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

Optimal. Leaf size=62 \[ \frac {\tanh ^{-1}\left (\sqrt {e} x\right ) (a+b \log (c x))}{\sqrt {e}}+\frac {b \text {Li}_2\left (-\sqrt {e} x\right )}{2 \sqrt {e}}-\frac {b \text {Li}_2\left (\sqrt {e} x\right )}{2 \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 12

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

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 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*}

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Mathematica [A]  time = 0.03, size = 68, normalized size = 1.10 \[ \frac {-\left (\left (\log \left (1-\sqrt {e} x\right )-\log \left (\sqrt {e} x+1\right )\right ) (a+b \log (c x))\right )+b \text {Li}_2\left (-\sqrt {e} x\right )-b \text {Li}_2\left (\sqrt {e} x\right )}{2 \sqrt {e}} \]

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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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \log \left (c x\right ) + a}{e x^{2} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \log \left (c x\right ) + a}{e x^{2} - 1}\,{d x} \]

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)

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maple [B]  time = 0.06, size = 103, normalized size = 1.66 \[ \frac {b \ln \left (c x \right ) \ln \left (\frac {c \sqrt {e}\, x +c}{c}\right )}{2 \sqrt {e}}-\frac {b \ln \left (c x \right ) \ln \left (-\frac {c \sqrt {e}\, x -c}{c}\right )}{2 \sqrt {e}}+\frac {a \arctanh \left (\sqrt {e}\, x \right )}{\sqrt {e}}+\frac {b \dilog \left (\frac {c \sqrt {e}\, x +c}{c}\right )}{2 \sqrt {e}}-\frac {b \dilog \left (-\frac {c \sqrt {e}\, x -c}{c}\right )}{2 \sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {b \log \left (c x\right ) + a}{e x^{2} - 1}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int -\frac {a+b\,\ln \left (c\,x\right )}{e\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*log(c*x))/(e*x^2 - 1),x)

[Out]

int(-(a + b*log(c*x))/(e*x^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a}{e x^{2} - 1}\, dx - \int \frac {b \log {\left (c x \right )}}{e x^{2} - 1}\, dx \]

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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