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

Optimal. Leaf size=184 \[ -\frac {\sqrt {1-a^2 x^2}}{4 a^3 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^3 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

1/8*(-a^2*x^2+1)^(1/2)/a^3/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)-1/4*(-a^2*x^2+1)^(1/2)/a^3/c^2/(-a*x+1)/(-a^2*c
*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a^3/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/8*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/
a^3/c^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6153, 6150, 88, 207} \[ -\frac {\sqrt {1-a^2 x^2}}{4 a^3 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^3 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 84, normalized size = 0.46 \[ \frac {\sqrt {1-a^2 x^2} \left (a^2 x^2+3 a x-(a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2\right )}{8 a^3 c^2 (a x-1)^2 (a x+1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 457, normalized size = 2.48 \[ \left [\frac {{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} + 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{32 \, {\left (a^{8} c^{3} x^{5} - a^{7} c^{3} x^{4} - 2 \, a^{6} c^{3} x^{3} + 2 \, a^{5} c^{3} x^{2} + a^{4} c^{3} x - a^{3} c^{3}\right )}}, -\frac {{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{16 \, {\left (a^{8} c^{3} x^{5} - a^{7} c^{3} x^{4} - 2 \, a^{6} c^{3} x^{3} + 2 \, a^{5} c^{3} x^{2} + a^{4} c^{3} x - a^{3} c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*sqrt(c)*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x
^2 + 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 -
1)) - 4*(2*a^3*x^3 - a^2*x^2 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^3*x^5 - a^7*c^3*x^4 - 2*a
^6*c^3*x^3 + 2*a^5*c^3*x^2 + a^4*c^3*x - a^3*c^3), -1/16*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1
)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a^4*c*x^4 - c)) + 2*(2*a^3*x^3 - a^2
*x^2 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^3*x^5 - a^7*c^3*x^4 - 2*a^6*c^3*x^3 + 2*a^5*c^3*x^
2 + a^4*c^3*x - a^3*c^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 161, normalized size = 0.88 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x -1\right ) x^{3} a^{3}-a^{3} x^{3} \ln \left (a x +1\right )-\ln \left (a x -1\right ) x^{2} a^{2}+\ln \left (a x +1\right ) x^{2} a^{2}+2 a^{2} x^{2}-\ln \left (a x -1\right ) x a +a x \ln \left (a x +1\right )+6 a x +\ln \left (a x -1\right )-\ln \left (a x +1\right )-4\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} a^{3} \left (a x -1\right )^{2} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(ln(a*x-1)*x^3*a^3-a^3*x^3*ln(a*x+1)-ln(a*x-1)*x^2*a^2+ln(a*x+
1)*x^2*a^2+2*a^2*x^2-ln(a*x-1)*x*a+a*x*ln(a*x+1)+6*a*x+ln(a*x-1)-ln(a*x+1)-4)/(a^2*x^2-1)/c^3/a^3/(a*x-1)^2/(a
*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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