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

Optimal. Leaf size=64 \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{a^2 c (1-a x)^2}+\frac{2 \sqrt{1-a^2 x^2}}{a^2 c}-\frac{2 \sin ^{-1}(a x)}{a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.0782248, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {6128, 793, 665, 216} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{a^2 c (1-a x)^2}+\frac{2 \sqrt{1-a^2 x^2}}{a^2 c}-\frac{2 \sin ^{-1}(a x)}{a^2 c} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x)/(c - a*c*x),x]

[Out]

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

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]

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*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 \frac{e^{\tanh ^{-1}(a x)} x}{c-a c x} \, dx &=c \int \frac{x \sqrt{1-a^2 x^2}}{(c-a c x)^2} \, dx\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{a^2 c (1-a x)^2}-\frac{2 \int \frac{\sqrt{1-a^2 x^2}}{c-a c x} \, dx}{a}\\ &=\frac{2 \sqrt{1-a^2 x^2}}{a^2 c}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^2 c (1-a x)^2}-\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a c}\\ &=\frac{2 \sqrt{1-a^2 x^2}}{a^2 c}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a^2 c (1-a x)^2}-\frac{2 \sin ^{-1}(a x)}{a^2 c}\\ \end{align*}

Mathematica [A]  time = 0.0350603, size = 53, normalized size = 0.83 \[ \frac{\frac{\sqrt{a x+1} (3-a x)}{\sqrt{1-a x}}+4 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{a^2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 98, normalized size = 1.5 \begin{align*}{\frac{1}{{a}^{2}c}\sqrt{-{a}^{2}{x}^{2}+1}}-2\,{\frac{1}{ac\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-2\,{\frac{1}{{a}^{3}c}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-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)*x/(-a*c*x+c),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50456, size = 155, normalized size = 2.42 \begin{align*} \frac{3 \, a x + 4 \,{\left (a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt{-a^{2} x^{2} + 1}{\left (a x - 3\right )} - 3}{a^{3} c x - a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{x}{a x \sqrt{- a^{2} x^{2} + 1} - \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{2}}{a x \sqrt{- a^{2} x^{2} + 1} - \sqrt{- a^{2} x^{2} + 1}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23721, size = 105, normalized size = 1.64 \begin{align*} -\frac{2 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{a c{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a^{2} c} + \frac{4}{a c{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \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)*x/(-a*c*x+c),x, algorithm="giac")

[Out]

-2*arcsin(a*x)*sgn(a)/(a*c*abs(a)) + sqrt(-a^2*x^2 + 1)/(a^2*c) + 4/(a*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2
*x) - 1)*abs(a))