Integrand size = 22, antiderivative size = 50 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-i a x-\frac {i a \cot (c+d x)}{d}-\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d} \] Output:
-I*a*x-I*a*cot(d*x+c)/d-1/2*a*cot(d*x+c)^2/d-a*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 = 62, normalized size of antiderivative = 1.24 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {a \csc ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}-\frac {a \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]
Output:
-1/2*(a*Csc[c + d*x]^2)/d - (I*a*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1 /2, -Tan[c + d*x]^2])/d - (a*Log[Sin[c + d*x]])/d
Time = 0.43 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4012, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+i a \tan (c+d x)}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {a \cot ^2(c+d x)}{2 d}+\int \cot ^2(c+d x) (i a-a \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \cot ^2(c+d x)}{2 d}+\int \frac {i a-a \tan (c+d x)}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \int -\cot (c+d x) (i \tan (c+d x) a+a)dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot (c+d x) (i \tan (c+d x) a+a)dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {i \tan (c+d x) a+a}{\tan (c+d x)}dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -a \int \cot (c+d x)dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}-i a x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}-i a x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}-i a x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}-\frac {a \log (-\sin (c+d x))}{d}-i a x\) |
Input:
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]
Output:
(-I)*a*x - (I*a*Cot[c + d*x])/d - (a*Cot[c + d*x]^2)/(2*d) - (a*Log[-Sin[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[(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]
Time = 0.92 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(-\frac {a \left (2 i d x +2 i \cot \left (d x +c \right )+2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )+\cot \left (d x +c \right )^{2}\right )}{2 d}\) | \(50\) |
derivativedivides | \(\frac {a \left (-\frac {1}{2 \tan \left (d x +c \right )^{2}}-\frac {i}{\tan \left (d x +c \right )}-\ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
default | \(\frac {a \left (-\frac {1}{2 \tan \left (d x +c \right )^{2}}-\frac {i}{\tan \left (d x +c \right )}-\ln \left (\tan \left (d x +c \right )\right )+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
risch | \(\frac {2 i a c}{d}+\frac {2 a \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(60\) |
norman | \(\frac {-\frac {a}{2 d}-\frac {i a \tan \left (d x +c \right )}{d}-i a x \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2}}-\frac {a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(74\) |
Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/2*a*(2*I*d*x+2*I*cot(d*x+c)+2*ln(tan(d*x+c))-ln(sec(d*x+c)^2)+cot(d*x+c )^2)/d
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.66 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {4 \, a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (a e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 2 \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
(4*a*e^(2*I*d*x + 2*I*c) - (a*e^(4*I*d*x + 4*I*c) - 2*a*e^(2*I*d*x + 2*I*c ) + a)*log(e^(2*I*d*x + 2*I*c) - 1) - 2*a)/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^ (2*I*d*x + 2*I*c) + d)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.60 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=- \frac {a \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {4 a e^{2 i c} e^{2 i d x} - 2 a}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \] Input:
integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c)),x)
Output:
-a*log(exp(2*I*d*x) - exp(-2*I*c))/d + (4*a*exp(2*I*c)*exp(2*I*d*x) - 2*a) /(d*exp(4*I*c)*exp(4*I*d*x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.16 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {2 i \, {\left (d x + c\right )} a - a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a \log \left (\tan \left (d x + c\right )\right ) + \frac {2 i \, a \tan \left (d x + c\right ) + a}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*I*(d*x + c)*a - a*log(tan(d*x + c)^2 + 1) + 2*a*log(tan(d*x + c)) + (2*I*a*tan(d*x + c) + a)/tan(d*x + c)^2)/d
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {1}{2} i \, a {\left (\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {2 i \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {2 \, \tan \left (d x + c\right ) - i}{d \tan \left (d x + c\right )^{2}}\right )} \] Input:
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
-1/2*I*a*(2*I*log(tan(d*x + c) + I)/d - 2*I*log(abs(tan(d*x + c)))/d + (2* tan(d*x + c) - I)/(d*tan(d*x + c)^2))
Time = 0.89 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d}-\frac {\frac {a}{2}+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2} \] Input:
int(cot(c + d*x)^3*(a + a*tan(c + d*x)*1i),x)
Output:
- (a*atan(2*tan(c + d*x) + 1i)*2i)/d - (a/2 + a*tan(c + d*x)*1i)/(d*tan(c + d*x)^2)
Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.92 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \left (-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )^{2} d i x +\sin \left (d x +c \right )^{2}-2\right )}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x)
Output:
(a*( - 4*cos(c + d*x)*sin(c + d*x)*i + 4*log(tan((c + d*x)/2)**2 + 1)*sin( c + d*x)**2 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 4*sin(c + d*x)**2* d*i*x + sin(c + d*x)**2 - 2))/(4*sin(c + d*x)**2*d)