3.733 \(\int e^{-\coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {x^{m+1} \sqrt {c-a^2 c x^2}}{(m+2) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^m \sqrt {c-a^2 c x^2}}{a (m+1) \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 83, 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 = {6192, 6193, 43} \[ \frac {x^{m+1} \sqrt {c-a^2 c x^2}}{(m+2) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^m \sqrt {c-a^2 c x^2}}{a (m+1) \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^m*Sqrt[c - a^2*c*x^2])/(a*(1 + m)*Sqrt[1 - 1/(a^2*x^2)])) + (x^(1 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*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 6192

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 58, normalized size = 0.70 \[ \frac {x^m \sqrt {c-a^2 c x^2} (m (a x-1)+a x-2)}{a (m+1) (m+2) \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 75, normalized size = 0.90 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} {\left ({\left (a m + a\right )} x^{2} - {\left (m + 2\right )} x\right )} x^{m} \sqrt {\frac {a x - 1}{a x + 1}}}{m^{2} - {\left (a m^{2} + 3 \, a m + 2 \, a\right )} x + 3 \, m + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 64, normalized size = 0.77 \[ \frac {x^{1+m} \left (a m x +a x -m -2\right ) \sqrt {-a^{2} c \,x^{2}+c}\, \sqrt {\frac {a x -1}{a x +1}}}{\left (2+m \right ) \left (1+m \right ) \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 57, normalized size = 0.69 \[ \frac {{\left (a \sqrt {-c} {\left (m + 1\right )} x^{2} - \sqrt {-c} {\left (m + 2\right )} x\right )} {\left (a x - 1\right )} x^{m}}{{\left (m^{2} + 3 \, m + 2\right )} a x - m^{2} - 3 \, m - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.45, size = 94, normalized size = 1.13 \[ \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (\frac {x^m\,x^2\,\sqrt {c-a^2\,c\,x^2}\,\left (m+1\right )}{m^2+3\,m+2}-\frac {x\,x^m\,\sqrt {c-a^2\,c\,x^2}\,\left (m+2\right )}{a\,\left (m^2+3\,m+2\right )}\right )}{x-\frac {1}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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