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

Optimal. Leaf size=83 \[ -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a}+\frac {1}{4} c^2 x \left (1-a^2 x^2\right )^{3/2}+\frac {3}{8} c^2 x \sqrt {1-a^2 x^2}+\frac {3 c^2 \sin ^{-1}(a x)}{8 a} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 83, 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 = {6138, 641, 195, 216} \[ -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{5 a}+\frac {1}{4} c^2 x \left (1-a^2 x^2\right )^{3/2}+\frac {3}{8} c^2 x \sqrt {1-a^2 x^2}+\frac {3 c^2 \sin ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6138

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] && IGtQ[(n + 1)/2, 0] &&
!IntegerQ[p - n/2]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 75, normalized size = 0.90 \[ -\frac {c^2 \left (\sqrt {1-a^2 x^2} \left (8 a^4 x^4+10 a^3 x^3-16 a^2 x^2-25 a x+8\right )+30 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{40 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/40*(c^2*(Sqrt[1 - a^2*x^2]*(8 - 25*a*x - 16*a^2*x^2 + 10*a^3*x^3 + 8*a^4*x^4) + 30*ArcSin[Sqrt[1 - a*x]/Sqr
t[2]]))/a

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fricas [A]  time = 0.48, size = 92, normalized size = 1.11 \[ -\frac {30 \, c^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (8 \, a^{4} c^{2} x^{4} + 10 \, a^{3} c^{2} x^{3} - 16 \, a^{2} c^{2} x^{2} - 25 \, a c^{2} x + 8 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{40 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(30*c^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (8*a^4*c^2*x^4 + 10*a^3*c^2*x^3 - 16*a^2*c^2*x^2 - 25*a
*c^2*x + 8*c^2)*sqrt(-a^2*x^2 + 1))/a

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giac [A]  time = 0.21, size = 78, normalized size = 0.94 \[ \frac {3 \, c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, {\left | a \right |}} + \frac {1}{40} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (25 \, c^{2} + 2 \, {\left (8 \, a c^{2} - {\left (4 \, a^{3} c^{2} x + 5 \, a^{2} c^{2}\right )} x\right )} x\right )} x - \frac {8 \, c^{2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*c^2*arcsin(a*x)*sgn(a)/abs(a) + 1/40*sqrt(-a^2*x^2 + 1)*((25*c^2 + 2*(8*a*c^2 - (4*a^3*c^2*x + 5*a^2*c^2)*
x)*x)*x - 8*c^2/a)

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maple [A]  time = 0.04, size = 137, normalized size = 1.65 \[ -\frac {c^{2} a^{3} x^{4} \sqrt {-a^{2} x^{2}+1}}{5}+\frac {2 c^{2} a \,x^{2} \sqrt {-a^{2} x^{2}+1}}{5}-\frac {c^{2} \sqrt {-a^{2} x^{2}+1}}{5 a}-\frac {c^{2} a^{2} x^{3} \sqrt {-a^{2} x^{2}+1}}{4}+\frac {5 c^{2} x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 c^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*c^2*a^3*x^4*(-a^2*x^2+1)^(1/2)+2/5*c^2*a*x^2*(-a^2*x^2+1)^(1/2)-1/5*c^2*(-a^2*x^2+1)^(1/2)/a-1/4*c^2*a^2*
x^3*(-a^2*x^2+1)^(1/2)+5/8*c^2*x*(-a^2*x^2+1)^(1/2)+3/8*c^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 = 0.41, size = 118, normalized size = 1.42 \[ -\frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{2} x^{4} - \frac {1}{4} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{2} x^{3} + \frac {2}{5} \, \sqrt {-a^{2} x^{2} + 1} a c^{2} x^{2} + \frac {5}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{2} x + \frac {3 \, c^{2} \arcsin \left (a x\right )}{8 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{5 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.91, size = 82, normalized size = 0.99 \[ \frac {3\,c^2\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {c^2\,x\,{\left (1-a^2\,x^2\right )}^{3/2}}{4}-\frac {c^2\,{\left (1-a^2\,x^2\right )}^{5/2}}{5\,a}-\frac {3\,c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}{8\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 20.15, size = 143, normalized size = 1.72 \[ \begin {cases} \frac {- \frac {c^{2} \left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} + c^{2} \left (\begin {cases} \frac {a x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\operatorname {asin}{\left (a x \right )}}{2} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) - c^{2} \left (\begin {cases} - \frac {a x \left (- 2 a^{2} x^{2} + 1\right ) \sqrt {- a^{2} x^{2} + 1}}{8} + \frac {\operatorname {asin}{\left (a x \right )}}{8} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) - c^{2} \left (\begin {cases} \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {5}{2}}}{5} - \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right )}{a} & \text {for}\: a \neq 0 \\c^{2} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-c**2*(-a**2*x**2 + 1)**(3/2)/3 + c**2*Piecewise((a*x*sqrt(-a**2*x**2 + 1)/2 + asin(a*x)/2, (a*x >
-1) & (a*x < 1))) - c**2*Piecewise((-a*x*(-2*a**2*x**2 + 1)*sqrt(-a**2*x**2 + 1)/8 + asin(a*x)/8, (a*x > -1)
& (a*x < 1))) - c**2*Piecewise(((-a**2*x**2 + 1)**(5/2)/5 - (-a**2*x**2 + 1)**(3/2)/3, (a*x > -1) & (a*x < 1))
))/a, Ne(a, 0)), (c**2*x, True))

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