3.1.80 \(\int e^{\frac {5}{2} \tanh ^{-1}(a x)} x^m \, dx\) [80]

Optimal. Leaf size=31 \[ \frac {x^{1+m} F_1\left (1+m;\frac {5}{4},-\frac {5}{4};2+m;a x,-a x\right )}{1+m} \]

[Out]

x^(1+m)*AppellF1(1+m,5/4,-5/4,2+m,a*x,-a*x)/(1+m)

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Rubi [A]
time = 0.02, antiderivative size = 31, 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 = {6261, 138} \begin {gather*} \frac {x^{m+1} F_1\left (m+1;\frac {5}{4},-\frac {5}{4};m+2;a x,-a x\right )}{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*AppellF1[1 + m, 5/4, -5/4, 2 + m, a*x, -(a*x)])/(1 + m)

Rule 138

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

Rule 6261

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

Rubi steps

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

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Mathematica [F]
time = 0.39, size = 0, normalized size = 0.00 \begin {gather*} \int e^{\frac {5}{2} \tanh ^{-1}(a x)} x^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int x^m\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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