Integrand size = 24, antiderivative size = 58 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-2 i a^2 x-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a^2 \log (\sin (c+d x))}{d} \] Output:
-2*I*a^2*x-2*I*a^2*cot(d*x+c)/d-1/2*a^2*cot(d*x+c)^2/d-2*a^2*ln(sin(d*x+c) )/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 i a^2 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}-\frac {2 a^2 \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]
Output:
-1/2*(a^2*Csc[c + d*x]^2)/d - ((2*I)*a^2*Cot[c + d*x]*Hypergeometric2F1[-1 /2, 1, 1/2, -Tan[c + d*x]^2])/d - (2*a^2*Log[Sin[c + d*x]])/d
Time = 0.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 4025, 27, 3042, 4012, 25, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^2}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4025 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+\int 2 \cot ^2(c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \int \cot ^2(c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \int \frac {i a^2-a^2 \tan (c+d x)}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (\int -\cot (c+d x) \left (i \tan (c+d x) a^2+a^2\right )dx-\frac {i a^2 \cot (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (-\int \cot (c+d x) \left (i \tan (c+d x) a^2+a^2\right )dx-\frac {i a^2 \cot (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (-\int \frac {i \tan (c+d x) a^2+a^2}{\tan (c+d x)}dx-\frac {i a^2 \cot (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (-a^2 \int \cot (c+d x)dx-\frac {i a^2 \cot (c+d x)}{d}-i a^2 x\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (-a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {i a^2 \cot (c+d x)}{d}-i a^2 x\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {i a^2 \cot (c+d x)}{d}-i a^2 x\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {a^2 \cot ^2(c+d x)}{2 d}+2 \left (-\frac {i a^2 \cot (c+d x)}{d}-\frac {a^2 \log (-\sin (c+d x))}{d}-i a^2 x\right )\) |
Input:
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]
Output:
-1/2*(a^2*Cot[c + d*x]^2)/d + 2*((-I)*a^2*x - (I*a^2*Cot[c + d*x])/d - (a^ 2*Log[-Sin[c + d*x]])/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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (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 + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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]
Time = 1.32 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(-\frac {a^{2} \left (4 i d x -2 \ln \left (\sec \left (d x +c \right )^{2}\right )+4 \ln \left (\tan \left (d x +c \right )\right )+4 i \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}\right )}{2 d}\) | \(52\) |
derivativedivides | \(\frac {-a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(64\) |
default | \(\frac {-a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(64\) |
risch | \(\frac {4 i a^{2} c}{d}+\frac {2 a^{2} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(66\) |
norman | \(\frac {-\frac {a^{2}}{2 d}-\frac {2 i a^{2} \tan \left (d x +c \right )}{d}-2 i a^{2} x \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2}}+\frac {a^{2} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(83\) |
Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/2*a^2*(4*I*d*x-2*ln(sec(d*x+c)^2)+4*ln(tan(d*x+c))+4*I*cot(d*x+c)+cot(d *x+c)^2)/d
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.62 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, a^{2} - {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
Output:
2*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*a^2 - (a^2*e^(4*I*d*x + 4*I*c) - 2*a^2*e^ (2*I*d*x + 2*I*c) + a^2)*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 = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.50 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=- \frac {2 a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {6 a^{2} e^{2 i c} e^{2 i d x} - 4 a^{2}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \] Input:
integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**2,x)
Output:
-2*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (6*a**2*exp(2*I*c)*exp(2*I*d*x ) - 4*a**2)/(d*exp(4*I*c)*exp(4*I*d*x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {4 i \, {\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {4 i \, a^{2} \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*(4*I*(d*x + c)*a^2 - 2*a^2*log(tan(d*x + c)^2 + 1) + 4*a^2*log(tan(d* x + c)) + (4*I*a^2*tan(d*x + c) + a^2)/tan(d*x + c)^2)/d
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {2 \, a^{2} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {4 i \, a^{2} \tan \left (d x + c\right ) + a^{2}}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
2*a^2*log(tan(d*x + c) + I)/d - 2*a^2*log(abs(tan(d*x + c)))/d - 1/2*(4*I* a^2*tan(d*x + c) + a^2)/(d*tan(d*x + c)^2)
Time = 0.92 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {\frac {a^2}{2}+a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d} \] Input:
int(cot(c + d*x)^3*(a + a*tan(c + d*x)*1i)^2,x)
Output:
- (a^2*tan(c + d*x)*2i + a^2/2)/(d*tan(c + d*x)^2) - (a^2*atan(2*tan(c + d *x) + 1i)*4i)/d
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.69 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^{2} \left (-8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )^{2} d i x +\sin \left (d x +c \right )^{2}-2\right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x)
Output:
(a**2*( - 8*cos(c + d*x)*sin(c + d*x)*i + 8*log(tan((c + d*x)/2)**2 + 1)*s in(c + d*x)**2 - 8*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 8*sin(c + d*x)* *2*d*i*x + sin(c + d*x)**2 - 2))/(4*sin(c + d*x)**2*d)