\(\int \frac {1}{\sqrt {a+b x+c x^2} (d+e x+f x^2)} \, dx\) [150]

Optimal result

Integrand size = 27, antiderivative size = 374 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {2} f \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {\sqrt {2} f \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:

-2^(1/2)*f*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))/(-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)+2^(1/2)*f*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))/(-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 = 0.48 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=-\text {RootSum}\left [c^2 d-b c e+b^2 f+2 \sqrt {a} c e \text {$\#$1}-4 \sqrt {a} b f \text {$\#$1}-2 c d \text {$\#$1}^2+b e \text {$\#$1}^2+4 a f \text {$\#$1}^2-2 \sqrt {a} e \text {$\#$1}^3+d \text {$\#$1}^4\&,\frac {c \log (x)-c \log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^2+\log \left (-\sqrt {a}+\sqrt {a+b x+c x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {a} c e-2 \sqrt {a} b f-2 c d \text {$\#$1}+b e \text {$\#$1}+4 a f \text {$\#$1}-3 \sqrt {a} e \text {$\#$1}^2+2 d \text {$\#$1}^3}\&\right ] \] Input:

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

Output:

-RootSum[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*Log[x] 
- c*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] - Log[x]*#1^2 + Log[-Sqrt 
[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1^2)/(Sqrt[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) & ]
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1314, 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.

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

\(\Big \downarrow \) 1314

\(\displaystyle \frac {2 f \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}-\frac {2 f \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {4 f \int \frac {1}{4 \left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}-\frac {4 f \int \frac {1}{4 \left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

-((Sqrt[2]*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[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]]) 
) + (Sqrt[2]*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[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 219

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])
 

rule 1154

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]
 

rule 1314

Int[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*( 
x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[1/( 
(b - q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x], x] - Simp[2*(c/q)   Int[1/((b + 
 q + 2*c*x)*Sqrt[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] && NeQ[c*e - b*f, 0] && PosQ 
[b^2 - 4*a*c]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(760\) vs. \(2(330)=660\).

Time = 2.92 (sec) , antiderivative size = 761, normalized size of antiderivative = 2.03

method	result	size
default	\(-\frac {\sqrt {2}\, \ln \left (\frac {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 f b \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}+\frac {\sqrt {2}\, \ln \left (\frac {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 f b \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}\)	\(761\)

Input:

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

Output:

-1/(-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))))+1/(-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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11287 vs. \(2 (328) = 656\).

Time = 3.74 (sec) , antiderivative size = 11287, normalized size of antiderivative = 30.18 \[ \int \frac {1}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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