3.13.74 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\) [1274]

Optimal. Leaf size=45 \[ -\frac {c (1-a x)^4 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6278, 6275, 32} \begin {gather*} -\frac {c (1-a x)^4 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(c*(1 - a*x)^4*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - a^2*x^2])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6275

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

Rule 6278

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^IntPart[p]*((c + d*x^2)^Frac
Part[p]/(1 - a^2*x^2)^FracPart[p]), Int[(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x
] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 65, normalized size = 1.44 \begin {gather*} \frac {c x \sqrt {1-a^2 x^2} \sqrt {c-a^2 c x^2} \left (-4+6 a x-4 a^2 x^2+a^3 x^3\right )}{-4+4 a^2 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*x*Sqrt[1 - a^2*x^2]*Sqrt[c - a^2*c*x^2]*(-4 + 6*a*x - 4*a^2*x^2 + a^3*x^3))/(-4 + 4*a^2*x^2)

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Maple [A]
time = 0.06, size = 47, normalized size = 1.04

method	result	size
default	\(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right )^{3} c}{4 a \left (a x +1\right )}\)	\(47\)
gosper	\(\frac {x \left (a^{3} x^{3}-4 a^{2} x^{2}+6 a x -4\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{3} \left (a x +1\right )^{3}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 70, normalized size = 1.56 \begin {gather*} -\frac {{\left (a^{4} c^{\frac {3}{2}} x^{4} - 4 \, a^{3} c^{\frac {3}{2}} x^{3} + 6 \, a^{2} c^{\frac {3}{2}} x^{2} - 4 \, a c^{\frac {3}{2}} x + 4 \, c^{\frac {3}{2}}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}{4 \, {\left (a^{3} x^{2} - a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^4*c^(3/2)*x^4 - 4*a^3*c^(3/2)*x^3 + 6*a^2*c^(3/2)*x^2 - 4*a*c^(3/2)*x + 4*c^(3/2))*(a*x + 1)*(a*x - 1)
/(a^3*x^2 - a)

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Fricas [A]
time = 0.34, size = 67, normalized size = 1.49 \begin {gather*} \frac {{\left (a^{3} c x^{4} - 4 \, a^{2} c x^{3} + 6 \, a c x^{2} - 4 \, c x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{2} x^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^3*c*x^4 - 4*a^2*c*x^3 + 6*a*c*x^2 - 4*c*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c-a^2\,c\,x^2\right )}^{3/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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