Integrand size = 23, antiderivative size = 449 \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\frac {3 \cos \left (2 e-\frac {2 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {3 \cos \left (4 e-\frac {4 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}+\frac {\cos \left (6 e-\frac {6 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}-\frac {i \operatorname {CosIntegral}\left (\frac {6 c f}{d}+6 f x\right ) \sin \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \operatorname {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {3 i \operatorname {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}-\frac {3 \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}-\frac {i \cos \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d} \]
1/8*Ci(6*c*f/d+6*f*x)*cos(-6*e+6*c*f/d)/a^3/d+3/8*Ci(4*c*f/d+4*f*x)*cos(-4 *e+4*c*f/d)/a^3/d+3/8*Ci(2*c*f/d+2*f*x)*cos(-2*e+2*c*f/d)/a^3/d+1/8*ln(d*x +c)/a^3/d-3/8*I*cos(-2*e+2*c*f/d)*Si(2*c*f/d+2*f*x)/a^3/d-3/8*I*cos(-4*e+4 *c*f/d)*Si(4*c*f/d+4*f*x)/a^3/d-1/8*I*cos(-6*e+6*c*f/d)*Si(6*c*f/d+6*f*x)/ a^3/d+1/8*I*Ci(6*c*f/d+6*f*x)*sin(-6*e+6*c*f/d)/a^3/d+1/8*Si(6*c*f/d+6*f*x )*sin(-6*e+6*c*f/d)/a^3/d+3/8*I*Ci(4*c*f/d+4*f*x)*sin(-4*e+4*c*f/d)/a^3/d+ 3/8*Si(4*c*f/d+4*f*x)*sin(-4*e+4*c*f/d)/a^3/d+3/8*I*Ci(2*c*f/d+2*f*x)*sin( -2*e+2*c*f/d)/a^3/d+3/8*Si(2*c*f/d+2*f*x)*sin(-2*e+2*c*f/d)/a^3/d
Time = 1.14 (sec) , antiderivative size = 336, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\cos (3 e) \log (f (c+d x))+i \log (f (c+d x)) \sin (3 e)+\left (\cos \left (e-\frac {4 c f}{d}\right )-i \sin \left (e-\frac {4 c f}{d}\right )\right ) \left (3 \operatorname {CosIntegral}\left (\frac {4 f (c+d x)}{d}\right )+\cos \left (2 e-\frac {2 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {6 f (c+d x)}{d}\right )+3 \operatorname {CosIntegral}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cos \left (2 e-\frac {2 c f}{d}\right )+i \sin \left (2 e-\frac {2 c f}{d}\right )\right )-i \operatorname {CosIntegral}\left (\frac {6 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-3 i \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {6 f (c+d x)}{d}\right )-\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {6 f (c+d x)}{d}\right )\right )\right )}{8 d (a+i a \tan (e+f x))^3} \]
(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(Cos[3*e]*Log[f*(c + d*x)] + I*L og[f*(c + d*x)]*Sin[3*e] + (Cos[e - (4*c*f)/d] - I*Sin[e - (4*c*f)/d])*(3* CosIntegral[(4*f*(c + d*x))/d] + Cos[2*e - (2*c*f)/d]*CosIntegral[(6*f*(c + d*x))/d] + 3*CosIntegral[(2*f*(c + d*x))/d]*(Cos[2*e - (2*c*f)/d] + I*Si n[2*e - (2*c*f)/d]) - I*CosIntegral[(6*f*(c + d*x))/d]*Sin[2*e - (2*c*f)/d ] - (3*I)*Cos[2*e - (2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d] + 3*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d] - (3*I)*SinIntegral[(4*f*(c + d*x))/d] - I*Cos[2*e - (2*c*f)/d]*SinIntegral[(6*f*(c + d*x))/d] - Sin[2*e - (2*c*f)/d]*SinIntegral[(6*f*(c + d*x))/d])))/(8*d*(a + I*a*Tan[e + f*x] )^3)
Time = 1.98 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4211, 2009}
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.
\(\displaystyle \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3}dx\) |
\(\Big \downarrow \) 4211 |
\(\displaystyle \int \left (\frac {i \sin ^3(2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 \sin ^2(2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 i \sin (2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 \sin (2 e+2 f x) \sin (4 e+4 f x)}{16 a^3 (c+d x)}-\frac {3 i \sin (4 e+4 f x)}{8 a^3 (c+d x)}+\frac {\cos ^3(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \cos ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \cos (2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 i \sin (2 e+2 f x) \cos ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {1}{8 a^3 (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 i \operatorname {CosIntegral}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {i \operatorname {CosIntegral}\left (6 x f+\frac {6 c f}{d}\right ) \sin \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \operatorname {CosIntegral}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {3 \operatorname {CosIntegral}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}+\frac {3 \operatorname {CosIntegral}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {\operatorname {CosIntegral}\left (6 x f+\frac {6 c f}{d}\right ) \cos \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {\sin \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {i \cos \left (6 e-\frac {6 c f}{d}\right ) \text {Si}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}\) |
(3*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/(8*a^3*d) + (3*Cos [4*e - (4*c*f)/d]*CosIntegral[(4*c*f)/d + 4*f*x])/(8*a^3*d) + (Cos[6*e - ( 6*c*f)/d]*CosIntegral[(6*c*f)/d + 6*f*x])/(8*a^3*d) + Log[c + d*x]/(8*a^3* d) - ((I/8)*CosIntegral[(6*c*f)/d + 6*f*x]*Sin[6*e - (6*c*f)/d])/(a^3*d) - (((3*I)/8)*CosIntegral[(4*c*f)/d + 4*f*x]*Sin[4*e - (4*c*f)/d])/(a^3*d) - (((3*I)/8)*CosIntegral[(2*c*f)/d + 2*f*x]*Sin[2*e - (2*c*f)/d])/(a^3*d) - (((3*I)/8)*Cos[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(a^3*d) - (3*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(8*a^3*d) - (((3* I)/8)*Cos[4*e - (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(a^3*d) - (3*Si n[4*e - (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(8*a^3*d) - ((I/8)*Cos[ 6*e - (6*c*f)/d]*SinIntegral[(6*c*f)/d + 6*f*x])/(a^3*d) - (Sin[6*e - (6*c *f)/d]*SinIntegral[(6*c*f)/d + 6*f*x])/(8*a^3*d)
3.1.32.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(2*a) + Cos[2*e + 2*f*x]/( 2*a) + Sin[2*e + 2*f*x]/(2*b))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0] && ILtQ[m, 0] && ILtQ[n, 0]
Time = 0.93 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.36
method | result | size |
risch | \(\frac {\ln \left (d x +c \right )}{8 a^{3} d}-\frac {{\mathrm e}^{\frac {6 i \left (c f -d e \right )}{d}} \operatorname {Ei}_{1}\left (6 i f x +6 i e +\frac {6 i \left (c f -d e \right )}{d}\right )}{8 a^{3} d}-\frac {3 \,{\mathrm e}^{\frac {4 i \left (c f -d e \right )}{d}} \operatorname {Ei}_{1}\left (4 i f x +4 i e +\frac {4 i \left (c f -d e \right )}{d}\right )}{8 a^{3} d}-\frac {3 \,{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \operatorname {Ei}_{1}\left (2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{8 a^{3} d}\) | \(163\) |
1/8*ln(d*x+c)/a^3/d-1/8/a^3/d*exp(6*I*(c*f-d*e)/d)*Ei(1,6*I*f*x+6*I*e+6*I* (c*f-d*e)/d)-3/8/a^3/d*exp(4*I*(c*f-d*e)/d)*Ei(1,4*I*f*x+4*I*e+4*I*(c*f-d* e)/d)-3/8/a^3/d*exp(2*I*(c*f-d*e)/d)*Ei(1,2*I*f*x+2*I*e+2*I*(c*f-d*e)/d)
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.26 \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\frac {3 \, {\rm Ei}\left (-\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (i \, d e - i \, c f\right )}}{d}\right )} + 3 \, {\rm Ei}\left (-\frac {4 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {4 \, {\left (i \, d e - i \, c f\right )}}{d}\right )} + {\rm Ei}\left (-\frac {6 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {6 \, {\left (i \, d e - i \, c f\right )}}{d}\right )} + \log \left (\frac {d x + c}{d}\right )}{8 \, a^{3} d} \]
1/8*(3*Ei(-2*(I*d*f*x + I*c*f)/d)*e^(-2*(I*d*e - I*c*f)/d) + 3*Ei(-4*(I*d* f*x + I*c*f)/d)*e^(-4*(I*d*e - I*c*f)/d) + Ei(-6*(I*d*f*x + I*c*f)/d)*e^(- 6*(I*d*e - I*c*f)/d) + log((d*x + c)/d))/(a^3*d)
\[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {1}{c \tan ^{3}{\left (e + f x \right )} - 3 i c \tan ^{2}{\left (e + f x \right )} - 3 c \tan {\left (e + f x \right )} + i c + d x \tan ^{3}{\left (e + f x \right )} - 3 i d x \tan ^{2}{\left (e + f x \right )} - 3 d x \tan {\left (e + f x \right )} + i d x}\, dx}{a^{3}} \]
I*Integral(1/(c*tan(e + f*x)**3 - 3*I*c*tan(e + f*x)**2 - 3*c*tan(e + f*x) + I*c + d*x*tan(e + f*x)**3 - 3*I*d*x*tan(e + f*x)**2 - 3*d*x*tan(e + f*x ) + I*d*x), x)/a**3
Time = 0.31 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=-\frac {3 \, f \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) + 3 \, f \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) + f \cos \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (-\frac {6 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) + 3 i \, f E_{1}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 3 i \, f E_{1}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + i \, f E_{1}\left (-\frac {6 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) \sin \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) - f \log \left ({\left (f x + e\right )} d - d e + c f\right )}{8 \, a^{3} d f} \]
-1/8*(3*f*cos(-2*(d*e - c*f)/d)*exp_integral_e(1, -2*(-I*(f*x + e)*d + I*d *e - I*c*f)/d) + 3*f*cos(-4*(d*e - c*f)/d)*exp_integral_e(1, -4*(-I*(f*x + e)*d + I*d*e - I*c*f)/d) + f*cos(-6*(d*e - c*f)/d)*exp_integral_e(1, -6*( -I*(f*x + e)*d + I*d*e - I*c*f)/d) + 3*I*f*exp_integral_e(1, -2*(-I*(f*x + e)*d + I*d*e - I*c*f)/d)*sin(-2*(d*e - c*f)/d) + 3*I*f*exp_integral_e(1, -4*(-I*(f*x + e)*d + I*d*e - I*c*f)/d)*sin(-4*(d*e - c*f)/d) + I*f*exp_int egral_e(1, -6*(-I*(f*x + e)*d + I*d*e - I*c*f)/d)*sin(-6*(d*e - c*f)/d) - f*log((f*x + e)*d - d*e + c*f))/(a^3*d*f)
Time = 0.58 (sec) , antiderivative size = 810, normalized size of antiderivative = 1.80 \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\text {Too large to display} \]
1/8*(3*cos(2*e)^2*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/d) + cos(2*e) ^3*log(d*x + c) + 6*I*cos(2*e)*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/ d)*sin(2*e) + 3*I*cos(2*e)^2*log(d*x + c)*sin(2*e) - 3*cos(2*c*f/d)*cos_in tegral(-2*(d*f*x + c*f)/d)*sin(2*e)^2 - 3*cos(2*e)*log(d*x + c)*sin(2*e)^2 - I*log(d*x + c)*sin(2*e)^3 + 3*I*cos(2*e)^2*cos_integral(-2*(d*f*x + c*f )/d)*sin(2*c*f/d) - 6*cos(2*e)*cos_integral(-2*(d*f*x + c*f)/d)*sin(2*e)*s in(2*c*f/d) - 3*I*cos_integral(-2*(d*f*x + c*f)/d)*sin(2*e)^2*sin(2*c*f/d) - 3*I*cos(2*e)^2*cos(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + 6*cos(2*e )*cos(2*c*f/d)*sin(2*e)*sin_integral(2*(d*f*x + c*f)/d) + 3*I*cos(2*c*f/d) *sin(2*e)^2*sin_integral(2*(d*f*x + c*f)/d) + 3*cos(2*e)^2*sin(2*c*f/d)*si n_integral(2*(d*f*x + c*f)/d) + 6*I*cos(2*e)*sin(2*e)*sin(2*c*f/d)*sin_int egral(2*(d*f*x + c*f)/d) - 3*sin(2*e)^2*sin(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + 3*cos(2*e)*cos(4*c*f/d)*cos_integral(-4*(d*f*x + c*f)/d) + 3* I*cos(4*c*f/d)*cos_integral(-4*(d*f*x + c*f)/d)*sin(2*e) + 3*I*cos(2*e)*co s_integral(-4*(d*f*x + c*f)/d)*sin(4*c*f/d) - 3*cos_integral(-4*(d*f*x + c *f)/d)*sin(2*e)*sin(4*c*f/d) - 3*I*cos(2*e)*cos(4*c*f/d)*sin_integral(4*(d *f*x + c*f)/d) + 3*cos(4*c*f/d)*sin(2*e)*sin_integral(4*(d*f*x + c*f)/d) + 3*cos(2*e)*sin(4*c*f/d)*sin_integral(4*(d*f*x + c*f)/d) + 3*I*sin(2*e)*si n(4*c*f/d)*sin_integral(4*(d*f*x + c*f)/d) + cos(6*c*f/d)*cos_integral(-6* (d*f*x + c*f)/d) + I*cos_integral(-6*(d*f*x + c*f)/d)*sin(6*c*f/d) - I*...
Timed out. \[ \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^3} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,\left (c+d\,x\right )} \,d x \]