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

Optimal. Leaf size=71 \[ \frac{8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt{c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.0607667, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6127, 657, 649} \[ \frac{8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6127

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

Rule 657

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*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

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)} (c-a c x)^{3/2} \, dx &=c \int \sqrt{c-a c x} \sqrt{1-a^2 x^2} \, dx\\ &=\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt{c-a c x}}+\frac{1}{5} \left (4 c^2\right ) \int \frac{\sqrt{1-a^2 x^2}}{\sqrt{c-a c x}} \, dx\\ &=\frac{8 c^3 \left (1-a^2 x^2\right )^{3/2}}{15 a (c-a c x)^{3/2}}+\frac{2 c^2 \left (1-a^2 x^2\right )^{3/2}}{5 a \sqrt{c-a c x}}\\ \end{align*}

Mathematica [A]  time = 0.0253229, size = 44, normalized size = 0.62 \[ -\frac{2 c (a x+1)^{3/2} (3 a x-7) \sqrt{c-a c x}}{15 a \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c*(1 + a*x)^(3/2)*(-7 + 3*a*x)*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - a*x])

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Maple [A]  time = 0.03, size = 47, normalized size = 0.7 \begin{align*}{\frac{2\, \left ( 3\,ax-7 \right ) \left ( ax+1 \right ) ^{2}}{15\, \left ( ax-1 \right ) a} \left ( -acx+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*c*x+c)^(3/2),x)

[Out]

2/15*(a*x+1)^2*(3*a*x-7)*(-a*c*x+c)^(3/2)/a/(a*x-1)/(-a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.03068, size = 111, normalized size = 1.56 \begin{align*} -\frac{2 \,{\left (a^{3} c^{\frac{3}{2}} x^{3} - 2 \, a^{2} c^{\frac{3}{2}} x^{2} + 3 \, a c^{\frac{3}{2}} x + 6 \, c^{\frac{3}{2}}\right )}}{5 \, \sqrt{a x + 1} a} - \frac{2 \,{\left (a^{2} c^{\frac{3}{2}} x^{2} - 4 \, a c^{\frac{3}{2}} x - 5 \, c^{\frac{3}{2}}\right )}}{3 \, \sqrt{a x + 1} 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*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78191, size = 113, normalized size = 1.59 \begin{align*} \frac{2 \,{\left (3 \, a^{2} c x^{2} - 4 \, a c x - 7 \, c\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{15 \,{\left (a^{2} x - a\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*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

2/15*(3*a^2*c*x^2 - 4*a*c*x - 7*c)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a)

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

[Out]

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

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Giac [A]  time = 1.22043, size = 63, normalized size = 0.89 \begin{align*} -\frac{2 \,{\left (8 \, \sqrt{2} \sqrt{c} + \frac{3 \,{\left (a c x + c\right )}^{\frac{5}{2}} - 10 \,{\left (a c x + c\right )}^{\frac{3}{2}} c}{c^{2}}\right )} c^{2}}{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)*(-a*c*x+c)^(3/2),x, algorithm="giac")

[Out]

-2/15*(8*sqrt(2)*sqrt(c) + (3*(a*c*x + c)^(5/2) - 10*(a*c*x + c)^(3/2)*c)/c^2)*c^2/(a*abs(c))