3.985 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^3 (c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=345 \[ \frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

-1/2*(-a^2*x^2+1)^(1/2)/c^2/x^2/(-a^2*c*x^2+c)^(1/2)-a*(-a^2*x^2+1)^(1/2)/c^2/x/(-a^2*c*x^2+c)^(1/2)+1/8*a^2*(
-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)+a^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*(-a^2*x^2+1)^(1/2)/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+3*a^2*ln(x)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)
^(1/2)-39/16*a^2*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)^(1/2)-9/16*a^2*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 = 0.26, antiderivative size = 345, 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} \[ \frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)^(5/2)),x]

[Out]

-Sqrt[1 - a^2*x^2]/(2*c^2*x^2*Sqrt[c - a^2*c*x^2]) - (a*Sqrt[1 - a^2*x^2])/(c^2*x*Sqrt[c - a^2*c*x^2]) + (a^2*
Sqrt[1 - a^2*x^2])/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + (a^2*Sqrt[1 - a^2*x^2])/(c^2*(1 - a*x)*Sqrt[c - a
^2*c*x^2]) + (a^2*Sqrt[1 - a^2*x^2])/(8*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (3*a^2*Sqrt[1 - a^2*x^2]*Log[x])/
(c^2*Sqrt[c - a^2*c*x^2]) - (39*a^2*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) - (9*a^2*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^3 \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^3 \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^3 (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^3}+\frac {a}{x^2}+\frac {3 a^2}{x}-\frac {a^3}{4 (-1+a x)^3}+\frac {a^3}{(-1+a x)^2}-\frac {39 a^3}{16 (-1+a x)}-\frac {a^3}{8 (1+a x)^2}-\frac {9 a^3}{16 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \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 = 0.12, size = 115, normalized size = 0.33 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {16 a^2}{1-a x}+\frac {2 a^2}{a x+1}+\frac {2 a^2}{(a x-1)^2}+48 a^2 \log (x)-39 a^2 \log (1-a x)-9 a^2 \log (a x+1)-\frac {16 a}{x}-\frac {8}{x^2}\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-8/x^2 - (16*a)/x + (16*a^2)/(1 - a*x) + (2*a^2)/(-1 + a*x)^2 + (2*a^2)/(1 + a*x) + 48*a^2
*Log[x] - 39*a^2*Log[1 - a*x] - 9*a^2*Log[1 + a*x]))/(16*c^2*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{a^{7} c^{3} x^{10} - a^{6} c^{3} x^{9} - 3 \, a^{5} c^{3} x^{8} + 3 \, a^{4} c^{3} x^{7} + 3 \, a^{3} c^{3} x^{6} - 3 \, a^{2} c^{3} x^{5} - a c^{3} x^{4} + c^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-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^10 - a^6*c^3*x^9 - 3*a^5*c^3*x^8 + 3*a^4*c^3*x^7 +
3*a^3*c^3*x^6 - 3*a^2*c^3*x^5 - a*c^3*x^4 + c^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.05, size = 242, normalized size = 0.70 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (48 a^{5} \ln \relax (x ) x^{5}-39 \ln \left (a x -1\right ) x^{5} a^{5}-9 \ln \left (a x +1\right ) x^{5} a^{5}-48 a^{4} \ln \relax (x ) x^{4}+39 \ln \left (a x -1\right ) x^{4} a^{4}+9 \ln \left (a x +1\right ) x^{4} a^{4}-30 x^{4} a^{4}-48 a^{3} \ln \relax (x ) x^{3}+39 \ln \left (a x -1\right ) x^{3} a^{3}+9 a^{3} x^{3} \ln \left (a x +1\right )+6 x^{3} a^{3}+48 a^{2} \ln \relax (x ) x^{2}-39 \ln \left (a x -1\right ) x^{2} a^{2}-9 \ln \left (a x +1\right ) x^{2} a^{2}+44 a^{2} x^{2}-8 a x -8\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} \left (a x -1\right )^{2} \left (a x +1\right ) x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-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)*(48*a^5*ln(x)*x^5-39*ln(a*x-1)*x^5*a^5-9*ln(a*x+1)*x^5*a^5-48*
a^4*ln(x)*x^4+39*ln(a*x-1)*x^4*a^4+9*ln(a*x+1)*x^4*a^4-30*x^4*a^4-48*a^3*ln(x)*x^3+39*ln(a*x-1)*x^3*a^3+9*a^3*
x^3*ln(a*x+1)+6*x^3*a^3+48*a^2*ln(x)*x^2-39*ln(a*x-1)*x^2*a^2-9*ln(a*x+1)*x^2*a^2+44*a^2*x^2-8*a*x-8)/(a^2*x^2
-1)/c^3/(a*x-1)^2/(a*x+1)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x + 1)/(x^3*(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 \[ \int \frac {a x + 1}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + 1)/(x**3*sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), x)

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