3.151 \(\int \frac{a+b \tanh ^{-1}(c x)}{x (d+e x)} \, dx\)

Optimal. Leaf size=148 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}-\frac{b \text{PolyLog}(2,-c x)}{2 d}+\frac{b \text{PolyLog}(2,c x)}{2 d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]

[Out]

(a*Log[x])/d + ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d - ((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e
)*(1 + c*x))])/d - (b*PolyLog[2, -(c*x)])/(2*d) + (b*PolyLog[2, c*x])/(2*d) - (b*PolyLog[2, 1 - 2/(1 + c*x)])/
(2*d) + (b*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d)

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Rubi [A]  time = 0.163711, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {5940, 5912, 5920, 2402, 2315, 2447} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}-\frac{b \text{PolyLog}(2,-c x)}{2 d}+\frac{b \text{PolyLog}(2,c x)}{2 d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x*(d + e*x)),x]

[Out]

(a*Log[x])/d + ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d - ((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e
)*(1 + c*x))])/d - (b*PolyLog[2, -(c*x)])/(2*d) + (b*PolyLog[2, c*x])/(2*d) - (b*PolyLog[2, 1 - 2/(1 + c*x)])/
(2*d) + (b*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d)

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rule 5920

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}(c x)}{x (d+e x)} \, dx &=\int \left (\frac{a+b \tanh ^{-1}(c x)}{d x}-\frac{e \left (a+b \tanh ^{-1}(c x)\right )}{d (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx}{d}-\frac{e \int \frac{a+b \tanh ^{-1}(c x)}{d+e x} \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \text{Li}_2(-c x)}{2 d}+\frac{b \text{Li}_2(c x)}{2 d}-\frac{(b c) \int \frac{\log \left (\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d}+\frac{(b c) \int \frac{\log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \text{Li}_2(-c x)}{2 d}+\frac{b \text{Li}_2(c x)}{2 d}+\frac{b \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}-\frac{b \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c x}\right )}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \text{Li}_2(-c x)}{2 d}+\frac{b \text{Li}_2(c x)}{2 d}-\frac{b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{b \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}\\ \end{align*}

Mathematica [C]  time = 1.63877, size = 294, normalized size = 1.99 \[ \frac{\frac{b \left (c d \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-c d \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+e \sqrt{1-\frac{c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )}+\frac{1}{2} i \pi c d \log \left (1-c^2 x^2\right )-2 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac{c d}{e}\right )-2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 c d \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+c d \tanh ^{-1}(c x)^2-i \pi c d \tanh ^{-1}(c x)+2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+i \pi c d \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-e \tanh ^{-1}(c x)^2\right )}{c}-2 a d \log (d+e x)+2 a d \log (x)}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])/(x*(d + e*x)),x]

[Out]

(2*a*d*Log[x] - 2*a*d*Log[d + e*x] + (b*((-I)*c*d*Pi*ArcTanh[c*x] - 2*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x] + c*d*
ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] + 2*c*d*ArcT
anh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d*ArcTanh[(c*d)/e]*Log[1 -
E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])
)] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d
*PolyLog[2, E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/c)/(2*d^2)

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Maple [A]  time = 0.119, size = 210, normalized size = 1.4 \begin{align*} -{\frac{a\ln \left ( cxe+cd \right ) }{d}}+{\frac{a\ln \left ( cx \right ) }{d}}-{\frac{b{\it Artanh} \left ( cx \right ) \ln \left ( cxe+cd \right ) }{d}}+{\frac{b{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) }{d}}+{\frac{b\ln \left ( cxe+cd \right ) }{2\,d}\ln \left ({\frac{cxe+e}{-cd+e}} \right ) }+{\frac{b}{2\,d}{\it dilog} \left ({\frac{cxe+e}{-cd+e}} \right ) }-{\frac{b\ln \left ( cxe+cd \right ) }{2\,d}\ln \left ({\frac{cxe-e}{-cd-e}} \right ) }-{\frac{b}{2\,d}{\it dilog} \left ({\frac{cxe-e}{-cd-e}} \right ) }-{\frac{b{\it dilog} \left ( cx \right ) }{2\,d}}-{\frac{b{\it dilog} \left ( cx+1 \right ) }{2\,d}}-{\frac{b\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x/(e*x+d),x)

[Out]

-a/d*ln(c*e*x+c*d)+a/d*ln(c*x)-b*arctanh(c*x)/d*ln(c*e*x+c*d)+b*arctanh(c*x)/d*ln(c*x)+1/2*b/d*ln((c*e*x+e)/(-
c*d+e))*ln(c*e*x+c*d)+1/2*b/d*dilog((c*e*x+e)/(-c*d+e))-1/2*b/d*ln((c*e*x-e)/(-c*d-e))*ln(c*e*x+c*d)-1/2*b/d*d
ilog((c*e*x-e)/(-c*d-e))-1/2*b/d*dilog(c*x)-1/2*b/d*dilog(c*x+1)-1/2*b/d*ln(c*x)*ln(c*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(e*x+d),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (c x\right ) + a}{e x^{2} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(e*x^2 + d*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{atanh}{\left (c x \right )}}{x \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x/(e*x+d),x)

[Out]

Integral((a + b*atanh(c*x))/(x*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c x\right ) + a}{{\left (e x + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)/((e*x + d)*x), x)