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

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

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 37, 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, 30} \[ \frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2} \]

Antiderivative was successfully verified.

[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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

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Mathematica [A]  time = 0.01, size = 37, normalized size = 1.00 \[ \frac {1}{2 b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 1.10, size = 59, normalized size = 1.59 \[ \frac {1}{2 \, {\left (b^{3} c \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b^{2} c \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2} b c\right )}} \]

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)))^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)

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giac [B]  time = 0.33, size = 85, normalized size = 2.30 \[ \frac {2}{b^{3} c \log \left (c x + 1\right )^{2} - 2 \, b^{3} c \log \left (c x + 1\right ) \log \left (-c x + 1\right ) + b^{3} c \log \left (-c x + 1\right )^{2} - 4 \, a b^{2} c \log \left (c x + 1\right ) + 4 \, a b^{2} c \log \left (-c x + 1\right ) + 4 \, a^{2} b c} \]

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)))^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)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-c^{2} x^{2}+1\right ) \left (b \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )+a \right )^{3}}\, dx \]

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)^3,x)

[Out]

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

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maxima [B]  time = 2.16, size = 80, normalized size = 2.16 \[ \frac {2}{b^{3} c \log \left (c x + 1\right )^{2} + b^{3} c \log \left (-c x + 1\right )^{2} - 4 \, a b^{2} c \log \left (c x + 1\right ) + 4 \, a^{2} b c - 2 \, {\left (b^{3} c \log \left (c x + 1\right ) - 2 \, a b^{2} c\right )} \log \left (-c x + 1\right )} \]

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)))^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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ -\int \frac {1}{{\left (a+b\,\ln \left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3\,\left (c^2\,x^2-1\right )} \,d x \]

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)))^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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)))**3,x)

[Out]

Timed out

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