\(\int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx\) [515]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx=-\frac {1}{3} \sqrt {5-2 x} \sqrt {2+3 x+x^2}-\frac {8 \sqrt {-2-3 x-x^2} E\left (\arcsin \left (\sqrt {2+x}\right )|\frac {2}{9}\right )}{\sqrt {2+3 x+x^2}}+\frac {7 \sqrt {-2-3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2+x}\right ),\frac {2}{9}\right )}{\sqrt {2+3 x+x^2}} \] Output:

-1/3*(5-2*x)^(1/2)*(x^2+3*x+2)^(1/2)-8*(-x^2-3*x-2)^(1/2)*EllipticE((2+x)^ 
(1/2),1/3*2^(1/2))/(x^2+3*x+2)^(1/2)+7*(-x^2-3*x-2)^(1/2)*EllipticF((2+x)^ 
(1/2),1/3*2^(1/2))/(x^2+3*x+2)^(1/2)
 

Mathematica [A] (verified)

Time = 27.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx=\frac {-\sqrt {5-2 x} \left (22+37 x+17 x^2+2 x^3\right )-24 (5-2 x)^2 \sqrt {\frac {1+x}{-5+2 x}} \sqrt {\frac {2+x}{-5+2 x}} E\left (\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right )|\frac {7}{9}\right )+3 (5-2 x)^2 \sqrt {\frac {1+x}{-5+2 x}} \sqrt {\frac {2+x}{-5+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right ),\frac {7}{9}\right )}{3 (-5+2 x) \sqrt {2+3 x+x^2}} \] Input:

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

Output:

(-(Sqrt[5 - 2*x]*(22 + 37*x + 17*x^2 + 2*x^3)) - 24*(5 - 2*x)^2*Sqrt[(1 + 
x)/(-5 + 2*x)]*Sqrt[(2 + x)/(-5 + 2*x)]*EllipticE[ArcSin[3/Sqrt[5 - 2*x]], 
 7/9] + 3*(5 - 2*x)^2*Sqrt[(1 + x)/(-5 + 2*x)]*Sqrt[(2 + x)/(-5 + 2*x)]*El 
lipticF[ArcSin[3/Sqrt[5 - 2*x]], 7/9])/(3*(-5 + 2*x)*Sqrt[2 + 3*x + x^2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1162, 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.

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

\(\Big \downarrow \) 1162

\(\displaystyle \frac {1}{6} \int \frac {16 x+23}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{6} \left (63 \int \frac {1}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx-8 \int \frac {\sqrt {5-2 x}}{\sqrt {x^2+3 x+2}}dx\right )-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

\(\Big \downarrow \) 1172

\(\displaystyle \frac {1}{6} \left (\frac {42 \sqrt {-x^2-3 x-2} \int \frac {3}{\sqrt {-x-1} \sqrt {9-2 (x+2)}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}-\frac {48 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{3 \sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {126 \sqrt {-x^2-3 x-2} \int \frac {1}{\sqrt {-x-1} \sqrt {9-2 (x+2)}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}-\frac {16 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{\sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{6} \left (\frac {42 \sqrt {-x^2-3 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x+2}\right ),\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}-\frac {16 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{\sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{6} \left (\frac {42 \sqrt {-x^2-3 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x+2}\right ),\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}-\frac {48 \sqrt {-x^2-3 x-2} E\left (\arcsin \left (\sqrt {x+2}\right )|\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}\right )-\frac {1}{3} \sqrt {5-2 x} \sqrt {x^2+3 x+2}\)

Input:

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

Output:

-1/3*(Sqrt[5 - 2*x]*Sqrt[2 + 3*x + x^2]) + ((-48*Sqrt[-2 - 3*x - x^2]*Elli 
pticE[ArcSin[Sqrt[2 + x]], 2/9])/Sqrt[2 + 3*x + x^2] + (42*Sqrt[-2 - 3*x - 
 x^2]*EllipticF[ArcSin[Sqrt[2 + x]], 2/9])/Sqrt[2 + 3*x + x^2])/6
 

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 1162

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p/(e*(m + 2*p + 1))   Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - 
b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x 
] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && 
!ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

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 = 1.38 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\sqrt {x^{2}+3 x +2}\, \sqrt {5-2 x}\, \left (\sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )+8 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )-4 x^{3}-2 x^{2}+22 x +20\right )}{12 x^{3}+6 x^{2}-66 x -60}\)	\(125\)
elliptic	\(\frac {\sqrt {-\left (-5+2 x \right ) \left (x^{2}+3 x +2\right )}\, \left (-\frac {\sqrt {-2 x^{3}-x^{2}+11 x +10}}{3}-\frac {23 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{42 \sqrt {-2 x^{3}-x^{2}+11 x +10}}-\frac {8 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \left (\frac {7 \operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{2}-\operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )\right )}{21 \sqrt {-2 x^{3}-x^{2}+11 x +10}}\right )}{\sqrt {5-2 x}\, \sqrt {x^{2}+3 x +2}}\)	\(186\)
risch	\(\frac {\left (-5+2 x \right ) \sqrt {x^{2}+3 x +2}\, \sqrt {\left (5-2 x \right ) \left (x^{2}+3 x +2\right )}}{3 \sqrt {-\left (-5+2 x \right ) \left (x^{2}+3 x +2\right )}\, \sqrt {5-2 x}}+\frac {\left (-\frac {23 \sqrt {35-14 x}\, \sqrt {2 x +4}\, \sqrt {14 x +14}\, \operatorname {EllipticF}\left (\frac {\sqrt {35-14 x}}{7}, \frac {\sqrt {7}}{3}\right )}{126 \sqrt {-2 x^{3}-x^{2}+11 x +10}}-\frac {8 \sqrt {35-14 x}\, \sqrt {2 x +4}\, \sqrt {14 x +14}\, \left (\frac {9 \operatorname {EllipticE}\left (\frac {\sqrt {35-14 x}}{7}, \frac {\sqrt {7}}{3}\right )}{2}-2 \operatorname {EllipticF}\left (\frac {\sqrt {35-14 x}}{7}, \frac {\sqrt {7}}{3}\right )\right )}{63 \sqrt {-2 x^{3}-x^{2}+11 x +10}}\right ) \sqrt {\left (5-2 x \right ) \left (x^{2}+3 x +2\right )}}{\sqrt {5-2 x}\, \sqrt {x^{2}+3 x +2}}\)	\(224\)

Input:

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

Output:

1/6*(x^2+3*x+2)^(1/2)*(5-2*x)^(1/2)*((5-2*x)^(1/2)*(14*x+14)^(1/2)*(2*x+4) 
^(1/2)*EllipticF(1/3*(5-2*x)^(1/2),3/7*7^(1/2))+8*(5-2*x)^(1/2)*(14*x+14)^ 
(1/2)*(2*x+4)^(1/2)*EllipticE(1/3*(5-2*x)^(1/2),3/7*7^(1/2))-4*x^3-2*x^2+2 
2*x+20)/(2*x^3+x^2-11*x-10)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx=-\frac {61}{18} \, \sqrt {-2} {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right ) + \frac {8}{3} \, \sqrt {-2} {\rm weierstrassZeta}\left (\frac {67}{3}, \frac {440}{27}, {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right )\right ) - \frac {1}{3} \, \sqrt {x^{2} + 3 \, x + 2} \sqrt {-2 \, x + 5} \] Input:

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

Output:

-61/18*sqrt(-2)*weierstrassPInverse(67/3, 440/27, x + 1/6) + 8/3*sqrt(-2)* 
weierstrassZeta(67/3, 440/27, weierstrassPInverse(67/3, 440/27, x + 1/6)) 
- 1/3*sqrt(x^2 + 3*x + 2)*sqrt(-2*x + 5)
 

Sympy [F]

\[ \int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx=\int \frac {\sqrt {\left (x + 1\right ) \left (x + 2\right )}}{\sqrt {5 - 2 x}}\, dx \] Input:

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

Output:

Integral(sqrt((x + 1)*(x + 2))/sqrt(5 - 2*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {2+3 x+x^2}}{\sqrt {5-2 x}} \, dx=-3 \sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}+8 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}\, x^{2}}{2 x^{3}+x^{2}-11 x -10}d x \right )-\frac {37 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}}{2 x^{3}+x^{2}-11 x -10}d x \right )}{2} \] Input:

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

Output:

( - 6*sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2) + 16*int((sqrt( - 2*x + 5)*sqr 
t(x**2 + 3*x + 2)*x**2)/(2*x**3 + x**2 - 11*x - 10),x) - 37*int((sqrt( - 2 
*x + 5)*sqrt(x**2 + 3*x + 2))/(2*x**3 + x**2 - 11*x - 10),x))/2