3.10.78 \(\int \frac {e^{\tanh ^{-1}(a x)} x^4}{(c-a^2 c x^2)^{5/2}} \, dx\) [978]

Optimal. Leaf size=232 \[ \frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{4 a^5 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (1+a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

1/8*(-a^2*x^2+1)^(1/2)/a^5/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)-3/4*(-a^2*x^2+1)^(1/2)/a^5/c^2/(-a*x+1)/(-a^2*c
*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a^5/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)-11/16*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/
a^5/c^2/(-a^2*c*x^2+c)^(1/2)-5/16*ln(a*x+1)*(-a^2*x^2+1)^(1/2)/a^5/c^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 232, 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 = {6288, 6285, 90} \begin {gather*} -\frac {3 \sqrt {1-a^2 x^2}}{4 a^5 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (a x+1)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]

[Out]

Sqrt[1 - a^2*x^2]/(8*a^5*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) - (3*Sqrt[1 - a^2*x^2])/(4*a^5*c^2*(1 - a*x)*Sqr
t[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a^5*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) - (11*Sqrt[1 - a^2*x^2]*Log[1
- a*x])/(16*a^5*c^2*Sqrt[c - a^2*c*x^2]) - (5*Sqrt[1 - a^2*x^2]*Log[1 + a*x])/(16*a^5*c^2*Sqrt[c - a^2*c*x^2])

Rule 90

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 6285

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 6288

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 )^{5/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{\tanh ^{-1}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {x^4}{(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^4 (-1+a x)^3}-\frac {3}{4 a^4 (-1+a x)^2}-\frac {11}{16 a^4 (-1+a x)}+\frac {1}{8 a^4 (1+a x)^2}-\frac {5}{16 a^4 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{4 a^5 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (1+a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 87, normalized size = 0.38 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (\frac {2 \left (-6+3 a x+5 a^2 x^2\right )}{(-1+a x)^2 (1+a x)}-11 \log (1-a x)-5 \log (1+a x)\right )}{16 a^5 c^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[1 - a^2*x^2]*((2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) - 11*Log[1 - a*x] - 5*Log[1 + a*x]))
/(16*a^5*c^2*Sqrt[c - a^2*c*x^2])

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Maple [A]
time = 0.07, size = 166, normalized size = 0.72

method	result	size
default	\(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (5 \ln \left (a x +1\right ) a^{3} x^{3}+11 \ln \left (a x -1\right ) a^{3} x^{3}-5 \ln \left (a x +1\right ) a^{2} x^{2}-11 \ln \left (a x -1\right ) a^{2} x^{2}-10 a^{2} x^{2}-5 \ln \left (a x +1\right ) a x -11 \ln \left (a x -1\right ) a x -6 a x +5 \ln \left (a x +1\right )+11 \ln \left (a x -1\right )+12\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} a^{5} \left (a x -1\right )^{2} \left (a x +1\right )}\)	\(166\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(5*ln(a*x+1)*a^3*x^3+11*ln(a*x-1)*a^3*x^3-5*ln(a*x+1)*a^2*x^2-1
1*ln(a*x-1)*a^2*x^2-10*a^2*x^2-5*ln(a*x+1)*a*x-11*ln(a*x-1)*a*x-6*a*x+5*ln(a*x+1)+11*ln(a*x-1)+12)/(a^2*x^2-1)
/c^3/a^5/(a*x-1)^2/(a*x+1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^4/(a^7*c^3*x^7 - a^6*c^3*x^6 - 3*a^5*c^3*x^5 + 3*a^4*c^3*x^
4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a*c^3*x + c^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}

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)**(5/2),x)

[Out]

Integral(x**4*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^4/((-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 = -0.00 \begin {gather*} \int \frac {x^4\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)

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