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

Optimal. Leaf size=41 \[ -\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6143, 6140, 31} \[ -\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 6140

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

Rule 6143

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.00 \[ -\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 227, normalized size = 5.54 \[ \left [\frac {\log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x + {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right )}{2 \, a \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} - 2 \, a^{3} c x^{3} - a^{2} c x^{2} + 2 \, a c x}\right )}{a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.03, size = 51, normalized size = 1.24 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \ln \left (a x -1\right )}{\left (a^{2} x^{2}-1\right ) c a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a\,x+1}{\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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