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

Optimal. Leaf size=34 \[ -\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{b c} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2512, 2302, 29} \[ -\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.00 \[ -\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 30, normalized size = 0.88 \[ -\frac {\log \left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 31, normalized size = 0.91 \[ -\frac {\log \left (-b \log \left (c x + 1\right ) + b \log \left (-c x + 1\right ) + 2 \, a\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-c^{2} x^{2}+1\right ) \left (b \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )+a \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.76, size = 36, normalized size = 1.06 \[ -\frac {\log \left (-\frac {b \log \left (c x + 1\right ) - b \log \left (-c x + 1\right ) - 2 \, a}{2 \, b}\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 23.96, size = 53, normalized size = 1.56 \[ \begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {\log {\left (x - \frac {1}{c} \right )}}{2 c} + \frac {\log {\left (x + \frac {1}{c} \right )}}{2 c}}{a} & \text {for}\: b = 0 \\\frac {x}{a} & \text {for}\: c = 0 \\- \frac {\log {\left (\frac {a}{b} + \frac {\log {\left (- c x + 1 \right )}}{2} - \frac {\log {\left (c x + 1 \right )}}{2} \right )}}{b c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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