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

Optimal. Leaf size=144 \[ \frac {a x^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {a^2 x^3 \log (x) \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}+\frac {a^3 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6160, 6150, 75} \[ \frac {a^3 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}+\frac {a x^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {a^2 x^3 \log (x) \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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

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Mathematica [A]  time = 0.04, size = 72, normalized size = 0.50 \[ -\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \left (2 a^3 x^3-3 a^2 x^2-2 a^2 x^2 \log (x)+2 a x-1\right )}{2 a^2 x \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 377, normalized size = 2.62 \[ \left [\frac {{\left (a^{3} c x^{3} - a c x\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} + {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) + {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + 2 \, a - 1\right )} c x^{2} + 2 \, a c x - c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}, -\frac {2 \, {\left (a^{3} c x^{3} - a c x\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) - {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + 2 \, a - 1\right )} c x^{2} + 2 \, a c x - c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 70, normalized size = 0.49 \[ -\frac {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {3}{2}} x \sqrt {-a^{2} x^{2}+1}\, \left (-2 x^{3} a^{3}+2 a^{2} \ln \relax (x ) x^{2}-2 a x +1\right )}{2 \left (a^{2} x^{2}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2}\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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