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

Optimal result

Integrand size = 40, antiderivative size = 34 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2573, 6816} \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{b c} \]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{b c} \]

[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))

Maple [F]

\[\int \frac {1}{\left (-x^{2} c^{2}+1\right ) \left (a +b \ln \left (\frac {\sqrt {-x c +1}}{\sqrt {x c +1}}\right )\right )}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}{b c} \]

[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)

Sympy [A] (verification not implemented)

Time = 7.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {\log {\left (x - \frac {1}{c} \right )}}{2 c} + \frac {\log {\left (x + \frac {1}{c} \right )}}{2 c}}{a} & \text {for}\: b = 0 \\\frac {x}{a} & \text {for}\: c = 0 \\- \frac {\log {\left (\frac {a}{b} + \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )} \right )}}{b c} & \text {otherwise} \end {cases} \]

[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(sqrt(-c*x + 1)/sqrt(c*x + 1)))/(b*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (-\frac {b \log \left (c x + 1\right ) - b \log \left (-c x + 1\right ) - 2 \, a}{2 \, b}\right )}{b c} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\frac {\log \left (-b \log \left (c x + 1\right ) + b \log \left (-c x + 1\right ) + 2 \, a\right )}{b c} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \, dx=-\int \frac {1}{\left (a+b\,\ln \left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )\,\left (c^2\,x^2-1\right )} \,d x \]

[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)