3.288 \(\int e^{\tanh ^{-1}(a x)} x^2 (c-a c x) \, dx\)

Optimal. Leaf size=58 \[ \frac {c \sin ^{-1}(a x)}{8 a^3}-\frac {c x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {1}{4} c x^3 \sqrt {1-a^2 x^2} \]

[Out]

1/8*c*arcsin(a*x)/a^3-1/8*c*x*(-a^2*x^2+1)^(1/2)/a^2+1/4*c*x^3*(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6128, 279, 321, 216} \[ \frac {1}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {c x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {c \sin ^{-1}(a x)}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^2*(c - a*c*x),x]

[Out]

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

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^2 (c-a c x) \, dx &=c \int x^2 \sqrt {1-a^2 x^2} \, dx\\ &=\frac {1}{4} c x^3 \sqrt {1-a^2 x^2}+\frac {1}{4} c \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {1}{4} c x^3 \sqrt {1-a^2 x^2}+\frac {c \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}\\ &=-\frac {c x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {1}{4} c x^3 \sqrt {1-a^2 x^2}+\frac {c \sin ^{-1}(a x)}{8 a^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 40, normalized size = 0.69 \[ \frac {c \left (a x \sqrt {1-a^2 x^2} \left (2 a^2 x^2-1\right )+\sin ^{-1}(a x)\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^2*(c - a*c*x),x]

[Out]

(c*(a*x*Sqrt[1 - a^2*x^2]*(-1 + 2*a^2*x^2) + ArcSin[a*x]))/(8*a^3)

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 60, normalized size = 1.03 \[ -\frac {2 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{3} c x^{3} - a c x\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^3*c*x^3 - a*c*x)*sqrt(-a^2*x^2 + 1))/a^3

________________________________________________________________________________________

giac [A]  time = 0.19, size = 45, normalized size = 0.78 \[ \frac {1}{8} \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, c x^{2} - \frac {c}{a^{2}}\right )} x + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, a^{2} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 70, normalized size = 1.21 \[ \frac {c \,x^{3} \sqrt {-a^{2} x^{2}+1}}{4}-\frac {c x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c*x^3*(-a^2*x^2+1)^(1/2)-1/8*c*x*(-a^2*x^2+1)^(1/2)/a^2+1/8*c/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x
^2+1)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 48, normalized size = 0.83 \[ \frac {1}{4} \, \sqrt {-a^{2} x^{2} + 1} c x^{3} - \frac {\sqrt {-a^{2} x^{2} + 1} c x}{8 \, a^{2}} + \frac {c \arcsin \left (a x\right )}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(-a^2*x^2 + 1)*c*x^3 - 1/8*sqrt(-a^2*x^2 + 1)*c*x/a^2 + 1/8*c*arcsin(a*x)/a^3

________________________________________________________________________________________

mupad [B]  time = 0.78, size = 61, normalized size = 1.05 \[ \frac {c\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {c\,x\,\sqrt {1-a^2\,x^2}}{8\,a^2}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^2\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^3*(1 - a^2*x^2)^(1/2))/4 - (c*x*(1 - a^2*x^2)^(1/2))/(8*a^2) + (c*asinh(x*(-a^2)^(1/2)))/(8*a^2*(-a^2)^(1
/2))

________________________________________________________________________________________

sympy [A]  time = 4.88, size = 150, normalized size = 2.59 \[ c \left (\begin {cases} \frac {i a^{2} x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {3 i x^{3}}{8 \sqrt {a^{2} x^{2} - 1}} + \frac {i x}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{8 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {3 x^{3}}{8 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{8 a^{3}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*Piecewise((I*a**2*x**5/(4*sqrt(a**2*x**2 - 1)) - 3*I*x**3/(8*sqrt(a**2*x**2 - 1)) + I*x/(8*a**2*sqrt(a**2*x*
*2 - 1)) - I*acosh(a*x)/(8*a**3), Abs(a**2*x**2) > 1), (-a**2*x**5/(4*sqrt(-a**2*x**2 + 1)) + 3*x**3/(8*sqrt(-
a**2*x**2 + 1)) - x/(8*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(8*a**3), True))

________________________________________________________________________________________