3.967 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=218 \[ \frac {\sqrt {1-a^2 x^2}}{2 a^5 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {7 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^5 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 a^5 c \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {1-a^2 x^2}}{a^4 c \sqrt {c-a^2 c x^2}}+\frac {x^2 \sqrt {1-a^2 x^2}}{2 a^3 c \sqrt {c-a^2 c x^2}} \]

[Out]

x*(-a^2*x^2+1)^(1/2)/a^4/c/(-a^2*c*x^2+c)^(1/2)+1/2*x^2*(-a^2*x^2+1)^(1/2)/a^3/c/(-a^2*c*x^2+c)^(1/2)+1/2*(-a^
2*x^2+1)^(1/2)/a^5/c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+7/4*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/a^5/c/(-a^2*c*x^2+c)^(1/2
)+1/4*ln(a*x+1)*(-a^2*x^2+1)^(1/2)/a^5/c/(-a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 88} \[ \frac {x^2 \sqrt {1-a^2 x^2}}{2 a^3 c \sqrt {c-a^2 c x^2}}+\frac {x \sqrt {1-a^2 x^2}}{a^4 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 a^5 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {7 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^5 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 a^5 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 77, normalized size = 0.35 \[ \frac {\sqrt {1-a^2 x^2} \left (2 \left (a^2 x^2+2 a x+\frac {1}{1-a x}\right )+7 \log (1-a x)+\log (a x+1)\right )}{4 a^5 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]

[Out]

(Sqrt[1 - a^2*x^2]*(2*(2*a*x + a^2*x^2 + (1 - a*x)^(-1)) + 7*Log[1 - a*x] + Log[1 + a*x]))/(4*a^5*c*Sqrt[c - a
^2*c*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{4}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^4/(a^5*c^2*x^5 - a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 2*a^2*c^2*x
^2 + a*c^2*x - c^2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^4/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)), x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 110, normalized size = 0.50 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2 x^{3} a^{3}+2 a^{2} x^{2}+7 \ln \left (a x -1\right ) x a +a x \ln \left (a x +1\right )-4 a x -7 \ln \left (a x -1\right )-\ln \left (a x +1\right )-2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a^{5} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x)

[Out]

-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(2*x^3*a^3+2*a^2*x^2+7*ln(a*x-1)*x*a+a*x*ln(a*x+1)-4*a*x-7*ln(a
*x-1)-ln(a*x+1)-2)/(a^2*x^2-1)/c^2/a^5/(a*x-1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)*x^4/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4/(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(x**4*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(3/2)), x)

________________________________________________________________________________________