\(\int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx\) [583]

Optimal result

Integrand size = 29, antiderivative size = 45 \[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=-\frac {\sqrt {-2-3 x-x^2} \operatorname {EllipticPi}\left (\frac {3}{2},\arcsin \left (\sqrt {2+x}\right ),\frac {2}{9}\right )}{3 \sqrt {2+3 x+x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 21.70 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.93 \[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=-\frac {4 (1+x) \sqrt {\frac {2+x}{-5+2 x}} \left (\operatorname {EllipticF}\left (\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right ),\frac {7}{9}\right )-\operatorname {EllipticPi}\left (\frac {23}{27},\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right ),\frac {7}{9}\right )\right )}{69 \sqrt {\frac {1+x}{-5+2 x}} \sqrt {2+3 x+x^2}} \] Input:

Integrate[1/(Sqrt[5 - 2*x]*(4 + 3*x)*Sqrt[2 + 3*x + x^2]),x]
 

Output:

(-4*(1 + x)*Sqrt[(2 + x)/(-5 + 2*x)]*(EllipticF[ArcSin[3/Sqrt[5 - 2*x]], 7 
/9] - EllipticPi[23/27, ArcSin[3/Sqrt[5 - 2*x]], 7/9]))/(69*Sqrt[(1 + x)/( 
-5 + 2*x)]*Sqrt[2 + 3*x + x^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1279, 27, 186, 27, 412}

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.

\(\displaystyle \int \frac {1}{\sqrt {5-2 x} (3 x+4) \sqrt {x^2+3 x+2}} \, dx\)

\(\Big \downarrow \) 1279

\(\displaystyle \frac {2 \sqrt {x+1} \sqrt {x+2} \int \frac {1}{2 \sqrt {5-2 x} \sqrt {x+1} \sqrt {x+2} (3 x+4)}dx}{\sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x+2} \int \frac {1}{\sqrt {5-2 x} \sqrt {x+1} \sqrt {x+2} (3 x+4)}dx}{\sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 186

\(\displaystyle -\frac {2 \sqrt {x+1} \sqrt {x+2} \int \frac {2}{(23-3 (5-2 x)) \sqrt {2 x+2} \sqrt {2 x+4}}d\sqrt {5-2 x}}{\sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {4 \sqrt {x+1} \sqrt {x+2} \int \frac {1}{(23-3 (5-2 x)) \sqrt {2 x+2} \sqrt {2 x+4}}d\sqrt {5-2 x}}{\sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 412

\(\displaystyle -\frac {4 \sqrt {x+1} \sqrt {x+2} \operatorname {EllipticPi}\left (\frac {27}{23},\arcsin \left (\frac {1}{3} \sqrt {5-2 x}\right ),\frac {9}{7}\right )}{23 \sqrt {7} \sqrt {x^2+3 x+2}}\)

Input:

Int[1/(Sqrt[5 - 2*x]*(4 + 3*x)*Sqrt[2 + 3*x + x^2]),x]
 

Output:

(-4*Sqrt[1 + x]*Sqrt[2 + x]*EllipticPi[27/23, ArcSin[Sqrt[5 - 2*x]/3], 9/7 
])/(23*Sqrt[7]*Sqrt[2 + 3*x + 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 186

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_ 
)]*Sqrt[(g_.) + (h_.)*(x_)]), x_] :> Simp[-2   Subst[Int[1/(Simp[b*c - a*d 
- b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + f*(x^2/d), x]]*Sqrt[Simp[(d*g - c*h)/ 
d + h*(x^2/d), x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, 
g, h}, x] && GtQ[(d*e - c*f)/d, 0]
 

rule 412

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* 
(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && S 
implerSqrtQ[-f/e, -d/c])
 

rule 1279

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[b 
 - q + 2*c*x]*(Sqrt[b + q + 2*c*x]/Sqrt[a + b*x + c*x^2])   Int[1/((d + e*x 
)*Sqrt[f + g*x]*Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x]), x], x]] /; FreeQ[ 
{a, b, c, d, e, f, g}, x]
 

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {2 \operatorname {EllipticPi}\left (\frac {\sqrt {5-2 x}}{3}, \frac {27}{23}, \frac {3 \sqrt {7}}{7}\right ) \sqrt {2 x +4}\, \sqrt {14 x +14}}{161 \sqrt {x^{2}+3 x +2}}\)	\(43\)
elliptic	\(-\frac {2 \sqrt {-\left (-5+2 x \right ) \left (x^{2}+3 x +2\right )}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticPi}\left (\frac {\sqrt {5-2 x}}{3}, \frac {27}{23}, \frac {3 \sqrt {7}}{7}\right )}{161 \sqrt {x^{2}+3 x +2}\, \sqrt {-2 x^{3}-x^{2}+11 x +10}}\)	\(77\)

Input:

int(1/(5-2*x)^(1/2)/(3*x+4)/(x^2+3*x+2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/161*EllipticPi(1/3*(5-2*x)^(1/2),27/23,3/7*7^(1/2))*(2*x+4)^(1/2)*(14*x 
+14)^(1/2)/(x^2+3*x+2)^(1/2)
 

Fricas [F]

\[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} + 3 \, x + 2} {\left (3 \, x + 4\right )} \sqrt {-2 \, x + 5}} \,d x } \] Input:

integrate(1/(5-2*x)^(1/2)/(4+3*x)/(x^2+3*x+2)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(x^2 + 3*x + 2)*sqrt(-2*x + 5)/(6*x^4 + 11*x^3 - 29*x^2 - 74 
*x - 40), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=\int \frac {1}{\sqrt {\left (x + 1\right ) \left (x + 2\right )} \sqrt {5 - 2 x} \left (3 x + 4\right )}\, dx \] Input:

integrate(1/(5-2*x)**(1/2)/(4+3*x)/(x**2+3*x+2)**(1/2),x)
 

Output:

Integral(1/(sqrt((x + 1)*(x + 2))*sqrt(5 - 2*x)*(3*x + 4)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} + 3 \, x + 2} {\left (3 \, x + 4\right )} \sqrt {-2 \, x + 5}} \,d x } \] Input:

integrate(1/(5-2*x)^(1/2)/(4+3*x)/(x^2+3*x+2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(x^2 + 3*x + 2)*(3*x + 4)*sqrt(-2*x + 5)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} + 3 \, x + 2} {\left (3 \, x + 4\right )} \sqrt {-2 \, x + 5}} \,d x } \] Input:

integrate(1/(5-2*x)^(1/2)/(4+3*x)/(x^2+3*x+2)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(x^2 + 3*x + 2)*(3*x + 4)*sqrt(-2*x + 5)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=\int \frac {1}{\sqrt {5-2\,x}\,\left (3\,x+4\right )\,\sqrt {x^2+3\,x+2}} \,d x \] Input:

int(1/((5 - 2*x)^(1/2)*(3*x + 4)*(3*x + x^2 + 2)^(1/2)),x)
 

Output:

int(1/((5 - 2*x)^(1/2)*(3*x + 4)*(3*x + x^2 + 2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {5-2 x} (4+3 x) \sqrt {2+3 x+x^2}} \, dx=-\left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}}{6 x^{4}+11 x^{3}-29 x^{2}-74 x -40}d x \right ) \] Input:

int(1/(5-2*x)^(1/2)/(4+3*x)/(x^2+3*x+2)^(1/2),x)
 

Output:

 - int((sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2))/(6*x**4 + 11*x**3 - 29*x**2 
 - 74*x - 40),x)