3.3.81 \(\int \frac {a+b \log (\frac {2 e}{e+f x})}{e^2-f^2 x^2} \, dx\) [281]

Optimal. Leaf size=41 \[ \frac {a \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f}+\frac {b \text {Li}_2\left (1-\frac {2 e}{e+f x}\right )}{2 e f} \]

[Out]

a*arctanh(f*x/e)/e/f+1/2*b*polylog(2,1-2*e/(f*x+e))/e/f

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Rubi [A]
time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2450, 214, 2449, 2352} \begin {gather*} \frac {b \text {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f}+\frac {a \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[(2*e)/(e + f*x)])/(e^2 - f^2*x^2),x]

[Out]

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

Rule 214

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

Rule 2352

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

Rule 2449

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 2450

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 82, normalized size = 2.00 \begin {gather*} \frac {-\left (\left (a+b \log \left (\frac {2 e}{e+f x}\right )\right ) \left (a+2 b \log \left (\frac {e-f x}{2 e}\right )+b \log \left (\frac {2 e}{e+f x}\right )\right )\right )+2 b^2 \text {Li}_2\left (\frac {e+f x}{2 e}\right )}{4 b e f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*Log[(2*e)/(e + f*x)])*(a + 2*b*Log[(e - f*x)/(2*e)] + b*Log[(2*e)/(e + f*x)])) + 2*b^2*PolyLog[2, (e
+ f*x)/(2*e)])/(4*b*e*f)

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Maple [A]
time = 0.84, size = 44, normalized size = 1.07

method	result	size
derivativedivides	\(-\frac {2 e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{4 e^{2}}-\frac {b \dilog \left (\frac {2 e}{f x +e}\right )}{4 e^{2}}\right )}{f}\)	\(44\)
default	\(-\frac {2 e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{4 e^{2}}-\frac {b \dilog \left (\frac {2 e}{f x +e}\right )}{4 e^{2}}\right )}{f}\)	\(44\)
risch	\(-\frac {a \ln \left (f x -e \right )}{2 e f}+\frac {a \ln \left (f x +e \right )}{2 e f}+\frac {b \dilog \left (\frac {2 e}{f x +e}\right )}{2 f e}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(2*e/(f*x+e)))/(-f^2*x^2+e^2),x,method=_RETURNVERBOSE)

[Out]

-2/f*e*(1/4/e^2*a*ln(2*e/(f*x+e)-1)-1/4/e^2*b*dilog(2*e/(f*x+e)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 46, normalized size = 1.12 \begin {gather*} \frac {{\left (b {\rm Li}_2\left (-\frac {2 \, e}{f x + e} + 1\right ) + a \log \left (f x + e\right ) - a \log \left (f x - e\right )\right )} e^{\left (-1\right )}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*dilog(-2*e/(f*x + e) + 1) + a*log(f*x + e) - a*log(f*x - e))*e^(-1)/f

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {b \log {\left (2 \right )}}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {b \log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a/(-e**2 + f**2*x**2), x) - Integral(b*log(2)/(-e**2 + f**2*x**2), x) - Integral(b*log(e/(e + f*x))/
(-e**2 + f**2*x**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.34, size = 43, normalized size = 1.05 \begin {gather*} -\frac {a\,\ln \left (f\,x-e\right )-b\,{\mathrm {Li}}_{\mathrm {2}}\left (\frac {2\,e}{e+f\,x}\right )+a\,\ln \left (\frac {1}{e+f\,x}\right )}{2\,e\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log((2*e)/(e + f*x)))/(e^2 - f^2*x^2),x)

[Out]

-(a*log(f*x - e) - b*dilog((2*e)/(e + f*x)) + a*log(1/(e + f*x)))/(2*e*f)

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