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]

2/5*c^2*(a*x+1)^5*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-1/6*c^2*(a*x+1)^6*(-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 = 93, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [A]  time = 0.04, 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]

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

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fricas [A]  time = 0.60, size = 98, normalized size = 1.05 \[ \frac {{\left (5 \, a^{5} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{30 \, {\left (a^{2} x^{2} - 1\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 81, normalized size = 0.87 \[ \frac {x \left (5 x^{5} a^{5}+18 x^{4} a^{4}+15 x^{3} a^{3}-20 a^{2} x^{2}-45 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x -1\right ) \left (a x +1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}} \]

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 = 0.39, size = 237, normalized size = 2.55 \[ -\frac {1}{3} \, a^{2} c^{\frac {5}{2}} x^{3} + \frac {1}{12} \, {\left (\frac {2 \, a^{4} c^{3} x^{8}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {5 \, a^{2} c^{3} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {3 \, c^{3} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}}\right )} a^{3} + c^{\frac {5}{2}} x - \frac {1}{5} \, {\left (3 \, a^{2} c^{\frac {5}{2}} x^{5} - 5 \, c^{\frac {5}{2}} x^{3}\right )} a^{2} + \frac {3}{4} \, {\left (\frac {a^{4} c^{3} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {3 \, a^{2} c^{3} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {2 \, c^{3}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2}}\right )} a \]

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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mupad [B]  time = 1.08, size = 85, normalized size = 0.91 \[ \frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {a^5\,c^2\,x^6}{6}-\frac {3\,a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}+\frac {2\,a^2\,c^2\,x^3}{3}+\frac {3\,a\,c^2\,x^2}{2}+c^2\,x\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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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