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

Optimal. Leaf size=42 \[ -\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^{n+1}}{b c (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 42, 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} \[ -\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^{n+1}}{b c (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 1.00 \[ -\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^{n+1}}{b c (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 56, normalized size = 1.33 \[ -\frac {{\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )} {\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{n}}{b c n + b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{n}}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(b*log(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^n/(c^2*x^2 - 1), x)

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )+a \right )^{n}}{-c^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{n}}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((b*log(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^n/(c^2*x^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {{\left (a+b\,\ln \left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^n}{c^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 133.77, size = 94, normalized size = 2.24 \[ \begin {cases} - \frac {a^{n} \operatorname {atan}{\left (\frac {x}{\sqrt {- \frac {1}{c^{2}}}} \right )}}{c^{2} \sqrt {- \frac {1}{c^{2}}}} & \text {for}\: b = 0 \\a^{n} x & \text {for}\: c = 0 \\- \frac {\begin {cases} \frac {\left (a + b \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}\right )^{n + 1}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (a + b \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )} \right )} & \text {otherwise} \end {cases}}{b c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**n*atan(x/sqrt(-1/c**2))/(c**2*sqrt(-1/c**2)), Eq(b, 0)), (a**n*x, Eq(c, 0)), (-Piecewise(((a +
b*log(sqrt(-c*x + 1)/sqrt(c*x + 1)))**(n + 1)/(n + 1), Ne(n, -1)), (log(a + b*log(sqrt(-c*x + 1)/sqrt(c*x + 1)
)), True))/(b*c), True))

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