3.14.37 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\) [1337]

Optimal. Leaf size=104 \[ -\frac {2^{\frac {1+n}{2}} (1-a x)^{\frac {1-n}{2}} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a (1-n) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6278, 6275, 71} \begin {gather*} -\frac {2^{\frac {n+1}{2}} \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a (1-n) \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

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

Rule 6275

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

Rule 6278

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 101, normalized size = 0.97 \begin {gather*} \frac {2^{\frac {1+n}{2}} (1-a x)^{\frac {1}{2}-\frac {n}{2}} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )}{a (-1+n) \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctanh(a*x))/(-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(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((-(a*x + 1)/(a*x - 1))^(1/2*n)/sqrt(-a^2*c*x^2 + c), 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(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{n \operatorname {atanh}{\left (a x \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(exp(n*atanh(a*x))/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(exp(n*atanh(a*x))/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(exp(n*arctanh(a*x))/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{\sqrt {c-a^2\,c\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^(1/2),x)

[Out]

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^(1/2), x)

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