3.1342 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac {2^{\frac {n+1}{2}} \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^3 c (1-n) \sqrt {c-a^2 c x^2}}-\frac {(n-a x) e^{n \tanh ^{-1}(a x)}}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}} \]

[Out]

-exp(n*arctanh(a*x))*(-a*x+n)/a^3/c/(-n^2+1)/(-a^2*c*x^2+c)^(1/2)+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^3/c/(1-n)/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6147, 6143, 6140, 69} \[ \frac {2^{\frac {n+1}{2}} \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^3 c (1-n) \sqrt {c-a^2 c x^2}}-\frac {(n-a x) e^{n \tanh ^{-1}(a x)}}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(n*ArcTanh[a*x])*x^2)/(c - a^2*c*x^2)^(3/2),x]

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 6140

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 6143

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])

Rule 6147

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^2*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((n + 2*(p + 1)*a*x)*(c
+ d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]))/(a*d*(n^2 - 4*(p + 1)^2)), x] + Dist[(n^2 + 2*(p + 1))/(d*(n^2 - 4*(p +
1)^2)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] && L
tQ[p, -1] &&  !IntegerQ[n] && NeQ[n^2 - 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 155, normalized size = 1.01 \[ \frac {\sqrt {1-a^2 x^2} (1-a x)^{-\frac {n}{2}-\frac {1}{2}} \left (2^{\frac {n+1}{2}} (n+1) (a x-1) \sqrt {a x+1} \, _2F_1\left (\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )+(n-a x) (a x+1)^{n/2}\right )}{a^3 c (n-1) (n+1) \sqrt {a x+1} \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(n*ArcTanh[a*x])*x^2)/(c - a^2*c*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x^2*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate(x^2*((a*x + 1)/(a*x - 1))^(1/2*n)/(-a^2*c*x^2 + c)^(3/2), x)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{2}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x)

[Out]

int(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^2*((a*x + 1)/(a*x - 1))^(1/2*n)/(-a^2*c*x^2 + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(3/2),x)

[Out]

int((x^2*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*x**2/(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(x**2*exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(3/2), x)

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