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

Optimal. Leaf size=69 \[ \frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac{2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt{c-a c x}} \]

[Out]

(2*c^2*(1 - a^2*x^2)^(3/2))/(15*a^2*(c - a*c*x)^(3/2)) - (2*c*(1 - a^2*x^2)^(3/2))/(5*a^2*Sqrt[c - a*c*x])

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Rubi [A]  time = 0.0897471, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {6128, 795, 649} \[ \frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac{2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(1 - a^2*x^2)^(3/2))/(15*a^2*(c - a*c*x)^(3/2)) - (2*c*(1 - a^2*x^2)^(3/2))/(5*a^2*Sqrt[c - a*c*x])

Rule 6128

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

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a x)} x \sqrt{c-a c x} \, dx &=c \int \frac{x \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}} \, dx\\ &=-\frac{2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt{c-a c x}}+\frac{c \int \frac{\sqrt{1-a^2 x^2}}{\sqrt{c-a c x}} \, dx}{5 a}\\ &=\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac{2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt{c-a c x}}\\ \end{align*}

Mathematica [A]  time = 0.0227393, size = 43, normalized size = 0.62 \[ \frac{2 (a x+1)^{3/2} (3 a x-2) \sqrt{c-a c x}}{15 a^2 \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(1 + a*x)^(3/2)*(-2 + 3*a*x)*Sqrt[c - a*c*x])/(15*a^2*Sqrt[1 - a*x])

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Maple [A]  time = 0.033, size = 40, normalized size = 0.6 \begin{align*}{\frac{2\, \left ( ax+1 \right ) ^{2} \left ( 3\,ax-2 \right ) }{15\,{a}^{2}}\sqrt{-acx+c}{\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)*x*(-a*c*x+c)^(1/2),x)

[Out]

2/15*(a*x+1)^2*(3*a*x-2)*(-a*c*x+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.03025, size = 112, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (3 \, a^{3} \sqrt{c} x^{3} - a^{2} \sqrt{c} x^{2} + 4 \, a \sqrt{c} x + 8 \, \sqrt{c}\right )}}{15 \, \sqrt{a x + 1} a^{2}} + \frac{2 \,{\left (a^{2} \sqrt{c} x^{2} - a \sqrt{c} x - 2 \, \sqrt{c}\right )}}{3 \, \sqrt{a x + 1} a^{2}} \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*c*x+c)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*a^3*sqrt(c)*x^3 - a^2*sqrt(c)*x^2 + 4*a*sqrt(c)*x + 8*sqrt(c))/(sqrt(a*x + 1)*a^2) + 2/3*(a^2*sqrt(c)*
x^2 - a*sqrt(c)*x - 2*sqrt(c))/(sqrt(a*x + 1)*a^2)

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Fricas [A]  time = 1.83253, size = 107, normalized size = 1.55 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} x^{2} + a x - 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{15 \,{\left (a^{3} x - a^{2}\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*c*x+c)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*a^2*x^2 + a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^3*x - a^2)

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

[Out]

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

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Giac [A]  time = 1.36497, size = 70, normalized size = 1.01 \begin{align*} -\frac{2 \, c{\left (\frac{2 \, \sqrt{2} \sqrt{c}}{a} - \frac{3 \,{\left (a c x + c\right )}^{\frac{5}{2}} - 5 \,{\left (a c x + c\right )}^{\frac{3}{2}} c}{a c^{2}}\right )}}{15 \, a{\left | 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*c*x+c)^(1/2),x, algorithm="giac")

[Out]

-2/15*c*(2*sqrt(2)*sqrt(c)/a - (3*(a*c*x + c)^(5/2) - 5*(a*c*x + c)^(3/2)*c)/(a*c^2))/(a*abs(c))