3.142 \(\int e^{3 \tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=151 \[ -\frac {3 x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2}+\frac {4 a x^{m+2} \, _2F_1\left (\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.88, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6124, 6742, 364, 850, 808} \[ -\frac {3 x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2}+\frac {4 a x^{m+2} \, _2F_1\left (\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*x^m,x]

[Out]

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

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 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 850

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

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 92, normalized size = 0.61 \[ \frac {\sqrt {-a x-1} \sqrt {1-a x} x^{m+1} \left (F_1\left (m+1;-\frac {1}{2},\frac {1}{2};m+2;-a x,a x\right )-2 F_1\left (m+1;-\frac {1}{2},\frac {3}{2};m+2;-a x,a x\right )\right )}{(m+1) \sqrt {a x-1} \sqrt {a x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a*x])*x^m,x]

[Out]

(x^(1 + m)*Sqrt[-1 - a*x]*Sqrt[1 - a*x]*(AppellF1[1 + m, -1/2, 1/2, 2 + m, -(a*x), a*x] - 2*AppellF1[1 + m, -1
/2, 3/2, 2 + m, -(a*x), a*x]))/((1 + m)*Sqrt[-1 + a*x]*Sqrt[1 + a*x])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + 1\right )} x^{m}}{a^{2} x^{2} - 2 \, a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.33, size = 139, normalized size = 0.92 \[ \frac {x^{1+m} \hypergeom \left (\left [\frac {3}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{1+m}+\frac {a^{3} x^{4+m} \hypergeom \left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [3+\frac {m}{2}\right ], a^{2} x^{2}\right )}{4+m}+\frac {3 a^{2} x^{3+m} \hypergeom \left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {5}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{3+m}+\frac {3 a \,x^{2+m} \hypergeom \left (\left [\frac {3}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], a^{2} x^{2}\right )}{2+m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x)

[Out]

x^(1+m)*hypergeom([3/2,1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+a^3/(4+m)*x^(4+m)*hypergeom([3/2,2+1/2*m],[3+1/2*
m],a^2*x^2)+3*a^2/(3+m)*x^(3+m)*hypergeom([3/2,3/2+1/2*m],[5/2+1/2*m],a^2*x^2)+3*a*x^(2+m)*hypergeom([3/2,1+1/
2*m],[2+1/2*m],a^2*x^2)/(2+m)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)

[Out]

int((x^m*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*x**m,x)

[Out]

Integral(x**m*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**(3/2), x)

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