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

Optimal result

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

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
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 a^4 \log (i+\tan (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {7 a^4 \tan ^4(c+d x)}{4 d}-\frac {4 i a^4 \tan ^5(c+d x)}{5 d}+\frac {a^4 \tan ^6(c+d x)}{6 d} \] Input:

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

Output:

(-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)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4039, 27, 3042, 4077, 3042, 4075, 3042, 4011, 3042, 4011, 3042, 4008, 3042, 3956}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

\(\displaystyle \frac {1}{6} a \int 2 \tan ^3(c+d x) (i \tan (c+d x) a+a)^2 (7 i \tan (c+d x) a+5 a)dx-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} a \int \tan ^3(c+d x) (i \tan (c+d x) a+a)^2 (7 i \tan (c+d x) a+5 a)dx-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \int \tan (c+d x)^3 (i \tan (c+d x) a+a)^2 (7 i \tan (c+d x) a+5 a)dx-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \int \tan ^3(c+d x) (i \tan (c+d x) a+a) \left (67 i \tan (c+d x) a^2+53 a^2\right )dx-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \int \tan (c+d x)^3 (i \tan (c+d x) a+a) \left (67 i \tan (c+d x) a^2+53 a^2\right )dx-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\int \tan ^3(c+d x) \left (120 i \tan (c+d x) a^3+120 a^3\right )dx\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\int \tan (c+d x)^3 \left (120 i \tan (c+d x) a^3+120 a^3\right )dx\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (\int \tan ^2(c+d x) \left (120 a^3 \tan (c+d x)-120 i a^3\right )dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (\int \tan (c+d x)^2 \left (120 a^3 \tan (c+d x)-120 i a^3\right )dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (\int \tan (c+d x) \left (-120 i \tan (c+d x) a^3-120 a^3\right )dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (\int \tan (c+d x) \left (-120 i \tan (c+d x) a^3-120 a^3\right )dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 4008

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-120 a^3 \int \tan (c+d x)dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}-\frac {120 i a^3 \tan (c+d x)}{d}+120 i a^3 x\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-120 a^3 \int \tan (c+d x)dx-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}-\frac {120 i a^3 \tan (c+d x)}{d}+120 i a^3 x\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}-\frac {120 i a^3 \tan (c+d x)}{d}+\frac {120 a^3 \log (\cos (c+d x))}{d}+120 i a^3 x\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\)

Input:

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

Output:

-1/6*(Tan[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (a*((-7*Tan[c + d*x 
]^4*(a^3 + I*a^3*Tan[c + d*x]))/(5*d) + ((120*I)*a^3*x + (120*a^3*Log[Cos[ 
c + d*x]])/d - ((120*I)*a^3*Tan[c + d*x])/d + (60*a^3*Tan[c + d*x]^2)/d + 
((40*I)*a^3*Tan[c + d*x]^3)/d - (67*a^3*Tan[c + d*x]^4)/(4*d))/5))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4008

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

rule 4011

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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[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] || IntegersQ[2*m, 2*n])
 

rule 4075

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 4077

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

Maple [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(\frac {a^{4} \left (-8 i \tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {4 i \tan \left (d x +c \right )^{5}}{5}-\frac {7 \tan \left (d x +c \right )^{4}}{4}+\frac {8 i \tan \left (d x +c \right )^{3}}{3}+4 \tan \left (d x +c \right )^{2}-4 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(93\)
default	\(\frac {a^{4} \left (-8 i \tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {4 i \tan \left (d x +c \right )^{5}}{5}-\frac {7 \tan \left (d x +c \right )^{4}}{4}+\frac {8 i \tan \left (d x +c \right )^{3}}{3}+4 \tan \left (d x +c \right )^{2}-4 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+8 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(93\)
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	\(-\frac {48 i a^{4} \tan \left (d x +c \right )^{5}-10 \tan \left (d x +c \right )^{6} a^{4}-160 i a^{4} \tan \left (d x +c \right )^{3}+105 \tan \left (d x +c \right )^{4} a^{4}-480 i a^{4} x d +480 i a^{4} \tan \left (d x +c \right )-240 a^{4} \tan \left (d x +c \right )^{2}+240 a^{4} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{60 d}\)	\(110\)
norman	\(\frac {4 a^{4} \tan \left (d x +c \right )^{2}}{d}-\frac {7 a^{4} \tan \left (d x +c \right )^{4}}{4 d}+\frac {a^{4} \tan \left (d x +c \right )^{6}}{6 d}+8 i a^{4} x -\frac {8 i a^{4} \tan \left (d x +c \right )}{d}+\frac {8 i a^{4} \tan \left (d x +c \right )^{3}}{3 d}-\frac {4 i a^{4} \tan \left (d x +c \right )^{5}}{5 d}-\frac {4 a^{4} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(125\)
parts	\(\frac {a^{4} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{4} \left (\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{4}}{4}+\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}-\frac {4 i a^{4} \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {4 i a^{4} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {6 a^{4} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}\)	\(206\)

Input:

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

Output:

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)

Time = 0.09 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.59 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 \, {\left (270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 86 \, a^{4} + 30 \, {\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

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 + 1 
0*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 = 0.93 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.54 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {1080 a^{4} e^{10 i c} e^{10 i d x} + 3420 a^{4} e^{8 i c} e^{8 i d x} + 5400 a^{4} e^{6 i c} e^{6 i d x} + 4500 a^{4} e^{4 i c} e^{4 i d x} + 1944 a^{4} e^{2 i c} e^{2 i d x} + 344 a^{4}}{15 d e^{12 i c} e^{12 i d x} + 90 d e^{10 i c} e^{10 i d x} + 225 d e^{8 i c} e^{8 i d x} + 300 d e^{6 i c} e^{6 i d x} + 225 d e^{4 i c} e^{4 i d x} + 90 d e^{2 i c} e^{2 i d x} + 15 d} \] Input:

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

Output:

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)

Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} \tan \left (d x + c\right )^{6} - 48 i \, a^{4} \tan \left (d x + c\right )^{5} - 105 \, a^{4} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} \tan \left (d x + c\right )^{3} + 240 \, a^{4} \tan \left (d x + c\right )^{2} + 480 i \, {\left (d x + c\right )} a^{4} - 240 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 480 i \, a^{4} \tan \left (d x + c\right )}{60 \, d} \] Input:

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

Output:

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 + 240*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 [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 \, a^{4} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {10 \, a^{4} d^{5} \tan \left (d x + c\right )^{6} - 48 i \, a^{4} d^{5} \tan \left (d x + c\right )^{5} - 105 \, a^{4} d^{5} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} d^{5} \tan \left (d x + c\right )^{3} + 240 \, a^{4} d^{5} \tan \left (d x + c\right )^{2} - 480 i \, a^{4} d^{5} \tan \left (d x + c\right )}{60 \, d^{6}} \] Input:

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

Output:

-8*a^4*log(tan(d*x + c) + I)/d + 1/60*(10*a^4*d^5*tan(d*x + c)^6 - 48*I*a^ 
4*d^5*tan(d*x + c)^5 - 105*a^4*d^5*tan(d*x + c)^4 + 160*I*a^4*d^5*tan(d*x 
+ c)^3 + 240*a^4*d^5*tan(d*x + c)^2 - 480*I*a^4*d^5*tan(d*x + c))/d^6
 

Mupad [B] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-4\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,8{}\mathrm {i}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}}{3}+\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}}{5}}{d} \] Input:

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

Output:

-(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
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (-240 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+10 \tan \left (d x +c \right )^{6}-48 \tan \left (d x +c \right )^{5} i -105 \tan \left (d x +c \right )^{4}+160 \tan \left (d x +c \right )^{3} i +240 \tan \left (d x +c \right )^{2}-480 \tan \left (d x +c \right ) i +480 d i x \right )}{60 d} \] Input:

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

Output:

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