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

Optimal. Leaf size=55 \[ -\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {c \sin ^{-1}(a x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6138, 641, 195, 216} \[ -\frac {c \left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} c x \sqrt {1-a^2 x^2}+\frac {c \sin ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - a^2*c*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.07, size = 57, normalized size = 1.04 \[ \frac {c \left (\sqrt {1-a^2 x^2} \left (2 a^2 x^2+3 a x-2\right )-6 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{6 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 63, normalized size = 1.15 \[ -\frac {6 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (2 \, a^{2} c x^{2} + 3 \, a c x - 2 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, a} \]

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),x, algorithm="fricas")

[Out]

-1/6*(6*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (2*a^2*c*x^2 + 3*a*c*x - 2*c)*sqrt(-a^2*x^2 + 1))/a

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giac [A]  time = 0.21, size = 46, normalized size = 0.84 \[ \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, {\left | a \right |}} + \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a c x + 3 \, c\right )} x - \frac {2 \, c}{a}\right )} \]

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),x, algorithm="giac")

[Out]

1/2*c*arcsin(a*x)*sgn(a)/abs(a) + 1/6*sqrt(-a^2*x^2 + 1)*((2*a*c*x + 3*c)*x - 2*c/a)

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maple [A]  time = 0.04, size = 83, normalized size = 1.51 \[ \frac {c a \,x^{2} \sqrt {-a^{2} x^{2}+1}}{3}-\frac {c \sqrt {-a^{2} x^{2}+1}}{3 a}+\frac {c x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 64, normalized size = 1.16 \[ \frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} a c x^{2} + \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} c x + \frac {c \arcsin \left (a x\right )}{2 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{3 \, a} \]

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.88, size = 80, normalized size = 1.45 \[ \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {a\,c}{3\,\sqrt {-a^2}}+\frac {c\,x\,\sqrt {-a^2}}{2}-\frac {a^3\,c\,x^2}{3\,\sqrt {-a^2}}\right )}{\sqrt {-a^2}}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.82, size = 53, normalized size = 0.96 \[ \begin {cases} \frac {- \frac {c \left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} + c \left (\begin {cases} \frac {a x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\operatorname {asin}{\left (a x \right )}}{2} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right )}{a} & \text {for}\: a \neq 0 \\c x & \text {otherwise} \end {cases} \]

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),x)

[Out]

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

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