\(\int \frac {1}{(1-c^2 x^2) (a+b \log (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^3} \, dx\) [49]

Optimal result

Integrand size = 40, antiderivative size = 37 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, 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, 6818} \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2} \]

[In]

Int[1/((1 - c^2*x^2)*(a + b*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3),x]

[Out]

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

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 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 )^3} \, dx,\sqrt {\frac {1-c x}{1+c x}},\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right ) \\ & = \frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \]

[In]

Integrate[1/((1 - c^2*x^2)*(a + b*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3),x]

[Out]

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

Maple [F]

\[\int \frac {1}{\left (-x^{2} c^{2}+1\right ) \left (a +b \ln \left (\frac {\sqrt {-x c +1}}{\sqrt {x c +1}}\right )\right )^{3}}d x\]

[In]

int(1/(-c^2*x^2+1)/(a+b*ln((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3,x)

[Out]

int(1/(-c^2*x^2+1)/(a+b*ln((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {1}{2 \, {\left (b^{3} c \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b^{2} c \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2} b c\right )}} \]

[In]

integrate(1/(-c^2*x^2+1)/(a+b*log((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3,x, algorithm="fricas")

[Out]

1/2/(b^3*c*log(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 2*a*b^2*c*log(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^2*b*c)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(1/(-c**2*x**2+1)/(a+b*ln((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (31) = 62\).

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.16 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {2}{b^{3} c \log \left (c x + 1\right )^{2} + b^{3} c \log \left (-c x + 1\right )^{2} - 4 \, a b^{2} c \log \left (c x + 1\right ) + 4 \, a^{2} b c - 2 \, {\left (b^{3} c \log \left (c x + 1\right ) - 2 \, a b^{2} c\right )} \log \left (-c x + 1\right )} \]

[In]

integrate(1/(-c^2*x^2+1)/(a+b*log((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3,x, algorithm="maxima")

[Out]

2/(b^3*c*log(c*x + 1)^2 + b^3*c*log(-c*x + 1)^2 - 4*a*b^2*c*log(c*x + 1) + 4*a^2*b*c - 2*(b^3*c*log(c*x + 1) -
2*a*b^2*c)*log(-c*x + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (31) = 62\).

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=\frac {2}{b^{3} c \log \left (c x + 1\right )^{2} - 2 \, b^{3} c \log \left (c x + 1\right ) \log \left (-c x + 1\right ) + b^{3} c \log \left (-c x + 1\right )^{2} - 4 \, a b^{2} c \log \left (c x + 1\right ) + 4 \, a b^{2} c \log \left (-c x + 1\right ) + 4 \, a^{2} b c} \]

[In]

integrate(1/(-c^2*x^2+1)/(a+b*log((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3,x, algorithm="giac")

[Out]

2/(b^3*c*log(c*x + 1)^2 - 2*b^3*c*log(c*x + 1)*log(-c*x + 1) + b^3*c*log(-c*x + 1)^2 - 4*a*b^2*c*log(c*x + 1)
+ 4*a*b^2*c*log(-c*x + 1) + 4*a^2*b*c)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3} \, dx=-\int \frac {1}{{\left (a+b\,\ln \left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3\,\left (c^2\,x^2-1\right )} \,d x \]

[In]

int(-1/((a + b*log((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3*(c^2*x^2 - 1)),x)

[Out]

-int(1/((a + b*log((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3*(c^2*x^2 - 1)), x)