3.491 \(\int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx\)

Optimal. Leaf size=124 \[ \frac {c x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 \sqrt {c-\frac {c}{a x}}}+\frac {c x \sqrt {1-\frac {1}{a^2 x^2}}}{4 a \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{4 a^2} \]

[Out]

-1/4*arctanh(c^(1/2)*(1-1/a^2/x^2)^(1/2)/(c-c/a/x)^(1/2))*c^(1/2)/a^2+1/4*c*x*(1-1/a^2/x^2)^(1/2)/a/(c-c/a/x)^
(1/2)+1/2*c*x^2*(1-1/a^2/x^2)^(1/2)/(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6178, 863, 873, 875, 208} \[ \frac {c x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 \sqrt {c-\frac {c}{a x}}}+\frac {c x \sqrt {1-\frac {1}{a^2 x^2}}}{4 a \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[1 - 1/(a^2*x^2)]*x)/(4*a*Sqrt[c - c/(a*x)]) + (c*Sqrt[1 - 1/(a^2*x^2)]*x^2)/(2*Sqrt[c - c/(a*x)]) - (S
qrt[c]*ArcTanh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(4*a^2)

Rule 208

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

Rule 863

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

Rule 873

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

Rule 875

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 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx &=-\left (c \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{x^3 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )\right )\\ &=\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 \sqrt {c-\frac {c}{a x}}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{4 a}\\ &=\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x}{4 a \sqrt {c-\frac {c}{a x}}}+\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 \sqrt {c-\frac {c}{a x}}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{8 a^2}\\ &=\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x}{4 a \sqrt {c-\frac {c}{a x}}}+\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 \sqrt {c-\frac {c}{a x}}}+\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{a^2}+\frac {c^2 x^2}{a^2}} \, dx,x,\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{4 a^4}\\ &=\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x}{4 a \sqrt {c-\frac {c}{a x}}}+\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{4 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 148, normalized size = 1.19 \[ \frac {2 a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}} (2 a x+1) \sqrt {c-\frac {c}{a x}}+\sqrt {c} (1-a x) \log \left (2 a^2 \sqrt {c} x^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+c \left (2 a^2 x^2-a x-1\right )\right )+\sqrt {c} (a x-1) \log (1-a x)}{8 a^2 (a x-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*x^2*(1 + 2*a*x) + Sqrt[c]*(-1 + a*x)*Log[1 - a*x] + Sqrt[c]*(1
- a*x)*Log[2*a^2*Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*x^2 + c*(-1 - a*x + 2*a^2*x^2)])/(8*a^2*(-1 +
a*x))

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fricas [A]  time = 0.67, size = 317, normalized size = 2.56 \[ \left [\frac {{\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{16 \, {\left (a^{3} x - a^{2}\right )}}, \frac {{\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{8 \, {\left (a^{3} x - a^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((a*x - 1)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x - 1)/
(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt((a*x - 1)/(a*x + 1))
*sqrt((a*c*x - c)/(a*x)))/(a^3*x - a^2), 1/8*((a*x - 1)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x -
1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt((a*x
- 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a^3*x - a^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} x}{\sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 102, normalized size = 0.82 \[ -\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (-4 a^{\frac {3}{2}} x \sqrt {\left (a x +1\right ) x}-2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+\ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right )\right )}{8 \sqrt {\frac {a x -1}{a x +1}}\, a^{\frac {3}{2}} \sqrt {\left (a x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/((a*x-1)/(a*x+1))^(1/2)*(c*(a*x-1)/a/x)^(1/2)*x/a^(3/2)*(-4*a^(3/2)*x*((a*x+1)*x)^(1/2)-2*((a*x+1)*x)^(1/
2)*a^(1/2)+ln(1/2*(2*((a*x+1)*x)^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2)))/((a*x+1)*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} x}{\sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\sqrt {c-\frac {c}{a\,x}}}{\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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