3.282 \(\int \frac {a+b \log (\frac {e}{e+f x})}{e^2-f^2 x^2} \, dx\)

Optimal. Leaf size=47 \[ \frac {(a-b \log (2)) \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f}+\frac {b \text {Li}_2\left (1-\frac {2 e}{e+f x}\right )}{2 e f} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2403, 208, 2402, 2315} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f}+\frac {(a-b \log (2)) \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

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 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 2403

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 {e}{e+f x}\right )}{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+(a-b \log (2)) \int \frac {1}{e^2-f^2 x^2} \, dx\\ &=\frac {\tanh ^{-1}\left (\frac {f x}{e}\right ) (a-b \log (2))}{e f}+\frac {b \operatorname {Subst}\left (\int \frac {\log (2 e x)}{1-2 e x} \, dx,x,\frac {1}{e+f x}\right )}{f}\\ &=\frac {\tanh ^{-1}\left (\frac {f x}{e}\right ) (a-b \log (2))}{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 = 0.03, size = 80, normalized size = 1.70 \[ \frac {2 b^2 \text {Li}_2\left (\frac {e+f x}{2 e}\right )-\left (a+b \log \left (\frac {e}{e+f x}\right )\right ) \left (a+2 b \log \left (\frac {e-f x}{2 e}\right )+b \log \left (\frac {e}{e+f x}\right )\right )}{4 b e f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \log \left (\frac {e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 109, normalized size = 2.32 \[ -\frac {b \ln \left (\frac {e}{f x +e}\right ) \ln \left (-\frac {2 e}{f x +e}+1\right )}{2 e f}+\frac {b \ln \left (\frac {2 e}{f x +e}\right ) \ln \left (-\frac {2 e}{f x +e}+1\right )}{2 e f}-\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{2 e f}+\frac {b \dilog \left (\frac {2 e}{f x +e}\right )}{2 e f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a/(-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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