\(\int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\) [205]

Optimal result

Integrand size = 25, antiderivative size = 125 \[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {10 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {10 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {1+x} \sqrt {2+3 x}}+\frac {2 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \] Output:

-10/3*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)+10/3*2^(1/2)*(3*x^2+5*x+2)^(1/2) 
*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))/(1+x)^(1/2)/(2+3*x)^(1/2)+2* 
2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2)*InverseJacobiAM(arctan(x^(1/2)),1/2*I*2^ 
(1/2))/(3*x^2+5*x+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.20 \[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {2 x^{3/2} \left (5 \left (3+\frac {2}{x^2}+\frac {5}{x}\right )+\frac {5 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )}{\sqrt {x}}-\frac {8 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{\sqrt {x}}\right )}{3 \sqrt {2+5 x+3 x^2}} \] Input:

Integrate[(2 - 5*x)/(Sqrt[x]*Sqrt[2 + 5*x + 3*x^2]),x]
 

Output:

(-2*x^(3/2)*(5*(3 + 2/x^2 + 5/x) + ((5*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 
+ 2/x]*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/Sqrt[x] - ((8*I)*Sqrt 
[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 
 3/2])/Sqrt[x]))/(3*Sqrt[2 + 5*x + 3*x^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1240, 1503, 1413, 1456}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {2-5 x}{\sqrt {x} \sqrt {3 x^2+5 x+2}} \, dx\)

\(\Big \downarrow \) 1240

\(\displaystyle 2 \int \frac {2-5 x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\)

\(\Big \downarrow \) 1503

\(\displaystyle 2 \left (2 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-5 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\)

\(\Big \downarrow \) 1413

\(\displaystyle 2 \left (\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}-5 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\)

\(\Big \downarrow \) 1456

\(\displaystyle 2 \left (\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}-5 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\)

Input:

Int[(2 - 5*x)/(Sqrt[x]*Sqrt[2 + 5*x + 3*x^2]),x]
 

Output:

2*(-5*((Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sq 
rt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 
3*x^2])) + (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[ 
x]], -1/2])/Sqrt[2 + 5*x + 3*x^2])
 

Defintions of rubi rules used

rule 1240

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), 
x_Symbol] :> Simp[2   Subst[Int[(f + g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, 
 Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x]
 

rule 1413

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q)*x^2)/(2*a + 
(b - q)*x^2)]/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF 
[ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; 
 FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1456

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[x*((b - q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 
])), x] - Simp[Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q 
)*x^2)/(2*a + (b - q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan 
[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[e   Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) 
/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \left (21 \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-5 \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{9 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(77\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (\frac {2 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {5 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(159\)

Input:

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

Output:

1/9/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^( 
1/2)*(21*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-5*EllipticE(1/2*(6*x+4)^(1 
/2),I*2^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.21 \[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=\frac {86}{27} \, \sqrt {3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) + \frac {10}{3} \, \sqrt {3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) \] Input:

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

Output:

86/27*sqrt(3)*weierstrassPInverse(28/27, 80/729, x + 5/9) + 10/3*sqrt(3)*w 
eierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9))
 

Sympy [F]

\[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=- \int \left (- \frac {2}{\sqrt {x} \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 \sqrt {x}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx \] Input:

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

Output:

-Integral(-2/(sqrt(x)*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5*sqrt(x)/sqr 
t(3*x**2 + 5*x + 2), x)
 

Maxima [F]

\[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {5 \, x - 2}{\sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}} \,d x } \] Input:

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

Output:

-integrate((5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*sqrt(x)), x)
 

Giac [F]

\[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=\int { -\frac {5 \, x - 2}{\sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}} \,d x } \] Input:

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

Output:

integrate(-(5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*sqrt(x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {5\,x-2}{\sqrt {x}\,\sqrt {3\,x^2+5\,x+2}} \,d x \] Input:

int(-(5*x - 2)/(x^(1/2)*(5*x + 3*x^2 + 2)^(1/2)),x)
 

Output:

-int((5*x - 2)/(x^(1/2)*(5*x + 3*x^2 + 2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {2-5 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx=2 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{3}+5 x^{2}+2 x}d x \right )-5 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{2}+5 x +2}d x \right ) \] Input:

int((2-5*x)/x^(1/2)/(3*x^2+5*x+2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

2*int((sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**3 + 5*x**2 + 2*x),x) - 5*int( 
(sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**2 + 5*x + 2),x)