3.2.33 \(\int e^{3 \coth ^{-1}(a x)} x^m \, dx\) [133]

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

[Out]

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

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Rubi [A]
time = 0.93, 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 = {6307, 6874, 371, 864, 822} \begin {gather*} -\frac {3 x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-m-1);\frac {1-m}{2};\frac {1}{a^2 x^2}\right )}{m+1}+\frac {4 x^{m+1} \, _2F_1\left (\frac {3}{2},\frac {1}{2} (-m-1);\frac {1-m}{2};\frac {1}{a^2 x^2}\right )}{m+1}-\frac {x^m \, _2F_1\left (\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{a^2 x^2}\right )}{a m}+\frac {4 x^m \, _2F_1\left (\frac {3}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{a^2 x^2}\right )}{a m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

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 864

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 6307

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

Rule 6874

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.22, size = 228, normalized size = 1.51 \begin {gather*} \frac {x^{1+m} \left (3 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1-a x} \sqrt {\frac {1+a x}{a^2}} F_1\left (m;-\frac {1}{2},\frac {1}{2};1+m;-a x,a x\right )-2 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1-a x} \sqrt {\frac {1+a x}{a^2}} F_1\left (m;-\frac {1}{2},\frac {3}{2};1+m;-a x,a x\right )+m \sqrt {-1+a x} \sqrt {1+a x} \sqrt {-\frac {1}{a^2}+x^2} \, _2F_1\left (-\frac {1}{2},-\frac {1}{2}-\frac {m}{2};\frac {1}{2}-\frac {m}{2};\frac {1}{a^2 x^2}\right )\right )}{m (1+m) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {-\frac {1}{a^2}+x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^(1 + m)*(3*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 - a*x]*Sqrt[(1 + a*x)/a^2]*AppellF1[m, -1/2, 1/2, 1 + m, -(
a*x), a*x] - 2*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 - a*x]*Sqrt[(1 + a*x)/a^2]*AppellF1[m, -1/2, 3/2, 1 + m, -
(a*x), a*x] + m*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*Sqrt[-a^(-2) + x^2]*Hypergeometric2F1[-1/2, -1/2 - m/2, 1/2 - m/2
, 1/(a^2*x^2)]))/(m*(1 + m)*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*Sqrt[-a^(-2) + x^2])

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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