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

Optimal. Leaf size=148 \[ -\frac {3 a \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}}+\frac {4 a^2 x \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 a^2 x \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{\sqrt {1-a^2 x^2}} \]

[Out]

-3*a*(c-c/a^2/x^2)^(1/2)/(-a^2*x^2+1)^(1/2)-1/2*(c-c/a^2/x^2)^(1/2)/x/(-a^2*x^2+1)^(1/2)+4*a^2*x*ln(x)*(c-c/a^
2/x^2)^(1/2)/(-a^2*x^2+1)^(1/2)-4*a^2*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 = 0.25, antiderivative size = 148, 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} \[ -\frac {3 a \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}}+\frac {4 a^2 x \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {4 a^2 x \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Sqrt[c - c/(a^2*x^2)])/Sqrt[1 - a^2*x^2] - Sqrt[c - c/(a^2*x^2)]/(2*x*Sqrt[1 - a^2*x^2]) + (4*a^2*Sqrt[c
- c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] - (4*a^2*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^2} \, 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^3} \, 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^3 (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^3}+\frac {3 a}{x^2}+\frac {4 a^2}{x}-\frac {4 a^3}{-1+a x}\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {3 a \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}}+\frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}} x \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^2 \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 = 0.04, size = 64, normalized size = 0.43 \[ \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^2 \log (x)-4 a^2 \log (1-a x)-\frac {3 a}{x}-\frac {1}{2 x^2}\right )}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 492, normalized size = 3.32 \[ \left [\frac {4 \, {\left (a^{3} x^{3} - a x\right )} \sqrt {-c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x - {\left (4 \, a^{4} x^{4} - 6 \, a^{3} x^{3} - {\left (4 \, a^{4} - 6 \, a^{3} + 4 \, a^{2} - a\right )} x^{5} + 4 \, a^{2} x^{2} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a + 1\right )} x^{2} - 6 \, a x - 1\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{2} x^{3} - x\right )}}, \frac {8 \, {\left (a^{3} x^{3} - a x\right )} \sqrt {c} \arctan \left (\frac {{\left (2 \, a^{2} x^{2} - {\left (2 \, a^{3} - 2 \, a^{2} + a\right )} x^{3} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a + 1\right )} x^{2} - 6 \, a x - 1\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{2} x^{3} - x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 0.53 \[ -\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {-a^{2} x^{2}+1}\, \left (8 a^{2} \ln \relax (x ) x^{2}-8 \ln \left (a x -1\right ) x^{2} a^{2}-6 a x -1\right )}{2 x \left (a^{2} x^{2}-1\right )} \]

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

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maxima [C]  time = 0.45, size = 159, normalized size = 1.07 \[ -\frac {1}{2} \, a^{3} {\left (\frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a^{2}} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a^{2}}\right )} - \frac {3}{2} \, a^{2} {\left (-\frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a} + \frac {2 i \, \sqrt {c} \log \relax (x)}{a}\right )} + \frac {1}{2} i \, a \sqrt {c} \log \left (a x + 1\right ) + \frac {1}{2} i \, a \sqrt {c} \log \left (a x - 1\right ) - i \, a \sqrt {c} \log \relax (x) - \frac {3}{2} \, {\left (i \, \sqrt {c} \log \left (a x + 1\right ) - i \, \sqrt {c} \log \left (a x - 1\right ) - \frac {2 i \, \sqrt {c}}{a x}\right )} a + \frac {i \, \sqrt {c}}{2 \, a x^{2}} \]

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^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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