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

Optimal. Leaf size=35 \[ \frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{5 a (c-a c x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6127, 649} \[ \frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{5 a (c-a c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{3/2}} \, dx\\ &=\frac {2 c^4 \left (1-a^2 x^2\right )^{5/2}}{5 a (c-a c x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 1.06 \[ \frac {2 (a x+1)^{5/2} (c-a c x)^{3/2}}{5 a (1-a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 49, normalized size = 1.40 \[ -\frac {2 \, {\left (a^{2} c x^{2} + 2 \, a c x + c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{5 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.02, size = 34, normalized size = 0.97 \[ \frac {2 \left (a x +1\right )^{4} \left (-a c x +c \right )^{\frac {3}{2}}}{5 a \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 121, normalized size = 3.46 \[ -\frac {2 \, c^{\frac {3}{2}}}{\sqrt {a x + 1} a} + \frac {2 \, {\left (a^{3} c^{\frac {3}{2}} x^{3} - 2 \, a^{2} c^{\frac {3}{2}} x^{2} + 8 \, a c^{\frac {3}{2}} x + 16 \, c^{\frac {3}{2}}\right )}}{5 \, \sqrt {a x + 1} a} + \frac {2 \, {\left (a^{2} c^{\frac {3}{2}} x^{2} - 4 \, a c^{\frac {3}{2}} x - 8 \, c^{\frac {3}{2}}\right )}}{\sqrt {a x + 1} a} + \frac {6 \, {\left (a c^{\frac {3}{2}} x + 2 \, c^{\frac {3}{2}}\right )}}{\sqrt {a x + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c^(3/2)/(sqrt(a*x + 1)*a) + 2/5*(a^3*c^(3/2)*x^3 - 2*a^2*c^(3/2)*x^2 + 8*a*c^(3/2)*x + 16*c^(3/2))/(sqrt(a*
x + 1)*a) + 2*(a^2*c^(3/2)*x^2 - 4*a*c^(3/2)*x - 8*c^(3/2))/(sqrt(a*x + 1)*a) + 6*(a*c^(3/2)*x + 2*c^(3/2))/(s
qrt(a*x + 1)*a)

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mupad [B]  time = 0.92, size = 49, normalized size = 1.40 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {6\,c\,x}{5}+\frac {2\,c}{5\,a}+\frac {2\,a^2\,c\,x^3}{5}+\frac {6\,a\,c\,x^2}{5}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (a x - 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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