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

Optimal. Leaf size=31 \[ \frac{2 (a x+1) (c-a c x)^{3/2} e^{3 \coth ^{-1}(a x)}}{5 a} \]

[Out]

(2*E^(3*ArcCoth[a*x])*(1 + a*x)*(c - a*c*x)^(3/2))/(5*a)

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Rubi [A]  time = 0.0365437, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {6174} \[ \frac{2 (a x+1) (c-a c x)^{3/2} e^{3 \coth ^{-1}(a x)}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(3*ArcCoth[a*x])*(1 + a*x)*(c - a*c*x)^(3/2))/(5*a)

Rule 6174

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

Rubi steps

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

Mathematica [A]  time = 0.0311314, size = 43, normalized size = 1.39 \[ \frac{2 x \left (\frac{1}{a x}+1\right )^{5/2} (c-a c x)^{3/2}}{5 \left (1-\frac{1}{a x}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.115, size = 35, normalized size = 1.1 \begin{align*}{\frac{2\,ax+2}{5\,a} \left ( -acx+c \right ) ^{{\frac{3}{2}}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06897, size = 55, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (a^{2} \sqrt{-c} c x^{2} - \sqrt{-c} c\right )}{\left (a x + 1\right )}^{\frac{3}{2}}}{5 \,{\left (a x - 1\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47355, size = 136, normalized size = 4.39 \begin{align*} -\frac{2 \,{\left (a^{3} c x^{3} + 3 \, a^{2} c x^{2} + 3 \, a c x + c\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{5 \,{\left (a^{2} x - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1924, size = 76, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left (\frac{4 \, \sqrt{2} \sqrt{-c} c}{\mathrm{sgn}\left (c\right )} + \frac{{\left (a c x + c\right )}^{2} \sqrt{-a c x - c}}{c \mathrm{sgn}\left (-a c x - c\right )}\right )}}{5 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(4*sqrt(2)*sqrt(-c)*c/sgn(c) + (a*c*x + c)^2*sqrt(-a*c*x - c)/(c*sgn(-a*c*x - c)))/a