3.1.19 \(\int \frac {1}{(c+d x) (a+a \coth (e+f x))} \, dx\) [19]

Optimal. Leaf size=157 \[ -\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}+\frac {\log (c+d x)}{2 a d}+\frac {\text {Chi}\left (\frac {2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{2 a d}+\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d}-\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a d} \]

[Out]

-1/2*Chi(2*c*f/d+2*f*x)*cosh(-2*e+2*c*f/d)/a/d+1/2*ln(d*x+c)/a/d+1/2*cosh(-2*e+2*c*f/d)*Shi(2*c*f/d+2*f*x)/a/d
-1/2*Chi(2*c*f/d+2*f*x)*sinh(-2*e+2*c*f/d)/a/d+1/2*Shi(2*c*f/d+2*f*x)*sinh(-2*e+2*c*f/d)/a/d

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Rubi [A]
time = 0.20, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3807, 3384, 3379, 3382} \begin {gather*} \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 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 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 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 d}+\frac {\log (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(a*d) + Log[c + d*x]/(2*a*d) + (CoshIntegral[(2*c
*f)/d + 2*f*x]*Sinh[2*e - (2*c*f)/d])/(2*a*d) + (Cosh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a*d
) - (Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a*d)

Rule 3379

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 3382

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 3384

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 3807

Int[1/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Log[c + d*x]/(2*a*d), x
] + (Dist[1/(2*a), Int[Cos[2*e + 2*f*x]/(c + d*x), x], x] + Dist[1/(2*b), Int[Sin[2*e + 2*f*x]/(c + d*x), x],
x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(c+d x) (a+a \coth (e+f x))} \, dx &=\frac {\log (c+d x)}{2 a d}+\frac {i \int \frac {\sin \left (2 \left (i e+\frac {\pi }{2}\right )+2 i f x\right )}{c+d x} \, dx}{2 a}+\frac {\int \frac {\cos \left (2 \left (i e+\frac {\pi }{2}\right )+2 i f x\right )}{c+d x} \, dx}{2 a}\\ &=\frac {\log (c+d x)}{2 a d}-\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}+\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}+\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}-\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}\\ &=-\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 d}+\frac {\log (c+d x)}{2 a 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 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 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 d}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 122, normalized size = 0.78 \begin {gather*} \frac {\text {csch}(e+f x) (\cosh (f x)+\sinh (f x)) \left (\log (f (c+d x)) (\cosh (e)+\sinh (e))+\text {Chi}\left (\frac {2 f (c+d x)}{d}\right ) \left (-\cosh \left (e-\frac {2 c f}{d}\right )+\sinh \left (e-\frac {2 c f}{d}\right )\right )+\left (\cosh \left (e-\frac {2 c f}{d}\right )-\sinh \left (e-\frac {2 c f}{d}\right )\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )\right )}{2 a d (1+\coth (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csch[e + f*x]*(Cosh[f*x] + Sinh[f*x])*(Log[f*(c + d*x)]*(Cosh[e] + Sinh[e]) + CoshIntegral[(2*f*(c + d*x))/d]
*(-Cosh[e - (2*c*f)/d] + Sinh[e - (2*c*f)/d]) + (Cosh[e - (2*c*f)/d] - Sinh[e - (2*c*f)/d])*SinhIntegral[(2*f*
(c + d*x))/d]))/(2*a*d*(1 + Coth[e + f*x]))

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Maple [A]
time = 10.85, size = 61, normalized size = 0.39

method	result	size
risch	\(\frac {\ln \left (d x +c \right )}{2 a d}+\frac {{\mathrm e}^{\frac {2 c f -2 d e}{d}} \expIntegral \left (1, 2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{2 a d}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x+c)/(a+a*coth(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(d*x+c)/a/d+1/2/a/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 = 0.40, size = 49, normalized size = 0.31 \begin {gather*} \frac {e^{\left (\frac {2 \, c f}{d} - 2 \, e\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{2 \, a d} + \frac {\log \left (d x + c\right )}{2 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 88, normalized size = 0.56 \begin {gather*} -\frac {{\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {2 \, {\left (c f - d \cosh \left (1\right ) - d \sinh \left (1\right )\right )}}{d}\right ) - {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {2 \, {\left (c f - d \cosh \left (1\right ) - d \sinh \left (1\right )\right )}}{d}\right ) - \log \left (d x + c\right )}{2 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{c \coth {\left (e + f x \right )} + c + d x \coth {\left (e + f x \right )} + d x}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 48, normalized size = 0.31 \begin {gather*} -\frac {{\left ({\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {2 \, c f}{d}\right )} - e^{\left (2 \, e\right )} \log \left (d x + c\right )\right )} e^{\left (-2 \, e\right )}}{2 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (a+a\,\mathrm {coth}\left (e+f\,x\right )\right )\,\left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*coth(e + f*x))*(c + d*x)),x)

[Out]

int(1/((a + a*coth(e + f*x))*(c + d*x)), x)

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