {4 a^4 \tan ^2(c+d x)}{d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {8 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x \]

Antiderivative was successfully verified.

[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 3475

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

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 3528

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 3556

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 3592

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 3594

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*} \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\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 [B]  time = 2.40, size = 349, normalized size = 2.18 \[ \frac {a^4 \sec (c) \sec ^6(c+d x) \left (-780 i \sin (c+2 d x)+510 i \sin (3 c+2 d x)-366 i \sin (3 c+4 d x)+150 i \sin (5 c+4 d x)-86 i \sin (5 c+6 d x)+450 i d x \cos (3 c+2 d x)+345 \cos (3 c+2 d x)+180 i d x \cos (3 c+4 d x)+120 \cos (3 c+4 d x)+180 i d x \cos (5 c+4 d x)+120 \cos (5 c+4 d x)+30 i d x \cos (5 c+6 d x)+30 i d x \cos (7 c+6 d x)+225 \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+90 \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )+90 \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+15 \cos (5 c+6 d x) \log \left (\cos ^2(c+d x)\right )+15 \cos (7 c+6 d x) \log \left (\cos ^2(c+d x)\right )+15 \cos (c+2 d x) \left (15 \log \left (\cos ^2(c+d x)\right )+30 i d x+23\right )+10 \cos (c) \left (30 \log \left (\cos ^2(c+d x)\right )+60 i d x+49\right )+860 i \sin (c)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Sec[c]*Sec[c + d*x]^6*(345*Cos[3*c + 2*d*x] + (450*I)*d*x*Cos[3*c + 2*d*x] + 120*Cos[3*c + 4*d*x] + (180*
I)*d*x*Cos[3*c + 4*d*x] + 120*Cos[5*c + 4*d*x] + (180*I)*d*x*Cos[5*c + 4*d*x] + (30*I)*d*x*Cos[5*c + 6*d*x] +
(30*I)*d*x*Cos[7*c + 6*d*x] + 225*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] + 90*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]^
2] + 90*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 15*Cos[5*c + 6*d*x]*Log[Cos[c + d*x]^2] + 15*Cos[7*c + 6*d*x]*L
og[Cos[c + d*x]^2] + 15*Cos[c + 2*d*x]*(23 + (30*I)*d*x + 15*Log[Cos[c + d*x]^2]) + 10*Cos[c]*(49 + (60*I)*d*x
+ 30*Log[Cos[c + d*x]^2]) + (860*I)*Sin[c] - (780*I)*Sin[c + 2*d*x] + (510*I)*Sin[3*c + 2*d*x] - (366*I)*Sin[
3*c + 4*d*x] + (150*I)*Sin[5*c + 4*d*x] - (86*I)*Sin[5*c + 6*d*x]))/(240*d)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 254, normalized size = 1.59 \[ \frac {4 \, {\left (270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 86 \, a^{4} + 30 \, {\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [B]  time = 9.97, size = 326, normalized size = 2.04 \[ \frac {4 \, {\left (30 \, a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 600 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 86 \, a^{4}\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 134, normalized size = 0.84 \[ -\frac {8 i a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}-\frac {4 i a^{4} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {7 a^{4} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {8 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {4 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {8 i a^{4} \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.65, size = 108, normalized size = 0.68 \[ \frac {10 \, a^{4} \tan \left (d x + c\right )^{6} - 48 i \, a^{4} \tan \left (d x + c\right )^{5} - 105 \, a^{4} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} \tan \left (d x + c\right )^{3} + 240 \, a^{4} \tan \left (d x + c\right )^{2} + 480 i \, {\left (d x + c\right )} a^{4} - 240 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 480 i \, a^{4} \tan \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 3.76, size = 100, normalized size = 0.62 \[ -\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-4\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,8{}\mathrm {i}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}}{3}+\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 4.80, size = 250, normalized size = 1.56 \[ \frac {8 a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 1080 a^{4} e^{10 i c} e^{10 i d x} - 3420 a^{4} e^{8 i c} e^{8 i d x} - 5400 a^{4} e^{6 i c} e^{6 i d x} - 4500 a^{4} e^{4 i c} e^{4 i d x} - 1944 a^{4} e^{2 i c} e^{2 i d x} - 344 a^{4}}{- 15 d e^{12 i c} e^{12 i d x} - 90 d e^{10 i c} e^{10 i d x} - 225 d e^{8 i c} e^{8 i d x} - 300 d e^{6 i c} e^{6 i d x} - 225 d e^{4 i c} e^{4 i d x} - 90 d e^{2 i c} e^{2 i d x} - 15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________