3.11.1 \(\int \frac {e^{\tanh ^{-1}(a x)} x^m}{\sqrt {c-a^2 c x^2}} \, dx\) [1001]

Optimal. Leaf size=51 \[ \frac {x^{1+m} \sqrt {1-a^2 x^2} \, _2F_1(1,1+m;2+m;a x)}{(1+m) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6288, 6285, 66} \begin {gather*} \frac {\sqrt {1-a^2 x^2} x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{(m+1) \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 6285

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

Rule 6288

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 51, normalized size = 1.00 \begin {gather*} \frac {x^{1+m} \sqrt {1-a^2 x^2} \, _2F_1(1,1+m;2+m;a x)}{(1+m) \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \sqrt {-a^{2} c \,x^{2}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^m/(a^3*c*x^3 - a^2*c*x^2 - a*c*x + c), x)

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

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**2*c*x**2+c)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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