Integrand size = 24, antiderivative size = 160 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {8 a^4 \log (\cos (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {67 a^4 \tan ^4(c+d x)}{60 d}-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}-\frac {7 \tan ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{15 d} \]
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
Time = 0.46 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 a^4 \log (i+\tan (c+d x))}{d}-\frac {8 i a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tan ^2(c+d x)}{d}+\frac {8 i a^4 \tan ^3(c+d x)}{3 d}-\frac {7 a^4 \tan ^4(c+d x)}{4 d}-\frac {4 i a^4 \tan ^5(c+d x)}{5 d}+\frac {a^4 \tan ^6(c+d x)}{6 d} \]
(-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)
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 |
\(\displaystyle \frac {1}{3} a \left (\frac {1}{5} \left (-\frac {67 a^3 \tan ^4(c+d x)}{4 d}+\frac {40 i a^3 \tan ^3(c+d x)}{d}+\frac {60 a^3 \tan ^2(c+d x)}{d}-\frac {120 i a^3 \tan (c+d x)}{d}+\frac {120 a^3 \log (\cos (c+d x))}{d}+120 i a^3 x\right )-\frac {7 \tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )-\frac {\tan ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{6 d}\) |
-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
3.1.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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]
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]
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])
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]
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]
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 | \(\frac {a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{4} \left (\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}-\frac {4 i a^{4} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {4 i a^{4} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {6 a^{4} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\) | \(206\) |
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)))
Time = 0.24 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.59 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 \, {\left (270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 86 \, a^{4} + 30 \, {\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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)
Time = 0.82 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.54 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {1080 a^{4} e^{10 i c} e^{10 i d x} + 3420 a^{4} e^{8 i c} e^{8 i d x} + 5400 a^{4} e^{6 i c} e^{6 i d x} + 4500 a^{4} e^{4 i c} e^{4 i d x} + 1944 a^{4} e^{2 i c} e^{2 i d x} + 344 a^{4}}{15 d e^{12 i c} e^{12 i d x} + 90 d e^{10 i c} e^{10 i d x} + 225 d e^{8 i c} e^{8 i d x} + 300 d e^{6 i c} e^{6 i d x} + 225 d e^{4 i c} e^{4 i d x} + 90 d e^{2 i c} e^{2 i d x} + 15 d} \]
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)
Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} \tan \left (d x + c\right )^{6} - 48 i \, a^{4} \tan \left (d x + c\right )^{5} - 105 \, a^{4} \tan \left (d x + c\right )^{4} + 160 i \, a^{4} \tan \left (d x + c\right )^{3} + 240 \, a^{4} \tan \left (d x + c\right )^{2} + 480 i \, {\left (d x + c\right )} a^{4} - 240 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 480 i \, a^{4} \tan \left (d x + c\right )}{60 \, d} \]
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
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 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 \, {\left (30 \, a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 600 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 450 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 180 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 270 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 855 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 1350 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1125 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 486 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 86 \, a^{4}\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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) + 3 0*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)
Time = 4.88 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-4\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,8{}\mathrm {i}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}}{3}+\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}}{5}}{d} \]