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

Optimal. Leaf size=57 \[ \frac {2 x^{m+1} \sqrt {c-\frac {c}{a x}} \, _2F_1\left (-\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};-a x\right )}{(2 m+1) \sqrt {1-a x}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6134, 6128, 848, 64} \[ \frac {2 x^{m+1} \sqrt {c-\frac {c}{a x}} \, _2F_1\left (-\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};-a x\right )}{(2 m+1) \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])))

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, 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] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

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 6134

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.98 \[ \frac {x^{m+1} \sqrt {c-\frac {c}{a x}} \, _2F_1\left (-\frac {1}{2},m+\frac {1}{2};m+\frac {3}{2};-a x\right )}{\left (m+\frac {1}{2}\right ) \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{m} \sqrt {\frac {a c x - c}{a x}}}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{m} \sqrt {c -\frac {c}{a x}}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^m\,\sqrt {c-\frac {c}{a\,x}}\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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