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

Optimal. Leaf size=89 \[ \frac {2 c (a x+1)^3 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-a^2 x^2}}-\frac {c (a x+1)^4 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}} \]

[Out]

2/3*c*(a*x+1)^3*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-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 = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6143, 6140, 43} \[ \frac {2 c (a x+1)^3 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-a^2 x^2}}-\frac {c (a x+1)^4 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6140

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 6143

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^{\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^{\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) (1+a x)^2 \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\left (c \sqrt {c-a^2 c x^2}\right ) \int \left (2 (1+a x)^2-(1+a x)^3\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {2 c (1+a x)^3 \sqrt {c-a^2 c x^2}}{3 a \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 = 57, normalized size = 0.64 \[ -\frac {c x \left (3 a^3 x^3+4 a^2 x^2-6 a x-12\right ) \sqrt {c-a^2 c x^2}}{12 \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*(c*x*Sqrt[c - a^2*c*x^2]*(-12 - 6*a*x + 4*a^2*x^2 + 3*a^3*x^3))/Sqrt[1 - a^2*x^2]

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fricas [A]  time = 0.81, size = 68, normalized size = 0.76 \[ \frac {{\left (3 \, a^{3} c x^{4} + 4 \, a^{2} c x^{3} - 6 \, a c x^{2} - 12 \, c x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{12 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^3*c*x^4 + 4*a^2*c*x^3 - 6*a*c*x^2 - 12*c*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 65, normalized size = 0.73 \[ \frac {x \left (3 x^{3} a^{3}+4 a^{2} x^{2}-6 a x -12\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{12 \left (a x -1\right ) \left (a x +1\right ) \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x*(3*a^3*x^3+4*a^2*x^2-6*a*x-12)*(-a^2*c*x^2+c)^(3/2)/(a*x-1)/(a*x+1)/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.33, size = 66, normalized size = 0.74 \[ -\frac {1}{3} \, a^{2} c^{\frac {3}{2}} x^{3} + c^{\frac {3}{2}} x + \frac {1}{4} \, {\left (a^{2} c^{\frac {3}{2}} x^{4} + 2 \, c^{\frac {3}{2}} x^{2} - \frac {4 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} c}{a^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.06, size = 55, normalized size = 0.62 \[ \frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {c\,a^3\,x^4}{4}-\frac {c\,a^2\,x^3}{3}+\frac {c\,a\,x^2}{2}+c\,x\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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