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

Optimal. Leaf size=122 \[ -\frac {\sqrt {1-a^2 x^2}}{a \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 x (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-a^2 x^2} \log (a x+1)}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 122, 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, 77} \[ -\frac {\sqrt {1-a^2 x^2}}{a \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 x (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-a^2 x^2} \log (a x+1)}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.05, size = 58, normalized size = 0.48 \[ \frac {\sqrt {1-a^2 x^2} \left (-a x+\frac {2}{a x+1}+3 \log (a x+1)\right )}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

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

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fricas [A]  time = 1.73, size = 439, normalized size = 3.60 \[ \left [-\frac {3 \, {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {-c} \log \left (\frac {a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x + {\left (a^{5} x^{5} + 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} + 4 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) + 2 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} c x^{3} + a^{3} c x^{2} - a^{2} c x - a c\right )}}, \frac {3 \, {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {c} \arctan \left (\frac {{\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} + 2 \, a^{2} c x^{2} - a c x - 2 \, c}\right ) - {\left (a^{3} x^{3} + 3 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{4} c x^{3} + a^{3} c x^{2} - a^{2} c x - a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 0.64 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} x^{2}+3 a x \ln \left (a x +1\right )-a x +3 \ln \left (a x +1\right )+2\right )}{\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \,a^{2} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.34, size = 53, normalized size = 0.43 \[ -\frac {i \, a^{2} \sqrt {c} x^{2} + i \, a \sqrt {c} x - 2 i \, \sqrt {c}}{a^{2} c x + a c} + \frac {3 i \, \log \left (a x + 1\right )}{a \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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