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

Optimal. Leaf size=63 \[ -\frac {1-a x}{a c^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^2}-\frac {\sin ^{-1}(a x)}{a c^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6131, 6128, 797, 641, 216, 637} \[ -\frac {1-a x}{a c^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^2}-\frac {\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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 6131

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 46, normalized size = 0.73 \[ -\frac {\sqrt {1-a^2 x^2} (a x+2)+(a x+1) \sin ^{-1}(a x)}{a c^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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fricas [A]  time = 0.54, size = 71, normalized size = 1.13 \[ -\frac {2 \, a x - 2 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 2\right )} + 2}{a^{2} c^{2} x + a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 73, normalized size = 1.16 \[ -\frac {\arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{c^{2} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a c^{2}} + \frac {2}{c^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 336, normalized size = 5.33 \[ -\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{8 a^{3} c^{2} \left (x -\frac {1}{a}\right )^{2}}-\frac {5 \left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{48 a \,c^{2}}+\frac {5 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}\, x}{32 c^{2}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{32 c^{2} \sqrt {a^{2}}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{4 a^{4} c^{2} \left (x +\frac {1}{a}\right )^{3}}-\frac {3 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{4 a^{3} c^{2} \left (x +\frac {1}{a}\right )^{2}}-\frac {37 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{48 a \,c^{2}}-\frac {37 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{32 c^{2}}-\frac {37 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{32 c^{2} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a^3/c^2/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-5/48/a/c^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+5/32/c
^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x+5/32/c^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))
^(1/2))-1/4/a^4/c^2/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-3/4/a^3/c^2/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+
1/a))^(5/2)-37/48/a/c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-37/32/c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-37/3
2/c^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 89, normalized size = 1.41 \[ \frac {\sqrt {1-a^2\,x^2}}{c^2\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \left (\int \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + a^{4} x^{4} - 2 a^{3} x^{3} - 2 a^{2} x^{2} + a x + 1}\, dx + \int \left (- \frac {a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + a^{4} x^{4} - 2 a^{3} x^{3} - 2 a^{2} x^{2} + a x + 1}\right )\, dx\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(x**2*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + a**4*x**4 - 2*a**3*x**3 - 2*a**2*x**2 + a*x + 1), x) + I
ntegral(-a**2*x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + a**4*x**4 - 2*a**3*x**3 - 2*a**2*x**2 + a*x + 1), x))/c**
2

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