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

Optimal. Leaf size=255 \[ \frac {a^2 \sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2 \sqrt {c-a^2 c x^2}}+\frac {2 a^2 \sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {7 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {a^2 \sqrt {1-a^2 x^2} \log (a x+1)}{4 c \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 255, 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} \[ \frac {a^2 \sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2 \sqrt {c-a^2 c x^2}}+\frac {2 a^2 \sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {7 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {a^2 \sqrt {1-a^2 x^2} \log (a x+1)}{4 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.07, size = 91, normalized size = 0.36 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (a x+1)-\frac {4 a}{x}-\frac {2}{x^2}\right )}{4 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8*a^2*Log[x] - 7*a^2*Log[1 - a*x] - a^2*Log[1 + a*x
]))/(4*c*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{a^{5} c^{2} x^{8} - a^{4} c^{2} x^{7} - 2 \, a^{3} c^{2} x^{6} + 2 \, a^{2} c^{2} x^{5} + a c^{2} x^{4} - c^{2} x^{3}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 142, normalized size = 0.56 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (8 a^{3} \ln \relax (x ) x^{3}-7 \ln \left (a x -1\right ) x^{3} a^{3}-a^{3} x^{3} \ln \left (a x +1\right )-8 a^{2} \ln \relax (x ) x^{2}+7 \ln \left (a x -1\right ) x^{2} a^{2}+\ln \left (a x +1\right ) x^{2} a^{2}-6 a^{2} x^{2}+2 a x +2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} \left (a x -1\right ) x^{2}} \]

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)^(3/2),x)

[Out]

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

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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