\(\int \frac {2-5 x}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx\) [206]

Optimal result

Integrand size = 25, antiderivative size = 142 \[ \int \frac {2-5 x}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx=\frac {2 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}-\frac {2 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {1+x} \sqrt {2+3 x}}-\frac {5 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \] Output:

2*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)-2*(3*x^2+5*x+2)^(1/2)/x^(1/2)-2*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)-5*2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2)*InverseJacobiAM(a 
rctan(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 = 21.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.63 \[ \int \frac {2-5 x}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx=\frac {i \sqrt {2+\frac {2}{x}} \sqrt {3+\frac {2}{x}} x \left (2 E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-7 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )\right )}{\sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

(I*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x*(2*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]] 
, 3/2] - 7*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2]))/Sqrt[2 + 5*x + 3 
*x^2]
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1237, 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.

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

\(\Big \downarrow \) 1237

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

\(\Big \downarrow \) 1240

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

\(\Big \downarrow \) 1503

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

\(\Big \downarrow \) 1413

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

\(\Big \downarrow \) 1456

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1237

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* 
x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) 
*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ 
(c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m 
+ 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 
] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 = 1.02 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {8 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-\sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+18 x^{2}+30 x +12}{3 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(108\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {2 \left (3 x^{2}+5 x +2\right )}{\sqrt {\left (3 x^{2}+5 x +2\right ) x}}-\frac {5 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {\sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(184\)

Input:

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

Output:

-1/3*(8*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+ 
4)^(1/2),I*2^(1/2))-(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ellipti 
cE(1/2*(6*x+4)^(1/2),I*2^(1/2))+18*x^2+30*x+12)/x^(1/2)/(3*x^2+5*x+2)^(1/2 
)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.35 \[ \int \frac {2-5 x}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {2 \, {\left (20 \, \sqrt {3} x {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) + 9 \, \sqrt {3} x {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) + 9 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{9 \, x} \] Input:

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

Output:

-2/9*(20*sqrt(3)*x*weierstrassPInverse(28/27, 80/729, x + 5/9) + 9*sqrt(3) 
*x*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5 
/9)) + 9*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/x
 

Sympy [F]

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

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

Output:

-Integral(-2/(x**(3/2)*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5/(sqrt(x)*s 
qrt(3*x**2 + 5*x + 2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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