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

Optimal. Leaf size=123 \[ \frac {\sqrt {1-a^2 x^2}}{a \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 x (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-a^2 x^2} \log (1-a x)}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

[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 = 123, 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} \[ \frac {\sqrt {1-a^2 x^2}}{a \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 x (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-a^2 x^2} \log (1-a x)}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[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 = 59, normalized size = 0.48 \[ \frac {\sqrt {1-a^2 x^2} \left (a x+\frac {2}{1-a x}+3 \log (1-a x)\right )}{a^2 x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[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 = 2.05, size = 440, normalized size = 3.58 \[ \left [-\frac {3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {-c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x + {\left (a^{5} x^{5} - 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - 4 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) - 2 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} c x^{3} - a^{3} c x^{2} - a^{2} c x + a c\right )}}, \frac {3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {c} \arctan \left (\frac {{\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} - 2 \, a^{2} c x^{2} - a c x + 2 \, c}\right ) + {\left (a^{3} x^{3} - 3 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{4} c x^{3} - a^{3} c x^{2} - a^{2} c x + a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 \[ \int \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c - \frac {c}{a^{2} x^{2}}}}\,{d x} \]

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, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 77, normalized size = 0.63 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \left (a^{2} x^{2}+3 \ln \left (a x -1\right ) x a -a x -3 \ln \left (a x -1\right )-2\right )}{\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \,a^{2} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*ln(a*x-1)*x*a-a*x-3*ln(a*x-1)-2)/(a*x-1)

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

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, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

[Out]

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

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