\(\int \frac {2-5 x}{\sqrt {x} (2+5 x+3 x^2)^{3/2}} \, dx\) [213]

Optimal result

Integrand size = 25, antiderivative size = 147 \[ \int \frac {2-5 x}{\sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {30 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}+\frac {2 \sqrt {x} (38+45 x)}{\sqrt {2+5 x+3 x^2}}+\frac {30 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {1+x} \sqrt {2+3 x}}-\frac {37 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \] Output:

-30*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)+2*x^(1/2)*(38+45*x)/(3*x^2+5*x+2)^ 
(1/2)+30*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)-37*2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2)*Inv 
erseJacobiAM(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 = 21.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \frac {2-5 x}{\sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {-60-74 x-30 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-7 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

(-60 - 74*x - (30*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*Ellipt 
icE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (7*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sq 
rt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(Sqrt[x] 
*Sqrt[2 + 5*x + 3*x^2])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 161, 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 = {1235, 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.

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

\(\Big \downarrow \) 1235

\(\displaystyle \frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-\int \frac {45 x+37}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\)

\(\Big \downarrow \) 1240

\(\displaystyle \frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-2 \int \frac {45 x+37}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-2 \left (37 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+45 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-2 \left (45 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {37 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-2 \left (\frac {37 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+45 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\)

Input:

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

Output:

(2*Sqrt[x]*(38 + 45*x))/Sqrt[2 + 5*x + 3*x^2] - 2*(45*((Sqrt[x]*(2 + 3*x)) 
/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*Elli 
pticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (37*(1 + x)*Sqr 
t[(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 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a 
+ b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] 
 + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^m 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -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 = 0.99 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {8 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-15 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+270 x^{2}+228 x}{3 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(107\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {2 x \left (-\frac {38}{3}-15 x \right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {37 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {15 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(182\)

Input:

int((2-5*x)/x^(1/2)/(3*x^2+5*x+2)^(3/2),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))-15*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ellip 
ticE(1/2*(6*x+4)^(1/2),I*2^(1/2))+270*x^2+228*x)/x^(1/2)/(3*x^2+5*x+2)^(1/ 
2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.56 \[ \int \frac {2-5 x}{\sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 15 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (45 \, x + 38\right )} \sqrt {x}\right )}}{3 \, x^{2} + 5 \, x + 2} \] Input:

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

Output:

-2*(4*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassPInverse(28/27, 80/729, x + 5/9 
) - 15*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassZeta(28/27, 80/729, weierstras 
sPInverse(28/27, 80/729, x + 5/9)) - sqrt(3*x^2 + 5*x + 2)*(45*x + 38)*sqr 
t(x))/(3*x^2 + 5*x + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {2-5 x}{\sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=2 \left (\int \frac {\sqrt {3 x^{2}+5 x +2}}{9 \sqrt {x}\, x^{4}+30 \sqrt {x}\, x^{3}+37 \sqrt {x}\, x^{2}+20 \sqrt {x}\, x +4 \sqrt {x}}d x \right )-5 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{9 x^{4}+30 x^{3}+37 x^{2}+20 x +4}d x \right ) \] Input:

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

Output:

2*int(sqrt(3*x**2 + 5*x + 2)/(9*sqrt(x)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x 
)*x**2 + 20*sqrt(x)*x + 4*sqrt(x)),x) - 5*int((sqrt(x)*sqrt(3*x**2 + 5*x + 
 2))/(9*x**4 + 30*x**3 + 37*x**2 + 20*x + 4),x)