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

Optimal. Leaf size=86 \[ \frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (a x+1)^3}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (a x+1)}+\frac {9 \sqrt {1-a^2 x^2}}{2 a^2}+\frac {9 \sin ^{-1}(a x)}{2 a^2} \]

[Out]

3/2*(-a^2*x^2+1)^(3/2)/a^2/(a*x+1)+(-a^2*x^2+1)^(5/2)/a^2/(a*x+1)^3+9/2*arcsin(a*x)/a^2+9/2*(-a^2*x^2+1)^(1/2)
/a^2

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Rubi [A]  time = 0.36, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6124, 1633, 1593, 12, 793, 665, 216} \[ \frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (a x+1)^3}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (a x+1)}+\frac {9 \sqrt {1-a^2 x^2}}{2 a^2}+\frac {9 \sin ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*Sqrt[1 - a^2*x^2])/(2*a^2) + (3*(1 - a^2*x^2)^(3/2))/(2*a^2*(1 + a*x)) + (1 - a^2*x^2)^(5/2)/(a^2*(1 + a*x)
^3) + (9*ArcSin[a*x])/(2*a^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 665

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

Rule 793

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

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]

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} x \, dx &=\int \frac {x (1-a x)^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx\\ &=a \int \frac {\left (\frac {x}{a}-x^2\right ) \sqrt {1-a^2 x^2}}{(1+a x)^2} \, dx\\ &=a \int \frac {\left (\frac {1}{a}-x\right ) x \sqrt {1-a^2 x^2}}{(1+a x)^2} \, dx\\ &=a^2 \int \frac {x \left (1-a^2 x^2\right )^{3/2}}{a^2 (1+a x)^3} \, dx\\ &=\int \frac {x \left (1-a^2 x^2\right )^{3/2}}{(1+a x)^3} \, dx\\ &=\frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac {3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(1+a x)^2} \, dx}{a}\\ &=\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac {9 \int \frac {\sqrt {1-a^2 x^2}}{1+a x} \, dx}{2 a}\\ &=\frac {9 \sqrt {1-a^2 x^2}}{2 a^2}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac {9 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a}\\ &=\frac {9 \sqrt {1-a^2 x^2}}{2 a^2}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{2 a^2 (1+a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^2 (1+a x)^3}+\frac {9 \sin ^{-1}(a x)}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 44, normalized size = 0.51 \[ \frac {\sqrt {1-a^2 x^2} \left (-a x+\frac {8}{a x+1}+6\right )+9 \sin ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(6 - a*x + 8/(1 + a*x)) + 9*ArcSin[a*x])/(2*a^2)

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fricas [A]  time = 0.56, size = 75, normalized size = 0.87 \[ \frac {14 \, a x - 18 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} x^{2} - 5 \, a x - 14\right )} \sqrt {-a^{2} x^{2} + 1} + 14}{2 \, {\left (a^{3} x + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(14*a*x - 18*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*x^2 - 5*a*x - 14)*sqrt(-a^2*x^2 + 1)
+ 14)/(a^3*x + a^2)

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giac [A]  time = 0.22, size = 78, normalized size = 0.91 \[ -\frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {x}{a} - \frac {6}{a^{2}}\right )} + \frac {9 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, a {\left | a \right |}} - \frac {8}{a {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-a^2*x^2 + 1)*(x/a - 6/a^2) + 9/2*arcsin(a*x)*sgn(a)/(a*abs(a)) - 8/(a*((sqrt(-a^2*x^2 + 1)*abs(a) +
a)/(a^2*x) + 1)*abs(a))

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maple [B]  time = 0.04, size = 169, normalized size = 1.97 \[ \frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{5} \left (x +\frac {1}{a}\right )^{3}}+\frac {3 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{4} \left (x +\frac {1}{a}\right )^{2}}+\frac {3 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a^{2}}+\frac {9 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{2 a}+\frac {9 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 a \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+3/a^4/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+3/a^2*(-
a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+9/2/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x+9/2/a/(a^2)^(1/2)*arctan((a^2)^(1/
2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))

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maxima [A]  time = 0.44, size = 110, normalized size = 1.28 \[ -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4} x^{2} + 2 \, a^{3} x + a^{2}} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (a^{3} x + a^{2}\right )}} + \frac {6 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3} x + a^{2}} + \frac {9 \, \arcsin \left (a x\right )}{2 \, a^{2}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2 + 1)^(3/2)/(a^4*x^2 + 2*a^3*x + a^2) + 1/2*(-a^2*x^2 + 1)^(3/2)/(a^3*x + a^2) + 6*sqrt(-a^2*x^2 + 1
)/(a^3*x + a^2) + 9/2*arcsin(a*x)/a^2 + 3/2*sqrt(-a^2*x^2 + 1)/a^2

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mupad [B]  time = 0.79, size = 101, normalized size = 1.17 \[ -\frac {\left (\frac {3}{\sqrt {-a^2}}+\frac {x\,\sqrt {-a^2}}{2\,a}\right )\,\sqrt {1-a^2\,x^2}-\frac {9\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a}+\frac {4\,\sqrt {1-a^2\,x^2}}{a\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )}}{\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((3/(-a^2)^(1/2) + (x*(-a^2)^(1/2))/(2*a))*(1 - a^2*x^2)^(1/2) - (9*asinh(x*(-a^2)^(1/2)))/(2*a) + (4*(1 - a^
2*x^2)^(1/2))/(a*(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)))/(-a^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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