\(\int \frac {5-x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx\) [1066]

Optimal result

Integrand size = 29, antiderivative size = 107 \[ \int \frac {5-x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {\sqrt {-2-5 x-3 x^2} E\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{\sqrt {3} \sqrt {2+5 x+3 x^2}}+\frac {13 \sqrt {-2-5 x-3 x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {2+5 x+3 x^2}} \] Output:

-1/3*(-3*x^2-5*x-2)^(1/2)*EllipticE((1+x)^(1/2)*3^(1/2),1/3*I*6^(1/2))*3^( 
1/2)/(3*x^2+5*x+2)^(1/2)+13/3*(-3*x^2-5*x-2)^(1/2)*EllipticF((1+x)^(1/2)*3 
^(1/2),1/3*I*6^(1/2))*3^(1/2)/(3*x^2+5*x+2)^(1/2)
 

Mathematica [A] (verified)

Time = 31.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.65 \[ \int \frac {5-x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx=-\frac {20+50 x+30 x^2+5 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{3/2} \sqrt {\frac {2+3 x}{3+2 x}} E\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right )|\frac {3}{5}\right )+34 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{3/2} \sqrt {\frac {2+3 x}{3+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right ),\frac {3}{5}\right )}{15 \sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

-1/15*(20 + 50*x + 30*x^2 + 5*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(3 
/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3 
/5] + 34*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(3/2)*Sqrt[(2 + 3*x)/(3 
 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(Sqrt[3 + 2*x]*S 
qrt[2 + 5*x + 3*x^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1269, 1172, 27, 321, 327}

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.

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

\(\Big \downarrow \) 1269

\(\displaystyle \frac {13}{2} \int \frac {1}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx-\frac {1}{2} \int \frac {\sqrt {2 x+3}}{\sqrt {3 x^2+5 x+2}}dx\)

\(\Big \downarrow \) 1172

\(\displaystyle \frac {13 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {3}}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {3} \sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {13 \sqrt {-3 x^2-5 x-2} \int \frac {1}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3 x^2+5 x+2}}-\frac {\sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {13 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {13 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {-3 x^2-5 x-2} E\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\)

Input:

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

Output:

-((Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(S 
qrt[3]*Sqrt[2 + 5*x + 3*x^2])) + (13*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcS 
in[Sqrt[3]*Sqrt[1 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])
 

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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 1172

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy 
mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 
)/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e 
*Rt[b^2 - 4*a*c, 2])))^m))   Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* 
c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 
2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[m^2, 1/4]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(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] &&  !IGtQ[m, 0]
 

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63

method	result	size
default	\(-\frac {\left (18 \operatorname {EllipticF}\left (\frac {\sqrt {-30 x -20}}{5}, \frac {\sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-30 x -20}}{5}, \frac {\sqrt {10}}{2}\right )\right ) \sqrt {15}\, \sqrt {3+3 x}\, \sqrt {-30 x -20}}{45 \sqrt {3 x^{2}+5 x +2}}\)	\(67\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) \left (2 x +3\right )}\, \left (-\frac {\sqrt {-30 x -20}\, \sqrt {3+3 x}\, \sqrt {30 x +45}\, \operatorname {EllipticF}\left (\frac {\sqrt {-30 x -20}}{5}, \frac {\sqrt {10}}{2}\right )}{3 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}+\frac {\sqrt {-30 x -20}\, \sqrt {3+3 x}\, \sqrt {30 x +45}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {-30 x -20}}{5}, \frac {\sqrt {10}}{2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {-30 x -20}}{5}, \frac {\sqrt {10}}{2}\right )\right )}{15 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}\right )}{\sqrt {2 x +3}\, \sqrt {3 x^{2}+5 x +2}}\)	\(170\)

Input:

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

Output:

-1/45*(18*EllipticF(1/5*(-30*x-20)^(1/2),1/2*10^(1/2))-EllipticE(1/5*(-30* 
x-20)^(1/2),1/2*10^(1/2)))*15^(1/2)*(3+3*x)^(1/2)*(-30*x-20)^(1/2)/(3*x^2+ 
5*x+2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.24 \[ \int \frac {5-x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx=\frac {109}{54} \, \sqrt {6} {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right ) + \frac {1}{3} \, \sqrt {6} {\rm weierstrassZeta}\left (\frac {19}{27}, -\frac {28}{729}, {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right )\right ) \] Input:

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

Output:

109/54*sqrt(6)*weierstrassPInverse(19/27, -28/729, x + 19/18) + 1/3*sqrt(6 
)*weierstrassZeta(19/27, -28/729, weierstrassPInverse(19/27, -28/729, x + 
19/18))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {5-x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx=-\left (\int \frac {\sqrt {2 x +3}\, \sqrt {3 x^{2}+5 x +2}\, x}{6 x^{3}+19 x^{2}+19 x +6}d x \right )+5 \left (\int \frac {\sqrt {2 x +3}\, \sqrt {3 x^{2}+5 x +2}}{6 x^{3}+19 x^{2}+19 x +6}d x \right ) \] Input:

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

Output:

 - int((sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2)*x)/(6*x**3 + 19*x**2 + 19*x + 
 6),x) + 5*int((sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2))/(6*x**3 + 19*x**2 + 
19*x + 6),x)