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

Optimal. Leaf size=131 \[ \frac{2 \sqrt{\frac{1}{a x}+1} x^{m+1} \sqrt{c-a c x}}{(2 m+3) \sqrt{1-\frac{1}{a x}}}-\frac{2 (4 m+5) x^m \sqrt{c-a c x} \text{Hypergeometric2F1}\left (\frac{1}{2},-m-\frac{1}{2},\frac{1}{2}-m,-\frac{1}{a x}\right )}{a (2 m+1) (2 m+3) \sqrt{1-\frac{1}{a x}}} \]

[Out]

(2*Sqrt[1 + 1/(a*x)]*x^(1 + m)*Sqrt[c - a*c*x])/((3 + 2*m)*Sqrt[1 - 1/(a*x)]) - (2*(5 + 4*m)*x^m*Sqrt[c - a*c*
x]*Hypergeometric2F1[1/2, -1/2 - m, 1/2 - m, -(1/(a*x))])/(a*(1 + 2*m)*(3 + 2*m)*Sqrt[1 - 1/(a*x)])

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Rubi [A]  time = 0.231738, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6176, 6181, 79, 64} \[ \frac{2 \sqrt{\frac{1}{a x}+1} x^{m+1} \sqrt{c-a c x}}{(2 m+3) \sqrt{1-\frac{1}{a x}}}-\frac{2 (4 m+5) x^m \sqrt{c-a c x} \, _2F_1\left (\frac{1}{2},-m-\frac{1}{2};\frac{1}{2}-m;-\frac{1}{a x}\right )}{a (2 m+1) (2 m+3) \sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 + 1/(a*x)]*x^(1 + m)*Sqrt[c - a*c*x])/((3 + 2*m)*Sqrt[1 - 1/(a*x)]) - (2*(5 + 4*m)*x^m*Sqrt[c - a*c*
x]*Hypergeometric2F1[1/2, -1/2 - m, 1/2 - m, -(1/(a*x))])/(a*(1 + 2*m)*(3 + 2*m)*Sqrt[1 - 1/(a*x)])

Rule 6176

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

Rule 6181

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

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int e^{-\coth ^{-1}(a x)} x^m \sqrt{c-a c x} \, dx &=\frac{\sqrt{c-a c x} \int e^{-\coth ^{-1}(a x)} \sqrt{1-\frac{1}{a x}} x^{\frac{1}{2}+m} \, dx}{\sqrt{1-\frac{1}{a x}} \sqrt{x}}\\ &=-\frac{\left (\left (\frac{1}{x}\right )^{\frac{1}{2}+m} x^m \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{x^{-\frac{5}{2}-m} \left (1-\frac{x}{a}\right )}{\sqrt{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )}{\sqrt{1-\frac{1}{a x}}}\\ &=\frac{2 \sqrt{1+\frac{1}{a x}} x^{1+m} \sqrt{c-a c x}}{(3+2 m) \sqrt{1-\frac{1}{a x}}}+\frac{\left ((5+4 m) \left (\frac{1}{x}\right )^{\frac{1}{2}+m} x^m \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{x^{-\frac{3}{2}-m}}{\sqrt{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )}{a (3+2 m) \sqrt{1-\frac{1}{a x}}}\\ &=\frac{2 \sqrt{1+\frac{1}{a x}} x^{1+m} \sqrt{c-a c x}}{(3+2 m) \sqrt{1-\frac{1}{a x}}}-\frac{2 (5+4 m) x^m \sqrt{c-a c x} \, _2F_1\left (\frac{1}{2},-\frac{1}{2}-m;\frac{1}{2}-m;-\frac{1}{a x}\right )}{a (1+2 m) (3+2 m) \sqrt{1-\frac{1}{a x}}}\\ \end{align*}

Mathematica [A]  time = 0.0705771, size = 102, normalized size = 0.78 \[ \frac{2 x^m \sqrt{c-a c x} \left (a (2 m+1) x \sqrt{\frac{1}{a x}+1}-(4 m+5) \text{Hypergeometric2F1}\left (\frac{1}{2},-m-\frac{1}{2},\frac{1}{2}-m,-\frac{1}{a x}\right )\right )}{a (2 m+1) (2 m+3) \sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^m*Sqrt[c - a*c*x]*(a*(1 + 2*m)*Sqrt[1 + 1/(a*x)]*x - (5 + 4*m)*Hypergeometric2F1[1/2, -1/2 - m, 1/2 - m,
-(1/(a*x))]))/(a*(1 + 2*m)*(3 + 2*m)*Sqrt[1 - 1/(a*x)])

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Maple [F]  time = 0.357, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sqrt{-acx+c}\sqrt{{\frac{ax-1}{ax+1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a c x + c} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-a c x + c} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a c x + c} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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