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

Optimal. Leaf size=91 \[ \frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{5}{8} c^3 x \sqrt{1-a^2 x^2}+\frac{5 c^3 \sin ^{-1}(a x)}{8 a} \]

[Out]

(5*c^3*x*Sqrt[1 - a^2*x^2])/8 + (5*c^3*(1 - a^2*x^2)^(3/2))/(12*a) + (c^3*(1 - a*x)*(1 - a^2*x^2)^(3/2))/(4*a)
+ (5*c^3*ArcSin[a*x])/(8*a)

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Rubi [A]  time = 0.057276, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6127, 671, 641, 195, 216} \[ \frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{5}{8} c^3 x \sqrt{1-a^2 x^2}+\frac{5 c^3 \sin ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a*c*x)^3,x]

[Out]

(5*c^3*x*Sqrt[1 - a^2*x^2])/8 + (5*c^3*(1 - a^2*x^2)^(3/2))/(12*a) + (c^3*(1 - a*x)*(1 - a^2*x^2)^(3/2))/(4*a)
+ (5*c^3*ArcSin[a*x])/(8*a)

Rule 6127

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

Rule 671

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[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

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 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]

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a x)} (c-a c x)^3 \, dx &=c \int (c-a c x)^2 \sqrt{1-a^2 x^2} \, dx\\ &=\frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{1}{4} \left (5 c^2\right ) \int (c-a c x) \sqrt{1-a^2 x^2} \, dx\\ &=\frac{5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{1}{4} \left (5 c^3\right ) \int \sqrt{1-a^2 x^2} \, dx\\ &=\frac{5}{8} c^3 x \sqrt{1-a^2 x^2}+\frac{5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{1}{8} \left (5 c^3\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{5}{8} c^3 x \sqrt{1-a^2 x^2}+\frac{5 c^3 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac{c^3 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}+\frac{5 c^3 \sin ^{-1}(a x)}{8 a}\\ \end{align*}

Mathematica [A]  time = 0.0833492, size = 67, normalized size = 0.74 \[ \frac{c^3 \left (\sqrt{1-a^2 x^2} \left (6 a^3 x^3-16 a^2 x^2+9 a x+16\right )-30 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{24 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(Sqrt[1 - a^2*x^2]*(16 + 9*a*x - 16*a^2*x^2 + 6*a^3*x^3) - 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(24*a)

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Maple [A]  time = 0.036, size = 114, normalized size = 1.3 \begin{align*}{\frac{{c}^{3}{a}^{2}{x}^{3}}{4}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3\,{c}^{3}x}{8}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{5\,{c}^{3}}{8}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{2\,{c}^{3}a{x}^{2}}{3}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{2\,{c}^{3}}{3\,a}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44562, size = 140, normalized size = 1.54 \begin{align*} \frac{1}{4} \, \sqrt{-a^{2} x^{2} + 1} a^{2} c^{3} x^{3} - \frac{2}{3} \, \sqrt{-a^{2} x^{2} + 1} a c^{3} x^{2} + \frac{3}{8} \, \sqrt{-a^{2} x^{2} + 1} c^{3} x + \frac{5 \, c^{3} \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{8 \, \sqrt{a^{2}}} + \frac{2 \, \sqrt{-a^{2} x^{2} + 1} c^{3}}{3 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6324, size = 178, normalized size = 1.96 \begin{align*} -\frac{30 \, c^{3} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (6 \, a^{3} c^{3} x^{3} - 16 \, a^{2} c^{3} x^{2} + 9 \, a c^{3} x + 16 \, c^{3}\right )} \sqrt{-a^{2} x^{2} + 1}}{24 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(30*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (6*a^3*c^3*x^3 - 16*a^2*c^3*x^2 + 9*a*c^3*x + 16*c^3)*s
qrt(-a^2*x^2 + 1))/a

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Sympy [A]  time = 6.46426, size = 136, normalized size = 1.49 \begin{align*} \begin{cases} - \frac{- 2 c^{3} \sqrt{- a^{2} x^{2} + 1} - 2 c^{3} \left (\begin{cases} \frac{\left (- a^{2} x^{2} + 1\right )^{\frac{3}{2}}}{3} - \sqrt{- a^{2} x^{2} + 1} & \text{for}\: a x > -1 \wedge a x < 1 \end{cases}\right ) + c^{3} \left (\begin{cases} \frac{a x \left (- 2 a^{2} x^{2} + 1\right ) \sqrt{- a^{2} x^{2} + 1}}{8} - \frac{a x \sqrt{- a^{2} x^{2} + 1}}{2} + \frac{3 \operatorname{asin}{\left (a x \right )}}{8} & \text{for}\: a x > -1 \wedge a x < 1 \end{cases}\right ) - c^{3} \operatorname{asin}{\left (a x \right )}}{a} & \text{for}\: a \neq 0 \\c^{3} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(-2*c**3*sqrt(-a**2*x**2 + 1) - 2*c**3*Piecewise(((-a**2*x**2 + 1)**(3/2)/3 - sqrt(-a**2*x**2 + 1)
, (a*x > -1) & (a*x < 1))) + c**3*Piecewise((a*x*(-2*a**2*x**2 + 1)*sqrt(-a**2*x**2 + 1)/8 - a*x*sqrt(-a**2*x*
*2 + 1)/2 + 3*asin(a*x)/8, (a*x > -1) & (a*x < 1))) - c**3*asin(a*x))/a, Ne(a, 0)), (c**3*x, True))

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Giac [A]  time = 1.3918, size = 89, normalized size = 0.98 \begin{align*} \frac{5 \, c^{3} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{8 \,{\left | a \right |}} + \frac{1}{24} \, \sqrt{-a^{2} x^{2} + 1}{\left (\frac{16 \, c^{3}}{a} +{\left (9 \, c^{3} + 2 \,{\left (3 \, a^{2} c^{3} x - 8 \, a c^{3}\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/24*sqrt(-a^2*x^2 + 1)*(16*c^3/a + (9*c^3 + 2*(3*a^2*c^3*x - 8*a*c^3)*x)*
x)