Integrand size = 30, antiderivative size = 523 \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d}+\frac {\left (c d \left (e-\sqrt {e^2-4 d f}\right )-f \left (2 b d-a \left (e+\sqrt {e^2-4 d f}\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )}}-\frac {\left (c d \left (e+\sqrt {e^2-4 d f}\right )-f \left (2 b d-a \left (e-\sqrt {e^2-4 d f}\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )}} \]
-arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))*a^(1/2)/d+1/2*arctanh( 1/4*(4*a*f+2*x*(b*f-c*(e-(-4*d*f+e^2)^(1/2)))-b*(e-(-4*d*f+e^2)^(1/2)))*2^ (1/2)/(c*x^2+b*x+a)^(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*d*(e-(-4*d*f+e^2)^(1/2))-f*(2*b*d-a*(e+(-4*d*f+e^2)^ (1/2))))/d*2^(1/2)/(-4*d*f+e^2)^(1/2)/(f*(2*a*f-b*(e-(-4*d*f+e^2)^(1/2)))+ c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)-1/2*arctanh(1/4*(4*a*f-b*(e+(-4* d*f+e^2)^(1/2))+2*x*(b*f-c*(e+(-4*d*f+e^2)^(1/2))))*2^(1/2)/(c*x^2+b*x+a)^ (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))*( -f*(2*b*d-a*(e-(-4*d*f+e^2)^(1/2)))+c*d*(e+(-4*d*f+e^2)^(1/2)))/d*2^(1/2)/ (-4*d*f+e^2)^(1/2)/(c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2))+f*(2*a*f-b*(e+(-4*d *f+e^2)^(1/2))))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.50 (sec) , antiderivative size = 467, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b^2 d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a b e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a^2 f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{d} \]
(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
Time = 2.78 (sec) , antiderivative size = 521, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle -\frac {\left (-a f \left (\sqrt {e^2-4 d f}+e\right )+2 b d f-c d \left (e-\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (-a f \left (e-\sqrt {e^2-4 d f}\right )+2 b d f-c d \left (\sqrt {e^2-4 d f}+e\right )\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (\sqrt {e^2-4 d f}+e\right )\right )+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d}\) |
-((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]))])
3.2.12.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1690\) vs. \(2(460)=920\).
Time = 1.05 (sec) , antiderivative size = 1691, normalized size of antiderivative = 3.23
-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*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c+4/f*(-c*(-4*d*f+e^2 )^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-b*f*(-4*d*f+e^2)^(1/ 2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)+1/2/f*(- c*(-4*d*f+e^2)^(1/2)+b*f-c*e)*ln((1/2/f*(-c*(-4*d*f+e^2)^(1/2)+b*f-c*e)+c* (x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))/c^(1/2)+((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f )^2*c+1/f*(-c*(-4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f) +1/2*(-b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f +c*e^2)/f^2)^(1/2))/c^(1/2)-1/2*(-b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2 )*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2*2^(1/2)/((-b*f*(-4*d*f+e^2)^(1/2)+( -4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-b*f*(- 4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2+1 /f*(-c*(-4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^ (1/2)*((-b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d *f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c+4/f*(-c*(-4*d *f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-b*f*(-4*d*f+e^ 2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2))/( x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)))+2*f/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e...
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {\sqrt {a + b x + c x^{2}}}{x \left (d + e x + f x^{2}\right )}\, dx \]
\[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} + e x + d\right )} x} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{x\,\left (f\,x^2+e\,x+d\right )} \,d x \]