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

Optimal. Leaf size=65 \[ -\frac {(1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2}}{a c}-\frac {2 \sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6131, 6128, 789, 641, 216} \[ -\frac {(1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2}}{a c}-\frac {2 \sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 789

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*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 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)}}{c-\frac {c}{a x}} \, dx &=-\frac {a \int \frac {e^{-3 \tanh ^{-1}(a x)} x}{1-a x} \, dx}{c}\\ &=-\frac {a \int \frac {x (1-a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac {(1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {2 \int \frac {1-a x}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {(1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2}}{a c}-\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {(1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {1-a^2 x^2}}{a c}-\frac {2 \sin ^{-1}(a x)}{a c}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 67, normalized size = 1.03 \[ \frac {a^2 x^2+4 \sqrt {1-a^2 x^2} \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )+2 a x-3}{a c \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 67, normalized size = 1.03 \[ -\frac {3 \, a x - 4 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 3\right )} + 3}{a^{2} c x + a c} \]

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),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^2*c*x + a*
c)

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giac [A]  time = 1.82, size = 73, normalized size = 1.12 \[ -\frac {2 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{c {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a c} + \frac {4}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]

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),x, algorithm="giac")

[Out]

-2*arcsin(a*x)*sgn(a)/(c*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c) + 4/(c*((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 = 292, normalized size = 4.49 \[ \frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{24 a c}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}\, x}{16 c}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{16 c \sqrt {a^{2}}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{2 a^{4} c \left (x +\frac {1}{a}\right )^{3}}-\frac {5 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{4 c \,a^{3} \left (x +\frac {1}{a}\right )^{2}}-\frac {31 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{24 c a}-\frac {31 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{16 c}-\frac {31 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{16 c \sqrt {a^{2}}} \]

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),x)

[Out]

1/24/a/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-1/16/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x-1/16/c/(a^2)^(1/2)*arc
tan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))-1/2/a^4/c/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-5
/4/c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-31/24/c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-31/16/c*(-a
^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-31/16/c/(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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}}\,{d x} \]

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 90, normalized size = 1.38 \[ \frac {2\,\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c}-\frac {2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x)

[Out]

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

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