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

Optimal. Leaf size=113 \[ \frac {x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 113, 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 = {6197, 6193, 43} \[ \frac {x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[c - c/(a^2*x^2)]*x)/(a*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[c - c/(a^2*x^2)]*x^2)/(2*Sqrt[1 - 1/(a^2*x^2)])
+ (4*Sqrt[c - c/(a^2*x^2)]*Log[1 - a*x])/(a^2*Sqrt[1 - 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 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rule 6197

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d/x^2
)^FracPart[p])/(1 - 1/(a^2*x^2))^FracPart[p], Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a
, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[n/2] &&  !(IntegerQ[p] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.48 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} (a x (a x+6)+8 \log (1-a x))}{2 a^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*(a*x*(6 + a*x) + 8*Log[1 - a*x]))/(2*a^2*Sqrt[1 - 1/(a^2*x^2)])

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fricas [A]  time = 0.69, size = 32, normalized size = 0.28 \[ \frac {{\left (a^{2} x^{2} + 6 \, a x + 8 \, \log \left (a x - 1\right )\right )} \sqrt {a^{2} c}}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2 + 6*a*x + 8*log(a*x - 1))*sqrt(a^2*c)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 73, normalized size = 0.65 \[ \frac {\left (a^{2} x^{2}+6 a x +8 \ln \left (a x -1\right )\right ) x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x -1\right )}{2 a \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2+6*a*x+8*ln(a*x-1))*x*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(a*x-1)/a/(a*x+1)^2/((a*x-1)/(a*x+1))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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