3.1141 \(\int \frac{e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x} \, dx\)

Optimal. Leaf size=66 \[ -\frac{1}{2} a c x \sqrt{1-a^2 x^2}-3 c \sqrt{1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{7}{2} c \sin ^{-1}(a x) \]

[Out]

-3*c*Sqrt[1 - a^2*x^2] - (a*c*x*Sqrt[1 - a^2*x^2])/2 + (7*c*ArcSin[a*x])/2 - c*ArcTanh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.19502, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6148, 1809, 844, 216, 266, 63, 208} \[ -\frac{1}{2} a c x \sqrt{1-a^2 x^2}-3 c \sqrt{1-a^2 x^2}-c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{7}{2} c \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3*c*Sqrt[1 - a^2*x^2] - (a*c*x*Sqrt[1 - a^2*x^2])/2 + (7*c*ArcSin[a*x])/2 - c*ArcTanh[Sqrt[1 - a^2*x^2]]

Rule 6148

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{e^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )}{x} \, dx &=c \int \frac{(1+a x)^3}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{1}{2} a c x \sqrt{1-a^2 x^2}-\frac{c \int \frac{-2 a^2-7 a^3 x-6 a^4 x^2}{x \sqrt{1-a^2 x^2}} \, dx}{2 a^2}\\ &=-3 c \sqrt{1-a^2 x^2}-\frac{1}{2} a c x \sqrt{1-a^2 x^2}+\frac{c \int \frac{2 a^4+7 a^5 x}{x \sqrt{1-a^2 x^2}} \, dx}{2 a^4}\\ &=-3 c \sqrt{1-a^2 x^2}-\frac{1}{2} a c x \sqrt{1-a^2 x^2}+c \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx+\frac{1}{2} (7 a c) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx\\ &=-3 c \sqrt{1-a^2 x^2}-\frac{1}{2} a c x \sqrt{1-a^2 x^2}+\frac{7}{2} c \sin ^{-1}(a x)+\frac{1}{2} c \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-3 c \sqrt{1-a^2 x^2}-\frac{1}{2} a c x \sqrt{1-a^2 x^2}+\frac{7}{2} c \sin ^{-1}(a x)-\frac{c \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{a^2}\\ &=-3 c \sqrt{1-a^2 x^2}-\frac{1}{2} a c x \sqrt{1-a^2 x^2}+\frac{7}{2} c \sin ^{-1}(a x)-c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0448069, size = 49, normalized size = 0.74 \[ -\frac{1}{2} c \left (\sqrt{1-a^2 x^2} (a x+6)+2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-7 \sin ^{-1}(a x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*((6 + a*x)*Sqrt[1 - a^2*x^2] - 7*ArcSin[a*x] + 2*ArcTanh[Sqrt[1 - a^2*x^2]]))/2

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Maple [B]  time = 0.043, size = 121, normalized size = 1.8 \begin{align*}{\frac{{a}^{3}c{x}^{3}}{2}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-{\frac{acx}{2}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+{\frac{7\,ac}{2}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+3\,{\frac{{a}^{2}c{x}^{2}}{\sqrt{-{a}^{2}{x}^{2}+1}}}-3\,{\frac{c}{\sqrt{-{a}^{2}{x}^{2}+1}}}-c{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*a^3*x^3/(-a^2*x^2+1)^(1/2)-1/2*c*a*x/(-a^2*x^2+1)^(1/2)+7/2*c*a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x
^2+1)^(1/2))+3*c*a^2*x^2/(-a^2*x^2+1)^(1/2)-3*c/(-a^2*x^2+1)^(1/2)-c*arctanh(1/(-a^2*x^2+1)^(1/2))

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Maxima [B]  time = 1.47956, size = 167, normalized size = 2.53 \begin{align*} \frac{a^{3} c x^{3}}{2 \, \sqrt{-a^{2} x^{2} + 1}} + \frac{3 \, a^{2} c x^{2}}{\sqrt{-a^{2} x^{2} + 1}} - \frac{a c x}{2 \, \sqrt{-a^{2} x^{2} + 1}} + \frac{7 \, a c \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{2 \, \sqrt{a^{2}}} - c \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - \frac{3 \, c}{\sqrt{-a^{2} x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*c*x^3/sqrt(-a^2*x^2 + 1) + 3*a^2*c*x^2/sqrt(-a^2*x^2 + 1) - 1/2*a*c*x/sqrt(-a^2*x^2 + 1) + 7/2*a*c*arc
sin(a^2*x/sqrt(a^2))/sqrt(a^2) - c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - 3*c/sqrt(-a^2*x^2 + 1)

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Fricas [A]  time = 2.64576, size = 162, normalized size = 2.45 \begin{align*} -7 \, c \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + c \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - \frac{1}{2} \, \sqrt{-a^{2} x^{2} + 1}{\left (a c x + 6 \, c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + c*log((sqrt(-a^2*x^2 + 1) - 1)/x) - 1/2*sqrt(-a^2*x^2 + 1)*(a*c*
x + 6*c)

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Sympy [C]  time = 10.2942, size = 197, normalized size = 2.98 \begin{align*} a^{3} c \left (\begin{cases} - \frac{i x \sqrt{a^{2} x^{2} - 1}}{2 a^{2}} - \frac{i \operatorname{acosh}{\left (a x \right )}}{2 a^{3}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac{x^{3}}{2 \sqrt{- a^{2} x^{2} + 1}} - \frac{x}{2 a^{2} \sqrt{- a^{2} x^{2} + 1}} + \frac{\operatorname{asin}{\left (a x \right )}}{2 a^{3}} & \text{otherwise} \end{cases}\right ) + 3 a^{2} c \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: a^{2} = 0 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{a^{2}} & \text{otherwise} \end{cases}\right ) + 3 a c \left (\begin{cases} \sqrt{\frac{1}{a^{2}}} \operatorname{asin}{\left (x \sqrt{a^{2}} \right )} & \text{for}\: a^{2} > 0 \\\sqrt{- \frac{1}{a^{2}}} \operatorname{asinh}{\left (x \sqrt{- a^{2}} \right )} & \text{for}\: a^{2} < 0 \end{cases}\right ) + c \left (\begin{cases} - \operatorname{acosh}{\left (\frac{1}{a x} \right )} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname{asin}{\left (\frac{1}{a x} \right )} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2*a**3), Abs(a**2*x**2) > 1), (x**3/(2*sqr
t(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a**3), True)) + 3*a**2*c*Piecewise((x**2/2
, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) + 3*a*c*Piecewise((sqrt(a**(-2))*asin(x*sqrt(a**2)), a**2
> 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) + c*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (
I*asin(1/(a*x)), True))

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Giac [A]  time = 1.18742, size = 103, normalized size = 1.56 \begin{align*} \frac{7 \, a c \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \,{\left | a \right |}} - \frac{a c \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{{\left | a \right |}} - \frac{1}{2} \, \sqrt{-a^{2} x^{2} + 1}{\left (a c x + 6 \, c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/2*a*c*arcsin(a*x)*sgn(a)/abs(a) - a*c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) -
1/2*sqrt(-a^2*x^2 + 1)*(a*c*x + 6*c)