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

Optimal. Leaf size=101 \[ \frac{64 c^2 \sqrt{1-a^2 x^2}}{15 a \sqrt{c-a c x}}+\frac{16 c \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{15 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{5 a} \]

[Out]

(64*c^2*Sqrt[1 - a^2*x^2])/(15*a*Sqrt[c - a*c*x]) + (16*c*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(15*a) + (2*(c -
a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(5*a)

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Rubi [A]  time = 0.0867293, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6127, 657, 649} \[ \frac{64 c^2 \sqrt{1-a^2 x^2}}{15 a \sqrt{c-a c x}}+\frac{16 c \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{15 a}+\frac{2 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*c^2*Sqrt[1 - a^2*x^2])/(15*a*Sqrt[c - a*c*x]) + (16*c*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(15*a) + (2*(c -
a*c*x)^(3/2)*Sqrt[1 - a^2*x^2])/(5*a)

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 &=\frac{\int \frac{(c-a c x)^{5/2}}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=\frac{2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{5 a}+\frac{8}{5} \int \frac{(c-a c x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{16 c \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{15 a}+\frac{2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{5 a}+\frac{1}{15} (32 c) \int \frac{\sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{64 c^2 \sqrt{1-a^2 x^2}}{15 a \sqrt{c-a c x}}+\frac{16 c \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{15 a}+\frac{2 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.028437, size = 49, normalized size = 0.49 \[ \frac{2 c^2 \sqrt{1-a^2 x^2} \left (3 a^2 x^2-14 a x+43\right )}{15 a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*Sqrt[1 - a^2*x^2]*(43 - 14*a*x + 3*a^2*x^2))/(15*a*Sqrt[c - a*c*x])

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Maple [A]  time = 0.03, size = 48, normalized size = 0.5 \begin{align*}{\frac{6\,{a}^{2}{x}^{2}-28\,ax+86}{15\, \left ( ax-1 \right ) ^{2}a}\sqrt{-{a}^{2}{x}^{2}+1} \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)*(-a^2*x^2+1)^(1/2),x)

[Out]

2/15*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(3/2)*(3*a^2*x^2-14*a*x+43)/(a*x-1)^2/a

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Maxima [A]  time = 0.997573, size = 66, normalized size = 0.65 \begin{align*} \frac{2 \,{\left (3 \, a^{2} c^{\frac{3}{2}} x^{2} - 14 \, a c^{\frac{3}{2}} x + 43 \, c^{\frac{3}{2}}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{15 \,{\left (a^{2} x - 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)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*a^2*c^(3/2)*x^2 - 14*a*c^(3/2)*x + 43*c^(3/2))*sqrt(a*x + 1)*(a*x - 1)/(a^2*x - a)

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Fricas [A]  time = 2.23512, size = 117, normalized size = 1.16 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} c x^{2} - 14 \, a c x + 43 \, 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*c*x+c)^(3/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*a^2*c*x^2 - 14*a*c*x + 43*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}} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22211, size = 80, normalized size = 0.79 \begin{align*} -\frac{64 \, \sqrt{2} \sqrt{c}{\left | c \right |}}{15 \, a} + \frac{2 \,{\left (3 \,{\left (a c x + c\right )}^{\frac{5}{2}} - 20 \,{\left (a c x + c\right )}^{\frac{3}{2}} c + 60 \, \sqrt{a c x + c} c^{2}\right )}{\left | c \right |}}{15 \, 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)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-64/15*sqrt(2)*sqrt(c)*abs(c)/a + 2/15*(3*(a*c*x + c)^(5/2) - 20*(a*c*x + c)^(3/2)*c + 60*sqrt(a*c*x + c)*c^2)
*abs(c)/(a*c^2)