3.962 \(\int \frac{e^{\tanh ^{-1}(a x)}}{x \sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=71 \[ \frac{\sqrt{1-a^2 x^2} \log (x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.181653, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6153, 6150, 36, 29, 31} \[ \frac{\sqrt{1-a^2 x^2} \log (x)}{\sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x*Sqrt[c - a^2*c*x^2]),x]

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0172875, size = 42, normalized size = 0.59 \[ \frac{\sqrt{1-a^2 x^2} (\log (x)-\log (1-a x))}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x*Sqrt[c - a^2*c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.087, size = 53, normalized size = 0.8 \begin{align*}{\frac{-\ln \left ( x \right ) +\ln \left ( ax-1 \right ) }{c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.629, size = 662, normalized size = 9.32 \begin{align*} \left [\frac{\log \left (-\frac{4 \, a^{5} c x^{5} -{\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} -{\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x +{\left (4 \, a^{3} x^{3} -{\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right )}{2 \, \sqrt{c}}, -\frac{\sqrt{-c} \arctan \left (-\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}{\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt{-c}}{2 \, a^{3} c x^{3} -{\left (2 \, a^{3} - a^{2}\right )} c x^{4} -{\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right )}{c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log(-(4*a^5*c*x^5 - (2*a^6 - 4*a^5 + 6*a^4 - 4*a^3 + a^2)*c*x^6 - (4*a^4 + 4*a^3 - 6*a^2 + 4*a - 1)*c*x^4
+ 5*a^2*c*x^2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 - 6*a^2 + 4*a - 1)*x^4 - 6*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*c*x^2
+ c)*sqrt(-a^2*x^2 + 1)*sqrt(c) + c)/(a^4*x^6 - 2*a^3*x^5 + 2*a*x^3 - x^2))/sqrt(c), -sqrt(-c)*arctan(-sqrt(-
a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 - 2*a + 1)*x^2 - 2*a*x + 1)*sqrt(-c)/(2*a^3*c*x^3 - (2*a^3 - a^2)*c*
x^4 - (a^2 - 2*a + 1)*c*x^2 - 2*a*c*x + c))/c]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{x \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + 1)/(x*sqrt(-(a*x - 1)*(a*x + 1))*sqrt(-c*(a*x - 1)*(a*x + 1))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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