3.5.90 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [490]

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

[Out]

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

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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 = {6266, 6263, 1653, 12, 807, 222} \begin {gather*} \frac {\sqrt {1-a^2 x^2}}{a c^2 (1-a x)}+\frac {\sqrt {1-a^2 x^2}}{a c^2}-\frac {\text {ArcSin}(a x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 222

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 807

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 1653

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

Rule 6263

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 6266

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

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Mathematica [A]
time = 0.05, size = 47, normalized size = 0.75 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{c^2}-\frac {1}{c^2 (-1+a x)}\right )}{a}-\frac {\text {ArcSin}(a x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(270\) vs. \(2(59)=118\).
time = 0.77, size = 271, normalized size = 4.30

method	result	size
risch	\(-\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{2}}+\frac {\left (-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{4} \left (x -\frac {1}{a}\right )}\right ) a^{2}}{c^{2}}\)	\(110\)
default	\(\frac {a^{2} \left (\frac {\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{2 a^{4}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{4 a^{3}}+\frac {\frac {3 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{4}-\frac {3 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{4 \sqrt {a^{2}}}}{a^{3}}\right )}{c^{2}}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^2,x,method=_RETURNVERBOSE)

[Out]

a^2/c^2*(1/2/a^4*(1/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+a*((-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-a/(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^3*((-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2
)+a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)))+3/4/a^3*((-a^2*(x-1/a)^2-2*a*(x-1/a)
)^(1/2)-a/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 71, normalized size = 1.13 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {a^{2} \int \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} - a^{2} x^{2} - a x + 1}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 72, normalized size = 1.14 \begin {gather*} -\frac {\arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(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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Mupad [B]
time = 0.86, size = 89, normalized size = 1.41 \begin {gather*} \frac {\sqrt {1-a^2\,x^2}}{a\,c^2}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}}{c^2\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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