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

Optimal. Leaf size=172 \[ -\frac{c (2 m+3) \sqrt{1-a^2 x^2} x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{(m+1) (m+2) \sqrt{c-a^2 c x^2}}-\frac{2 a c \sqrt{1-a^2 x^2} x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{(m+2) \sqrt{c-a^2 c x^2}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{m+2} \]

[Out]

(x^(1 + m)*Sqrt[c - a^2*c*x^2])/(2 + m) - (c*(3 + 2*m)*x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*(2 + m)*Sqrt[c - a^2*c*x^2]) - (2*a*c*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hyperge
ometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.386524, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6167, 6151, 1809, 808, 365, 364} \[ -\frac{c (2 m+3) \sqrt{1-a^2 x^2} x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{(m+1) (m+2) \sqrt{c-a^2 c x^2}}-\frac{2 a c \sqrt{1-a^2 x^2} x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{(m+2) \sqrt{c-a^2 c x^2}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[c - a^2*c*x^2])/(2 + m) - (c*(3 + 2*m)*x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*(2 + m)*Sqrt[c - a^2*c*x^2]) - (2*a*c*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hyperge
ometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[c - a^2*c*x^2])

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6151

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [C]  time = 0.209297, size = 129, normalized size = 0.75 \[ \frac{x^{m+1} \left (\frac{\sqrt{c-a^2 c x^2} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{\sqrt{1-a^2 x^2}}+\frac{2 \sqrt{1-a x} \sqrt{-c (a x+1)} F_1\left (m+1;\frac{1}{2},-\frac{1}{2};m+2;a x,-a x\right )}{\sqrt{a x-1} \sqrt{a x+1}}\right )}{m+1} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^(1 + m)*((2*Sqrt[1 - a*x]*Sqrt[-(c*(1 + a*x))]*AppellF1[1 + m, 1/2, -1/2, 2 + m, a*x, -(a*x)])/(Sqrt[-1 + a
*x]*Sqrt[1 + a*x]) + (Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[-1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/Sqrt[1 - a^2
*x^2]))/(1 + m)

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Maple [F]  time = 0.477, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ){x}^{m}}{ax-1}\sqrt{-{a}^{2}c{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x + 1\right )} x^{m}}{a x - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x + 1\right )} x^{m}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x + 1\right )} x^{m}}{a x - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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