3.1257 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=93 \[ \frac {c \sqrt {1-a^2 x^2} (1-a x)^2}{3 a}+\frac {5 c \sqrt {1-a^2 x^2} (1-a x)}{6 a}+\frac {5 c \sqrt {1-a^2 x^2}}{2 a}+\frac {5 c \sin ^{-1}(a x)}{2 a} \]

[Out]

5/2*c*arcsin(a*x)/a+5/2*c*(-a^2*x^2+1)^(1/2)/a+5/6*c*(-a*x+1)*(-a^2*x^2+1)^(1/2)/a+1/3*c*(-a*x+1)^2*(-a^2*x^2+
1)^(1/2)/a

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Rubi [A]  time = 0.06, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6139, 671, 641, 216} \[ \frac {c \sqrt {1-a^2 x^2} (1-a x)^2}{3 a}+\frac {5 c \sqrt {1-a^2 x^2} (1-a x)}{6 a}+\frac {5 c \sqrt {1-a^2 x^2}}{2 a}+\frac {5 c \sin ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)/E^(3*ArcTanh[a*x]),x]

[Out]

(5*c*Sqrt[1 - a^2*x^2])/(2*a) + (5*c*(1 - a*x)*Sqrt[1 - a^2*x^2])/(6*a) + (c*(1 - a*x)^2*Sqrt[1 - a^2*x^2])/(3
*a) + (5*c*ArcSin[a*x])/(2*a)

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 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 6139

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

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right ) \, dx &=c \int \frac {(1-a x)^3}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {c (1-a x)^2 \sqrt {1-a^2 x^2}}{3 a}+\frac {1}{3} (5 c) \int \frac {(1-a x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5 c (1-a x) \sqrt {1-a^2 x^2}}{6 a}+\frac {c (1-a x)^2 \sqrt {1-a^2 x^2}}{3 a}+\frac {1}{2} (5 c) \int \frac {1-a x}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5 c \sqrt {1-a^2 x^2}}{2 a}+\frac {5 c (1-a x) \sqrt {1-a^2 x^2}}{6 a}+\frac {c (1-a x)^2 \sqrt {1-a^2 x^2}}{3 a}+\frac {1}{2} (5 c) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5 c \sqrt {1-a^2 x^2}}{2 a}+\frac {5 c (1-a x) \sqrt {1-a^2 x^2}}{6 a}+\frac {c (1-a x)^2 \sqrt {1-a^2 x^2}}{3 a}+\frac {5 c \sin ^{-1}(a x)}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 70, normalized size = 0.75 \[ \frac {c \left (\frac {\sqrt {a x+1} \left (-2 a^3 x^3+11 a^2 x^2-31 a x+22\right )}{\sqrt {1-a x}}-30 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{6 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a^2*c*x^2)/E^(3*ArcTanh[a*x]),x]

[Out]

(c*((Sqrt[1 + a*x]*(22 - 31*a*x + 11*a^2*x^2 - 2*a^3*x^3))/Sqrt[1 - a*x] - 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/
(6*a)

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fricas [A]  time = 0.61, size = 63, normalized size = 0.68 \[ -\frac {30 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{2} c x^{2} - 9 \, a c x + 22 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(30*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^2*c*x^2 - 9*a*c*x + 22*c)*sqrt(-a^2*x^2 + 1))/a

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giac [A]  time = 0.27, size = 46, normalized size = 0.49 \[ \frac {5 \, c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, {\left | a \right |}} + \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a c x - 9 \, c\right )} x + \frac {22 \, c}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*c*arcsin(a*x)*sgn(a)/abs(a) + 1/6*sqrt(-a^2*x^2 + 1)*((2*a*c*x - 9*c)*x + 22*c/a)

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maple [A]  time = 0.04, size = 133, normalized size = 1.43 \[ \frac {2 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{3} \left (x +\frac {1}{a}\right )^{2}}+\frac {5 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3 a}+\frac {5 c \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{2}+\frac {5 c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.50, size = 122, normalized size = 1.31 \[ -\frac {1}{2} \, \sqrt {a^{2} x^{2} + 4 \, a x + 3} c x + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{a^{2} x + a} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{3 \, a} + \frac {i \, c \arcsin \left (a x + 2\right )}{2 \, a} + \frac {3 \, c \arcsin \left (a x\right )}{a} - \frac {\sqrt {a^{2} x^{2} + 4 \, a x + 3} c}{a} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} c}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 74, normalized size = 0.80 \[ \frac {11\,c\,\sqrt {1-a^2\,x^2}}{3\,a}-\frac {3\,c\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {5\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}+\frac {a\,c\,x^2\,\sqrt {1-a^2\,x^2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c \left (\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx + \int \left (- \frac {2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\right )\, dx + \int \frac {a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(Integral(sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1), x) + Integral(-2*a**2*x**2*sqrt(-a**2*
x**2 + 1)/(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1), x) + Integral(a**4*x**4*sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3*a
**2*x**2 + 3*a*x + 1), x))

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