\(\int \frac {1}{x \sqrt {a+b x+c x^2} (d+e x+f x^2)} \, dx\) [145]

Optimal result

Integrand size = 30, antiderivative size = 451 \[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d}+\frac {f \left (e+\sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}-\frac {f \left (e-\sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:

-arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(1/2)/d+1/2*f*(e+(-4 
*d*f+e^2)^(1/2))*arctanh(1/4*(4*a*f-b*(e-(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e-( 
-4*d*f+e^2)^(1/2)))*x)*2^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4 
*d*f+e^2)^(1/2))^(1/2)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/d/(-4*d*f+e^2)^(1/2)/( 
c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2)-1/2*f*(e- 
(-4*d*f+e^2)^(1/2))*arctanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*(b*f-c*( 
e+(-4*d*f+e^2)^(1/2)))*x)*2^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)* 
(-4*d*f+e^2)^(1/2))^(1/2)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/d/(-4*d*f+e^2)^(1/2 
)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.57 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 \sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-2 b \sqrt {c} d+a \sqrt {c} e+4 c d \text {$\#$1}+b e \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{d} \] Input:

Integrate[1/(x*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
 

Output:

((2*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]])/Sqrt[a] - RootSu 
m[b^2*d - a*b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 
 + b*e*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (b*e*Log[-(Sqrt[c 
]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + 
 c*x^2] - #1] - 2*Sqrt[c]*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] 
*#1 + f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(-2*b*Sqrt[c] 
*d + a*Sqrt[c]*e + 4*c*d*#1 + b*e*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*# 
1^3) & ])/d
 

Rubi [A] (verified)

Time = 1.84 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {7279, 2009}

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.

\(\displaystyle \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {-e-f x}{d \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )}+\frac {1}{d x \sqrt {a+b x+c x^2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f \left (\sqrt {e^2-4 d f}+e\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {f \left (e-\sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d}\)

Input:

Int[1/(x*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
 

Output:

-(ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])]/(Sqrt[a]*d)) + (f 
*(e + Sqrt[e^2 - 4*d*f])*ArcTanh[(4*a*f - b*(e - Sqrt[e^2 - 4*d*f]) + 2*(b 
*f - c*(e - Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f 
 + 2*a*f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/(Sqrt 
[2]*d*Sqrt[e^2 - 4*d*f]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 - (c*e - b* 
f)*Sqrt[e^2 - 4*d*f]]) - (f*(e - Sqrt[e^2 - 4*d*f])*ArcTanh[(4*a*f - b*(e 
+ Sqrt[e^2 - 4*d*f]) + 2*(b*f - c*(e + Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt[2]*S 
qrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqr 
t[a + b*x + c*x^2])])/(Sqrt[2]*d*Sqrt[e^2 - 4*d*f]*Sqrt[c*e^2 - 2*c*d*f - 
b*e*f + 2*a*f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(858\) vs. \(2(396)=792\).

Time = 2.79 (sec) , antiderivative size = 859, normalized size of antiderivative = 1.90

method	result	size
default	\(\frac {4 f \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {a}}-\frac {2 f \sqrt {2}\, \ln \left (\frac {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 f b \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}-\frac {2 f \sqrt {2}\, \ln \left (\frac {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 f b \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}\)	\(859\)

Input:

int(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

Output:

4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))/a^(1/2)*ln((2*a+b*x+2*a 
^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-2*f/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1 
/2)*2^(1/2)/((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f- 
2*d*f*c+c*e^2)/f^2)^(1/2)*ln(((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c 
*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2+(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x-1/ 
2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+ 
e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x-1/2/f*(-e+( 
-4*d*f+e^2)^(1/2)))^2+4*(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x-1/2/f*(-e+(-4* 
d*f+e^2)^(1/2)))+2*(f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2- 
b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2))/(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2))))-2*f/(e 
+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-f*b*(-4*d*f+e^2)^(1/2)+ 
(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*ln(((-f*b*( 
-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2+ 
1/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2 
^(1/2)*((-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d* 
f*c+c*e^2)/f^2)^(1/2)*(4*c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2+4/f*(-c*(-4* 
d*f+e^2)^(1/2)+f*b-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-f*b*(-4*d*f+e 
^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2))/ 
(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x \sqrt {a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:

integrate(1/x/(c*x**2+b*x+a)**(1/2)/(f*x**2+e*x+d),x)
 

Output:

Integral(1/(x*sqrt(a + b*x + c*x**2)*(d + e*x + f*x**2)), x)
 

Maxima [F]

\[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (f x^{2} + e x + d\right )} x} \,d x } \] Input:

integrate(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(c*x^2 + b*x + a)*(f*x^2 + e*x + d)*x), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionNot implemented, e.g. for multivariate mod/approx polynomi 
alsError:
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x\,\sqrt {c\,x^2+b\,x+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:

int(1/(x*(a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)),x)
 

Output:

int(1/(x*(a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)), x)
 

Reduce [F]

\[ \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x \sqrt {c \,x^{2}+b x +a}\, \left (f \,x^{2}+e x +d \right )}d x \] Input:

int(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)
 

Output:

int(1/x/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)