3.763 \(\int \frac{e^{3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}}}{x^4} \, dx\)

Optimal. Leaf size=222 \[ -\frac{4 a^3 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{2 a^2 \sqrt{c-\frac{c}{a^2 x^2}}}{x \sqrt{1-a^2 x^2}}-\frac{a \sqrt{c-\frac{c}{a^2 x^2}}}{x^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{4 x^3 \sqrt{1-a^2 x^2}}+\frac{4 a^4 x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{4 a^4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{\sqrt{1-a^2 x^2}} \]

[Out]

(-4*a^3*Sqrt[c - c/(a^2*x^2)])/Sqrt[1 - a^2*x^2] - Sqrt[c - c/(a^2*x^2)]/(4*x^3*Sqrt[1 - a^2*x^2]) - (a*Sqrt[c
- c/(a^2*x^2)])/(x^2*Sqrt[1 - a^2*x^2]) - (2*a^2*Sqrt[c - c/(a^2*x^2)])/(x*Sqrt[1 - a^2*x^2]) + (4*a^4*Sqrt[c
- c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] - (4*a^4*Sqrt[c - c/(a^2*x^2)]*x*Log[1 - a*x])/Sqrt[1 - a^2*x^2]

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Rubi [A]  time = 0.267187, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6160, 6150, 88} \[ -\frac{4 a^3 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{2 a^2 \sqrt{c-\frac{c}{a^2 x^2}}}{x \sqrt{1-a^2 x^2}}-\frac{a \sqrt{c-\frac{c}{a^2 x^2}}}{x^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{c-\frac{c}{a^2 x^2}}}{4 x^3 \sqrt{1-a^2 x^2}}+\frac{4 a^4 x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{4 a^4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)])/x^4,x]

[Out]

(-4*a^3*Sqrt[c - c/(a^2*x^2)])/Sqrt[1 - a^2*x^2] - Sqrt[c - c/(a^2*x^2)]/(4*x^3*Sqrt[1 - a^2*x^2]) - (a*Sqrt[c
- c/(a^2*x^2)])/(x^2*Sqrt[1 - a^2*x^2]) - (2*a^2*Sqrt[c - c/(a^2*x^2)])/(x*Sqrt[1 - a^2*x^2]) + (4*a^4*Sqrt[c
- c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] - (4*a^4*Sqrt[c - c/(a^2*x^2)]*x*Log[1 - a*x])/Sqrt[1 - a^2*x^2]

Rule 6160

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/
(1 + (c*x^2)/d)^p, Int[(u*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[c + a^2*d, 0] &&  !IntegerQ[p] &&  !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 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]))

Rubi steps

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

Mathematica [A]  time = 0.0548248, size = 80, normalized size = 0.36 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (-\frac{2 a^2}{x^2}-\frac{4 a^3}{x}+4 a^4 \log (x)-4 a^4 \log (1-a x)-\frac{a}{x^3}-\frac{1}{4 x^4}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)])/x^4,x]

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(-1/(4*x^4) - a/x^3 - (2*a^2)/x^2 - (4*a^3)/x + 4*a^4*Log[x] - 4*a^4*Log[1 - a*x]))/S
qrt[1 - a^2*x^2]

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Maple [A]  time = 0.154, size = 94, normalized size = 0.4 \begin{align*} -{\frac{16\,{a}^{4}\ln \left ( x \right ){x}^{4}-16\,\ln \left ( ax-1 \right ){a}^{4}{x}^{4}-16\,{x}^{3}{a}^{3}-8\,{a}^{2}{x}^{2}-4\,ax-1}{4\,{x}^{3} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)/x^4,x)

[Out]

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

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Maxima [C]  time = 1.36656, size = 282, normalized size = 1.27 \begin{align*} \frac{1}{2} i \, a^{3} \sqrt{c} \log \left (a x + 1\right ) + \frac{1}{2} i \, a^{3} \sqrt{c} \log \left (a x - 1\right ) - i \, a^{3} \sqrt{c} \log \left (x\right ) - \frac{1}{2} \,{\left (i \, \sqrt{c} \log \left (a x + 1\right ) - i \, \sqrt{c} \log \left (a x - 1\right ) - \frac{2 i \, \sqrt{c}}{a x}\right )} a^{3} - \frac{3}{2} \,{\left (-i \, a \sqrt{c} \log \left (a x + 1\right ) - i \, a \sqrt{c} \log \left (a x - 1\right ) + 2 i \, a \sqrt{c} \log \left (x\right ) - \frac{i \, \sqrt{c}}{a x^{2}}\right )} a^{2} - \frac{1}{2} \,{\left (3 i \, a^{2} \sqrt{c} \log \left (a x + 1\right ) - 3 i \, a^{2} \sqrt{c} \log \left (a x - 1\right ) - \frac{2 \,{\left (3 i \, a^{2} \sqrt{c} x^{2} + i \, \sqrt{c}\right )}}{a x^{3}}\right )} a + \frac{2 i \, a^{2} \sqrt{c} x^{2} + i \, \sqrt{c}}{4 \, a x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

1/2*I*a^3*sqrt(c)*log(a*x + 1) + 1/2*I*a^3*sqrt(c)*log(a*x - 1) - I*a^3*sqrt(c)*log(x) - 1/2*(I*sqrt(c)*log(a*
x + 1) - I*sqrt(c)*log(a*x - 1) - 2*I*sqrt(c)/(a*x))*a^3 - 3/2*(-I*a*sqrt(c)*log(a*x + 1) - I*a*sqrt(c)*log(a*
x - 1) + 2*I*a*sqrt(c)*log(x) - I*sqrt(c)/(a*x^2))*a^2 - 1/2*(3*I*a^2*sqrt(c)*log(a*x + 1) - 3*I*a^2*sqrt(c)*l
og(a*x - 1) - 2*(3*I*a^2*sqrt(c)*x^2 + I*sqrt(c))/(a*x^3))*a + 1/4*(2*I*a^2*sqrt(c)*x^2 + I*sqrt(c))/(a*x^4)

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Fricas [A]  time = 2.68817, size = 1154, normalized size = 5.2 \begin{align*} \left [\frac{8 \,{\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^{4} x^{4} - 6 \, a^{3} x^{3} -{\left (4 \, a^{4} - 6 \, a^{3} + 4 \, a^{2} - a\right )} x^{5} + 4 \, a^{2} x^{2} - a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) +{\left (16 \, a^{3} x^{3} -{\left (16 \, a^{3} + 8 \, a^{2} + 4 \, a + 1\right )} x^{4} + 8 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \,{\left (a^{2} x^{5} - x^{3}\right )}}, \frac{16 \,{\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt{c} \arctan \left (\frac{{\left (2 \, a^{2} x^{2} -{\left (2 \, a^{3} - 2 \, a^{2} + a\right )} x^{3} - a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{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 ) +{\left (16 \, a^{3} x^{3} -{\left (16 \, a^{3} + 8 \, a^{2} + 4 \, a + 1\right )} x^{4} + 8 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \,{\left (a^{2} x^{5} - x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/4*(8*(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^4*x^4 - 6*a^3*x^3 - (4*a^4 - 6*a^3 + 4*a^2 - a
)*x^5 + 4*a^2*x^2 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) + c)/(a^4*x^6 - 2*a^3*x^5
+ 2*a*x^3 - x^2)) + (16*a^3*x^3 - (16*a^3 + 8*a^2 + 4*a + 1)*x^4 + 8*a^2*x^2 + 4*a*x + 1)*sqrt(-a^2*x^2 + 1)*
sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^5 - x^3), 1/4*(16*(a^5*x^5 - a^3*x^3)*sqrt(c)*arctan((2*a^2*x^2 - (2*a
^3 - 2*a^2 + a)*x^3 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(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)) + (16*a^3*x^3 - (16*a^3 + 8*a^2 + 4*a + 1)*x^4 + 8*a^2*x^2
+ 4*a*x + 1)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^5 - x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (-1 + \frac{1}{a x}\right ) \left (1 + \frac{1}{a x}\right )} \left (a x + 1\right )^{3}}{x^{4} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(c-c/a**2/x**2)**(1/2)/x**4,x)

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))*(a*x + 1)**3/(x**4*(-(a*x - 1)*(a*x + 1))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{3} \sqrt{c - \frac{c}{a^{2} x^{2}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate((a*x + 1)^3*sqrt(c - c/(a^2*x^2))/((-a^2*x^2 + 1)^(3/2)*x^4), x)