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

Optimal. Leaf size=37 \[ \frac {\text {Li}_2\left (1-\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b} \]

[Out]

1/2*polylog(2,1+(-a*(1-c)-b*(1+c)*x)/(b*x+a))/a/b

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2447} \[ \frac {\text {PolyLog}\left (2,1-\frac {a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 0.20, size = 252, normalized size = 6.81 \[ \frac {2 \text {Li}_2\left (\frac {(c+1) (a-b x)}{2 a}\right )-2 \text {Li}_2\left (\frac {(c+1) (a+b x)}{2 a c}\right )+\log ^2\left (\frac {2 a c}{(c+1) (a+b x)}\right )+2 \log \left (-\frac {a (-c)+a+b (c+1) x}{2 a c}\right ) \log \left (\frac {2 a c}{(c+1) (a+b x)}\right )-2 \log \left (\frac {a (-c)+a+b (c+1) x}{a+b x}\right ) \log \left (\frac {2 a c}{(c+1) (a+b x)}\right )+2 \log (a-b x) \log \left (\frac {a (-c)+a+b (c+1) x}{2 a}\right )-2 \log (a-b x) \log \left (\frac {a (-c)+a+b (c+1) x}{a+b x}\right )-2 \text {Li}_2\left (\frac {a-b x}{2 a}\right )-2 \log (a-b x) \log \left (\frac {a+b x}{2 a}\right )}{4 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[(2*a*c)/((1 + c)*(a + b*x))]^2 - 2*Log[a - b*x]*Log[(a + b*x)/(2*a)] + 2*Log[a - b*x]*Log[(a - a*c + b*(1
+ c)*x)/(2*a)] + 2*Log[(2*a*c)/((1 + c)*(a + b*x))]*Log[-1/2*(a - a*c + b*(1 + c)*x)/(a*c)] - 2*Log[a - b*x]*
Log[(a - a*c + b*(1 + c)*x)/(a + b*x)] - 2*Log[(2*a*c)/((1 + c)*(a + b*x))]*Log[(a - a*c + b*(1 + c)*x)/(a + b
*x)] - 2*PolyLog[2, (a - b*x)/(2*a)] + 2*PolyLog[2, ((1 + c)*(a - b*x))/(2*a)] - 2*PolyLog[2, ((1 + c)*(a + b*
x))/(2*a*c)])/(4*a*b)

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fricas [A]  time = 0.43, size = 34, normalized size = 0.92 \[ \frac {{\rm Li}_2\left (\frac {a c - {\left (b c + b\right )} x - a}{b x + a} + 1\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*dilog((a*c - (b*c + b)*x - a)/(b*x + a) + 1)/(a*b)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.06, size = 24, normalized size = 0.65 \[ \frac {\dilog \left (-\frac {2 a c}{b x +a}+c +1\right )}{2 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*dilog(1+c-2*a*c/(b*x+a))/a

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maxima [B]  time = 0.50, size = 246, normalized size = 6.65 \[ \frac {1}{2} \, {\left (\frac {\log \left (b x + a\right )}{a b} - \frac {\log \left (b x - a\right )}{a b}\right )} \log \left (\frac {b {\left (c + 1\right )} x - a {\left (c - 1\right )}}{b x + a}\right ) + \frac {\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (b x - a\right )}{4 \, a b} + \frac {\log \left (b x - a\right ) \log \left (\frac {b {\left (c + 1\right )} x - a {\left (c + 1\right )}}{2 \, a} + 1\right ) + {\rm Li}_2\left (-\frac {b {\left (c + 1\right )} x - a {\left (c + 1\right )}}{2 \, a}\right )}{2 \, a b} + \frac {\log \left (b x + a\right ) \log \left (-\frac {b x + a}{2 \, a} + 1\right ) + {\rm Li}_2\left (\frac {b x + a}{2 \, a}\right )}{2 \, a b} - \frac {\log \left (b x + a\right ) \log \left (-\frac {b {\left (c + 1\right )} x + a {\left (c + 1\right )}}{2 \, a c} + 1\right ) + {\rm Li}_2\left (\frac {b {\left (c + 1\right )} x + a {\left (c + 1\right )}}{2 \, a c}\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(b*x + a)/(a*b) - log(b*x - a)/(a*b))*log((b*(c + 1)*x - a*(c - 1))/(b*x + a)) + 1/4*(log(b*x + a)^2 -
2*log(b*x + a)*log(b*x - a))/(a*b) + 1/2*(log(b*x - a)*log(1/2*(b*(c + 1)*x - a*(c + 1))/a + 1) + dilog(-1/2*
(b*(c + 1)*x - a*(c + 1))/a))/(a*b) + 1/2*(log(b*x + a)*log(-1/2*(b*x + a)/a + 1) + dilog(1/2*(b*x + a)/a))/(a
*b) - 1/2*(log(b*x + a)*log(-1/2*(b*(c + 1)*x + a*(c + 1))/(a*c) + 1) + dilog(1/2*(b*(c + 1)*x + a*(c + 1))/(a
*c)))/(a*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\ln \left (-\frac {a\,\left (c-1\right )-b\,x\,\left (c+1\right )}{a+b\,x}\right )}{a^2-b^2\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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