3.965 \(\int \frac{e^{\tanh ^{-1}(a x)}}{x^4 \sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=187 \[ -\frac{a^2 \sqrt{1-a^2 x^2}}{x \sqrt{c-a^2 c x^2}}-\frac{a \sqrt{1-a^2 x^2}}{2 x^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{3 x^3 \sqrt{c-a^2 c x^2}}+\frac{a^3 \sqrt{1-a^2 x^2} \log (x)}{\sqrt{c-a^2 c x^2}}-\frac{a^3 \sqrt{1-a^2 x^2} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.19943, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 44} \[ -\frac{a^2 \sqrt{1-a^2 x^2}}{x \sqrt{c-a^2 c x^2}}-\frac{a \sqrt{1-a^2 x^2}}{2 x^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{3 x^3 \sqrt{c-a^2 c x^2}}+\frac{a^3 \sqrt{1-a^2 x^2} \log (x)}{\sqrt{c-a^2 c x^2}}-\frac{a^3 \sqrt{1-a^2 x^2} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0403855, size = 72, normalized size = 0.39 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{a^2}{x}+a^3 \log (x)-a^3 \log (1-a x)-\frac{a}{2 x^2}-\frac{1}{3 x^3}\right )}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 84, normalized size = 0.5 \begin{align*} -{\frac{6\,{a}^{3}\ln \left ( x \right ){x}^{3}-6\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-6\,{a}^{2}{x}^{2}-3\,ax-2}{ \left ( 6\,{a}^{2}{x}^{2}-6 \right ) c{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

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)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} x^{4}}\,{d x} \end{align*}

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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.56926, size = 1026, normalized size = 5.49 \begin{align*} \left [\frac{3 \,{\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt{c} \log \left (-\frac{4 \, a^{5} c x^{5} -{\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} -{\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x +{\left (4 \, a^{3} x^{3} -{\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (6 \, a^{2} x^{2} -{\left (6 \, a^{2} + 3 \, a + 2\right )} x^{3} + 3 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{2} c x^{5} - c x^{3}\right )}}, -\frac{6 \,{\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt{-c} \arctan \left (-\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}{\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt{-c}}{2 \, a^{3} c x^{3} -{\left (2 \, a^{3} - a^{2}\right )} c x^{4} -{\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) - \sqrt{-a^{2} c x^{2} + c}{\left (6 \, a^{2} x^{2} -{\left (6 \, a^{2} + 3 \, a + 2\right )} x^{3} + 3 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{2} c x^{5} - c x^{3}\right )}}\right ] \end{align*}

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)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^5*x^5 - a^3*x^3)*sqrt(c)*log(-(4*a^5*c*x^5 - (2*a^6 - 4*a^5 + 6*a^4 - 4*a^3 + a^2)*c*x^6 - (4*a^4 +
4*a^3 - 6*a^2 + 4*a - 1)*c*x^4 + 5*a^2*c*x^2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 - 6*a^2 + 4*a - 1)*x^4 - 6*a^2*x
^2 + 4*a*x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) + c)/(a^4*x^6 - 2*a^3*x^5 + 2*a*x^3 - x^2)) +
sqrt(-a^2*c*x^2 + c)*(6*a^2*x^2 - (6*a^2 + 3*a + 2)*x^3 + 3*a*x + 2)*sqrt(-a^2*x^2 + 1))/(a^2*c*x^5 - c*x^3),
-1/6*(6*(a^5*x^5 - a^3*x^3)*sqrt(-c)*arctan(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 - 2*a + 1)*x^2 -
2*a*x + 1)*sqrt(-c)/(2*a^3*c*x^3 - (2*a^3 - a^2)*c*x^4 - (a^2 - 2*a + 1)*c*x^2 - 2*a*c*x + c)) - sqrt(-a^2*c*x
^2 + c)*(6*a^2*x^2 - (6*a^2 + 3*a + 2)*x^3 + 3*a*x + 2)*sqrt(-a^2*x^2 + 1))/(a^2*c*x^5 - c*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{x^{4} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

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)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} x^{4}}\,{d x} \end{align*}

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)^(1/2),x, algorithm="giac")

[Out]

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