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

Optimal. Leaf size=45 \[ \frac {c \left (1-a^2 x^2\right )^{5/2}}{5 a^4}-\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6128, 266, 43} \[ \frac {c \left (1-a^2 x^2\right )^{5/2}}{5 a^4}-\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^3*(c - a*c*x),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6128

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} x^3 (c-a c x) \, dx &=c \int x^3 \sqrt {1-a^2 x^2} \, dx\\ &=\frac {1}{2} c \operatorname {Subst}\left (\int x \sqrt {1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{2} c \operatorname {Subst}\left (\int \left (\frac {\sqrt {1-a^2 x}}{a^2}-\frac {\left (1-a^2 x\right )^{3/2}}{a^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^4}+\frac {c \left (1-a^2 x^2\right )^{5/2}}{5 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 0.71 \[ -\frac {c \left (1-a^2 x^2\right )^{3/2} \left (3 a^2 x^2+2\right )}{15 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(c*(1 - a^2*x^2)^(3/2)*(2 + 3*a^2*x^2))/a^4

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fricas [A]  time = 0.51, size = 39, normalized size = 0.87 \[ \frac {{\left (3 \, a^{4} c x^{4} - a^{2} c x^{2} - 2 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{15 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a^4*c*x^4 - a^2*c*x^2 - 2*c)*sqrt(-a^2*x^2 + 1)/a^4

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giac [A]  time = 0.16, size = 47, normalized size = 1.04 \[ \frac {3 \, {\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} c - 5 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{15 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*c - 5*(-a^2*x^2 + 1)^(3/2)*c)/a^4

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maple [A]  time = 0.03, size = 43, normalized size = 0.96 \[ -\frac {\left (a x -1\right )^{2} \left (a x +1\right )^{2} \left (3 a^{2} x^{2}+2\right ) c}{15 a^{4} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(a*x-1)^2*(a*x+1)^2*(3*a^2*x^2+2)*c/a^4/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.40, size = 58, normalized size = 1.29 \[ \frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} c x^{4} - \frac {\sqrt {-a^{2} x^{2} + 1} c x^{2}}{15 \, a^{2}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c}{15 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(-a^2*x^2 + 1)*c*x^4 - 1/15*sqrt(-a^2*x^2 + 1)*c*x^2/a^2 - 2/15*sqrt(-a^2*x^2 + 1)*c/a^4

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mupad [B]  time = 0.04, size = 36, normalized size = 0.80 \[ -\frac {5\,c\,{\left (1-a^2\,x^2\right )}^{3/2}-3\,c\,{\left (1-a^2\,x^2\right )}^{5/2}}{15\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*c*(1 - a^2*x^2)^(3/2) - 3*c*(1 - a^2*x^2)^(5/2))/(15*a^4)

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sympy [A]  time = 0.70, size = 66, normalized size = 1.47 \[ \begin {cases} \frac {c x^{4} \sqrt {- a^{2} x^{2} + 1}}{5} - \frac {c x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{2}} - \frac {2 c \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} & \text {for}\: a \neq 0 \\\frac {c x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*x**4*sqrt(-a**2*x**2 + 1)/5 - c*x**2*sqrt(-a**2*x**2 + 1)/(15*a**2) - 2*c*sqrt(-a**2*x**2 + 1)/(1
5*a**4), Ne(a, 0)), (c*x**4/4, True))

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