3.957 \(\int \frac{e^{\tanh ^{-1}(a x)} x^4}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=194 \[ -\frac{x^4 \sqrt{1-a^2 x^2}}{4 a \sqrt{c-a^2 c x^2}}-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{x^2 \sqrt{1-a^2 x^2}}{2 a^3 \sqrt{c-a^2 c x^2}}-\frac{x \sqrt{1-a^2 x^2}}{a^4 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \log (1-a x)}{a^5 \sqrt{c-a^2 c x^2}} \]

[Out]

-((x*Sqrt[1 - a^2*x^2])/(a^4*Sqrt[c - a^2*c*x^2])) - (x^2*Sqrt[1 - a^2*x^2])/(2*a^3*Sqrt[c - a^2*c*x^2]) - (x^
3*Sqrt[1 - a^2*x^2])/(3*a^2*Sqrt[c - a^2*c*x^2]) - (x^4*Sqrt[1 - a^2*x^2])/(4*a*Sqrt[c - a^2*c*x^2]) - (Sqrt[1
- a^2*x^2]*Log[1 - a*x])/(a^5*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.212941, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 43} \[ -\frac{x^4 \sqrt{1-a^2 x^2}}{4 a \sqrt{c-a^2 c x^2}}-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{x^2 \sqrt{1-a^2 x^2}}{2 a^3 \sqrt{c-a^2 c x^2}}-\frac{x \sqrt{1-a^2 x^2}}{a^4 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \log (1-a x)}{a^5 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/Sqrt[c - a^2*c*x^2],x]

[Out]

-((x*Sqrt[1 - a^2*x^2])/(a^4*Sqrt[c - a^2*c*x^2])) - (x^2*Sqrt[1 - a^2*x^2])/(2*a^3*Sqrt[c - a^2*c*x^2]) - (x^
3*Sqrt[1 - a^2*x^2])/(3*a^2*Sqrt[c - a^2*c*x^2]) - (x^4*Sqrt[1 - a^2*x^2])/(4*a*Sqrt[c - a^2*c*x^2]) - (Sqrt[1
- a^2*x^2]*Log[1 - a*x])/(a^5*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 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 \frac{e^{\tanh ^{-1}(a x)} x^4}{\sqrt{c-a^2 c x^2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)} x^4}{\sqrt{1-a^2 x^2}} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x^4}{1-a x} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \left (-\frac{1}{a^4}-\frac{x}{a^3}-\frac{x^2}{a^2}-\frac{x^3}{a}-\frac{1}{a^4 (-1+a x)}\right ) \, dx}{\sqrt{c-a^2 c x^2}}\\ &=-\frac{x \sqrt{1-a^2 x^2}}{a^4 \sqrt{c-a^2 c x^2}}-\frac{x^2 \sqrt{1-a^2 x^2}}{2 a^3 \sqrt{c-a^2 c x^2}}-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{x^4 \sqrt{1-a^2 x^2}}{4 a \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \log (1-a x)}{a^5 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0451362, size = 71, normalized size = 0.37 \[ -\frac{\sqrt{1-a^2 x^2} \left (a x \left (3 a^3 x^3+4 a^2 x^2+6 a x+12\right )+12 \log (1-a x)\right )}{12 a^5 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/Sqrt[c - a^2*c*x^2],x]

[Out]

-(Sqrt[1 - a^2*x^2]*(a*x*(12 + 6*a*x + 4*a^2*x^2 + 3*a^3*x^3) + 12*Log[1 - a*x]))/(12*a^5*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.085, size = 83, normalized size = 0.4 \begin{align*}{\frac{3\,{x}^{4}{a}^{4}+4\,{x}^{3}{a}^{3}+6\,{a}^{2}{x}^{2}+12\,ax+12\,\ln \left ( ax-1 \right ) }{ \left ( 12\,{a}^{2}{x}^{2}-12 \right ) c{a}^{5}}\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^4/(-a^2*c*x^2+c)^(1/2),x)

[Out]

1/12*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(3*x^4*a^4+4*x^3*a^3+6*a^2*x^2+12*a*x+12*ln(a*x-1))/(a^2*x^2-1)
/c/a^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{4}}{\sqrt{-a^{2} c x^{2} + c} \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)*x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)*x^4/(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)), x)

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Fricas [A]  time = 2.46467, size = 828, normalized size = 4.27 \begin{align*} \left [\frac{6 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \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 ) +{\left (3 \, a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 12 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{12 \,{\left (a^{7} c x^{2} - a^{5} c\right )}}, -\frac{12 \,{\left (a^{2} x^{2} - 1\right )} \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 ) -{\left (3 \, a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 12 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{12 \,{\left (a^{7} c x^{2} - a^{5} c\right )}}\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^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(6*(a^2*x^2 - 1)*sqrt(c)*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)) + (3*a^4*x^4 + 4*a^3*x^3 + 6*a^2*x^2 + 12*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^7*c*x^2
- a^5*c), -1/12*(12*(a^2*x^2 - 1)*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)) - (3*a^4*x^4 + 4*a^3*x^3 + 6*a^2*x^2 + 12*a*x)*sq
rt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^7*c*x^2 - a^5*c)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (a x + 1\right )}{\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**4/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**4*(a*x + 1)/(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{{\left (a x + 1\right )} x^{4}}{\sqrt{-a^{2} c x^{2} + c} \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)*x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^4/(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)), x)