\(\int \frac {\sqrt {a+b x+c x^2}}{x (d+e x+f x^2)} \, dx\) [133]

Optimal result

Integrand size = 30, antiderivative size = 493 \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d}-\frac {\left (2 (b d-a e) f-(c d-a f) \left (e-\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {\left (2 (b d-a e) f-(c d-a f) \left (e+\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:

-a^(1/2)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/d-1/2*(2*(-a*e 
+b*d)*f-(-a*f+c*d)*(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*(2*(-a*e+b*d)*f-(-a*f+c*d)*(e+(-4*d*f+e^2)^(1/2)))*ar 
ctanh(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 = 0.69 (sec) , antiderivative size = 467, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b^2 d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a b e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a^2 f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{d} \] Input:

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

Output:

(2*Sqrt[a]*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] - RootSum[ 
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^2*d*Log[-(Sqrt[c 
]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x 
 + c*x^2] - #1] - a*b*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a 
^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[c]*d*Log[-( 
Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*e*Log[-(Sqrt[c]* 
x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x 
+ c*x^2] - #1]*#1^2 - a*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 = 3.06 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.06, 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.

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((Sqrt[a]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/d) - (( 
2*b*d*f - c*d*(e - Sqrt[e^2 - 4*d*f]) - a*f*(e + Sqrt[e^2 - 4*d*f]))*ArcTa 
nh[(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*d*f - e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e - Sqrt[e^2 - 4*d*f] 
))]) + ((2*b*d*f - a*f*(e - Sqrt[e^2 - 4*d*f]) - c*d*(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*d*f + e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e + 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1690\) vs. \(2(438)=876\).

Time = 2.73 (sec) , antiderivative size = 1691, normalized size of antiderivative = 3.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1691\)

Input:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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