3.41 \(\int \frac{1}{(c+d x) (a+a \tanh (e+f x))^2} \, dx\)

Optimal. Leaf size=297 \[ -\frac{\text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \sinh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\text{Chi}\left (4 x f+\frac{4 c f}{d}\right ) \sinh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \cosh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{\text{Chi}\left (4 x f+\frac{4 c f}{d}\right ) \cosh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{\sinh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\log (c+d x)}{4 a^2 d} \]

[Out]

(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) + (Cosh[4*e - (4*c*f)/d]*CoshIntegral[(4*c*f
)/d + 4*f*x])/(4*a^2*d) + Log[c + d*x]/(4*a^2*d) - (CoshIntegral[(4*c*f)/d + 4*f*x]*Sinh[4*e - (4*c*f)/d])/(4*
a^2*d) - (CoshIntegral[(2*c*f)/d + 2*f*x]*Sinh[2*e - (2*c*f)/d])/(2*a^2*d) - (Cosh[2*e - (2*c*f)/d]*SinhIntegr
al[(2*c*f)/d + 2*f*x])/(2*a^2*d) + (Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) - (Cosh[4
*e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(4*a^2*d) + (Sinh[4*e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4
*f*x])/(4*a^2*d)

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Rubi [A]  time = 0.770409, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3728, 3303, 3298, 3301, 3312} \[ -\frac{\text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \sinh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\text{Chi}\left (4 x f+\frac{4 c f}{d}\right ) \sinh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \cosh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{\text{Chi}\left (4 x f+\frac{4 c f}{d}\right ) \cosh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}+\frac{\sinh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (4 x f+\frac{4 c f}{d}\right )}{4 a^2 d}+\frac{\log (c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + d*x)*(a + a*Tanh[e + f*x])^2),x]

[Out]

(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) + (Cosh[4*e - (4*c*f)/d]*CoshIntegral[(4*c*f
)/d + 4*f*x])/(4*a^2*d) + Log[c + d*x]/(4*a^2*d) - (CoshIntegral[(4*c*f)/d + 4*f*x]*Sinh[4*e - (4*c*f)/d])/(4*
a^2*d) - (CoshIntegral[(2*c*f)/d + 2*f*x]*Sinh[2*e - (2*c*f)/d])/(2*a^2*d) - (Cosh[2*e - (2*c*f)/d]*SinhIntegr
al[(2*c*f)/d + 2*f*x])/(2*a^2*d) + (Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) - (Cosh[4
*e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(4*a^2*d) + (Sinh[4*e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4
*f*x])/(4*a^2*d)

Rule 3728

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]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin{align*} \int \frac{1}{(c+d x) (a+a \tanh (e+f x))^2} \, dx &=\int \left (\frac{1}{4 a^2 (c+d x)}+\frac{\cosh (2 e+2 f x)}{2 a^2 (c+d x)}+\frac{\cosh ^2(2 e+2 f x)}{4 a^2 (c+d x)}-\frac{\sinh (2 e+2 f x)}{2 a^2 (c+d x)}+\frac{\sinh ^2(2 e+2 f x)}{4 a^2 (c+d x)}-\frac{\sinh (4 e+4 f x)}{4 a^2 (c+d x)}\right ) \, dx\\ &=\frac{\log (c+d x)}{4 a^2 d}+\frac{\int \frac{\cosh ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}+\frac{\int \frac{\sinh ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}-\frac{\int \frac{\sinh (4 e+4 f x)}{c+d x} \, dx}{4 a^2}+\frac{\int \frac{\cosh (2 e+2 f x)}{c+d x} \, dx}{2 a^2}-\frac{\int \frac{\sinh (2 e+2 f x)}{c+d x} \, dx}{2 a^2}\\ &=\frac{\log (c+d x)}{4 a^2 d}-\frac{\int \left (\frac{1}{2 (c+d x)}-\frac{\cosh (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}+\frac{\int \left (\frac{1}{2 (c+d x)}+\frac{\cosh (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \int \frac{\sinh \left (\frac{4 c f}{d}+4 f x\right )}{c+d x} \, dx}{4 a^2}+\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \int \frac{\cosh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \int \frac{\sinh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}-\frac{\sinh \left (4 e-\frac{4 c f}{d}\right ) \int \frac{\cosh \left (\frac{4 c f}{d}+4 f x\right )}{c+d x} \, dx}{4 a^2}-\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \int \frac{\cosh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \int \frac{\sinh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}\\ &=\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Chi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\log (c+d x)}{4 a^2 d}-\frac{\text{Chi}\left (\frac{4 c f}{d}+4 f x\right ) \sinh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\text{Chi}\left (\frac{2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (\frac{4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \frac{\int \frac{\cosh (4 e+4 f x)}{c+d x} \, dx}{8 a^2}\\ &=\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Chi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\log (c+d x)}{4 a^2 d}-\frac{\text{Chi}\left (\frac{4 c f}{d}+4 f x\right ) \sinh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\text{Chi}\left (\frac{2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (\frac{4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \int \frac{\cosh \left (\frac{4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}+\frac{\sinh \left (4 e-\frac{4 c f}{d}\right ) \int \frac{\sinh \left (\frac{4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}\right )\\ &=\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Chi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\log (c+d x)}{4 a^2 d}-\frac{\text{Chi}\left (\frac{4 c f}{d}+4 f x\right ) \sinh \left (4 e-\frac{4 c f}{d}\right )}{4 a^2 d}-\frac{\text{Chi}\left (\frac{2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac{2 c f}{d}\right )}{2 a^2 d}-\frac{\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac{\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (\frac{4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac{\cosh \left (4 e-\frac{4 c f}{d}\right ) \text{Chi}\left (\frac{4 c f}{d}+4 f x\right )}{8 a^2 d}+\frac{\sinh \left (4 e-\frac{4 c f}{d}\right ) \text{Shi}\left (\frac{4 c f}{d}+4 f x\right )}{8 a^2 d}\right )\\ \end{align*}

Mathematica [A]  time = 0.443738, size = 199, normalized size = 0.67 \[ \frac{\left (\cosh \left (2 e-\frac{2 c f}{d}\right )-\sinh \left (2 e-\frac{2 c f}{d}\right )\right ) \left (\text{Chi}\left (\frac{4 f (c+d x)}{d}\right ) \left (\cosh \left (2 e-\frac{2 c f}{d}\right )-\sinh \left (2 e-\frac{2 c f}{d}\right )\right )+2 \text{Chi}\left (\frac{2 f (c+d x)}{d}\right )+\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{4 f (c+d x)}{d}\right )-\cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{4 f (c+d x)}{d}\right )+\sinh \left (2 e-\frac{2 c f}{d}\right ) \log (f (c+d x))+\cosh \left (2 e-\frac{2 c f}{d}\right ) \log (f (c+d x))-2 \text{Shi}\left (\frac{2 f (c+d x)}{d}\right )\right )}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + d*x)*(a + a*Tanh[e + f*x])^2),x]

[Out]

((Cosh[2*e - (2*c*f)/d] - Sinh[2*e - (2*c*f)/d])*(2*CoshIntegral[(2*f*(c + d*x))/d] + Cosh[2*e - (2*c*f)/d]*Lo
g[f*(c + d*x)] + CoshIntegral[(4*f*(c + d*x))/d]*(Cosh[2*e - (2*c*f)/d] - Sinh[2*e - (2*c*f)/d]) + Log[f*(c +
d*x)]*Sinh[2*e - (2*c*f)/d] - 2*SinhIntegral[(2*f*(c + d*x))/d] - Cosh[2*e - (2*c*f)/d]*SinhIntegral[(4*f*(c +
d*x))/d] + Sinh[2*e - (2*c*f)/d]*SinhIntegral[(4*f*(c + d*x))/d]))/(4*a^2*d)

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Maple [A]  time = 0.3, size = 106, normalized size = 0.4 \begin{align*}{\frac{\ln \left ( dx+c \right ) }{4\,d{a}^{2}}}-{\frac{1}{4\,d{a}^{2}}{{\rm e}^{4\,{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,4\,fx+4\,e+4\,{\frac{cf-de}{d}} \right ) }-{\frac{1}{2\,d{a}^{2}}{{\rm e}^{2\,{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x+c)/(a+a*tanh(f*x+e))^2,x)

[Out]

1/4*ln(d*x+c)/d/a^2-1/4/a^2/d*exp(4*(c*f-d*e)/d)*Ei(1,4*f*x+4*e+4*(c*f-d*e)/d)-1/2/a^2/d*exp(2*(c*f-d*e)/d)*Ei
(1,2*f*x+2*e+2*(c*f-d*e)/d)

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Maxima [A]  time = 3.49823, size = 109, normalized size = 0.37 \begin{align*} -\frac{e^{\left (-4 \, e + \frac{4 \, c f}{d}\right )} E_{1}\left (\frac{4 \,{\left (d x + c\right )} f}{d}\right )}{4 \, a^{2} d} - \frac{e^{\left (-2 \, e + \frac{2 \, c f}{d}\right )} E_{1}\left (\frac{2 \,{\left (d x + c\right )} f}{d}\right )}{2 \, a^{2} d} + \frac{\log \left (d x + c\right )}{4 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+a*tanh(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/4*e^(-4*e + 4*c*f/d)*exp_integral_e(1, 4*(d*x + c)*f/d)/(a^2*d) - 1/2*e^(-2*e + 2*c*f/d)*exp_integral_e(1,
2*(d*x + c)*f/d)/(a^2*d) + 1/4*log(d*x + c)/(a^2*d)

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Fricas [A]  time = 2.09417, size = 302, normalized size = 1.02 \begin{align*} \frac{2 \,{\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 ) +{\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 ) + 2 \,{\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 ) +{\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 ) + \log \left (d x + c\right )}{4 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+a*tanh(f*x+e))^2,x, algorithm="fricas")

[Out]

1/4*(2*Ei(-2*(d*f*x + c*f)/d)*cosh(-2*(d*e - c*f)/d) + Ei(-4*(d*f*x + c*f)/d)*cosh(-4*(d*e - c*f)/d) + 2*Ei(-2
*(d*f*x + c*f)/d)*sinh(-2*(d*e - c*f)/d) + Ei(-4*(d*f*x + c*f)/d)*sinh(-4*(d*e - c*f)/d) + log(d*x + c))/(a^2*
d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{c \tanh ^{2}{\left (e + f x \right )} + 2 c \tanh{\left (e + f x \right )} + c + d x \tanh ^{2}{\left (e + f x \right )} + 2 d x \tanh{\left (e + f x \right )} + d x}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+a*tanh(f*x+e))**2,x)

[Out]

Integral(1/(c*tanh(e + f*x)**2 + 2*c*tanh(e + f*x) + c + d*x*tanh(e + f*x)**2 + 2*d*x*tanh(e + f*x) + d*x), x)
/a**2

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Giac [A]  time = 1.23541, size = 105, normalized size = 0.35 \begin{align*} \frac{{\left ({\rm Ei}\left (-\frac{4 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac{4 \, c f}{d}\right )} + 2 \,{\rm Ei}\left (-\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac{2 \, c f}{d} + 2 \, e\right )} + e^{\left (4 \, e\right )} \log \left (d x + c\right )\right )} e^{\left (-4 \, e\right )}}{4 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+a*tanh(f*x+e))^2,x, algorithm="giac")

[Out]

1/4*(Ei(-4*(d*f*x + c*f)/d)*e^(4*c*f/d) + 2*Ei(-2*(d*f*x + c*f)/d)*e^(2*c*f/d + 2*e) + e^(4*e)*log(d*x + c))*e
^(-4*e)/(a^2*d)