Integrand size = 27, antiderivative size = 431 \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f}-\frac {\sqrt {c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )} \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}}+\frac {\sqrt {c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )} \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}} \]
arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*c^(1/2)/f-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))*(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)/f*2^(1/2)/(-4*d*f+e^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))*(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)/f*2^(1/2)/(-4*d*f+e^2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )+\text {RootSum}\left [c^2 d-b c e+b^2 f+2 \sqrt {a} c e \text {$\#$1}-4 \sqrt {a} b f \text {$\#$1}-2 c d \text {$\#$1}^2+b e \text {$\#$1}^2+4 a f \text {$\#$1}^2-2 \sqrt {a} e \text {$\#$1}^3+d \text {$\#$1}^4\&,\frac {-c^2 d \log (x)+b c e \log (x)-b^2 f \log (x)+a c f \log (x)+c^2 d \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )-b c e \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )+b^2 f \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )-a c f \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )-2 \sqrt {a} c e \log (x) \text {$\#$1}+2 \sqrt {a} b f \log (x) \text {$\#$1}+2 \sqrt {a} c e \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}-2 \sqrt {a} b f \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}+c d \log (x) \text {$\#$1}^2-a f \log (x) \text {$\#$1}^2-c d \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2+a f \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-\sqrt {a} c e+2 \sqrt {a} b f+2 c d \text {$\#$1}-b e \text {$\#$1}-4 a f \text {$\#$1}+3 \sqrt {a} e \text {$\#$1}^2-2 d \text {$\#$1}^3}\&\right ]}{f} \]
(2*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])] + RootS um[c^2*d - b*c*e + b^2*f + 2*Sqrt[a]*c*e*#1 - 4*Sqrt[a]*b*f*#1 - 2*c*d*#1^ 2 + b*e*#1^2 + 4*a*f*#1^2 - 2*Sqrt[a]*e*#1^3 + d*#1^4 & , (-(c^2*d*Log[x]) + b*c*e*Log[x] - b^2*f*Log[x] + a*c*f*Log[x] + c^2*d*Log[-Sqrt[a] + Sqrt[ a + b*x + c*x^2] - x*#1] - b*c*e*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x* #1] + b^2*f*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] - a*c*f*Log[-Sqrt [a] + Sqrt[a + b*x + c*x^2] - x*#1] - 2*Sqrt[a]*c*e*Log[x]*#1 + 2*Sqrt[a]* b*f*Log[x]*#1 + 2*Sqrt[a]*c*e*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] *#1 - 2*Sqrt[a]*b*f*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1 + c*d* Log[x]*#1^2 - a*f*Log[x]*#1^2 - c*d*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1^2 + a*f*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1^2)/(-(Sq rt[a]*c*e) + 2*Sqrt[a]*b*f + 2*c*d*#1 - b*e*#1 - 4*a*f*#1 + 3*Sqrt[a]*e*#1 ^2 - 2*d*#1^3) & ])/f
Time = 0.67 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1320, 1092, 219, 1365, 1154, 219}
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.
\(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx\) |
\(\Big \downarrow \) 1320 |
\(\displaystyle \frac {c \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{f}-\frac {\int \frac {c d-a f+(c e-b f) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {2 c \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{f}-\frac {\int \frac {c d-a f+(c e-b f) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f}-\frac {\int \frac {c d-a f+(c e-b f) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\) |
\(\Big \downarrow \) 1365 |
\(\displaystyle \frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f}-\frac {\frac {\left (2 f (c d-a f)-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f)\right ) \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}-\frac {\left (2 f (c d-a f)-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f)\right ) \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}}{f}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f}-\frac {\frac {2 \left (2 f (c d-a f)-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f)\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}-\frac {2 \left (2 f (c d-a f)-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f)\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f}-\frac {\frac {\left (2 f (c d-a f)-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f)\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\left (2 f (c d-a f)-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f)\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}}{f}\) |
(Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/f - (-((( 2*f*(c*d - a*f) - (c*e - b*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]*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]])) + ((2*f*(c*d - a*f) - (c*e - b*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]*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
3.2.11.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^ 2), x_Symbol] :> Simp[c/f Int[1/Sqrt[a + b*x + c*x^2], x], x] - Simp[1/f Int[(c*d - a*f + (c*e - b*f)*x)/(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)) , x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + ( e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Sim p[(2*c*g - h*(b - q))/q Int[1/((b - q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x] , x] - Simp[(2*c*g - h*(b + q))/q Int[1/((b + q + 2*c*x)*Sqrt[d + e*x + f *x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0 ] && NeQ[e^2 - 4*d*f, 0] && PosQ[b^2 - 4*a*c]
Leaf count of result is larger than twice the leaf count of optimal. \(1546\) vs. \(2(376)=752\).
Time = 0.94 (sec) , antiderivative size = 1547, normalized size of antiderivative = 3.59
-1/(-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)))+1/(-4*d*f+e^2)^(1/2)*(1/2*(4 *(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2*c+4*(c*(-4*d*f+e^2)^(1/2)+b*f-c*e)/f* (x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+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*(c*(-4*d*f+e^2)^(1...
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\int \frac {\sqrt {a + b x + c x^{2}}}{d + e x + f x^{2}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for more deta
Exception generated. \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{f\,x^2+e\,x+d} \,d x \]