3.1272 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{7/2} \, dx\)

Optimal. Leaf size=142 \[ -\frac{c^3 (1-a x)^8 \sqrt{c-a^2 c x^2}}{8 a \sqrt{1-a^2 x^2}}+\frac{4 c^3 (1-a x)^7 \sqrt{c-a^2 c x^2}}{7 a \sqrt{1-a^2 x^2}}-\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}} \]

[Out]

(-2*c^3*(1 - a*x)^6*Sqrt[c - a^2*c*x^2])/(3*a*Sqrt[1 - a^2*x^2]) + (4*c^3*(1 - a*x)^7*Sqrt[c - a^2*c*x^2])/(7*
a*Sqrt[1 - a^2*x^2]) - (c^3*(1 - a*x)^8*Sqrt[c - a^2*c*x^2])/(8*a*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.103736, antiderivative size = 142, 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} \[ -\frac{c^3 (1-a x)^8 \sqrt{c-a^2 c x^2}}{8 a \sqrt{1-a^2 x^2}}+\frac{4 c^3 (1-a x)^7 \sqrt{c-a^2 c x^2}}{7 a \sqrt{1-a^2 x^2}}-\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^(7/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(-2*c^3*(1 - a*x)^6*Sqrt[c - a^2*c*x^2])/(3*a*Sqrt[1 - a^2*x^2]) + (4*c^3*(1 - a*x)^7*Sqrt[c - a^2*c*x^2])/(7*
a*Sqrt[1 - a^2*x^2]) - (c^3*(1 - a*x)^8*Sqrt[c - a^2*c*x^2])/(8*a*Sqrt[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 )^{7/2} \, dx &=\frac{\left (c^3 \sqrt{c-a^2 c x^2}\right ) \int e^{-3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{7/2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^3 \sqrt{c-a^2 c x^2}\right ) \int (1-a x)^5 (1+a x)^2 \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^3 \sqrt{c-a^2 c x^2}\right ) \int \left (4 (1-a x)^5-4 (1-a x)^6+(1-a x)^7\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{3 a \sqrt{1-a^2 x^2}}+\frac{4 c^3 (1-a x)^7 \sqrt{c-a^2 c x^2}}{7 a \sqrt{1-a^2 x^2}}-\frac{c^3 (1-a x)^8 \sqrt{c-a^2 c x^2}}{8 a \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0494028, size = 60, normalized size = 0.42 \[ -\frac{c^3 (a x-1)^6 \left (21 a^2 x^2+54 a x+37\right ) \sqrt{c-a^2 c x^2}}{168 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^(7/2)/E^(3*ArcTanh[a*x]),x]

[Out]

-(c^3*(-1 + a*x)^6*(37 + 54*a*x + 21*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(168*a*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.03, size = 97, normalized size = 0.7 \begin{align*}{\frac{x \left ( 21\,{a}^{7}{x}^{7}-72\,{x}^{6}{a}^{6}+28\,{x}^{5}{a}^{5}+168\,{x}^{4}{a}^{4}-210\,{x}^{3}{a}^{3}-56\,{a}^{2}{x}^{2}+252\,ax-168 \right ) }{168\, \left ( ax+1 \right ) ^{5} \left ( ax-1 \right ) ^{5}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{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^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

1/168*x*(21*a^7*x^7-72*a^6*x^6+28*a^5*x^5+168*a^4*x^4-210*a^3*x^3-56*a^2*x^2+252*a*x-168)*(-a^2*c*x^2+c)^(7/2)
*(-a^2*x^2+1)^(3/2)/(a*x+1)^5/(a*x-1)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{7}{2}}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a^2*c*x^2 + c)^(7/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)

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Fricas [A]  time = 3.16564, size = 261, normalized size = 1.84 \begin{align*} \frac{{\left (21 \, a^{7} c^{3} x^{8} - 72 \, a^{6} c^{3} x^{7} + 28 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} - 210 \, a^{3} c^{3} x^{4} - 56 \, a^{2} c^{3} x^{3} + 252 \, a c^{3} x^{2} - 168 \, c^{3} x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{168 \,{\left (a^{2} x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

1/168*(21*a^7*c^3*x^8 - 72*a^6*c^3*x^7 + 28*a^5*c^3*x^6 + 168*a^4*c^3*x^5 - 210*a^3*c^3*x^4 - 56*a^2*c^3*x^3 +
252*a*c^3*x^2 - 168*c^3*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**(7/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Timed out

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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{7}{2}}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(7/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)