3.302 \(\int \frac{e^{\tanh ^{-1}(a x)} (c-a c x)^2}{x^4} \, dx\)

Optimal. Leaf size=75 \[ \frac{a c^2 \sqrt{1-a^2 x^2}}{2 x^2}-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}-\frac{1}{2} a^3 c^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0927444, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {6128, 807, 266, 47, 63, 208} \[ \frac{a c^2 \sqrt{1-a^2 x^2}}{2 x^2}-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}-\frac{1}{2} a^3 c^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x)^2)/x^4,x]

[Out]

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

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.0285044, size = 91, normalized size = 1.21 \[ -\frac{c^2 \left (2 a^4 x^4+3 a^3 x^3-4 a^2 x^2+3 a^3 x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-3 a x+2\right )}{6 x^3 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^2)/x^4,x]

[Out]

-(c^2*(2 - 3*a*x - 4*a^2*x^2 + 3*a^3*x^3 + 2*a^4*x^4 + 3*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])
)/(6*x^3*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.039, size = 100, normalized size = 1.3 \begin{align*}{c}^{2} \left ({\frac{{a}^{2}}{3\,x}\sqrt{-{a}^{2}{x}^{2}+1}}-{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -a \left ( -{\frac{1}{2\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) } \right ) -{\frac{1}{3\,{x}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^2/x^4,x)

[Out]

c^2*(1/3*a^2*(-a^2*x^2+1)^(1/2)/x-a^3*arctanh(1/(-a^2*x^2+1)^(1/2))-a*(-1/2*(-a^2*x^2+1)^(1/2)/x^2-1/2*a^2*arc
tanh(1/(-a^2*x^2+1)^(1/2)))-1/3*(-a^2*x^2+1)^(1/2)/x^3)

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Maxima [A]  time = 1.44382, size = 134, normalized size = 1.79 \begin{align*} -\frac{1}{2} \, a^{3} c^{2} \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-a^{2} x^{2} + 1} a^{2} c^{2}}{3 \, x} + \frac{\sqrt{-a^{2} x^{2} + 1} a c^{2}}{2 \, x^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^2/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51902, size = 154, normalized size = 2.05 \begin{align*} \frac{3 \, a^{3} c^{2} x^{3} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (2 \, a^{2} c^{2} x^{2} + 3 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^2/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 7.48187, size = 270, normalized size = 3.6 \begin{align*} a^{3} c^{2} \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 ) - a^{2} c^{2} \left (\begin{cases} - \frac{i \sqrt{a^{2} x^{2} - 1}}{x} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{x} & \text{otherwise} \end{cases}\right ) - a c^{2} \left (\begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} - \frac{a \sqrt{-1 + \frac{1}{a^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac{i a^{2} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a}{2 x \sqrt{1 - \frac{1}{a^{2} x^{2}}}} + \frac{i}{2 a x^{3} \sqrt{1 - \frac{1}{a^{2} x^{2}}}} & \text{otherwise} \end{cases}\right ) + c^{2} \left (\begin{cases} - \frac{2 i a^{2} \sqrt{a^{2} x^{2} - 1}}{3 x} - \frac{i \sqrt{a^{2} x^{2} - 1}}{3 x^{3}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{2 a^{2} \sqrt{- a^{2} x^{2} + 1}}{3 x} - \frac{\sqrt{- a^{2} x^{2} + 1}}{3 x^{3}} & \text{otherwise} \end{cases}\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)*(-a*c*x+c)**2/x**4,x)

[Out]

a**3*c**2*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) - a**2*c**2*Piecewise((-
I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) - a*c**2*Piecewise((-a**2*acosh
(1/(a*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a/(2*x*sqrt
(1 - 1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True)) + c**2*Piecewise((-2*I*a**2*sqrt(a**2*x**2
- 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-a**2*x**2 + 1)/(3*x) - sqrt(
-a**2*x**2 + 1)/(3*x**3), True))

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Giac [B]  time = 1.38843, size = 315, normalized size = 4.2 \begin{align*} \frac{{\left (a^{4} c^{2} - \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{2} c^{2}}{x} - \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{2}}{x^{2}}\right )} a^{6} x^{3}}{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left | a \right |}} - \frac{a^{4} c^{2} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \,{\left | a \right |}} + \frac{\frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4} c^{2}}{x} + \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2} c^{2}}{x^{2}} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{2}}{x^{3}}}{24 \, a^{2}{\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)*(-a*c*x+c)^2/x^4,x, algorithm="giac")

[Out]

1/24*(a^4*c^2 - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2*c^2/x - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^2/x^2)*a^6
*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) - 1/2*a^4*c^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/
(a^2*abs(x)))/abs(a) + 1/24*(3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^2/x + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2
*a^2*c^2/x^2 - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^2/x^3)/(a^2*abs(a))