3.13.40 \(\int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx\) [1240]

Optimal. Leaf size=78 \[ -\sqrt {c-a^2 c x^2}-2 \sqrt {c} \text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6287, 1823, 858, 223, 209, 272, 65, 214} \begin {gather*} -2 \sqrt {c} \text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\sqrt {c-a^2 c x^2}-\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 214

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 272

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 858

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 1823

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 6287

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c^(n/2), Int[x^m*
((c + d*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
] || GtQ[c, 0]) && ILtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 99, normalized size = 1.27 \begin {gather*} -\sqrt {c-a^2 c x^2}+2 \sqrt {c} \text {ArcTan}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )+\sqrt {c} \log (x)-\sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 120, normalized size = 1.54

method	result	size
default	\(-2 \sqrt {-c \,a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a c \left (x +\frac {1}{a}\right )}-\frac {2 a c \arctan \left (\frac {\sqrt {c \,a^{2}}\, x}{\sqrt {-c \,a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a c \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {c \,a^{2}}}+\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.33, size = 196, normalized size = 2.51 \begin {gather*} \left [2 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + \frac {1}{2} \, \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - \sqrt {-a^{2} c x^{2} + c}, -\sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - \sqrt {-a^{2} c x^{2} + c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + 1/2*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt(-a^
2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - sqrt(-a^2*c*x^2 + c), -sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c
*x^2 - c)) + sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - sqrt(-a^2*c*x^2 + c)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 124, normalized size = 1.59 \begin {gather*} -{\left (\frac {{\left (a x + 1\right )} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{a^{2}} - \frac {2 \, c \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{a^{2} \sqrt {-c}} - \frac {4 \, \sqrt {c} \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{a^{2}}\right )} a {\left | a \right |} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*x + 1)*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a)/a^2 - 2*c*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(-c
))*sgn(1/(a*x + 1))*sgn(a)/(a^2*sqrt(-c)) - 4*sqrt(c)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a*x + 1)
)*sgn(a)/a^2)*a*abs(a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )}{x\,{\left (a\,x+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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