3.100 \(\int e^{-\frac {3}{2} \tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=31 \[ \frac {x^{m+1} F_1\left (m+1;-\frac {3}{4},\frac {3}{4};m+2;a x,-a x\right )}{m+1} \]

[Out]

x^(1+m)*AppellF1(1+m,-3/4,3/4,2+m,a*x,-a*x)/(1+m)

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Rubi [A]  time = 0.03, 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 = {6126, 133} \[ \frac {x^{m+1} F_1\left (m+1;-\frac {3}{4},\frac {3}{4};m+2;a x,-a x\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^((3*ArcTanh[a*x])/2),x]

[Out]

(x^(1 + m)*AppellF1[1 + m, -3/4, 3/4, 2 + m, a*x, -(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 6126

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 {3}{2} \tanh ^{-1}(a x)} x^m \, dx &=\int \frac {x^m (1-a x)^{3/4}}{(1+a x)^{3/4}} \, dx\\ &=\frac {x^{1+m} F_1\left (1+m;-\frac {3}{4},\frac {3}{4};2+m;a x,-a x\right )}{1+m}\\ \end {align*}

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Mathematica [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int e^{-\frac {3}{2} \tanh ^{-1}(a x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^m/E^((3*ArcTanh[a*x])/2),x]

[Out]

Integrate[x^m/E^((3*ArcTanh[a*x])/2), x]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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