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

Optimal. Leaf size=22 \[ -\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6128, 261} \[ -\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x*(c - a*c*x),x]

[Out]

-(c*(1 - a^2*x^2)^(3/2))/(3*a^2)

Rule 261

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

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 (c-a c x) \, dx &=c \int x \sqrt {1-a^2 x^2} \, dx\\ &=-\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ -\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(c*(1 - a^2*x^2)^(3/2))/a^2

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fricas [A]  time = 0.43, size = 29, normalized size = 1.32 \[ \frac {{\left (a^{2} c x^{2} - c\right )} \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 18, normalized size = 0.82 \[ -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{3 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-a^2*x^2 + 1)^(3/2)*c/a^2

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maple [A]  time = 0.03, size = 33, normalized size = 1.50 \[ -\frac {\left (a x -1\right )^{2} \left (a x +1\right )^{2} c}{3 a^{2} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 37, normalized size = 1.68 \[ \frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} c x^{2} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{3 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 18, normalized size = 0.82 \[ -\frac {c\,{\left (1-a^2\,x^2\right )}^{3/2}}{3\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*(1 - a^2*x^2)^(3/2))/(3*a^2)

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sympy [A]  time = 0.41, size = 42, normalized size = 1.91 \[ \begin {cases} \frac {c x^{2} \sqrt {- a^{2} x^{2} + 1}}{3} - \frac {c \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} & \text {for}\: a \neq 0 \\\frac {c x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*x**2*sqrt(-a**2*x**2 + 1)/3 - c*sqrt(-a**2*x**2 + 1)/(3*a**2), Ne(a, 0)), (c*x**2/2, True))

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