Integrand size = 20, antiderivative size = 437 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\frac {3 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {3 \cosh \left (4 e-\frac {4 c f}{d}\right ) \text {Chi}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}+\frac {\cosh \left (6 e-\frac {6 c f}{d}\right ) \text {Chi}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}-\frac {\text {Chi}\left (\frac {6 c f}{d}+6 f x\right ) \sinh \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \text {Chi}\left (\frac {4 c f}{d}+4 f x\right ) \sinh \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {3 \text {Chi}\left (\frac {2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}+\frac {3 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{8 a^3 d}-\frac {3 \cosh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}+\frac {3 \sinh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^3 d}-\frac {\cosh \left (6 e-\frac {6 c f}{d}\right ) \text {Shi}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d}+\frac {\sinh \left (6 e-\frac {6 c f}{d}\right ) \text {Shi}\left (\frac {6 c f}{d}+6 f x\right )}{8 a^3 d} \]
1/8*Chi(6*c*f/d+6*f*x)*cosh(-6*e+6*c*f/d)/a^3/d+3/8*Chi(4*c*f/d+4*f*x)*cos h(-4*e+4*c*f/d)/a^3/d+3/8*Chi(2*c*f/d+2*f*x)*cosh(-2*e+2*c*f/d)/a^3/d+1/8* ln(d*x+c)/a^3/d-3/8*cosh(-2*e+2*c*f/d)*Shi(2*c*f/d+2*f*x)/a^3/d-3/8*cosh(- 4*e+4*c*f/d)*Shi(4*c*f/d+4*f*x)/a^3/d-1/8*cosh(-6*e+6*c*f/d)*Shi(6*c*f/d+6 *f*x)/a^3/d+1/8*Chi(6*c*f/d+6*f*x)*sinh(-6*e+6*c*f/d)/a^3/d-1/8*Shi(6*c*f/ d+6*f*x)*sinh(-6*e+6*c*f/d)/a^3/d+3/8*Chi(4*c*f/d+4*f*x)*sinh(-4*e+4*c*f/d )/a^3/d-3/8*Shi(4*c*f/d+4*f*x)*sinh(-4*e+4*c*f/d)/a^3/d+3/8*Chi(2*c*f/d+2* f*x)*sinh(-2*e+2*c*f/d)/a^3/d-3/8*Shi(2*c*f/d+2*f*x)*sinh(-2*e+2*c*f/d)/a^ 3/d
Time = 0.96 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\frac {\text {sech}^3(e+f x) (\cosh (f x)+\sinh (f x))^3 \left (\cosh (3 e) \log (f (c+d x))+\log (f (c+d x)) \sinh (3 e)+\left (\cosh \left (e-\frac {4 c f}{d}\right )-\sinh \left (e-\frac {4 c f}{d}\right )\right ) \left (3 \text {Chi}\left (\frac {4 f (c+d x)}{d}\right )+\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {6 f (c+d x)}{d}\right )-\text {Chi}\left (\frac {6 f (c+d x)}{d}\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )+3 \text {Chi}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cosh \left (2 e-\frac {2 c f}{d}\right )+\sinh \left (2 e-\frac {2 c f}{d}\right )\right )-3 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )-3 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )-3 \text {Shi}\left (\frac {4 f (c+d x)}{d}\right )-\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {6 f (c+d x)}{d}\right )+\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {6 f (c+d x)}{d}\right )\right )\right )}{8 a^3 d (1+\tanh (e+f x))^3} \]
(Sech[e + f*x]^3*(Cosh[f*x] + Sinh[f*x])^3*(Cosh[3*e]*Log[f*(c + d*x)] + L og[f*(c + d*x)]*Sinh[3*e] + (Cosh[e - (4*c*f)/d] - Sinh[e - (4*c*f)/d])*(3 *CoshIntegral[(4*f*(c + d*x))/d] + Cosh[2*e - (2*c*f)/d]*CoshIntegral[(6*f *(c + d*x))/d] - CoshIntegral[(6*f*(c + d*x))/d]*Sinh[2*e - (2*c*f)/d] + 3 *CoshIntegral[(2*f*(c + d*x))/d]*(Cosh[2*e - (2*c*f)/d] + Sinh[2*e - (2*c* f)/d]) - 3*Cosh[2*e - (2*c*f)/d]*SinhIntegral[(2*f*(c + d*x))/d] - 3*Sinh[ 2*e - (2*c*f)/d]*SinhIntegral[(2*f*(c + d*x))/d] - 3*SinhIntegral[(4*f*(c + d*x))/d] - Cosh[2*e - (2*c*f)/d]*SinhIntegral[(6*f*(c + d*x))/d] + Sinh[ 2*e - (2*c*f)/d]*SinhIntegral[(6*f*(c + d*x))/d])))/(8*a^3*d*(1 + Tanh[e + f*x])^3)
Time = 2.01 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.
| \(\displaystyle \int \frac {1}{(c+d x) (a \tanh (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(c+d x) (a-i a \tan (i e+i f x))^3}dx\) |
| \(\Big \downarrow \) 4211 |
| \(\displaystyle \int \left (-\frac {\sinh ^3(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \sinh ^2(2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 \sinh (2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \sinh (2 e+2 f x) \sinh (4 e+4 f x)}{16 a^3 (c+d x)}-\frac {3 \sinh (4 e+4 f x)}{8 a^3 (c+d x)}+\frac {\cosh ^3(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \cosh ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {3 \cosh (2 e+2 f x)}{8 a^3 (c+d x)}-\frac {3 \sinh (2 e+2 f x) \cosh ^2(2 e+2 f x)}{8 a^3 (c+d x)}+\frac {1}{8 a^3 (c+d x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {\text {Chi}\left (6 x f+\frac {6 c f}{d}\right ) \sinh \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \text {Chi}\left (4 x f+\frac {4 c f}{d}\right ) \sinh \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{8 a^3 d}+\frac {3 \text {Chi}\left (4 x f+\frac {4 c f}{d}\right ) \cosh \left (4 e-\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {\text {Chi}\left (6 x f+\frac {6 c f}{d}\right ) \cosh \left (6 e-\frac {6 c f}{d}\right )}{8 a^3 d}+\frac {3 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}+\frac {3 \sinh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}+\frac {\sinh \left (6 e-\frac {6 c f}{d}\right ) \text {Shi}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}-\frac {3 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{8 a^3 d}-\frac {3 \cosh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (4 x f+\frac {4 c f}{d}\right )}{8 a^3 d}-\frac {\cosh \left (6 e-\frac {6 c f}{d}\right ) \text {Shi}\left (6 x f+\frac {6 c f}{d}\right )}{8 a^3 d}+\frac {\log (c+d x)}{8 a^3 d}\) |
(3*Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(8*a^3*d) + (3*C osh[4*e - (4*c*f)/d]*CoshIntegral[(4*c*f)/d + 4*f*x])/(8*a^3*d) + (Cosh[6* e - (6*c*f)/d]*CoshIntegral[(6*c*f)/d + 6*f*x])/(8*a^3*d) + Log[c + d*x]/( 8*a^3*d) - (CoshIntegral[(6*c*f)/d + 6*f*x]*Sinh[6*e - (6*c*f)/d])/(8*a^3* d) - (3*CoshIntegral[(4*c*f)/d + 4*f*x]*Sinh[4*e - (4*c*f)/d])/(8*a^3*d) - (3*CoshIntegral[(2*c*f)/d + 2*f*x]*Sinh[2*e - (2*c*f)/d])/(8*a^3*d) - (3* Cosh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(8*a^3*d) + (3*Sinh [2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(8*a^3*d) - (3*Cosh[4*e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(8*a^3*d) + (3*Sinh[4*e - ( 4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(8*a^3*d) - (Cosh[6*e - (6*c*f) /d]*SinhIntegral[(6*c*f)/d + 6*f*x])/(8*a^3*d) + (Sinh[6*e - (6*c*f)/d]*Si nhIntegral[(6*c*f)/d + 6*f*x])/(8*a^3*d)
3.1.46.3.1 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 = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {\ln \left (d x +c \right )}{8 a^{3} d}-\frac {{\mathrm e}^{\frac {6 c f -6 d e}{d}} \operatorname {Ei}_{1}\left (6 f x +6 e +\frac {6 c f -6 d e}{d}\right )}{8 a^{3} d}-\frac {3 \,{\mathrm e}^{\frac {4 c f -4 d e}{d}} \operatorname {Ei}_{1}\left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right )}{8 a^{3} d}-\frac {3 \,{\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {Ei}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{8 a^{3} d}\) | \(151\) |
1/8*ln(d*x+c)/a^3/d-1/8/a^3/d*exp(6*(c*f-d*e)/d)*Ei(1,6*f*x+6*e+6*(c*f-d*e )/d)-3/8/a^3/d*exp(4*(c*f-d*e)/d)*Ei(1,4*f*x+4*e+4*(c*f-d*e)/d)-3/8/a^3/d* exp(2*(c*f-d*e)/d)*Ei(1,2*f*x+2*e+2*(c*f-d*e)/d)
Time = 0.26 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\frac {3 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 3 \, {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + {\rm Ei}\left (-\frac {6 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) + 3 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 3 \, {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + {\rm Ei}\left (-\frac {6 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {6 \, {\left (d e - c f\right )}}{d}\right ) + \log \left (d x + c\right )}{8 \, a^{3} d} \]
1/8*(3*Ei(-2*(d*f*x + c*f)/d)*cosh(-2*(d*e - c*f)/d) + 3*Ei(-4*(d*f*x + c* f)/d)*cosh(-4*(d*e - c*f)/d) + Ei(-6*(d*f*x + c*f)/d)*cosh(-6*(d*e - c*f)/ d) + 3*Ei(-2*(d*f*x + c*f)/d)*sinh(-2*(d*e - c*f)/d) + 3*Ei(-4*(d*f*x + c* f)/d)*sinh(-4*(d*e - c*f)/d) + Ei(-6*(d*f*x + c*f)/d)*sinh(-6*(d*e - c*f)/ d) + log(d*x + c))/(a^3*d)
\[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\frac {\int \frac {1}{c \tanh ^{3}{\left (e + f x \right )} + 3 c \tanh ^{2}{\left (e + f x \right )} + 3 c \tanh {\left (e + f x \right )} + c + d x \tanh ^{3}{\left (e + f x \right )} + 3 d x \tanh ^{2}{\left (e + f x \right )} + 3 d x \tanh {\left (e + f x \right )} + d x}\, dx}{a^{3}} \]
Integral(1/(c*tanh(e + f*x)**3 + 3*c*tanh(e + f*x)**2 + 3*c*tanh(e + f*x) + c + d*x*tanh(e + f*x)**3 + 3*d*x*tanh(e + f*x)**2 + 3*d*x*tanh(e + f*x) + d*x), x)/a**3
Time = 2.49 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.26 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=-\frac {e^{\left (-6 \, e + \frac {6 \, c f}{d}\right )} E_{1}\left (\frac {6 \, {\left (d x + c\right )} f}{d}\right )}{8 \, a^{3} d} - \frac {3 \, e^{\left (-4 \, e + \frac {4 \, c f}{d}\right )} E_{1}\left (\frac {4 \, {\left (d x + c\right )} f}{d}\right )}{8 \, a^{3} d} - \frac {3 \, e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{8 \, a^{3} d} + \frac {\log \left (d x + c\right )}{8 \, a^{3} d} \]
-1/8*e^(-6*e + 6*c*f/d)*exp_integral_e(1, 6*(d*x + c)*f/d)/(a^3*d) - 3/8*e ^(-4*e + 4*c*f/d)*exp_integral_e(1, 4*(d*x + c)*f/d)/(a^3*d) - 3/8*e^(-2*e + 2*c*f/d)*exp_integral_e(1, 2*(d*x + c)*f/d)/(a^3*d) + 1/8*log(d*x + c)/ (a^3*d)
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.24 \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\frac {{\left (3 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (4 \, e + \frac {2 \, c f}{d}\right )} + 3 \, {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (2 \, e + \frac {4 \, c f}{d}\right )} + {\rm Ei}\left (-\frac {6 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {6 \, c f}{d}\right )} + e^{\left (6 \, e\right )} \log \left (d x + c\right )\right )} e^{\left (-6 \, e\right )}}{8 \, a^{3} d} \]
1/8*(3*Ei(-2*(d*f*x + c*f)/d)*e^(4*e + 2*c*f/d) + 3*Ei(-4*(d*f*x + c*f)/d) *e^(2*e + 4*c*f/d) + Ei(-6*(d*f*x + c*f)/d)*e^(6*c*f/d) + e^(6*e)*log(d*x + c))*e^(-6*e)/(a^3*d)
Timed out. \[ \int \frac {1}{(c+d x) (a+a \tanh (e+f x))^3} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tanh}\left (e+f\,x\right )\right )}^3\,\left (c+d\,x\right )} \,d x \]