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

Optimal. Leaf size=131 \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}+\frac{9 (2-3 a x) \sqrt{1-a^2 x^2}}{8 a^4}+\frac{27 \sqrt{1-a^2 x^2}}{4 a^4}+\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{51 \sin ^{-1}(a x)}{8 a^4} \]

[Out]

(1 - a*x)^3/(a^4*Sqrt[1 - a^2*x^2]) + (27*Sqrt[1 - a^2*x^2])/(4*a^4) + (x^2*Sqrt[1 - a^2*x^2])/a^2 - (x^3*Sqrt
[1 - a^2*x^2])/(4*a) + (9*(2 - 3*a*x)*Sqrt[1 - a^2*x^2])/(8*a^4) + (51*ArcSin[a*x])/(8*a^4)

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Rubi [A]  time = 0.689497, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {6124, 1633, 1593, 12, 852, 1635, 1815, 27, 743, 641, 216} \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}+\frac{9 (2-3 a x) \sqrt{1-a^2 x^2}}{8 a^4}+\frac{27 \sqrt{1-a^2 x^2}}{4 a^4}+\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{51 \sin ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - a*x)^3/(a^4*Sqrt[1 - a^2*x^2]) + (27*Sqrt[1 - a^2*x^2])/(4*a^4) + (x^2*Sqrt[1 - a^2*x^2])/a^2 - (x^3*Sqrt
[1 - a^2*x^2])/(4*a) + (9*(2 - 3*a*x)*Sqrt[1 - a^2*x^2])/(8*a^4) + (51*ArcSin[a*x])/(8*a^4)

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]

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 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 12

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

Rule 852

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

Rule 1635

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

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 743

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

Rule 641

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

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]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} x^3 \, dx &=\int \frac{x^3 (1-a x)^2}{(1+a x) \sqrt{1-a^2 x^2}} \, dx\\ &=a \int \frac{\sqrt{1-a^2 x^2} \left (\frac{x^3}{a}-x^4\right )}{(1+a x)^2} \, dx\\ &=a \int \frac{\left (\frac{1}{a}-x\right ) x^3 \sqrt{1-a^2 x^2}}{(1+a x)^2} \, dx\\ &=a^2 \int \frac{x^3 \left (1-a^2 x^2\right )^{3/2}}{a^2 (1+a x)^3} \, dx\\ &=\int \frac{x^3 \left (1-a^2 x^2\right )^{3/2}}{(1+a x)^3} \, dx\\ &=\int \frac{x^3 (1-a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}-\int \frac{(1-a x)^2 \left (-\frac{3}{a^3}+\frac{x}{a^2}-\frac{x^2}{a}\right )}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{\int \frac{\frac{12}{a}-28 x+27 a x^2-12 a^2 x^3}{\sqrt{1-a^2 x^2}} \, dx}{4 a^2}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{\int \frac{-36 a+108 a^2 x-81 a^3 x^2}{\sqrt{1-a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{\int -\frac{9 a (-2+3 a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{3 \int \frac{(-2+3 a x)^2}{\sqrt{1-a^2 x^2}} \, dx}{4 a^3}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{9 (2-3 a x) \sqrt{1-a^2 x^2}}{8 a^4}-\frac{3 \int \frac{-17 a^2+18 a^3 x}{\sqrt{1-a^2 x^2}} \, dx}{8 a^5}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{27 \sqrt{1-a^2 x^2}}{4 a^4}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{9 (2-3 a x) \sqrt{1-a^2 x^2}}{8 a^4}+\frac{51 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{8 a^3}\\ &=\frac{(1-a x)^3}{a^4 \sqrt{1-a^2 x^2}}+\frac{27 \sqrt{1-a^2 x^2}}{4 a^4}+\frac{x^2 \sqrt{1-a^2 x^2}}{a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{9 (2-3 a x) \sqrt{1-a^2 x^2}}{8 a^4}+\frac{51 \sin ^{-1}(a x)}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0554135, size = 70, normalized size = 0.53 \[ \sqrt{1-a^2 x^2} \left (\frac{x^2}{a^2}-\frac{19 x}{8 a^3}+\frac{4}{a^4 (a x+1)}+\frac{6}{a^4}-\frac{x^3}{4 a}\right )+\frac{51 \sin ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]*(6/a^4 - (19*x)/(8*a^3) + x^2/a^2 - x^3/(4*a) + 4/(a^4*(1 + a*x))) + (51*ArcSin[a*x])/(8*a^4
)

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Maple [B]  time = 0.054, size = 235, normalized size = 1.8 \begin{align*}{\frac{x}{4\,{a}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{3\,x}{8\,{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3}{8\,{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+5\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{6} \left ( x+{a}^{-1} \right ) ^{2}}}+4\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{3/2}}{{a}^{4}}}+6\,{\frac{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }x}{{a}^{3}}}+6\,{\frac{1}{{a}^{3}\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}} \right ) }+{\frac{1}{{a}^{7} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^3*x*(-a^2*x^2+1)^(3/2)+3/8/a^3*x*(-a^2*x^2+1)^(1/2)+3/8/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1
)^(1/2))+5/a^6/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+4/a^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+6/a^3*(-a
^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x+6/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+1
/a^7/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)

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Maxima [C]  time = 1.48586, size = 290, normalized size = 2.21 \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{6} x^{2} + 2 \, a^{5} x + a^{4}} + \frac{3 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{2 \,{\left (a^{5} x + a^{4}\right )}} + \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{5} x + a^{4}} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{4 \, a^{3}} - \frac{3 \, \sqrt{a^{2} x^{2} + 4 \, a x + 3} x}{2 \, a^{3}} + \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x}{8 \, a^{3}} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4}} + \frac{3 i \, \arcsin \left (a x + 2\right )}{2 \, a^{4}} + \frac{63 \, \arcsin \left (a x\right )}{8 \, a^{4}} - \frac{3 \, \sqrt{a^{2} x^{2} + 4 \, a x + 3}}{a^{4}} + \frac{9 \, \sqrt{-a^{2} x^{2} + 1}}{2 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2 + 1)^(3/2)/(a^6*x^2 + 2*a^5*x + a^4) + 3/2*(-a^2*x^2 + 1)^(3/2)/(a^5*x + a^4) + 6*sqrt(-a^2*x^2 + 1
)/(a^5*x + a^4) + 1/4*(-a^2*x^2 + 1)^(3/2)*x/a^3 - 3/2*sqrt(a^2*x^2 + 4*a*x + 3)*x/a^3 + 3/8*sqrt(-a^2*x^2 + 1
)*x/a^3 - (-a^2*x^2 + 1)^(3/2)/a^4 + 3/2*I*arcsin(a*x + 2)/a^4 + 63/8*arcsin(a*x)/a^4 - 3*sqrt(a^2*x^2 + 4*a*x
+ 3)/a^4 + 9/2*sqrt(-a^2*x^2 + 1)/a^4

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Fricas [A]  time = 2.05419, size = 216, normalized size = 1.65 \begin{align*} \frac{80 \, a x - 102 \,{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 11 \, a^{2} x^{2} - 29 \, a x - 80\right )} \sqrt{-a^{2} x^{2} + 1} + 80}{8 \,{\left (a^{5} x + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(80*a*x - 102*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^4*x^4 - 6*a^3*x^3 + 11*a^2*x^2 - 29*
a*x - 80)*sqrt(-a^2*x^2 + 1) + 80)/(a^5*x + a^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22591, size = 128, normalized size = 0.98 \begin{align*} -\frac{1}{8} \, \sqrt{-a^{2} x^{2} + 1}{\left ({\left (2 \, x{\left (\frac{x}{a} - \frac{4}{a^{2}}\right )} + \frac{19}{a^{3}}\right )} x - \frac{48}{a^{4}}\right )} + \frac{51 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{8 \, a^{3}{\left | a \right |}} - \frac{8}{a^{3}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-a^2*x^2 + 1)*((2*x*(x/a - 4/a^2) + 19/a^3)*x - 48/a^4) + 51/8*arcsin(a*x)*sgn(a)/(a^3*abs(a)) - 8/(
a^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))