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

Optimal. Leaf size=64 \[ \frac{2 c (a x+1) \sqrt{1-a^2 x^2} x^m (-a x)^{-m} \text{Hypergeometric2F1}\left (\frac{3}{2},-m,\frac{5}{2},a x+1\right )}{3 a \sqrt{c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.124062, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {6128, 892, 67, 65} \[ \frac{2 c (a x+1) \sqrt{1-a^2 x^2} x^m (-a x)^{-m} \, _2F_1\left (\frac{3}{2},-m;\frac{5}{2};a x+1\right )}{3 a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^m*Sqrt[c - a*c*x],x]

[Out]

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

Rule 6128

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

Rule 892

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + c*x^
2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (
c*x)/e)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !Int
egerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rule 67

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

Rule 65

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

Rubi steps

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

Mathematica [A]  time = 0.01767, size = 46, normalized size = 0.72 \[ \frac{x^{m+1} \sqrt{c-a c x} \text{Hypergeometric2F1}\left (-\frac{1}{2},m+1,m+2,-a x\right )}{(m+1) \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a*x]*x^m*Sqrt[c - a*c*x],x]

[Out]

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

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Maple [F]  time = 0.368, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ){x}^{m}\sqrt{-acx+c}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*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 )}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**m*sqrt(-c*(a*x - 1))*(a*x + 1)/sqrt(-(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 c x + c}{\left (a x + 1\right )} x^{m}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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