3.954 \(\int e^{\tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=89 \[ \frac{2 c (a x+1)^3 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}}-\frac{c (a x+1)^4 \sqrt{c-a^2 c x^2}}{4 a \sqrt{1-a^2 x^2}} \]

[Out]

(2*c*(1 + a*x)^3*Sqrt[c - a^2*c*x^2])/(3*a*Sqrt[1 - a^2*x^2]) - (c*(1 + a*x)^4*Sqrt[c - a^2*c*x^2])/(4*a*Sqrt[
1 - a^2*x^2])

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Rubi [A]  time = 0.0872342, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6143, 6140, 43} \[ \frac{2 c (a x+1)^3 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}}-\frac{c (a x+1)^4 \sqrt{c-a^2 c x^2}}{4 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2),x]

[Out]

(2*c*(1 + a*x)^3*Sqrt[c - a^2*c*x^2])/(3*a*Sqrt[1 - a^2*x^2]) - (c*(1 + a*x)^4*Sqrt[c - a^2*c*x^2])/(4*a*Sqrt[
1 - a^2*x^2])

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])

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx &=\frac{\left (c \sqrt{c-a^2 c x^2}\right ) \int e^{\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{3/2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c \sqrt{c-a^2 c x^2}\right ) \int (1-a x) (1+a x)^2 \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c \sqrt{c-a^2 c x^2}\right ) \int \left (2 (1+a x)^2-(1+a x)^3\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{2 c (1+a x)^3 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}}-\frac{c (1+a x)^4 \sqrt{c-a^2 c x^2}}{4 a \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0270392, size = 57, normalized size = 0.64 \[ -\frac{c x \left (3 a^3 x^3+4 a^2 x^2-6 a x-12\right ) \sqrt{c-a^2 c x^2}}{12 \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2),x]

[Out]

-(c*x*Sqrt[c - a^2*c*x^2]*(-12 - 6*a*x + 4*a^2*x^2 + 3*a^3*x^3))/(12*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.032, size = 65, normalized size = 0.7 \begin{align*}{\frac{x \left ( 3\,{x}^{3}{a}^{3}+4\,{a}^{2}{x}^{2}-6\,ax-12 \right ) }{ \left ( 12\,ax-12 \right ) \left ( ax+1 \right ) } \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

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)^(3/2),x)

[Out]

1/12*x*(3*a^3*x^3+4*a^2*x^2-6*a*x-12)*(-a^2*c*x^2+c)^(3/2)/(a*x-1)/(a*x+1)/(-a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.03253, size = 178, normalized size = 2. \begin{align*} -\frac{1}{3} \, a^{2} c^{\frac{3}{2}} x^{3} + c^{\frac{3}{2}} x + \frac{1}{4} \,{\left (\frac{2 \, a^{8} c^{4} \log \left (x^{2} - \frac{1}{a^{2}}\right )}{\left (a^{4} c\right )^{\frac{5}{2}}} + \frac{2 \, a^{6} c^{3} x^{2}}{\left (a^{4} c\right )^{\frac{3}{2}}} + \frac{a^{4} c^{2} x^{4}}{\sqrt{a^{4} c}} - 2 \, c^{2} \sqrt{\frac{1}{a^{4} c}} \log \left (x^{2} - \frac{1}{a^{2}}\right ) - \frac{4 \, \sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} c}{a^{2}}\right )} a \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

-1/3*a^2*c^(3/2)*x^3 + c^(3/2)*x + 1/4*(2*a^8*c^4*log(x^2 - 1/a^2)/(a^4*c)^(5/2) + 2*a^6*c^3*x^2/(a^4*c)^(3/2)
+ a^4*c^2*x^4/sqrt(a^4*c) - 2*c^2*sqrt(1/(a^4*c))*log(x^2 - 1/a^2) - 4*sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*c/a^
2)*a

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Fricas [A]  time = 2.08864, size = 147, normalized size = 1.65 \begin{align*} \frac{{\left (3 \, a^{3} c x^{4} + 4 \, a^{2} c x^{3} - 6 \, a c x^{2} - 12 \, c x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{12 \,{\left (a^{2} x^{2} - 1\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)*(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*a^3*c*x^4 + 4*a^2*c*x^3 - 6*a*c*x^2 - 12*c*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}} \left (a x + 1\right )}{\sqrt{- \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)*(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**(3/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (a x + 1\right )}}{\sqrt{-a^{2} x^{2} + 1}}\,{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)*(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(3/2)*(a*x + 1)/sqrt(-a^2*x^2 + 1), x)