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

Optimal. Leaf size=114 \[ -\frac {3 x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 x \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{a \sqrt {1-a^2 x^2}}-\frac {a x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6160, 6140, 43} \[ -\frac {a x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}}-\frac {3 x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 x \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, 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^{3 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx &=\frac {\left (\sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int e^{3 \tanh ^{-1}(a x)} \sqrt {1-a^2 x^2} \, 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}{1-a x} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \left (-3-a x+\frac {4}{1-a x}\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {3 \sqrt {c-\frac {c}{a^2 x^2}} x^2}{\sqrt {1-a^2 x^2}}-\frac {a \sqrt {c-\frac {c}{a^2 x^2}} x^3}{2 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1-a x)}{a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.48 \[ \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {a x^2}{2}-\frac {4 \log (1-a x)}{a}-3 x\right )}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.42, size = 387, normalized size = 3.39 \[ \left [\frac {4 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x - {\left (a^{5} x^{5} - 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - 4 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) + {\left (a^{3} x^{3} + 6 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{2} - a^{2}\right )}}, \frac {8 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {{\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} - 2 \, a^{2} c x^{2} - a c x + 2 \, c}\right ) + {\left (a^{3} x^{3} + 6 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{2} - a^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 69, normalized size = 0.61 \[ \frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (a^{2} x^{2}+6 a x +8 \ln \left (a x -1\right )\right )}{2 \left (a^{2} x^{2}-1\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(a^2*x^2+6*a*x+8*ln(a*x-1))/(a^2*x^2-1)/a

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maxima [C]  time = 0.47, size = 149, normalized size = 1.31 \[ -\frac {1}{2} \, a^{3} {\left (-\frac {i \, \sqrt {c} x^{2}}{a^{3}} - \frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a^{5}} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a^{5}}\right )} - \frac {3}{2} \, a^{2} {\left (-\frac {2 i \, \sqrt {c} x}{a^{3}} + \frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a^{4}} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a^{4}}\right )} - \frac {3}{2} \, a {\left (-\frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a^{3}} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a^{3}}\right )} - \frac {i \, \sqrt {c} \log \left (a x + 1\right )}{2 \, a^{2}} + \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^3*(-I*sqrt(c)*x^2/a^3 - I*sqrt(c)*log(a*x + 1)/a^5 - I*sqrt(c)*log(a*x - 1)/a^5) - 3/2*a^2*(-2*I*sqrt(c
)*x/a^3 + I*sqrt(c)*log(a*x + 1)/a^4 - I*sqrt(c)*log(a*x - 1)/a^4) - 3/2*a*(-I*sqrt(c)*log(a*x + 1)/a^3 - I*sq
rt(c)*log(a*x - 1)/a^3) - 1/2*I*sqrt(c)*log(a*x + 1)/a^2 + 1/2*I*sqrt(c)*log(a*x - 1)/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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