3.1209 \(\int e^{-\tanh ^{-1}(a x)} (c-a^2 c x^2)^{7/2} \, dx\)

Optimal. Leaf size=187 \[ \frac{c^3 (1-a x)^8 \sqrt{c-a^2 c x^2}}{8 a \sqrt{1-a^2 x^2}}-\frac{6 c^3 (1-a x)^7 \sqrt{c-a^2 c x^2}}{7 a \sqrt{1-a^2 x^2}}+\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{8 c^3 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}} \]

[Out]

(-8*c^3*(1 - a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) + (2*c^3*(1 - a*x)^6*Sqrt[c - a^2*c*x^2])/(a*
Sqrt[1 - a^2*x^2]) - (6*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])

________________________________________________________________________________________

Rubi [A]  time = 0.116011, antiderivative size = 187, 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} \[ \frac{c^3 (1-a x)^8 \sqrt{c-a^2 c x^2}}{8 a \sqrt{1-a^2 x^2}}-\frac{6 c^3 (1-a x)^7 \sqrt{c-a^2 c x^2}}{7 a \sqrt{1-a^2 x^2}}+\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{8 c^3 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^3*(1 - a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) + (2*c^3*(1 - a*x)^6*Sqrt[c - a^2*c*x^2])/(a*
Sqrt[1 - a^2*x^2]) - (6*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^{-\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^{-\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)^4 (1+a x)^3 \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^3 \sqrt{c-a^2 c x^2}\right ) \int \left (8 (1-a x)^4-12 (1-a x)^5+6 (1-a x)^6-(1-a x)^7\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{8 c^3 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}+\frac{2 c^3 (1-a x)^6 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{6 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 = 0.0481089, size = 68, normalized size = 0.36 \[ \frac{c^3 (a x-1)^5 \left (35 a^3 x^3+135 a^2 x^2+185 a x+93\right ) \sqrt{c-a^2 c x^2}}{280 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(-1 + a*x)^5*Sqrt[c - a^2*c*x^2]*(93 + 185*a*x + 135*a^2*x^2 + 35*a^3*x^3))/(280*a*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 97, normalized size = 0.5 \begin{align*}{\frac{x \left ( 35\,{a}^{7}{x}^{7}-40\,{x}^{6}{a}^{6}-140\,{x}^{5}{a}^{5}+168\,{x}^{4}{a}^{4}+210\,{x}^{3}{a}^{3}-280\,{a}^{2}{x}^{2}-140\,ax+280 \right ) }{280\, \left ( ax+1 \right ) ^{4} \left ( ax-1 \right ) ^{4}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}\sqrt{-{a}^{2}{x}^{2}+1}} \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)*(-a^2*x^2+1)^(1/2),x)

[Out]

1/280*x*(35*a^7*x^7-40*a^6*x^6-140*a^5*x^5+168*a^4*x^4+210*a^3*x^3-280*a^2*x^2-140*a*x+280)*(-a^2*c*x^2+c)^(7/
2)*(-a^2*x^2+1)^(1/2)/(a*x+1)^4/(a*x-1)^4

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.27553, size = 265, normalized size = 1.42 \begin{align*} -\frac{{\left (35 \, a^{7} c^{3} x^{8} - 40 \, a^{6} c^{3} x^{7} - 140 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} + 210 \, a^{3} c^{3} x^{4} - 280 \, a^{2} c^{3} x^{3} - 140 \, a c^{3} x^{2} + 280 \, c^{3} x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{280 \,{\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)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/280*(35*a^7*c^3*x^8 - 40*a^6*c^3*x^7 - 140*a^5*c^3*x^6 + 168*a^4*c^3*x^5 + 210*a^3*c^3*x^4 - 280*a^2*c^3*x^
3 - 140*a*c^3*x^2 + 280*c^3*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

________________________________________________________________________________________

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)*(-a**2*x**2+1)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{7}{2}} \sqrt{-a^{2} x^{2} + 1}}{a x + 1}\,{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)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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