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

Optimal. Leaf size=252 \[ \frac {\sqrt {1-a^2 x^2}}{2 c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 252, 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} \[ \frac {\sqrt {1-a^2 x^2}}{2 c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(2*c^2*(1 - a*x)*Sqrt[c - a^2*c*
x^2]) + Sqrt[1 - a^2*x^2]/(8*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*Log[x])/(c^2*Sqrt[c - a^2
*c*x^2]) - (11*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) - (5*Sqrt[1 - a^2*x^2]*Log[1 + a*x
])/(16*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 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 \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 \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 {1}{x (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}{x}-\frac {a}{4 (-1+a x)^3}+\frac {a}{2 (-1+a x)^2}-\frac {11 a}{16 (-1+a x)}-\frac {a}{8 (1+a x)^2}-\frac {5 a}{16 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 86, normalized size = 0.34 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {8}{1-a x}+\frac {2}{a x+1}+\frac {2}{(a x-1)^2}-11 \log (1-a x)-5 \log (a x+1)+16 \log (x)\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{a^{7} c^{3} x^{8} - a^{6} c^{3} x^{7} - 3 \, a^{5} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{5} + 3 \, a^{3} c^{3} x^{4} - 3 \, a^{2} c^{3} x^{3} - a c^{3} x^{2} + c^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-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)/(a^7*c^3*x^8 - a^6*c^3*x^7 - 3*a^5*c^3*x^6 + 3*a^4*c^3*x^5 +
3*a^3*c^3*x^4 - 3*a^2*c^3*x^3 - a*c^3*x^2 + c^3*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 193, normalized size = 0.77 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (16 a^{3} \ln \relax (x ) x^{3}-11 \ln \left (a x -1\right ) x^{3} a^{3}-5 a^{3} x^{3} \ln \left (a x +1\right )-16 a^{2} \ln \relax (x ) x^{2}+11 \ln \left (a x -1\right ) x^{2} a^{2}+5 \ln \left (a x +1\right ) x^{2} a^{2}-6 a^{2} x^{2}-16 a \ln \relax (x ) x +11 \ln \left (a x -1\right ) x a +5 a x \ln \left (a x +1\right )-2 a x +16 \ln \relax (x )-11 \ln \left (a x -1\right )-5 \ln \left (a x +1\right )+12\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} \left (a x -1\right )^{2} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x + 1)/(x*(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 \[ \int \frac {a x + 1}{x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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