3.9.98 \(\int e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} \, dx\) [898]

Optimal. Leaf size=109 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 109, 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 = {6332, 6328, 84} \begin {gather*} \frac {x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\log (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 \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), 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 6332

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

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Mathematica [A]
time = 0.04, size = 50, normalized size = 0.46 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} (a x-\log (x)+4 \log (1-a x))}{a \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 65, normalized size = 0.60

method	result	size
default	\(-\frac {\left (-a x +\ln \left (x \right )-4 \ln \left (a x -1\right )\right ) x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x -1\right )}{\left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a*x+ln(x)-4*ln(a*x-1))*x*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(a*x-1)/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 27, normalized size = 0.25 \begin {gather*} \frac {\sqrt {a^{2} c} {\left (a x + 4 \, \log \left (a x - 1\right ) - \log \left (x\right )\right )}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a^2*c)*(a*x + 4*log(a*x - 1) - log(x))/a^2

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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