3.1003 \(\int \frac {e^{\tanh ^{-1}(a x)} x^m}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=134 \[ \frac {\sqrt {1-a^2 x^2} x^{m+1} \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{c^2 (m+1) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} x^{m+2} \, _2F_1\left (3,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{c^2 (m+2) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6153, 6150, 82, 73, 364} \[ \frac {\sqrt {1-a^2 x^2} x^{m+1} \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{c^2 (m+1) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} x^{m+2} \, _2F_1\left (3,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{c^2 (m+2) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^m)/(c - a^2*c*x^2)^(5/2),x]

[Out]

(x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2])/(c^2*(1 + m)*Sqrt[c - a^2*c*
x^2]) + (a*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[3, (2 + m)/2, (4 + m)/2, a^2*x^2])/(c^2*(2 + m)*Sqrt[
c - a^2*c*x^2])

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 82

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[a, Int[(a + b*
x)^n*(c + d*x)^n*(f*x)^p, x], x] + Dist[b/f, Int[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b,
c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] &&  !IGtQ[m, 0] && NeQ[m +
n + p + 2, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 6150

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 6153

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}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{\tanh ^{-1}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {x^m}{(1-a x)^3 (1+a x)^2} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {x^m}{(1-a x)^3 (1+a x)^3} \, dx}{c^2 \sqrt {c-a^2 c x^2}}+\frac {\left (a \sqrt {1-a^2 x^2}\right ) \int \frac {x^{1+m}}{(1-a x)^3 (1+a x)^3} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {x^m}{\left (1-a^2 x^2\right )^3} \, dx}{c^2 \sqrt {c-a^2 c x^2}}+\frac {\left (a \sqrt {1-a^2 x^2}\right ) \int \frac {x^{1+m}}{\left (1-a^2 x^2\right )^3} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {x^{1+m} \sqrt {1-a^2 x^2} \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{c^2 (1+m) \sqrt {c-a^2 c x^2}}+\frac {a x^{2+m} \sqrt {1-a^2 x^2} \, _2F_1\left (3,\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{c^2 (2+m) \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 107, normalized size = 0.80 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {x^{m+1} \, _2F_1\left (3,\frac {m+1}{2};\frac {m+1}{2}+1;a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (3,\frac {m+2}{2};\frac {m+2}{2}+1;a^2 x^2\right )}{m+2}\right )}{c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^m)/(c - a^2*c*x^2)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{m}}{a^{7} c^{3} x^{7} - a^{6} c^{3} x^{6} - 3 \, a^{5} c^{3} x^{5} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} - a c^{3} x + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^m/(a^7*c^3*x^7 - a^6*c^3*x^6 - 3*a^5*c^3*x^5 + 3*a^4*c^3*x^
4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a*c^3*x + c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^m/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\, dx \]

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)^(5/2),x)

[Out]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)*x^m/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int((x^m*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**m/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x**m*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), x)

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