\(\int \frac {1}{(5+x) \sqrt {2 x+3 x^2+x^3}} \, dx\) [7]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int \frac {1}{(5+x) \sqrt {2 x+3 x^2+x^3}} \, dx=\frac {\sqrt {2} \sqrt {-x} \sqrt {-2-3 x-x^2} \operatorname {EllipticPi}\left (-\frac {1}{3},\arcsin \left (\sqrt {2+x}\right ),\frac {1}{2}\right )}{3 \sqrt {2 x+3 x^2+x^3}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 22.54 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(5+x) \sqrt {2 x+3 x^2+x^3}} \, dx=\frac {i \sqrt {2+\frac {2}{x}} x^{3/2} \sqrt {\frac {2+x}{x}} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {2}}{\sqrt {x}}\right ),\frac {1}{2}\right )-\operatorname {EllipticPi}\left (\frac {5}{2},i \text {arcsinh}\left (\frac {\sqrt {2}}{\sqrt {x}}\right ),\frac {1}{2}\right )\right )}{5 \sqrt {x \left (2+3 x+x^2\right )}} \] Input:

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

Output:

((I/5)*Sqrt[2 + 2/x]*x^(3/2)*Sqrt[(2 + x)/x]*(EllipticF[I*ArcSinh[Sqrt[2]/ 
Sqrt[x]], 1/2] - EllipticPi[5/2, I*ArcSinh[Sqrt[2]/Sqrt[x]], 1/2]))/Sqrt[x 
*(2 + 3*x + x^2)]
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(133\) vs. \(2(61)=122\).

Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2467, 1279, 27, 186, 25, 411, 320, 414}

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 1279

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 186

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 411

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

\(\Big \downarrow \) 320

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x+1} \sqrt {x+2} \left (\frac {1}{4} \int \frac {\sqrt {x+1}}{\sqrt {x+2} (x+5)}d\sqrt {x}-\frac {\sqrt {x+2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt {x+1} \sqrt {\frac {x+2}{x+1}}}\right )}{\sqrt {x^3+3 x^2+2 x}}\)

\(\Big \downarrow \) 414

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x+1} \sqrt {x+2} \left (\frac {\sqrt {x+2} \operatorname {EllipticPi}\left (\frac {4}{5},\arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{20 \sqrt {2} \sqrt {x+1} \sqrt {\frac {x+2}{x+1}}}-\frac {\sqrt {x+2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt {x+1} \sqrt {\frac {x+2}{x+1}}}\right )}{\sqrt {x^3+3 x^2+2 x}}\)

Input:

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

Output:

(-2*Sqrt[x]*Sqrt[1 + x]*Sqrt[2 + x]*(-1/4*(Sqrt[2 + x]*EllipticF[ArcTan[Sq 
rt[x]], 1/2])/(Sqrt[2]*Sqrt[1 + x]*Sqrt[(2 + x)/(1 + x)]) + (Sqrt[2 + x]*E 
llipticPi[4/5, ArcTan[Sqrt[x]], 1/2])/(20*Sqrt[2]*Sqrt[1 + x]*Sqrt[(2 + x) 
/(1 + x)])))/Sqrt[2*x + 3*x^2 + x^3]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 320

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

rule 411

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

rule 414

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

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\sqrt {2 x +4}\, \sqrt {-1-x}\, \sqrt {-2 x}\, \operatorname {EllipticPi}\left (\frac {\sqrt {2 x +4}}{2}, -\frac {2}{3}, \sqrt {2}\right )}{3 \sqrt {x^{3}+3 x^{2}+2 x}}\)	\(50\)
elliptic	\(\frac {\sqrt {2 x +4}\, \sqrt {-1-x}\, \sqrt {-2 x}\, \operatorname {EllipticPi}\left (\frac {\sqrt {2 x +4}}{2}, -\frac {2}{3}, \sqrt {2}\right )}{3 \sqrt {x^{3}+3 x^{2}+2 x}}\)	\(50\)

Input:

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

Output:

1/3*(2*x+4)^(1/2)*(-1-x)^(1/2)*(-2*x)^(1/2)/(x^3+3*x^2+2*x)^(1/2)*Elliptic 
Pi(1/2*(2*x+4)^(1/2),-2/3,2^(1/2))
 

Fricas [F]

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

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

Output:

integral(sqrt(x^3 + 3*x^2 + 2*x)/(x^4 + 8*x^3 + 17*x^2 + 10*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(5+x) \sqrt {2 x+3 x^2+x^3}} \, dx=-\frac {\sqrt {-x}\,\sqrt {-x-1}\,\sqrt {x+2}\,\Pi \left (\frac {1}{4};\mathrm {asin}\left (\sqrt {-x-1}\right )\middle |-1\right )}{2\,\sqrt {x^3+3\,x^2+2\,x}} \] Input:

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

Output:

-((-x)^(1/2)*(- x - 1)^(1/2)*(x + 2)^(1/2)*ellipticPi(1/4, asin((- x - 1)^ 
(1/2)), -1))/(2*(2*x + 3*x^2 + x^3)^(1/2))
 

Reduce [F]

\[ \int \frac {1}{(5+x) \sqrt {2 x+3 x^2+x^3}} \, dx=\int \frac {\sqrt {x}\, \sqrt {x^{2}+3 x +2}}{x^{4}+8 x^{3}+17 x^{2}+10 x}d x \] Input:

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

Output:

int((sqrt(x)*sqrt(x**2 + 3*x + 2))/(x**4 + 8*x**3 + 17*x**2 + 10*x),x)