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

Optimal. Leaf size=81 \[ \frac {x^{m+1} \sqrt {c-\frac {c}{a^2 x^2}}}{(m+1) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^m \sqrt {c-\frac {c}{a^2 x^2}}}{a m \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 81, 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 = {6197, 6193, 43} \[ \frac {x^{m+1} \sqrt {c-\frac {c}{a^2 x^2}}}{(m+1) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^m \sqrt {c-\frac {c}{a^2 x^2}}}{a m \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[c - c/(a^2*x^2)]*x^m)/(a*m*Sqrt[1 - 1/(a^2*x^2)])) + (Sqrt[c - c/(a^2*x^2)]*x^(1 + m))/((1 + 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 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^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^m \, 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^m \, dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\\ &=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \int x^{-1+m} (-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^{-1+m}+a x^m\right ) \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\\ &=-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^m}{a m \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^{1+m}}{(1+m) \sqrt {1-\frac {1}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.62 \[ \frac {x^m \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {a x}{m+1}-\frac {1}{m}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 73, normalized size = 0.90 \[ -\frac {{\left (a m x^{2} - {\left (m + 1\right )} x\right )} x^{m} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{m^{2} - {\left (a m^{2} + a m\right )} x + m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.04, size = 65, normalized size = 0.80 \[ \frac {x^{1+m} \left (a x m -m -1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {\frac {a x -1}{a x +1}}}{\left (1+m \right ) m \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 46, normalized size = 0.57 \[ \frac {{\left (a \sqrt {c} m x - \sqrt {c} {\left (m + 1\right )}\right )} {\left (a x - 1\right )} x^{m}}{{\left (m^{2} + m\right )} a^{2} x - {\left (m^{2} + m\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\sqrt {c-\frac {c}{a^2\,x^2}}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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