3.5.15 \(\int \frac {e^{-\tanh ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx\) [415]

Optimal. Leaf size=68 \[ -\frac {2 c \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6263, 895, 889, 214} \begin {gather*} -\frac {2 c \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 889

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 895

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e^2*(d +
e*x)^(m - 2)*(f + g*x)^(n + 1)*((a + c*x^2)^(p + 1)/(c*g*(n + p + 2))), x] - Dist[(e*f*(p + 1) - d*g*(2*n + p
+ 3))/(g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m,
n, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1]
&& IntegerQ[2*p]

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 44, normalized size = 0.65 \begin {gather*} -\frac {2 c \sqrt {1-a x} \left (\sqrt {1+a x}+\tanh ^{-1}\left (\sqrt {1+a x}\right )\right )}{\sqrt {c-a c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.08, size = 69, normalized size = 1.01

method	result	size
default	\(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {c}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}}{\sqrt {c}}\right )+\sqrt {\left (a x +1\right ) c}\right )}{\sqrt {\left (a x +1\right ) c}\, \left (a x -1\right )}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.37, size = 184, normalized size = 2.71 \begin {gather*} \left [\frac {{\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{a x - 1}, -\frac {2 \, {\left ({\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) - \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{a x - 1}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 82, normalized size = 1.21 \begin {gather*} 2 \, {\left (\frac {\arctan \left (\frac {\sqrt {a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {\sqrt {a c x + c}}{c}\right )} {\left | c \right |} - \frac {2 \, {\left (\sqrt {c} {\left | c \right |} \arctan \left (\frac {\sqrt {2} \sqrt {c}}{\sqrt {-c}}\right ) - \sqrt {2} \sqrt {-c} {\left | c \right |}\right )}}{\sqrt {-c} \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(arctan(sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) - sqrt(a*c*x + c)/c)*abs(c) - 2*(sqrt(c)*abs(c)*arctan(sqrt(2)*sq
rt(c)/sqrt(-c)) - sqrt(2)*sqrt(-c)*abs(c))/(sqrt(-c)*sqrt(c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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