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

Optimal. Leaf size=113 \[ \frac {c^2 \sin ^{-1}(a x)}{8 a^3}-\frac {c^2 x \left (1-a^2 x^2\right )^{3/2}}{4 a^2}+\frac {c^2 x \sqrt {1-a^2 x^2}}{8 a^2}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3} \]

[Out]

1/3*c^2*(-a^2*x^2+1)^(3/2)/a^3-1/4*c^2*x*(-a^2*x^2+1)^(3/2)/a^2-1/5*c^2*(-a^2*x^2+1)^(5/2)/a^3+1/8*c^2*arcsin(
a*x)/a^3+1/8*c^2*x*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6128, 797, 641, 195, 216} \[ -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {c^2 x \left (1-a^2 x^2\right )^{3/2}}{4 a^2}+\frac {c^2 x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {c^2 \sin ^{-1}(a x)}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x*Sqrt[1 - a^2*x^2])/(8*a^2) + (c^2*(1 - a^2*x^2)^(3/2))/(3*a^3) - (c^2*x*(1 - a^2*x^2)^(3/2))/(4*a^2) -
(c^2*(1 - a^2*x^2)^(5/2))/(5*a^3) + (c^2*ArcSin[a*x])/(8*a^3)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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)^2 \, dx &=c \int x^2 (c-a c x) \sqrt {1-a^2 x^2} \, dx\\ &=\frac {c \int (c-a c x) \sqrt {1-a^2 x^2} \, dx}{a^2}-\frac {c \int (c-a c x) \left (1-a^2 x^2\right )^{3/2} \, dx}{a^2}\\ &=\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \int \sqrt {1-a^2 x^2} \, dx}{a^2}-\frac {c^2 \int \left (1-a^2 x^2\right )^{3/2} \, dx}{a^2}\\ &=\frac {c^2 x \sqrt {1-a^2 x^2}}{2 a^2}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {c^2 x \left (1-a^2 x^2\right )^{3/2}}{4 a^2}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}-\frac {\left (3 c^2\right ) \int \sqrt {1-a^2 x^2} \, dx}{4 a^2}\\ &=\frac {c^2 x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {c^2 x \left (1-a^2 x^2\right )^{3/2}}{4 a^2}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \sin ^{-1}(a x)}{2 a^3}-\frac {\left (3 c^2\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}\\ &=\frac {c^2 x \sqrt {1-a^2 x^2}}{8 a^2}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {c^2 x \left (1-a^2 x^2\right )^{3/2}}{4 a^2}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {c^2 \sin ^{-1}(a x)}{8 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 75, normalized size = 0.66 \[ -\frac {c^2 \left (\sqrt {1-a^2 x^2} \left (24 a^4 x^4-30 a^3 x^3-8 a^2 x^2+15 a x-16\right )+30 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{120 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/120*(c^2*(Sqrt[1 - a^2*x^2]*(-16 + 15*a*x - 8*a^2*x^2 - 30*a^3*x^3 + 24*a^4*x^4) + 30*ArcSin[Sqrt[1 - a*x]/
Sqrt[2]]))/a^3

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fricas [A]  time = 0.44, size = 92, normalized size = 0.81 \[ -\frac {30 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (24 \, a^{4} c^{2} x^{4} - 30 \, a^{3} c^{2} x^{3} - 8 \, a^{2} c^{2} x^{2} + 15 \, a c^{2} x - 16 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{120 \, a^{3}} \]

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)^2,x, algorithm="fricas")

[Out]

-1/120*(30*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (24*a^4*c^2*x^4 - 30*a^3*c^2*x^3 - 8*a^2*c^2*x^2 + 15*
a*c^2*x - 16*c^2)*sqrt(-a^2*x^2 + 1))/a^3

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giac [A]  time = 0.25, size = 81, normalized size = 0.72 \[ -\frac {1}{120} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (3 \, {\left (4 \, a c^{2} x - 5 \, c^{2}\right )} x - \frac {4 \, c^{2}}{a}\right )} x + \frac {15 \, c^{2}}{a^{2}}\right )} x - \frac {16 \, c^{2}}{a^{3}}\right )} + \frac {c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, a^{2} {\left | a \right |}} \]

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)^2,x, algorithm="giac")

[Out]

-1/120*sqrt(-a^2*x^2 + 1)*((2*(3*(4*a*c^2*x - 5*c^2)*x - 4*c^2/a)*x + 15*c^2/a^2)*x - 16*c^2/a^3) + 1/8*c^2*ar
csin(a*x)*sgn(a)/(a^2*abs(a))

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maple [A]  time = 0.04, size = 140, normalized size = 1.24 \[ -\frac {c^{2} a \,x^{4} \sqrt {-a^{2} x^{2}+1}}{5}+\frac {c^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a}+\frac {2 c^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{3}}+\frac {c^{2} x^{3} \sqrt {-a^{2} x^{2}+1}}{4}-\frac {c^{2} x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}} \]

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)^2,x)

[Out]

-1/5*c^2*a*x^4*(-a^2*x^2+1)^(1/2)+1/15*c^2/a*x^2*(-a^2*x^2+1)^(1/2)+2/15*c^2/a^3*(-a^2*x^2+1)^(1/2)+1/4*c^2*x^
3*(-a^2*x^2+1)^(1/2)-1/8*c^2*x*(-a^2*x^2+1)^(1/2)/a^2+1/8*c^2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1
)^(1/2))

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maxima [A]  time = 0.52, size = 118, normalized size = 1.04 \[ -\frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} a c^{2} x^{4} + \frac {1}{4} \, \sqrt {-a^{2} x^{2} + 1} c^{2} x^{3} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x^{2}}{15 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2} x}{8 \, a^{2}} + \frac {c^{2} \arcsin \left (a x\right )}{8 \, a^{3}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{15 \, a^{3}} \]

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)^2,x, algorithm="maxima")

[Out]

-1/5*sqrt(-a^2*x^2 + 1)*a*c^2*x^4 + 1/4*sqrt(-a^2*x^2 + 1)*c^2*x^3 + 1/15*sqrt(-a^2*x^2 + 1)*c^2*x^2/a - 1/8*s
qrt(-a^2*x^2 + 1)*c^2*x/a^2 + 1/8*c^2*arcsin(a*x)/a^3 + 2/15*sqrt(-a^2*x^2 + 1)*c^2/a^3

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mupad [B]  time = 0.03, size = 131, normalized size = 1.16 \[ \frac {2\,c^2\,\sqrt {1-a^2\,x^2}}{15\,a^3}+\frac {c^2\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {c^2\,x\,\sqrt {1-a^2\,x^2}}{8\,a^2}-\frac {a\,c^2\,x^4\,\sqrt {1-a^2\,x^2}}{5}+\frac {c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^2\,\sqrt {-a^2}}+\frac {c^2\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^2*(1 - a^2*x^2)^(1/2))/(15*a^3) + (c^2*x^3*(1 - a^2*x^2)^(1/2))/4 - (c^2*x*(1 - a^2*x^2)^(1/2))/(8*a^2) -
(a*c^2*x^4*(1 - a^2*x^2)^(1/2))/5 + (c^2*asinh(x*(-a^2)^(1/2)))/(8*a^2*(-a^2)^(1/2)) + (c^2*x^2*(1 - a^2*x^2)
^(1/2))/(15*a)

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sympy [C]  time = 7.52, size = 374, normalized size = 3.31 \[ a^{3} c^{2} \left (\begin {cases} - \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} - \frac {8 \sqrt {- a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) - a^{2} c^{2} \left (\begin {cases} - \frac {i x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {3 i x}{8 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \operatorname {acosh}{\left (a x \right )}}{8 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8 a^{5}} & \text {otherwise} \end {cases}\right ) - a c^{2} \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) \]

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)**2,x)

[Out]

a**3*c**2*Piecewise((-x**4*sqrt(-a**2*x**2 + 1)/(5*a**2) - 4*x**2*sqrt(-a**2*x**2 + 1)/(15*a**4) - 8*sqrt(-a**
2*x**2 + 1)/(15*a**6), Ne(a, 0)), (x**6/6, True)) - a**2*c**2*Piecewise((-I*x**5/(4*sqrt(a**2*x**2 - 1)) - I*x
**3/(8*a**2*sqrt(a**2*x**2 - 1)) + 3*I*x/(8*a**4*sqrt(a**2*x**2 - 1)) - 3*I*acosh(a*x)/(8*a**5), Abs(a**2*x**2
) > 1), (x**5/(4*sqrt(-a**2*x**2 + 1)) + x**3/(8*a**2*sqrt(-a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 + 1)
) + 3*asin(a*x)/(8*a**5), True)) - a*c**2*Piecewise((-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 +
1)/(3*a**4), Ne(a, 0)), (x**4/4, True)) + c**2*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2
*a**3), Abs(a**2*x**2) > 1), (x**3/(2*sqrt(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a
**3), True))

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