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]

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

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.58, size = 61, normalized size = 1.97 \[ -\frac {2 \, {\left (a^{3} c x^{3} + 3 \, a^{2} c x^{2} + 3 \, a c x + c\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{5 \, {\left (a^{2} x - a\right )}} \]

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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giac [A]  time = 0.15, size = 56, normalized size = 1.81 \[ -\frac {2 \, {\left (\frac {4 \, \sqrt {2} \sqrt {-c} c}{\mathrm {sgn}\relax (c)} + \frac {{\left (a c x + c\right )}^{2} \sqrt {-a c x - c}}{c \mathrm {sgn}\left (-a c x - c\right )}\right )}}{5 \, a} \]

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

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maple [A]  time = 0.04, size = 35, normalized size = 1.13 \[ \frac {2 \left (a x +1\right ) \left (-a c x +c \right )^{\frac {3}{2}}}{5 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a} \]

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 = 0.34, size = 41, normalized size = 1.32 \[ -\frac {2 \, {\left (a^{2} \sqrt {-c} c x^{2} - \sqrt {-c} c\right )} {\left (a x + 1\right )}^{\frac {3}{2}}}{5 \, {\left (a x - 1\right )} a} \]

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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mupad [B]  time = 1.38, size = 81, normalized size = 2.61 \[ -\frac {2\,c\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (a^2\,x^2+4\,a\,x+7\right )}{5\,a}-\frac {16\,c\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{5\,a\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*c*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(4*a*x + a^2*x^2 + 7))/(5*a) - (16*c*(c - a*c*x)^(1/2)*((
a*x - 1)/(a*x + 1))^(1/2))/(5*a*(a*x - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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