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

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

[Out]

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

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Rubi [A]  time = 0.0379821, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {2512, 2301} \[ -\frac{\left (a+b \log \left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2512

Int[((a_.) + Log[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]]*(b_.))^(n_.)/((A_.) + (C_.)*(x_)^2
), x_Symbol] :> Dist[g/(C*f), Subst[Int[(a + b*Log[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x]], x] /; FreeQ
[{a, b, c, d, e, f, g, A, C, n}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0063933, size = 37, normalized size = 1. \[ -\frac{\left (a+b \log \left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.364, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-{c}^{2}{x}^{2}+1} \left ( a+b\ln \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.20672, size = 142, normalized size = 3.84 \begin{align*} \frac{1}{2} \, b{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \log \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + \frac{1}{2} \, a{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} + \frac{{\left (\log \left (c x + 1\right )^{2} - 2 \, \log \left (c x + 1\right ) \log \left (c x - 1\right ) + \log \left (c x - 1\right )^{2}\right )} b}{8 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*(log(c*x + 1)/c - log(c*x - 1)/c)*log(sqrt(-c*x + 1)/sqrt(c*x + 1)) + 1/2*a*(log(c*x + 1)/c - log(c*x -
1)/c) + 1/8*(log(c*x + 1)^2 - 2*log(c*x + 1)*log(c*x - 1) + log(c*x - 1)^2)*b/c

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Fricas [A]  time = 1.99544, size = 119, normalized size = 3.22 \begin{align*} -\frac{b \log \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{2} + 2 \, a \log \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 30.8092, size = 61, normalized size = 1.65 \begin{align*} \begin{cases} - \frac{a \operatorname{atan}{\left (\frac{x}{\sqrt{- \frac{1}{c^{2}}}} \right )}}{c^{2} \sqrt{- \frac{1}{c^{2}}}} & \text{for}\: b = 0 \\a x & \text{for}\: c = 0 \\- \frac{\left (a + b \log{\left (\frac{\sqrt{- c x + 1}}{\sqrt{c x + 1}} \right )}\right )^{2}}{2 b c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*atan(x/sqrt(-1/c**2))/(c**2*sqrt(-1/c**2)), Eq(b, 0)), (a*x, Eq(c, 0)), (-(a + b*log(sqrt(-c*x +
1)/sqrt(c*x + 1)))**2/(2*b*c), True))

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Giac [B]  time = 1.26335, size = 116, normalized size = 3.14 \begin{align*} -\frac{b \log \left (c x + 1\right )^{2}}{8 \, c} + \frac{b \log \left (c x - 1\right )^{2}}{8 \, c} + \frac{1}{4} \,{\left (\frac{b \log \left (c x + 1\right )}{c} - \frac{b \log \left (c x - 1\right )}{c}\right )} \log \left (-c x + 1\right ) + \frac{a \log \left (c x + 1\right )}{2 \, c} - \frac{a \log \left (c x - 1\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b*log(c*x + 1)^2/c + 1/8*b*log(c*x - 1)^2/c + 1/4*(b*log(c*x + 1)/c - b*log(c*x - 1)/c)*log(-c*x + 1) + 1
/2*a*log(c*x + 1)/c - 1/2*a*log(c*x - 1)/c