3.1.47 \(\int \frac {1}{(1-c^2 x^2) (a+b \log (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))} \, dx\) [47]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2573, 6816} \begin {gather*} -\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{b c} \end {gather*}

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 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {\text {Subst}\left (\int \frac {1}{x (a+b \log (x))} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {\text {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 = 0.03, size = 34, normalized size = 1.00 \begin {gather*} -\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{b c} \end {gather*}

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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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-c^{2} x^{2}+1\right ) \left (a +b \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 36, normalized size = 1.06 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.37, size = 30, normalized size = 0.88 \begin {gather*} -\frac {\log \left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}{b c} \end {gather*}

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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Sympy [A]
time = 8.29, size = 54, normalized size = 1.59 \begin {gather*} \begin {cases} \frac {x}{a} & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \\\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 {\log {\left (\frac {a}{b} + \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )} \right )}}{b c} & \text {otherwise} \end {cases} \end {gather*}

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

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Giac [A]
time = 5.08, size = 31, normalized size = 0.91 \begin {gather*} -\frac {\log \left (-b \log \left (c x + 1\right ) + b \log \left (-c x + 1\right ) + 2 \, a\right )}{b c} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\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 \end {gather*}

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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