Integrand size = 15, antiderivative size = 117 \[ \int (a+i a \tan (c+d x))^5 \, dx=16 a^5 x-\frac {16 i a^5 \log (\cos (c+d x))}{d}-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\frac {2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \] Output:
16*a^5*x-16*I*a^5*ln(cos(d*x+c))/d-8*a^5*tan(d*x+c)/d+2/3*I*a^2*(a+I*a*tan (d*x+c))^3/d+1/4*I*a*(a+I*a*tan(d*x+c))^4/d+2*I*a*(a^2+I*a^2*tan(d*x+c))^2 /d
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \left (35 i+192 i \log (i+\tan (c+d x))-180 \tan (c+d x)-66 i \tan ^2(c+d x)+20 \tan ^3(c+d x)+3 i \tan ^4(c+d x)\right )}{12 d} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^5,x]
Output:
(a^5*(35*I + (192*I)*Log[I + Tan[c + d*x]] - 180*Tan[c + d*x] - (66*I)*Tan [c + d*x]^2 + 20*Tan[c + d*x]^3 + (3*I)*Tan[c + d*x]^4))/(12*d)
Time = 0.55 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3959, 3042, 3959, 3042, 3959, 3042, 3958, 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.
\(\displaystyle \int (a+i a \tan (c+d x))^5 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^5dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^4dx+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^4dx+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^3dx+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^3dx+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3958 |
\(\displaystyle 2 a \left (2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle 2 a \left (2 a \left (2 a \left (-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\right )+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^5,x]
Output:
((I/4)*a*(a + I*a*Tan[c + d*x])^4)/d + 2*a*(((I/3)*a*(a + I*a*Tan[c + d*x] )^3)/d + 2*a*(((I/2)*a*(a + I*a*Tan[c + d*x])^2)/d + 2*a*(2*a^2*x - ((2*I) *a^2*Log[Cos[c + d*x]])/d - (a^2*Tan[c + d*x])/d)))
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[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2) *x, x] + (Simp[b^2*(Tan[c + d*x]/d), x] + Simp[2*a*b Int[Tan[c + d*x], x] , x]) /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a Int[(a + b*Tan[c + d* x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n , 1]
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{4}}{4}+\frac {5 \tan \left (d x +c \right )^{3}}{3}-\frac {11 i \tan \left (d x +c \right )^{2}}{2}+8 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
default | \(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{4}}{4}+\frac {5 \tan \left (d x +c \right )^{3}}{3}-\frac {11 i \tan \left (d x +c \right )^{2}}{2}+8 i \ln \left (1+\tan \left (d x +c \right )^{2}\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
parallelrisch | \(\frac {3 i a^{5} \tan \left (d x +c \right )^{4}-66 i a^{5} \tan \left (d x +c \right )^{2}+20 \tan \left (d x +c \right )^{3} a^{5}+96 i a^{5} \ln \left (1+\tan \left (d x +c \right )^{2}\right )+192 a^{5} x d -180 a^{5} \tan \left (d x +c \right )}{12 d}\) | \(83\) |
risch | \(-\frac {32 a^{5} c}{d}-\frac {4 i a^{5} \left (48 \,{\mathrm e}^{6 i \left (d x +c \right )}+108 \,{\mathrm e}^{4 i \left (d x +c \right )}+88 \,{\mathrm e}^{2 i \left (d x +c \right )}+25\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {16 i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(89\) |
norman | \(16 a^{5} x -\frac {15 a^{5} \tan \left (d x +c \right )}{d}+\frac {5 a^{5} \tan \left (d x +c \right )^{3}}{3 d}-\frac {11 i a^{5} \tan \left (d x +c \right )^{2}}{2 d}+\frac {i a^{5} \tan \left (d x +c \right )^{4}}{4 d}+\frac {8 i a^{5} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\) | \(92\) |
parts | \(a^{5} x -\frac {10 i a^{5} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {i a^{5} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {5 i a^{5} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}-\frac {10 a^{5} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {5 a^{5} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(161\) |
Input:
int((a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
1/d*a^5*(-15*tan(d*x+c)+1/4*I*tan(d*x+c)^4+5/3*tan(d*x+c)^3-11/2*I*tan(d*x +c)^2+8*I*ln(1+tan(d*x+c)^2)+16*arctan(tan(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.51 \[ \int (a+i a \tan (c+d x))^5 \, dx=-\frac {4 \, {\left (48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 108 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 25 i \, a^{5} + 12 \, {\left (i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^5,x, algorithm="fricas")
Output:
-4/3*(48*I*a^5*e^(6*I*d*x + 6*I*c) + 108*I*a^5*e^(4*I*d*x + 4*I*c) + 88*I* a^5*e^(2*I*d*x + 2*I*c) + 25*I*a^5 + 12*(I*a^5*e^(8*I*d*x + 8*I*c) + 4*I*a ^5*e^(6*I*d*x + 6*I*c) + 6*I*a^5*e^(4*I*d*x + 4*I*c) + 4*I*a^5*e^(2*I*d*x + 2*I*c) + I*a^5)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4 *d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.52 \[ \int (a+i a \tan (c+d x))^5 \, dx=- \frac {16 i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 192 i a^{5} e^{6 i c} e^{6 i d x} - 432 i a^{5} e^{4 i c} e^{4 i d x} - 352 i a^{5} e^{2 i c} e^{2 i d x} - 100 i a^{5}}{3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} + 3 d} \] Input:
integrate((a+I*a*tan(d*x+c))**5,x)
Output:
-16*I*a**5*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-192*I*a**5*exp(6*I*c)*exp (6*I*d*x) - 432*I*a**5*exp(4*I*c)*exp(4*I*d*x) - 352*I*a**5*exp(2*I*c)*exp (2*I*d*x) - 100*I*a**5)/(3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp (6*I*d*x) + 18*d*exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)
Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.41 \[ \int (a+i a \tan (c+d x))^5 \, dx=a^{5} x + \frac {5 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{5}}{3 \, d} + \frac {10 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{5}}{d} + \frac {i \, a^{5} {\left (\frac {4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{4 \, d} + \frac {5 i \, a^{5} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {5 i \, a^{5} \log \left (\sec \left (d x + c\right )\right )}{d} \] Input:
integrate((a+I*a*tan(d*x+c))^5,x, algorithm="maxima")
Output:
a^5*x + 5/3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^5/d + 10*(d* x + c - tan(d*x + c))*a^5/d + 1/4*I*a^5*((4*sin(d*x + c)^2 - 3)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 2*log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*(1/ (sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*log(sec(d*x + c))/d
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {16 i \, a^{5} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {-3 i \, a^{5} d^{3} \tan \left (d x + c\right )^{4} - 20 \, a^{5} d^{3} \tan \left (d x + c\right )^{3} + 66 i \, a^{5} d^{3} \tan \left (d x + c\right )^{2} + 180 \, a^{5} d^{3} \tan \left (d x + c\right )}{12 \, d^{4}} \] Input:
integrate((a+I*a*tan(d*x+c))^5,x, algorithm="giac")
Output:
16*I*a^5*log(tan(d*x + c) + I)/d - 1/12*(-3*I*a^5*d^3*tan(d*x + c)^4 - 20* a^5*d^3*tan(d*x + c)^3 + 66*I*a^5*d^3*tan(d*x + c)^2 + 180*a^5*d^3*tan(d*x + c))/d^4
Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}-15\,a^5\,\mathrm {tan}\left (c+d\,x\right )-\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\,11{}\mathrm {i}}{2}+\frac {5\,a^5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4}}{d} \] Input:
int((a + a*tan(c + d*x)*1i)^5,x)
Output:
(a^5*log(tan(c + d*x) + 1i)*16i - 15*a^5*tan(c + d*x) - (a^5*tan(c + d*x)^ 2*11i)/2 + (5*a^5*tan(c + d*x)^3)/3 + (a^5*tan(c + d*x)^4*1i)/4)/d
Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {a^{5} \left (96 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) i +3 \tan \left (d x +c \right )^{4} i +20 \tan \left (d x +c \right )^{3}-66 \tan \left (d x +c \right )^{2} i -180 \tan \left (d x +c \right )+192 d x \right )}{12 d} \] Input:
int((a+I*a*tan(d*x+c))^5,x)
Output:
(a**5*(96*log(tan(c + d*x)**2 + 1)*i + 3*tan(c + d*x)**4*i + 20*tan(c + d* x)**3 - 66*tan(c + d*x)**2*i - 180*tan(c + d*x) + 192*d*x))/(12*d)