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

Optimal. Leaf size=136 \[ \frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{8 (c-a c x)^{5/2}}{5 a c \sqrt{1-a^2 x^2}}+\frac{64 (c-a c x)^{3/2}}{5 a \sqrt{1-a^2 x^2}}-\frac{256 c \sqrt{c-a c x}}{5 a \sqrt{1-a^2 x^2}} \]

[Out]

(-256*c*Sqrt[c - a*c*x])/(5*a*Sqrt[1 - a^2*x^2]) + (64*(c - a*c*x)^(3/2))/(5*a*Sqrt[1 - a^2*x^2]) + (8*(c - a*
c*x)^(5/2))/(5*a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c*x)^(7/2))/(5*a*c^2*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.112988, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6127, 657, 649} \[ \frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{8 (c-a c x)^{5/2}}{5 a c \sqrt{1-a^2 x^2}}+\frac{64 (c-a c x)^{3/2}}{5 a \sqrt{1-a^2 x^2}}-\frac{256 c \sqrt{c-a c x}}{5 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*c*Sqrt[c - a*c*x])/(5*a*Sqrt[1 - a^2*x^2]) + (64*(c - a*c*x)^(3/2))/(5*a*Sqrt[1 - a^2*x^2]) + (8*(c - a*
c*x)^(5/2))/(5*a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c*x)^(7/2))/(5*a*c^2*Sqrt[1 - a^2*x^2])

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^{-3 \tanh ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=\frac{\int \frac{(c-a c x)^{9/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{12 \int \frac{(c-a c x)^{7/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{5 c^2}\\ &=\frac{8 (c-a c x)^{5/2}}{5 a c \sqrt{1-a^2 x^2}}+\frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{32 \int \frac{(c-a c x)^{5/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{5 c}\\ &=\frac{64 (c-a c x)^{3/2}}{5 a \sqrt{1-a^2 x^2}}+\frac{8 (c-a c x)^{5/2}}{5 a c \sqrt{1-a^2 x^2}}+\frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{128}{5} \int \frac{(c-a c x)^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{256 c \sqrt{c-a c x}}{5 a \sqrt{1-a^2 x^2}}+\frac{64 (c-a c x)^{3/2}}{5 a \sqrt{1-a^2 x^2}}+\frac{8 (c-a c x)^{5/2}}{5 a c \sqrt{1-a^2 x^2}}+\frac{2 (c-a c x)^{7/2}}{5 a c^2 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0341686, size = 61, normalized size = 0.45 \[ -\frac{2 c^2 \sqrt{1-a x} \left (a^3 x^3-7 a^2 x^2+43 a x+91\right )}{5 a \sqrt{a x+1} \sqrt{c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c^2*Sqrt[1 - a*x]*(91 + 43*a*x - 7*a^2*x^2 + a^3*x^3))/(5*a*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]  time = 0.033, size = 62, normalized size = 0.5 \begin{align*}{\frac{2\,{x}^{3}{a}^{3}-14\,{a}^{2}{x}^{2}+86\,ax+182}{5\, \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{3}a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}} \left ( -acx+c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(3/2)*(a^3*x^3-7*a^2*x^2+43*a*x+91)/(a*x+1)^2/(a*x-1)^3/a

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Maxima [A]  time = 1.02732, size = 82, normalized size = 0.6 \begin{align*} -\frac{2 \,{\left (a^{3} c^{\frac{3}{2}} x^{3} - 7 \, a^{2} c^{\frac{3}{2}} x^{2} + 43 \, a c^{\frac{3}{2}} x + 91 \, c^{\frac{3}{2}}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{5 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

-2/5*(a^3*c^(3/2)*x^3 - 7*a^2*c^(3/2)*x^2 + 43*a*c^(3/2)*x + 91*c^(3/2))*sqrt(a*x + 1)*(a*x - 1)/(a^3*x^2 - a)

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Fricas [A]  time = 1.99046, size = 134, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (a^{3} c x^{3} - 7 \, a^{2} c x^{2} + 43 \, a c x + 91 \, c\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{5 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

2/5*(a^3*c*x^3 - 7*a^2*c*x^2 + 43*a*c*x + 91*c)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^3*x^2 - 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 (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2811, size = 95, normalized size = 0.7 \begin{align*} \frac{128 \, \sqrt{2} \sqrt{c}{\left | c \right |}}{5 \, a} - \frac{2 \,{\left ({\left (a c x + c\right )}^{\frac{5}{2}} - 10 \,{\left (a c x + c\right )}^{\frac{3}{2}} c + 60 \, \sqrt{a c x + c} c^{2} + \frac{40 \, c^{3}}{\sqrt{a c x + c}}\right )}{\left | c \right |}}{5 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(3/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

128/5*sqrt(2)*sqrt(c)*abs(c)/a - 2/5*((a*c*x + c)^(5/2) - 10*(a*c*x + c)^(3/2)*c + 60*sqrt(a*c*x + c)*c^2 + 40
*c^3/sqrt(a*c*x + c))*abs(c)/(a*c^2)