3.24.64 \(\int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx\) [2364]

Optimal. Leaf size=88 \[ \sqrt {-2-3 x+5 x^2}+\sqrt {2} \tan ^{-1}\left (\frac {4+3 x}{2 \sqrt {2} \sqrt {-2-3 x+5 x^2}}\right )+\frac {3 \tanh ^{-1}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {-2-3 x+5 x^2}}\right )}{2 \sqrt {5}} \]

[Out]

arctan(1/4*(4+3*x)*2^(1/2)/(5*x^2-3*x-2)^(1/2))*2^(1/2)+3/10*arctanh(1/10*(3-10*x)*5^(1/2)/(5*x^2-3*x-2)^(1/2)
)*5^(1/2)+(5*x^2-3*x-2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {748, 857, 635, 212, 738, 210} \begin {gather*} \sqrt {2} \text {ArcTan}\left (\frac {3 x+4}{2 \sqrt {2} \sqrt {5 x^2-3 x-2}}\right )+\sqrt {5 x^2-3 x-2}+\frac {3 \tanh ^{-1}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {5 x^2-3 x-2}}\right )}{2 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-2 - 3*x + 5*x^2]/x,x]

[Out]

Sqrt[-2 - 3*x + 5*x^2] + Sqrt[2]*ArcTan[(4 + 3*x)/(2*Sqrt[2]*Sqrt[-2 - 3*x + 5*x^2])] + (3*ArcTanh[(3 - 10*x)/
(2*Sqrt[5]*Sqrt[-2 - 3*x + 5*x^2])])/(2*Sqrt[5])

Rule 210

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

Rule 212

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

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx &=\sqrt {-2-3 x+5 x^2}-\frac {1}{2} \int \frac {4+3 x}{x \sqrt {-2-3 x+5 x^2}} \, dx\\ &=\sqrt {-2-3 x+5 x^2}-\frac {3}{2} \int \frac {1}{\sqrt {-2-3 x+5 x^2}} \, dx-2 \int \frac {1}{x \sqrt {-2-3 x+5 x^2}} \, dx\\ &=\sqrt {-2-3 x+5 x^2}-3 \text {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-3+10 x}{\sqrt {-2-3 x+5 x^2}}\right )+4 \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,\frac {-4-3 x}{\sqrt {-2-3 x+5 x^2}}\right )\\ &=\sqrt {-2-3 x+5 x^2}+\sqrt {2} \tan ^{-1}\left (\frac {4+3 x}{2 \sqrt {2} \sqrt {-2-3 x+5 x^2}}\right )+\frac {3 \tanh ^{-1}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {-2-3 x+5 x^2}}\right )}{2 \sqrt {5}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 83, normalized size = 0.94 \begin {gather*} \sqrt {-2-3 x+5 x^2}+2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {-2-3 x+5 x^2}}{\sqrt {2} (-1+x)}\right )-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {5} \sqrt {-2-3 x+5 x^2}}{2+5 x}\right )}{\sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-2 - 3*x + 5*x^2]/x,x]

[Out]

Sqrt[-2 - 3*x + 5*x^2] + 2*Sqrt[2]*ArcTan[Sqrt[-2 - 3*x + 5*x^2]/(Sqrt[2]*(-1 + x))] - (3*ArcTanh[(Sqrt[5]*Sqr
t[-2 - 3*x + 5*x^2])/(2 + 5*x)])/Sqrt[5]

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Maple [A]
time = 0.69, size = 71, normalized size = 0.81

method	result	size
default	\(\sqrt {5 x^{2}-3 x -2}-\frac {3 \ln \left (\frac {\left (-\frac {3}{2}+5 x \right ) \sqrt {5}}{5}+\sqrt {5 x^{2}-3 x -2}\right ) \sqrt {5}}{10}-\sqrt {2}\, \arctan \left (\frac {\left (-3 x -4\right ) \sqrt {2}}{4 \sqrt {5 x^{2}-3 x -2}}\right )\)	\(71\)
trager	\(\sqrt {5 x^{2}-3 x -2}+\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {3 \RootOf \left (\textit {\_Z}^{2}+2\right ) x +4 \RootOf \left (\textit {\_Z}^{2}+2\right )-4 \sqrt {5 x^{2}-3 x -2}}{x}\right )+\frac {3 \RootOf \left (\textit {\_Z}^{2}-5\right ) \ln \left (-10 \RootOf \left (\textit {\_Z}^{2}-5\right ) x +3 \RootOf \left (\textit {\_Z}^{2}-5\right )+10 \sqrt {5 x^{2}-3 x -2}\right )}{10}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2-3*x-2)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

(5*x^2-3*x-2)^(1/2)-3/10*ln(1/5*(-3/2+5*x)*5^(1/2)+(5*x^2-3*x-2)^(1/2))*5^(1/2)-2^(1/2)*arctan(1/4*(-3*x-4)*2^
(1/2)/(5*x^2-3*x-2)^(1/2))

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Maxima [A]
time = 0.51, size = 60, normalized size = 0.68 \begin {gather*} \sqrt {2} \arcsin \left (\frac {3 \, x}{7 \, {\left | x \right |}} + \frac {4}{7 \, {\left | x \right |}}\right ) - \frac {3}{10} \, \sqrt {5} \log \left (2 \, \sqrt {5} \sqrt {5 \, x^{2} - 3 \, x - 2} + 10 \, x - 3\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2-3*x-2)^(1/2)/x,x, algorithm="maxima")

[Out]

sqrt(2)*arcsin(3/7*x/abs(x) + 4/7/abs(x)) - 3/10*sqrt(5)*log(2*sqrt(5)*sqrt(5*x^2 - 3*x - 2) + 10*x - 3) + sqr
t(5*x^2 - 3*x - 2)

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Fricas [A]
time = 3.03, size = 78, normalized size = 0.89 \begin {gather*} \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (3 \, x + 4\right )}}{4 \, \sqrt {5 \, x^{2} - 3 \, x - 2}}\right ) + \frac {3}{20} \, \sqrt {5} \log \left (-4 \, \sqrt {5} \sqrt {5 \, x^{2} - 3 \, x - 2} {\left (10 \, x - 3\right )} + 200 \, x^{2} - 120 \, x - 31\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2-3*x-2)^(1/2)/x,x, algorithm="fricas")

[Out]

sqrt(2)*arctan(1/4*sqrt(2)*(3*x + 4)/sqrt(5*x^2 - 3*x - 2)) + 3/20*sqrt(5)*log(-4*sqrt(5)*sqrt(5*x^2 - 3*x - 2
)*(10*x - 3) + 200*x^2 - 120*x - 31) + sqrt(5*x^2 - 3*x - 2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (x - 1\right ) \left (5 x + 2\right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2-3*x-2)**(1/2)/x,x)

[Out]

Integral(sqrt((x - 1)*(5*x + 2))/x, x)

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Giac [A]
time = 5.22, size = 77, normalized size = 0.88 \begin {gather*} -2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {5} x - \sqrt {5 \, x^{2} - 3 \, x - 2}\right )}\right ) + \frac {3}{10} \, \sqrt {5} \log \left ({\left | -10 \, \sqrt {5} x + 3 \, \sqrt {5} + 10 \, \sqrt {5 \, x^{2} - 3 \, x - 2} \right |}\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2-3*x-2)^(1/2)/x,x, algorithm="giac")

[Out]

-2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(5)*x - sqrt(5*x^2 - 3*x - 2))) + 3/10*sqrt(5)*log(abs(-10*sqrt(5)*x + 3*s
qrt(5) + 10*sqrt(5*x^2 - 3*x - 2))) + sqrt(5*x^2 - 3*x - 2)

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Mupad [B]
time = 0.21, size = 77, normalized size = 0.88 \begin {gather*} \sqrt {5\,x^2-3\,x-2}-\frac {3\,\sqrt {5}\,\ln \left (\sqrt {5\,x^2-3\,x-2}+\frac {\sqrt {5}\,\left (5\,x-\frac {3}{2}\right )}{5}\right )}{10}-\sqrt {2}\,\ln \left (-\frac {2}{x}-\frac {3}{2}+\frac {\sqrt {2}\,\sqrt {5\,x^2-3\,x-2}\,1{}\mathrm {i}}{x}\right )\,1{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2 - 3*x - 2)^(1/2)/x,x)

[Out]

(5*x^2 - 3*x - 2)^(1/2) - (3*5^(1/2)*log((5*x^2 - 3*x - 2)^(1/2) + (5^(1/2)*(5*x - 3/2))/5))/10 - 2^(1/2)*log(
(2^(1/2)*(5*x^2 - 3*x - 2)^(1/2)*1i)/x - 2/x - 3/2)*1i

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