3.245 \(\int \frac{a+b \log (c x)}{1-e x^2} \, dx\)

Optimal. Leaf size=62 \[ \frac{b \text{PolyLog}\left (2,-\sqrt{e} x\right )}{2 \sqrt{e}}-\frac{b \text{PolyLog}\left (2,\sqrt{e} x\right )}{2 \sqrt{e}}+\frac{\tanh ^{-1}\left (\sqrt{e} x\right ) (a+b \log (c x))}{\sqrt{e}} \]

[Out]

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

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Rubi [A]  time = 0.0387105, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {206, 2324, 12, 5912} \[ \frac{b \text{PolyLog}\left (2,-\sqrt{e} x\right )}{2 \sqrt{e}}-\frac{b \text{PolyLog}\left (2,\sqrt{e} x\right )}{2 \sqrt{e}}+\frac{\tanh ^{-1}\left (\sqrt{e} x\right ) (a+b \log (c x))}{\sqrt{e}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x])/(1 - e*x^2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

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

Mathematica [A]  time = 0.0295573, size = 68, normalized size = 1.1 \[ \frac{b \text{PolyLog}\left (2,-\sqrt{e} x\right )-b \text{PolyLog}\left (2,\sqrt{e} x\right )+\left (\log \left (1-\sqrt{e} x\right )-\log \left (\sqrt{e} x+1\right )\right ) (-(a+b \log (c x)))}{2 \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x])/(1 - e*x^2),x]

[Out]

(-((a + b*Log[c*x])*(Log[1 - Sqrt[e]*x] - Log[1 + Sqrt[e]*x])) + b*PolyLog[2, -(Sqrt[e]*x)] - b*PolyLog[2, Sqr
t[e]*x])/(2*Sqrt[e])

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Maple [B]  time = 0.133, size = 103, normalized size = 1.7 \begin{align*}{a{\it Artanh} \left ( x\sqrt{e} \right ){\frac{1}{\sqrt{e}}}}-{\frac{b\ln \left ( cx \right ) }{2}\ln \left ( -{\frac{1}{c} \left ( cx\sqrt{e}-c \right ) } \right ){\frac{1}{\sqrt{e}}}}+{\frac{b\ln \left ( cx \right ) }{2}\ln \left ({\frac{1}{c} \left ( cx\sqrt{e}+c \right ) } \right ){\frac{1}{\sqrt{e}}}}-{\frac{b}{2}{\it dilog} \left ( -{\frac{1}{c} \left ( cx\sqrt{e}-c \right ) } \right ){\frac{1}{\sqrt{e}}}}+{\frac{b}{2}{\it dilog} \left ({\frac{1}{c} \left ( cx\sqrt{e}+c \right ) } \right ){\frac{1}{\sqrt{e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x))/(-e*x^2+1),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(-e*x^2+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(-e*x^2+1),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x))/(-e*x**2+1),x)

[Out]

-Integral(a/(e*x**2 - 1), x) - Integral(b*log(c*x)/(e*x**2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(-e*x^2+1),x, algorithm="giac")

[Out]

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