3.11.62 \(\int \frac {2-5 x}{x^{5/2} \sqrt {2+5 x+3 x^2}} \, dx\) [1062]

3.11.62.1 Optimal result
3.11.62.2 Mathematica [C] (verified)
3.11.62.3 Rubi [A] (verified)
3.11.62.4 Maple [A] (verified)
3.11.62.5 Fricas [C] (verification not implemented)
3.11.62.6 Sympy [F]
3.11.62.7 Maxima [F]
3.11.62.8 Giac [F]
3.11.62.9 Mupad [F(-1)]

3.11.62.1 Optimal result

Integrand size = 25, antiderivative size = 175 \[ \int \frac {2-5 x}{x^{5/2} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {25 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {2 \sqrt {2+5 x+3 x^2}}{3 x^{3/2}}+\frac {25 \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}+\frac {25 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}}-\frac {\sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]

output
-25/3*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+25/3*(1+x)^(3/2)*(1/(1+x))^(1/2) 
*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2 
)/(3*x^2+5*x+2)^(1/2)-(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^ 
(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-2/3 
*(3*x^2+5*x+2)^(1/2)/x^(3/2)+25/3*(3*x^2+5*x+2)^(1/2)/x^(1/2)
 
3.11.62.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.85 \[ \int \frac {2-5 x}{x^{5/2} \sqrt {2+5 x+3 x^2}} \, dx=\frac {-2 \left (2+5 x+3 x^2\right )-25 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+22 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{3 x^{3/2} \sqrt {2+5 x+3 x^2}} \]

input
Integrate[(2 - 5*x)/(x^(5/2)*Sqrt[2 + 5*x + 3*x^2]),x]
 
output
(-2*(2 + 5*x + 3*x^2) - (25*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5 
/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (22*I)*Sqrt[2]*Sqrt[1 + 
 x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2 
])/(3*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])
 
3.11.62.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1237, 1237, 27, 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.

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

\(\Big \downarrow \) 1237

\(\displaystyle -\frac {1}{3} \int \frac {3 x+25}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 1237

\(\displaystyle \frac {1}{3} \left (\int -\frac {3 (25 x+2)}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {3}{2} \int \frac {25 x+2}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 1240

\(\displaystyle \frac {1}{3} \left (\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \int \frac {25 x+2}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{3} \left (\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (2 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+25 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {1}{3} \left (\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (25 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {1}{3} \left (\frac {25 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \left (\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+25 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )-\frac {2 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\)

input
Int[(2 - 5*x)/(x^(5/2)*Sqrt[2 + 5*x + 3*x^2]),x]
 
output
(-2*Sqrt[2 + 5*x + 3*x^2])/(3*x^(3/2)) + ((25*Sqrt[2 + 5*x + 3*x^2])/Sqrt[ 
x] - 3*(25*((Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + 
x)*Sqrt[(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]))/3
 

3.11.62.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 
3.11.62.4 Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.66

method result size
default \(\frac {69 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -25 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x +450 x^{3}+714 x^{2}+240 x -24}{18 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {3}{2}}}\) \(115\)
risch \(\frac {75 x^{3}+119 x^{2}+40 x -4}{3 x^{\frac {3}{2}} \sqrt {3 x^{2}+5 x +2}}-\frac {\left (\frac {\sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {25 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{6 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {x \left (3 x^{2}+5 x +2\right )}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) \(193\)
elliptic \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {2 \sqrt {3 x^{3}+5 x^{2}+2 x}}{3 x^{2}}+\frac {25 x^{2}+\frac {125}{3} x +\frac {50}{3}}{\sqrt {x \left (3 x^{2}+5 x +2\right )}}-\frac {\sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {25 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{6 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) \(206\)

input
int((2-5*x)/x^(5/2)/(3*x^2+5*x+2)^(1/2),x,method=_RETURNVERBOSE)
 
output
1/18*(69*(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))*x-25*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*E 
llipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x+450*x^3+714*x^2+240*x-24)/(3*x^2+5 
*x+2)^(1/2)/x^(3/2)
 
3.11.62.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.34 \[ \int \frac {2-5 x}{x^{5/2} \sqrt {2+5 x+3 x^2}} \, dx=\frac {107 \, \sqrt {3} x^{2} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) + 225 \, \sqrt {3} x^{2} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) + 9 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (25 \, x - 2\right )} \sqrt {x}}{27 \, x^{2}} \]

input
integrate((2-5*x)/x^(5/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")
 
output
1/27*(107*sqrt(3)*x^2*weierstrassPInverse(28/27, 80/729, x + 5/9) + 225*sq 
rt(3)*x^2*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729 
, x + 5/9)) + 9*sqrt(3*x^2 + 5*x + 2)*(25*x - 2)*sqrt(x))/x^2
 
3.11.62.6 Sympy [F]

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

input
integrate((2-5*x)/x**(5/2)/(3*x**2+5*x+2)**(1/2),x)
 
output
-Integral(-2/(x**(5/2)*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5/(x**(3/2)* 
sqrt(3*x**2 + 5*x + 2)), x)
 
3.11.62.7 Maxima [F]

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

input
integrate((2-5*x)/x^(5/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")
 
output
-integrate((5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*x^(5/2)), x)
 
3.11.62.8 Giac [F]

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

input
integrate((2-5*x)/x^(5/2)/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")
 
output
integrate(-(5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*x^(5/2)), x)
 
3.11.62.9 Mupad [F(-1)]

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

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