3.144 \(\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=41 \[ \frac {x^{m+1} F_1\left (-m-1;-\frac {5}{4},\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{m+1} \]

[Out]

x^(1+m)*AppellF1(-1-m,-5/4,5/4,-m,1/a/x,-1/a/x)/(1+m)

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6173, 133} \[ \frac {x^{m+1} F_1\left (-m-1;-\frac {5}{4},\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^((5*ArcCoth[a*x])/2),x]

[Out]

(x^(1 + m)*AppellF1[-1 - m, -5/4, 5/4, -m, 1/(a*x), -(1/(a*x))])/(1 + m)

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 6173

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

Rubi steps

\begin {align*} \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx &=-\left (\left (\left (\frac {1}{x}\right )^m x^m\right ) \operatorname {Subst}\left (\int \frac {x^{-2-m} \left (1-\frac {x}{a}\right )^{5/4}}{\left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )\right )\\ &=\frac {x^{1+m} F_1\left (-1-m;-\frac {5}{4},\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{1+m}\\ \end {align*}

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Mathematica [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m/E^((5*ArcCoth[a*x])/2),x]

[Out]

Integrate[x^m/E^((5*ArcCoth[a*x])/2), x]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*((a*x-1)/(a*x+1))^(5/4),x, algorithm="fricas")

[Out]

integral((a*x - 1)*x^m*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*((a*x-1)/(a*x+1))^(5/4),x, algorithm="giac")

[Out]

integrate(x^m*((a*x - 1)/(a*x + 1))^(5/4), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*((a*x-1)/(a*x+1))^(5/4),x)

[Out]

int(x^m*((a*x-1)/(a*x+1))^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*((a*x-1)/(a*x+1))^(5/4),x, algorithm="maxima")

[Out]

integrate(x^m*((a*x - 1)/(a*x + 1))^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^m\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

int(x^m*((a*x - 1)/(a*x + 1))^(5/4), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*((a*x-1)/(a*x+1))**(5/4),x)

[Out]

Timed out

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