\(\int \frac {(a+b \log (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^n}{1-c^2 x^2} \, dx\) [43]

Optimal result

Integrand size = 40, antiderivative size = 42 \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=-\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^{1+n}}{b c (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 42, 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} \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=-\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^{n+1}}{b c (n+1)} \]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=-\frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^{1+n}}{b c (1+n)} \]

[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)))

Maple [F]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=-\frac {{\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )} {\left (b \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{n}}{b c n + b c} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (32) = 64\).

Time = 78.37 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.24 \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=\begin {cases} - \frac {a^{n} \operatorname {atan}{\left (\frac {x}{\sqrt {- \frac {1}{c^{2}}}} \right )}}{c^{2} \sqrt {- \frac {1}{c^{2}}}} & \text {for}\: b = 0 \\a^{n} x & \text {for}\: c = 0 \\- \frac {\begin {cases} \frac {\left (a + b \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}\right )^{n + 1}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (a + b \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )} \right )} & \text {otherwise} \end {cases}}{b c} & \text {otherwise} \end {cases} \]

[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))

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^n}{1-c^2 x^2} \, dx=-\frac {{\left (-\frac {1}{2} \, b \log \left (c x + 1\right ) + \frac {1}{2} \, b \log \left (-c x + 1\right ) + a\right )}^{n + 1}}{b c {\left (n + 1\right )}} \]

[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]

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

Mupad [F(-1)]

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

[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)