3.1166 \(\int e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=93 \[ \frac{2 c^2 (a x+1)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}-\frac{c^2 (a x+1)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}} \]

[Out]

(2*c^2*(1 + a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) - (c^2*(1 + a*x)^6*Sqrt[c - a^2*c*x^2])/(6*a*S
qrt[1 - a^2*x^2])

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Rubi [A]  time = 0.0911275, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6143, 6140, 43} \[ \frac{2 c^2 (a x+1)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}-\frac{c^2 (a x+1)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2),x]

[Out]

(2*c^2*(1 + a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) - (c^2*(1 + a*x)^6*Sqrt[c - a^2*c*x^2])/(6*a*S
qrt[1 - a^2*x^2])

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])

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 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])

Rubi steps

\begin{align*} \int e^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx &=\frac{\left (c^2 \sqrt{c-a^2 c x^2}\right ) \int e^{3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{5/2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^2 \sqrt{c-a^2 c x^2}\right ) \int (1-a x) (1+a x)^4 \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^2 \sqrt{c-a^2 c x^2}\right ) \int \left (2 (1+a x)^4-(1+a x)^5\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{2 c^2 (1+a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}-\frac{c^2 (1+a x)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0379155, size = 52, normalized size = 0.56 \[ -\frac{c^2 (a x+1)^5 (5 a x-7) \sqrt{c-a^2 c x^2}}{30 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2),x]

[Out]

-(c^2*(1 + a*x)^5*(-7 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(30*a*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.028, size = 81, normalized size = 0.9 \begin{align*}{\frac{x \left ( 5\,{x}^{5}{a}^{5}+18\,{x}^{4}{a}^{4}+15\,{x}^{3}{a}^{3}-20\,{a}^{2}{x}^{2}-45\,ax-30 \right ) }{ \left ( 30\,ax-30 \right ) \left ( ax+1 \right ) } \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*x*(5*a^5*x^5+18*a^4*x^4+15*a^3*x^3-20*a^2*x^2-45*a*x-30)*(-a^2*c*x^2+c)^(5/2)/(a*x-1)/(a*x+1)/(-a^2*x^2+1
)^(3/2)

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Maxima [B]  time = 1.105, size = 320, normalized size = 3.44 \begin{align*} -\frac{1}{3} \, a^{2} c^{\frac{5}{2}} x^{3} + \frac{1}{12} \,{\left (\frac{2 \, a^{4} c^{3} x^{8}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac{5 \, a^{2} c^{3} x^{6}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac{3 \, c^{3} x^{4}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}}\right )} a^{3} + c^{\frac{5}{2}} x - \frac{1}{5} \,{\left (3 \, a^{2} c^{\frac{5}{2}} x^{5} - 5 \, c^{\frac{5}{2}} x^{3}\right )} a^{2} + \frac{3}{4} \,{\left (\frac{a^{4} c^{3} x^{6}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac{3 \, a^{2} c^{3} x^{4}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac{2 \, c^{3}}{\sqrt{a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a^2*c^(5/2)*x^3 + 1/12*(2*a^4*c^3*x^8/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 5*a^2*c^3*x^6/sqrt(a^4*c*x^4 -
2*a^2*c*x^2 + c) + 3*c^3*x^4/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c))*a^3 + c^(5/2)*x - 1/5*(3*a^2*c^(5/2)*x^5 - 5*c
^(5/2)*x^3)*a^2 + 3/4*(a^4*c^3*x^6/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 3*a^2*c^3*x^4/sqrt(a^4*c*x^4 - 2*a^2*c*
x^2 + c) + 2*c^3/(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*a^2))*a

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Fricas [A]  time = 2.51615, size = 207, normalized size = 2.23 \begin{align*} \frac{{\left (5 \, a^{5} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{30 \,{\left (a^{2} x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/30*(5*a^5*c^2*x^6 + 18*a^4*c^2*x^5 + 15*a^3*c^2*x^4 - 20*a^2*c^2*x^3 - 45*a*c^2*x^2 - 30*c^2*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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}}{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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