Integrand size = 24, antiderivative size = 71 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-4 i a^3 x-\frac {2 i a^3 \cot (c+d x)}{d}-\frac {4 a^3 \log (\sin (c+d x))}{d}-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \] Output:
-4*I*a^3*x-2*I*a^3*cot(d*x+c)/d-4*a^3*ln(sin(d*x+c))/d-1/2*a*cot(d*x+c)^2* (a+I*a*tan(d*x+c))^2/d
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=a^3 \left (-\frac {3 i \cot (c+d x)}{d}-\frac {\cot ^2(c+d x)}{2 d}-\frac {4 \log (\tan (c+d x))}{d}+\frac {4 \log (i+\tan (c+d x))}{d}\right ) \] Input:
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^3,x]
Output:
a^3*(((-3*I)*Cot[c + d*x])/d - Cot[c + d*x]^2/(2*d) - (4*Log[Tan[c + d*x]] )/d + (4*Log[I + Tan[c + d*x]])/d)
Time = 0.51 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4028, 3042, 4025, 27, 3042, 4014, 3042, 25, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4028 |
\(\displaystyle 2 i a \int \cot ^2(c+d x) (i \tan (c+d x) a+a)^2dx-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 i a \int \frac {(i \tan (c+d x) a+a)^2}{\tan (c+d x)^2}dx-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 4025 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+\int 2 \cot (c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \int \cot (c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \int \frac {i a^2-a^2 \tan (c+d x)}{\tan (c+d x)}dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+i a^2 \int \cot (c+d x)dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+i a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (a^2 (-x)-i a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle 2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+\frac {i a^2 \log (-\sin (c+d x))}{d}\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\) |
Input:
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^3,x]
Output:
(2*I)*a*(-((a^2*Cot[c + d*x])/d) + 2*(-(a^2*x) + (I*a^2*Log[-Sin[c + d*x]] )/d)) - (a*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^2)/(2*d)
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[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*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] && N eQ[a*c + b*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - b*d)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
Time = 1.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73
method | result | size |
parallelrisch | \(-\frac {a^{3} \left (8 i d x +8 \ln \left (\tan \left (d x +c \right )\right )-4 \ln \left (\sec \left (d x +c \right )^{2}\right )+6 i \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}\right )}{2 d}\) | \(52\) |
derivativedivides | \(-\frac {a^{3} \left (-2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+4 i \arctan \left (\tan \left (d x +c \right )\right )+\frac {3 i}{\tan \left (d x +c \right )}+\frac {1}{2 \tan \left (d x +c \right )^{2}}+4 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(63\) |
default | \(-\frac {a^{3} \left (-2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )+4 i \arctan \left (\tan \left (d x +c \right )\right )+\frac {3 i}{\tan \left (d x +c \right )}+\frac {1}{2 \tan \left (d x +c \right )^{2}}+4 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(63\) |
risch | \(\frac {8 i a^{3} c}{d}+\frac {2 a^{3} \left (4 \,{\mathrm e}^{2 i \left (d x +c \right )}-3\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(66\) |
norman | \(\frac {-\frac {a^{3}}{2 d}-4 i a^{3} x \tan \left (d x +c \right )^{2}-\frac {3 i a^{3} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {4 a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\) | \(84\) |
Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-1/2*a^3*(8*I*d*x+8*ln(tan(d*x+c))-4*ln(sec(d*x+c)^2)+6*I*cot(d*x+c)+cot(d *x+c)^2)/d
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2 \, {\left (4 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, a^{3} - 2 \, {\left (a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
2*(4*a^3*e^(2*I*d*x + 2*I*c) - 3*a^3 - 2*(a^3*e^(4*I*d*x + 4*I*c) - 2*a^3* e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(4*I*d*x + 4 *I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=- \frac {4 a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {8 a^{3} e^{2 i c} e^{2 i d x} - 6 a^{3}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \] Input:
integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**3,x)
Output:
-4*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (8*a**3*exp(2*I*c)*exp(2*I*d*x ) - 6*a**3)/(d*exp(4*I*c)*exp(4*I*d*x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {8 i \, {\left (d x + c\right )} a^{3} - 4 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 8 \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 i \, a^{3} \tan \left (d x + c\right ) + a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/2*(8*I*(d*x + c)*a^3 - 4*a^3*log(tan(d*x + c)^2 + 1) + 8*a^3*log(tan(d* x + c)) + (6*I*a^3*tan(d*x + c) + a^3)/tan(d*x + c)^2)/d
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4 \, a^{3} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {4 \, a^{3} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {6 i \, a^{3} \tan \left (d x + c\right ) + a^{3}}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
4*a^3*log(tan(d*x + c) + I)/d - 4*a^3*log(abs(tan(d*x + c)))/d - 1/2*(6*I* a^3*tan(d*x + c) + a^3)/(d*tan(d*x + c)^2)
Time = 0.94 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {\frac {a^3}{2}+a^3\,\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \] Input:
int(cot(c + d*x)^3*(a + a*tan(c + d*x)*1i)^3,x)
Output:
- (a^3*tan(c + d*x)*3i + a^3/2)/(d*tan(c + d*x)^2) - (a^3*atan(2*tan(c + d *x) + 1i)*8i)/d
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.38 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (-12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i +16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-16 \sin \left (d x +c \right )^{2} d i x +\sin \left (d x +c \right )^{2}-2\right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x)
Output:
(a**3*( - 12*cos(c + d*x)*sin(c + d*x)*i + 16*log(tan((c + d*x)/2)**2 + 1) *sin(c + d*x)**2 - 16*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 16*sin(c + d *x)**2*d*i*x + sin(c + d*x)**2 - 2))/(4*sin(c + d*x)**2*d)