3.10.5 \(\int e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx\) [905]

Optimal. Leaf size=76 \[ -\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^3}{3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 76, 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 = {6332, 6328, 45} \begin {gather*} \frac {x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 45, normalized size = 0.59 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (-3+2 a x)}{6 a \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x^2*(-3 + 2*a*x))/(6*a*Sqrt[1 - 1/(a^2*x^2)])

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 53, normalized size = 0.70

method	result	size
gosper	\(\frac {x^{3} \left (2 a x -3\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {\frac {a x -1}{a x +1}}}{6 a x -6}\)	\(53\)
default	\(\frac {x^{3} \left (2 a x -3\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {\frac {a x -1}{a x +1}}}{6 a x -6}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 24, normalized size = 0.32 \begin {gather*} \frac {{\left (2 \, a x^{3} - 3 \, x^{2}\right )} \sqrt {a^{2} c}}{6 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*x^3 - 3*x^2)*sqrt(a^2*c)/a^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 1.36, size = 46, normalized size = 0.61 \begin {gather*} \frac {x^3\,\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (2\,a\,x-3\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{6\,\left (a\,x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________