Integrand size = 22, antiderivative size = 60 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=4 i a^3 x+\frac {3 a^3 \log (\cos (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d}-\frac {a^3+i a^3 \tan (c+d x)}{d} \] Output:
4*I*a^3*x+3*a^3*ln(cos(d*x+c))/d+a^3*ln(sin(d*x+c))/d-(a^3+I*a^3*tan(d*x+c ))/d
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \log (\tan (c+d x))}{d}-\frac {4 a^3 \log (i+\tan (c+d x))}{d}-\frac {i a^3 \tan (c+d x)}{d} \] Input:
Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^3*Log[Tan[c + d*x]])/d - (4*a^3*Log[I + Tan[c + d*x]])/d - (I*a^3*Tan[c + d*x])/d
Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4039, 3042, 4072, 3042, 3956, 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.
\(\displaystyle \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 4039 |
\(\displaystyle a \int \cot (c+d x) (i \tan (c+d x) a+a) (3 i \tan (c+d x) a+a)dx-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {(i \tan (c+d x) a+a) (3 i \tan (c+d x) a+a)}{\tan (c+d x)}dx-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4072 |
\(\displaystyle a \left (-3 a^2 \int \tan (c+d x)dx+\int \cot (c+d x) \left (4 i \tan (c+d x) a^2+a^2\right )dx\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (-3 a^2 \int \tan (c+d x)dx+\int \frac {4 i \tan (c+d x) a^2+a^2}{\tan (c+d x)}dx\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle a \left (\frac {3 a^2 \log (\cos (c+d x))}{d}+\int \frac {4 i \tan (c+d x) a^2+a^2}{\tan (c+d x)}dx\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle a \left (a^2 \int \cot (c+d x)dx+\frac {3 a^2 \log (\cos (c+d x))}{d}+4 i a^2 x\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+\frac {3 a^2 \log (\cos (c+d x))}{d}+4 i a^2 x\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \left (-a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx+\frac {3 a^2 \log (\cos (c+d x))}{d}+4 i a^2 x\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle a \left (\frac {a^2 \log (-\sin (c+d x))}{d}+\frac {3 a^2 \log (\cos (c+d x))}{d}+4 i a^2 x\right )-\frac {a^3+i a^3 \tan (c+d x)}{d}\) |
Input:
Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^3,x]
Output:
a*((4*I)*a^2*x + (3*a^2*Log[Cos[c + d*x]])/d + (a^2*Log[-Sin[c + d*x]])/d) - (a^3 + I*a^3*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[((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_)])^(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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[1/b Int[Simp[A*b*c + (A*b*d + B*(b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d , e, f, A, B}, x] && NeQ[b*c - a*d, 0]
Time = 0.92 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68
method | result | size |
parallelrisch | \(\frac {a^{3} \left (4 i d x -i \tan \left (d x +c \right )+\ln \left (\tan \left (d x +c \right )\right )-2 \ln \left (\sec \left (d x +c \right )^{2}\right )\right )}{d}\) | \(41\) |
norman | \(4 i a^{3} x -\frac {i a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\) | \(57\) |
derivativedivides | \(\frac {a^{3} \left (-\frac {i}{\cot \left (d x +c \right )}+3 \ln \left (\cot \left (d x +c \right )\right )-2 \ln \left (\cot \left (d x +c \right )^{2}+1\right )-4 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\) | \(58\) |
default | \(\frac {a^{3} \left (-\frac {i}{\cot \left (d x +c \right )}+3 \ln \left (\cot \left (d x +c \right )\right )-2 \ln \left (\cot \left (d x +c \right )^{2}+1\right )-4 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\) | \(58\) |
risch | \(-\frac {8 i a^{3} c}{d}+\frac {2 a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(72\) |
Input:
int(cot(d*x+c)*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
a^3*(4*I*d*x-I*tan(d*x+c)+ln(tan(d*x+c))-2*ln(sec(d*x+c)^2))/d
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2 \, a^{3} + 3 \, {\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
(2*a^3 + 3*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*I*d*x + 2*I*c) + 1) + (a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(2*I*d* x + 2*I*c) + d)
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2 a^{3}}{d e^{2 i c} e^{2 i d x} + d} + \frac {a^{3} \left (\log {\left (e^{2 i d x} - e^{- 2 i c} \right )} + 3 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**3,x)
Output:
2*a**3/(d*exp(2*I*c)*exp(2*I*d*x) + d) + a**3*(log(exp(2*I*d*x) - exp(-2*I *c)) + 3*log(exp(2*I*d*x) + exp(-2*I*c)))/d
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4 i \, {\left (d x + c\right )} a^{3} - 2 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + a^{3} \log \left (\tan \left (d x + c\right )\right ) - i \, a^{3} \tan \left (d x + c\right )}{d} \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
(4*I*(d*x + c)*a^3 - 2*a^3*log(tan(d*x + c)^2 + 1) + a^3*log(tan(d*x + c)) - I*a^3*tan(d*x + c))/d
Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {4 \, a^{3} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {a^{3} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {i \, a^{3} \tan \left (d x + c\right )}{d} \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-4*a^3*log(tan(d*x + c) + I)/d + a^3*log(abs(tan(d*x + c)))/d - I*a^3*tan( d*x + c)/d
Time = 0.94 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.65 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\left (4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{d} \] Input:
int(cot(c + d*x)*(a + a*tan(c + d*x)*1i)^3,x)
Output:
-(a^3*(4*log(tan(c + d*x) + 1i) + tan(c + d*x)*1i - log(tan(c + d*x))))/d
Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.92 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (-4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \cos \left (d x +c \right ) d i x -\sin \left (d x +c \right ) i \right )}{\cos \left (d x +c \right ) d} \] Input:
int(cot(d*x+c)*(a+I*a*tan(d*x+c))^3,x)
Output:
(a**3*( - 4*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1) + 3*cos(c + d*x)*log (tan((c + d*x)/2) - 1) + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1) + cos(c + d*x)*log(tan((c + d*x)/2)) + 4*cos(c + d*x)*d*i*x - sin(c + d*x)*i))/(co s(c + d*x)*d)