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

Optimal. Leaf size=220 \[ -\frac {2 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-\frac {c}{a^2 x^2}}}{x^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 x^3 \sqrt {1-a^2 x^2}}+\frac {4 a^4 x \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 x \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{\sqrt {1-a^2 x^2}}+\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 220, normalized size of antiderivative = 1.00, 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} \[ \frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {2 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-\frac {c}{a^2 x^2}}}{x^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 x^3 \sqrt {1-a^2 x^2}}+\frac {4 a^4 x \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 x \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

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*}

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Mathematica [A]  time = 0.05, size = 78, normalized size = 0.35 \[ \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac {4 a^3}{x}-\frac {2 a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(-1/4*1/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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fricas [A]  time = 0.68, size = 551, normalized size = 2.50 \[ \left [\frac {8 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {-c} \log \left (\frac {4 \, a^{5} c x^{5} + {\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} + {\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x - {\left (4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - {\left (4 \, a^{4} + 6 \, a^{3} + 4 \, a^{2} + a\right )} x^{5} + 4 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) - {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{2} x^{5} - x^{3}\right )}}, -\frac {16 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {c} \arctan \left (-\frac {{\left (2 \, a^{2} x^{2} + {\left (2 \, a^{3} + 2 \, a^{2} + a\right )} x^{3} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + a^{2}\right )} c x^{4} + {\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) + {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{2} x^{5} - x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/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)*s
qrt((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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{{\left (a x + 1\right )}^{3} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 94, normalized size = 0.43 \[ -\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {-a^{2} x^{2}+1}\, \left (16 a^{4} \ln \relax (x ) x^{4}-16 \ln \left (a x +1\right ) x^{4} a^{4}+16 x^{3} a^{3}-8 a^{2} x^{2}+4 a x -1\right )}{4 x^{3} \left (a^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/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)*x^4*a^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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{{\left (a x + 1\right )}^{3} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^4\,{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{x^{4} \left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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