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

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

[Out]

51/8*arcsin(a*x)/a^4+(-a*x+1)^3/a^4/(-a^2*x^2+1)^(1/2)+27/4*(-a^2*x^2+1)^(1/2)/a^4+x^2*(-a^2*x^2+1)^(1/2)/a^2-
1/4*x^3*(-a^2*x^2+1)^(1/2)/a+9/8*(-3*a*x+2)*(-a^2*x^2+1)^(1/2)/a^4

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Rubi [A]  time = 0.69, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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 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 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 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 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^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*}

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Mathematica [A]  time = 0.06, size = 70, normalized size = 0.53 \[ \frac {51 \sin ^{-1}(a x)}{8 a^4}+\sqrt {1-a^2 x^2} \left (\frac {4}{a^4 (a x+1)}+\frac {6}{a^4}-\frac {19 x}{8 a^3}+\frac {x^2}{a^2}-\frac {x^3}{4 a}\right ) \]

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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fricas [A]  time = 0.49, size = 92, normalized size = 0.70 \[ \frac {80 \, a x - 102 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 11 \, a^{2} x^{2} - 29 \, a x - 80\right )} \sqrt {-a^{2} x^{2} + 1} + 80}{8 \, {\left (a^{5} x + a^{4}\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.05, size = 235, normalized size = 1.79 \[ \frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 a^{3}}+\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{3}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3} \sqrt {a^{2}}}+\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{7} \left (x +\frac {1}{a}\right )^{3}}+\frac {5 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{6} \left (x +\frac {1}{a}\right )^{2}}+\frac {4 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a^{4}}+\frac {6 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{a^{3}}+\frac {6 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{a^{3} \sqrt {a^{2}}} \]

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

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maxima [C]  time = 0.43, size = 215, normalized size = 1.64 \[ -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{6} x^{2} + 2 \, a^{5} x + a^{4}} + \frac {3 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (a^{5} x + a^{4}\right )}} + \frac {6 \, \sqrt {-a^{2} x^{2} + 1}}{a^{5} x + a^{4}} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, a^{3}} - \frac {3 \, \sqrt {a^{2} x^{2} + 4 \, a x + 3} x}{2 \, a^{3}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{3}} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}} + \frac {3 i \, \arcsin \left (a x + 2\right )}{2 \, a^{4}} + \frac {63 \, \arcsin \left (a x\right )}{8 \, a^{4}} - \frac {3 \, \sqrt {a^{2} x^{2} + 4 \, a x + 3}}{a^{4}} + \frac {9 \, \sqrt {-a^{2} x^{2} + 1}}{2 \, a^{4}} \]

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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mupad [B]  time = 0.07, size = 154, normalized size = 1.18 \[ \frac {51\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^3\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{{\left (-a^2\right )}^{3/2}}-\frac {4}{a^2\,\sqrt {-a^2}}-\frac {19\,x\,\sqrt {-a^2}}{8\,a^3}+\frac {a^2\,x^2}{{\left (-a^2\right )}^{3/2}}+\frac {x^3\,{\left (-a^2\right )}^{3/2}}{4\,a^3}\right )}{\sqrt {-a^2}}-\frac {4\,\sqrt {1-a^2\,x^2}}{a^3\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(51*asinh(x*(-a^2)^(1/2)))/(8*a^3*(-a^2)^(1/2)) + ((1 - a^2*x^2)^(1/2)*(2/(-a^2)^(3/2) - 4/(a^2*(-a^2)^(1/2))
- (19*x*(-a^2)^(1/2))/(8*a^3) + (a^2*x^2)/(-a^2)^(3/2) + (x^3*(-a^2)^(3/2))/(4*a^3)))/(-a^2)^(1/2) - (4*(1 - a
^2*x^2)^(1/2))/(a^3*(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 \[ \int \frac {x^{3} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**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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