\(\int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx\) [109]

Optimal result

Integrand size = 27, antiderivative size = 653 \[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=-\frac {\left (\sqrt {5} f+\sqrt {5 f^2-2 f g+3 g^2}\right ) \text {arctanh}\left (\frac {\sqrt {3} \sqrt {f+g x}}{\sqrt {f} \sqrt {3+2 x+5 x^2}}\right )}{\sqrt {15} \sqrt {f} \left (f+\sqrt {f^2-\frac {2 f g}{5}+\frac {3 g^2}{5}}\right )}-\frac {\sqrt [4]{5 f^2-2 f g+3 g^2} \sqrt {\frac {g^2 \left (3+2 x+5 x^2\right )}{\left (5 f^2-2 f g+3 g^2\right ) \left (1+\frac {\sqrt {5} (f+g x)}{\sqrt {5 f^2-2 f g+3 g^2}}\right )^2}} \left (1+\frac {\sqrt {5} (f+g x)}{\sqrt {5 f^2-2 f g+3 g^2}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{5} \sqrt {f+g x}}{\sqrt [4]{5 f^2-2 f g+3 g^2}}\right ),\frac {1}{10} \left (5+\frac {5 f-g}{\sqrt {f^2-\frac {2 f g}{5}+\frac {3 g^2}{5}}}\right )\right )}{\sqrt [4]{5} \left (f+\sqrt {f^2-\frac {2 f g}{5}+\frac {3 g^2}{5}}\right ) \sqrt {3+2 x+5 x^2}}+\frac {\left (\sqrt {5} f-\sqrt {5 f^2-2 f g+3 g^2}\right ) \sqrt {\frac {g^2 \left (3+2 x+5 x^2\right )}{\left (\sqrt {5 f^2-2 f g+3 g^2}+\sqrt {5} (f+g x)\right )^2}} \left (\sqrt {5 f^2-2 f g+3 g^2}+\sqrt {5} (f+g x)\right ) \operatorname {EllipticPi}\left (\frac {\left (\sqrt {5} f+\sqrt {5 f^2-2 f g+3 g^2}\right )^2}{4 \sqrt {5} f \sqrt {5 f^2-2 f g+3 g^2}},2 \arctan \left (\frac {\sqrt [4]{5} \sqrt {f+g x}}{\sqrt [4]{5 f^2-2 f g+3 g^2}}\right ),\frac {1}{10} \left (5+\frac {5 f-g}{\sqrt {f^2-\frac {2 f g}{5}+\frac {3 g^2}{5}}}\right )\right )}{2\ 5^{3/4} f \sqrt [4]{5 f^2-2 f g+3 g^2} \left (f+\sqrt {f^2-\frac {2 f g}{5}+\frac {3 g^2}{5}}\right ) \sqrt {3+2 x+5 x^2}} \] Output:

-1/15*(5^(1/2)*f+(5*f^2-2*f*g+3*g^2)^(1/2))*arctanh(3^(1/2)*(g*x+f)^(1/2)/ 
f^(1/2)/(5*x^2+2*x+3)^(1/2))*15^(1/2)/f^(1/2)/(f+1/5*(25*f^2-10*f*g+15*g^2 
)^(1/2))-1/5*(5*f^2-2*f*g+3*g^2)^(1/4)*(g^2*(5*x^2+2*x+3)/(5*f^2-2*f*g+3*g 
^2)/(1+5^(1/2)*(g*x+f)/(5*f^2-2*f*g+3*g^2)^(1/2))^2)^(1/2)*(1+5^(1/2)*(g*x 
+f)/(5*f^2-2*f*g+3*g^2)^(1/2))*InverseJacobiAM(2*arctan(5^(1/4)*(g*x+f)^(1 
/2)/(5*f^2-2*f*g+3*g^2)^(1/4)),1/2*(2+2*(5*f-g)/(25*f^2-10*f*g+15*g^2)^(1/ 
2))^(1/2))*5^(3/4)/(f+1/5*(25*f^2-10*f*g+15*g^2)^(1/2))/(5*x^2+2*x+3)^(1/2 
)+1/10*(5^(1/2)*f-(5*f^2-2*f*g+3*g^2)^(1/2))*(g^2*(5*x^2+2*x+3)/((5*f^2-2* 
f*g+3*g^2)^(1/2)+5^(1/2)*(g*x+f))^2)^(1/2)*((5*f^2-2*f*g+3*g^2)^(1/2)+5^(1 
/2)*(g*x+f))*EllipticPi(sin(2*arctan(5^(1/4)*(g*x+f)^(1/2)/(5*f^2-2*f*g+3* 
g^2)^(1/4))),1/20*(5^(1/2)*f+(5*f^2-2*f*g+3*g^2)^(1/2))^2*5^(1/2)/f/(5*f^2 
-2*f*g+3*g^2)^(1/2),1/2*(2+2*(5*f-g)/(25*f^2-10*f*g+15*g^2)^(1/2))^(1/2))* 
5^(1/4)/f/(5*f^2-2*f*g+3*g^2)^(1/4)/(f+1/5*(25*f^2-10*f*g+15*g^2)^(1/2))/( 
5*x^2+2*x+3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 22.78 (sec) , antiderivative size = 385, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=-\frac {2 i (f+g x) \sqrt {1+\frac {-5 f^2+2 f g-3 g^2}{\left (5 f+\left (-1-i \sqrt {14}\right ) g\right ) (f+g x)}} \sqrt {1+\frac {-5 f^2+2 f g-3 g^2}{\left (5 f+i \left (i+\sqrt {14}\right ) g\right ) (f+g x)}} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {-5 f^2+2 f g-3 g^2}{5 f+\left (-1-i \sqrt {14}\right ) g}}}{\sqrt {f+g x}}\right ),\frac {5 f-g-i \sqrt {14} g}{5 f-g+i \sqrt {14} g}\right )-\operatorname {EllipticPi}\left (\frac {f \left (5 f+\left (-1-i \sqrt {14}\right ) g\right )}{5 f^2-2 f g+3 g^2},i \text {arcsinh}\left (\frac {\sqrt {\frac {-5 f^2+2 f g-3 g^2}{5 f+\left (-1-i \sqrt {14}\right ) g}}}{\sqrt {f+g x}}\right ),\frac {5 f-g-i \sqrt {14} g}{5 f-g+i \sqrt {14} g}\right )\right )}{f \sqrt {\frac {-5 f^2+2 f g-3 g^2}{5 f+\left (-1-i \sqrt {14}\right ) g}} \sqrt {3+2 x+5 x^2}} \] Input:

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

Output:

((-2*I)*(f + g*x)*Sqrt[1 + (-5*f^2 + 2*f*g - 3*g^2)/((5*f + (-1 - I*Sqrt[1 
4])*g)*(f + g*x))]*Sqrt[1 + (-5*f^2 + 2*f*g - 3*g^2)/((5*f + I*(I + Sqrt[1 
4])*g)*(f + g*x))]*(EllipticF[I*ArcSinh[Sqrt[(-5*f^2 + 2*f*g - 3*g^2)/(5*f 
 + (-1 - I*Sqrt[14])*g)]/Sqrt[f + g*x]], (5*f - g - I*Sqrt[14]*g)/(5*f - g 
 + I*Sqrt[14]*g)] - EllipticPi[(f*(5*f + (-1 - I*Sqrt[14])*g))/(5*f^2 - 2* 
f*g + 3*g^2), I*ArcSinh[Sqrt[(-5*f^2 + 2*f*g - 3*g^2)/(5*f + (-1 - I*Sqrt[ 
14])*g)]/Sqrt[f + g*x]], (5*f - g - I*Sqrt[14]*g)/(5*f - g + I*Sqrt[14]*g) 
]))/(f*Sqrt[(-5*f^2 + 2*f*g - 3*g^2)/(5*f + (-1 - I*Sqrt[14])*g)]*Sqrt[3 + 
 2*x + 5*x^2])
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.74 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.47, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1279, 27, 187, 413, 413, 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.

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

\(\Big \downarrow \) 1279

\(\displaystyle \frac {2 \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \int \frac {1}{2 x \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \sqrt {f+g x}}dx}{\sqrt {5 x^2+2 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \int \frac {1}{x \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \sqrt {f+g x}}dx}{\sqrt {5 x^2+2 x+3}}\)

\(\Big \downarrow \) 187

\(\displaystyle -\frac {2 \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \int -\frac {1}{g x \sqrt {-\frac {5 f}{g}+\frac {5 (f+g x)}{g}-i \sqrt {14}+1} \sqrt {-\frac {5 f}{g}+\frac {5 (f+g x)}{g}+i \sqrt {14}+1}}d\sqrt {f+g x}}{\sqrt {5 x^2+2 x+3}}\)

\(\Big \downarrow \) 413

\(\displaystyle -\frac {2 \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \sqrt {1-\frac {5 (f+g x)}{5 f+i \sqrt {14} g-g}} \int -\frac {1}{g x \sqrt {-\frac {5 f}{g}+\frac {5 (f+g x)}{g}+i \sqrt {14}+1} \sqrt {1-\frac {5 (f+g x)}{5 f+i \sqrt {14} g-g}}}d\sqrt {f+g x}}{\sqrt {5 x^2+2 x+3} \sqrt {\frac {5 (f+g x)}{g}-\frac {5 f}{g}-i \sqrt {14}+1}}\)

\(\Big \downarrow \) 413

\(\displaystyle -\frac {2 \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \sqrt {1-\frac {5 (f+g x)}{5 f+i \sqrt {14} g-g}} \sqrt {1-\frac {5 (f+g x)}{5 f-\left (1+i \sqrt {14}\right ) g}} \int -\frac {1}{g x \sqrt {1-\frac {5 (f+g x)}{5 f+i \sqrt {14} g-g}} \sqrt {1-\frac {5 (f+g x)}{5 f-\left (1+i \sqrt {14}\right ) g}}}d\sqrt {f+g x}}{\sqrt {5 x^2+2 x+3} \sqrt {\frac {5 (f+g x)}{g}-\frac {5 f}{g}-i \sqrt {14}+1} \sqrt {\frac {5 (f+g x)}{g}-\frac {5 f}{g}+i \sqrt {14}+1}}\)

\(\Big \downarrow \) 412

\(\displaystyle -\frac {2 \sqrt {5 x-i \sqrt {14}+1} \sqrt {5 x+i \sqrt {14}+1} \sqrt {5 f+i \left (\sqrt {14}+i\right ) g} \sqrt {1-\frac {5 (f+g x)}{5 f+i \sqrt {14} g-g}} \sqrt {1-\frac {5 (f+g x)}{5 f-\left (1+i \sqrt {14}\right ) g}} \operatorname {EllipticPi}\left (\frac {5 f+i \sqrt {14} g-g}{5 f},\arcsin \left (\frac {\sqrt {5} \sqrt {f+g x}}{\sqrt {5 f+i \left (i+\sqrt {14}\right ) g}}\right ),\frac {5 f+i \sqrt {14} g-g}{5 f-i \sqrt {14} g-g}\right )}{\sqrt {5} f \sqrt {5 x^2+2 x+3} \sqrt {\frac {5 (f+g x)}{g}-\frac {5 f}{g}-i \sqrt {14}+1} \sqrt {\frac {5 (f+g x)}{g}-\frac {5 f}{g}+i \sqrt {14}+1}}\)

Input:

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

Output:

(-2*Sqrt[5*f + I*(I + Sqrt[14])*g]*Sqrt[1 - I*Sqrt[14] + 5*x]*Sqrt[1 + I*S 
qrt[14] + 5*x]*Sqrt[1 - (5*(f + g*x))/(5*f - g + I*Sqrt[14]*g)]*Sqrt[1 - ( 
5*(f + g*x))/(5*f - (1 + I*Sqrt[14])*g)]*EllipticPi[(5*f - g + I*Sqrt[14]* 
g)/(5*f), ArcSin[(Sqrt[5]*Sqrt[f + g*x])/Sqrt[5*f + I*(I + Sqrt[14])*g]], 
(5*f - g + I*Sqrt[14]*g)/(5*f - g - I*Sqrt[14]*g)])/(Sqrt[5]*f*Sqrt[3 + 2* 
x + 5*x^2]*Sqrt[1 - I*Sqrt[14] - (5*f)/g + (5*(f + g*x))/g]*Sqrt[1 + I*Sqr 
t[14] - (5*f)/g + (5*(f + g*x))/g])
 

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 187

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] &&  !SimplerQ[e + f*x, c + d*x] &&  !SimplerQ[g + h*x, c + d*x]
 

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 413

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

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 [C] (verified)

Result contains complex when optimal does not.

Time = 3.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {2 \left (i \sqrt {14}\, g -5 f +g \right ) \operatorname {EllipticPi}\left (\sqrt {-\frac {5 \left (g x +f \right )}{i \sqrt {14}\, g -5 f +g}}, -\frac {i \sqrt {14}\, g -5 f +g}{5 f}, \sqrt {-\frac {i \sqrt {14}\, g -5 f +g}{i \sqrt {14}\, g +5 f -g}}\right ) \sqrt {\frac {g \left (i \sqrt {14}+5 x +1\right )}{i \sqrt {14}\, g -5 f +g}}\, \sqrt {\frac {\left (i \sqrt {14}-5 x -1\right ) g}{i \sqrt {14}\, g +5 f -g}}\, \sqrt {-\frac {5 \left (g x +f \right )}{i \sqrt {14}\, g -5 f +g}}\, \sqrt {5 x^{2}+2 x +3}\, \sqrt {g x +f}}{5 \left (5 g \,x^{3}+5 f \,x^{2}+2 g \,x^{2}+2 f x +3 g x +3 f \right ) f}\)	\(225\)
elliptic	\(-\frac {2 \sqrt {\left (g x +f \right ) \left (5 x^{2}+2 x +3\right )}\, \left (\frac {f}{g}-\frac {1}{5}-\frac {i \sqrt {14}}{5}\right ) \sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {1}{5}-\frac {i \sqrt {14}}{5}}}\, \sqrt {\frac {x +\frac {1}{5}-\frac {i \sqrt {14}}{5}}{-\frac {f}{g}+\frac {1}{5}-\frac {i \sqrt {14}}{5}}}\, \sqrt {\frac {x +\frac {1}{5}+\frac {i \sqrt {14}}{5}}{-\frac {f}{g}+\frac {1}{5}+\frac {i \sqrt {14}}{5}}}\, g \operatorname {EllipticPi}\left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {1}{5}-\frac {i \sqrt {14}}{5}}}, -\frac {\left (-\frac {f}{g}+\frac {1}{5}+\frac {i \sqrt {14}}{5}\right ) g}{f}, \sqrt {\frac {-\frac {f}{g}+\frac {1}{5}+\frac {i \sqrt {14}}{5}}{-\frac {f}{g}+\frac {1}{5}-\frac {i \sqrt {14}}{5}}}\right )}{\sqrt {g x +f}\, \sqrt {5 x^{2}+2 x +3}\, \sqrt {5 g \,x^{3}+5 f \,x^{2}+2 g \,x^{2}+2 f x +3 g x +3 f}\, f}\)	\(249\)

Input:

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

Output:

2/5*(I*14^(1/2)*g-5*f+g)*EllipticPi((-5*(g*x+f)/(I*14^(1/2)*g-5*f+g))^(1/2 
),-1/5*(I*14^(1/2)*g-5*f+g)/f,(-(I*14^(1/2)*g-5*f+g)/(I*14^(1/2)*g+5*f-g)) 
^(1/2))*(g*(I*14^(1/2)+5*x+1)/(I*14^(1/2)*g-5*f+g))^(1/2)*((I*14^(1/2)-5*x 
-1)*g/(I*14^(1/2)*g+5*f-g))^(1/2)*(-5*(g*x+f)/(I*14^(1/2)*g-5*f+g))^(1/2)* 
(5*x^2+2*x+3)^(1/2)*(g*x+f)^(1/2)/(5*g*x^3+5*f*x^2+2*g*x^2+2*f*x+3*g*x+3*f 
)/f
 

Fricas [F]

\[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int { \frac {1}{\sqrt {g x + f} \sqrt {5 \, x^{2} + 2 \, x + 3} x} \,d x } \] Input:

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

Output:

integral(sqrt(g*x + f)*sqrt(5*x^2 + 2*x + 3)/(5*g*x^4 + (5*f + 2*g)*x^3 + 
(2*f + 3*g)*x^2 + 3*f*x), x)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int \frac {1}{x \sqrt {f + g x} \sqrt {5 x^{2} + 2 x + 3}}\, dx \] Input:

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

Output:

Integral(1/(x*sqrt(f + g*x)*sqrt(5*x**2 + 2*x + 3)), x)
 

Maxima [F]

\[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int { \frac {1}{\sqrt {g x + f} \sqrt {5 \, x^{2} + 2 \, x + 3} x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int { \frac {1}{\sqrt {g x + f} \sqrt {5 \, x^{2} + 2 \, x + 3} x} \,d x } \] Input:

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

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int \frac {1}{x\,\sqrt {f+g\,x}\,\sqrt {5\,x^2+2\,x+3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x \sqrt {f+g x} \sqrt {3+2 x+5 x^2}} \, dx=\int \frac {\sqrt {g x +f}\, \sqrt {5 x^{2}+2 x +3}}{5 g \,x^{4}+5 f \,x^{3}+2 g \,x^{3}+2 f \,x^{2}+3 g \,x^{2}+3 f x}d x \] Input:

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

Output:

int((sqrt(f + g*x)*sqrt(5*x**2 + 2*x + 3))/(5*f*x**3 + 2*f*x**2 + 3*f*x + 
5*g*x**4 + 2*g*x**3 + 3*g*x**2),x)