Integrand size = 22, antiderivative size = 62 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=-2 i a^2 x-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {i a^2 \tan (c+d x)}{d}+\frac {(a+i a \tan (c+d x))^2}{2 d} \] Output:
-2*I*a^2*x-2*a^2*ln(cos(d*x+c))/d+I*a^2*tan(d*x+c)/d+1/2*(a+I*a*tan(d*x+c) )^2/d
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \left (-4 i \arctan (\tan (c+d x))-4 \log (\cos (c+d x))-\sec ^2(c+d x)+4 i \tan (c+d x)\right )}{2 d} \] Input:
Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^2,x]
Output:
(a^2*((-4*I)*ArcTan[Tan[c + d*x]] - 4*Log[Cos[c + d*x]] - Sec[c + d*x]^2 + (4*I)*Tan[c + d*x]))/(2*d)
Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4010, 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.
\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {(a+i a \tan (c+d x))^2}{2 d}-i \int (i \tan (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a+i a \tan (c+d x))^2}{2 d}-i \int (i \tan (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3958 |
\(\displaystyle \frac {(a+i a \tan (c+d x))^2}{2 d}-i \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a+i a \tan (c+d x))^2}{2 d}-i \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(a+i a \tan (c+d x))^2}{2 d}-i \left (-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x\right )\) |
Input:
Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^2,x]
Output:
(a + I*a*Tan[c + d*x])^2/(2*d) - I*(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[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]
Time = 0.38 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {a^{2} \left (2 i \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (1+\tan \left (d x +c \right )^{2}\right )-2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(49\) |
default | \(\frac {a^{2} \left (2 i \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (1+\tan \left (d x +c \right )^{2}\right )-2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(49\) |
parallelrisch | \(-\frac {4 i a^{2} x d -4 i a^{2} \tan \left (d x +c \right )+\tan \left (d x +c \right )^{2} a^{2}-2 a^{2} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(55\) |
norman | \(-\frac {a^{2} \tan \left (d x +c \right )^{2}}{2 d}-2 i a^{2} x +\frac {2 i a^{2} \tan \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\) | \(58\) |
risch | \(\frac {4 i a^{2} c}{d}-\frac {2 a^{2} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(66\) |
parts | \(\frac {a^{2} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {2 i a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}\) | \(78\) |
Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*a^2*(2*I*tan(d*x+c)-1/2*tan(d*x+c)^2+ln(1+tan(d*x+c)^2)-2*I*arctan(tan (d*x+c)))
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.50 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a^{2} + {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \] Input:
integrate(tan(d*x+c)*(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 = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.42 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=- \frac {2 a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 6 a^{2} e^{2 i c} e^{2 i d x} - 4 a^{2}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} \] Input:
integrate(tan(d*x+c)*(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 = 55, normalized size of antiderivative = 0.89 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^{2} \tan \left (d x + c\right )^{2} + 4 i \, {\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 4 i \, a^{2} \tan \left (d x + c\right )}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*(a^2*tan(d*x + c)^2 + 4*I*(d*x + c)*a^2 - 2*a^2*log(tan(d*x + c)^2 + 1) - 4*I*a^2*tan(d*x + c))/d
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {a^{2} d \tan \left (d x + c\right )^{2} - 4 i \, a^{2} d \tan \left (d x + c\right )}{2 \, d^{2}} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
2*a^2*log(tan(d*x + c) + I)/d - 1/2*(a^2*d*tan(d*x + c)^2 - 4*I*a^2*d*tan( d*x + c))/d^2
Time = 0.83 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.65 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\left (4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}\right )}{2\,d} \] Input:
int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^2,x)
Output:
(a^2*(4*log(tan(c + d*x) + 1i) + tan(c + d*x)*4i - tan(c + d*x)^2))/(2*d)
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^{2} \left (2 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-\tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i -4 d i x \right )}{2 d} \] Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^2,x)
Output:
(a**2*(2*log(tan(c + d*x)**2 + 1) - tan(c + d*x)**2 + 4*tan(c + d*x)*i - 4 *d*i*x))/(2*d)