\(\int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+(c f^2-b f g+a g^2) x^4} \, dx\) [335]

Optimal result

Integrand size = 64, antiderivative size = 270 \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=-\frac {\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} x}{\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} (e f-d g)}+\frac {\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} \text {arctanh}\left (\frac {\sqrt {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g} x}{\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} (e f-d g)} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.20 \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\frac {\sqrt {2} \sqrt {c f^2+g (-b f+a g)} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {c f^2-b f g+a g^2} x}{\sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g}}\right )}{\sqrt {-2 c d f+b e f+\sqrt {b^2-4 a c} e f+b d g-\sqrt {b^2-4 a c} d g-2 a e g}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt {c f^2-b f g+a g^2} x}{\sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g}}\right )}{\sqrt {-2 c d f+b e f-\sqrt {b^2-4 a c} e f+b d g+\sqrt {b^2-4 a c} d g-2 a e g}}\right )}{\sqrt {b^2-4 a c} (-e f+d g)} \] Input:

Integrate[(c*d^2 - b*d*e + a*e^2 - (2*c*d*f - b*e*f - b*d*g + 2*a*e*g)*x^2 
 + (c*f^2 - b*f*g + a*g^2)*x^4)^(-1),x]
 

Output:

(Sqrt[2]*Sqrt[c*f^2 + g*(-(b*f) + a*g)]*(ArcTan[(Sqrt[2]*Sqrt[c*f^2 - b*f* 
g + a*g^2]*x)/Sqrt[-2*c*d*f + b*e*f + Sqrt[b^2 - 4*a*c]*e*f + b*d*g - Sqrt 
[b^2 - 4*a*c]*d*g - 2*a*e*g]]/Sqrt[-2*c*d*f + b*e*f + Sqrt[b^2 - 4*a*c]*e* 
f + b*d*g - Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g] - ArcTan[(Sqrt[2]*Sqrt[c*f^2 
- b*f*g + a*g^2]*x)/Sqrt[-2*c*d*f + b*e*f - Sqrt[b^2 - 4*a*c]*e*f + b*d*g 
+ Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g]]/Sqrt[-2*c*d*f + b*e*f - Sqrt[b^2 - 4*a 
*c]*e*f + b*d*g + Sqrt[b^2 - 4*a*c]*d*g - 2*a*e*g]))/(Sqrt[b^2 - 4*a*c]*(- 
(e*f) + d*g))
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {1406, 221}

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.

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

\(\Big \downarrow \) 1406

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

\(\Big \downarrow \) 221

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

Input:

Int[(c*d^2 - b*d*e + a*e^2 - (2*c*d*f - b*e*f - b*d*g + 2*a*e*g)*x^2 + (c* 
f^2 - b*f*g + a*g^2)*x^4)^(-1),x]
 

Output:

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

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1406

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 
2 - 4*a*c, 2]}, Simp[c/q   Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q   I 
nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c 
, 0] && PosQ[b^2 - 4*a*c]
 

Maple [C] (verified)

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

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.46

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a \,g^{2}-b f g +c \,f^{2}\right ) \textit {\_Z}^{4}+\left (-2 a e g +b d g +b e f -2 c d f \right ) \textit {\_Z}^{2}+a \,e^{2}-b d e +c \,d^{2}\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} a \,g^{2}-2 \textit {\_R}^{3} b f g +2 \textit {\_R}^{3} c \,f^{2}-2 \textit {\_R} a e g +\textit {\_R} b d g +\textit {\_R} b e f -2 \textit {\_R} c d f}\right )}{2}\)	\(124\)
default	\(\left (4 a \,g^{2}-4 b f g +4 c \,f^{2}\right ) \left (-\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 a \,g^{2}-2 b f g +2 c \,f^{2}\right ) x \sqrt {2}}{2 \sqrt {\left (a \,g^{2}-b f g +c \,f^{2}\right ) \left (-2 a e g +b d g +b e f -2 c d f +\sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\right )}}\right )}{4 \sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (a \,g^{2}-b f g +c \,f^{2}\right ) \left (-2 a e g +b d g +b e f -2 c d f +\sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\right )}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (-2 a \,g^{2}+2 b f g -2 c \,f^{2}\right ) x \sqrt {2}}{2 \sqrt {\left (2 a e g -b d g -b e f +2 c d f +\sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\right ) \left (a \,g^{2}-b f g +c \,f^{2}\right )}}\right )}{4 \sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\, \sqrt {\left (2 a e g -b d g -b e f +2 c d f +\sqrt {-\left (d g -e f \right )^{2} \left (4 a c -b^{2}\right )}\right ) \left (a \,g^{2}-b f g +c \,f^{2}\right )}}\right )\)	\(381\)

Input:

int(1/(c*d^2-b*d*e+a*e^2-(2*a*e*g-b*d*g-b*e*f+2*c*d*f)*x^2+(a*g^2-b*f*g+c* 
f^2)*x^4),x,method=_RETURNVERBOSE)
 

Output:

1/2*sum(1/(2*_R^3*a*g^2-2*_R^3*b*f*g+2*_R^3*c*f^2-2*_R*a*e*g+_R*b*d*g+_R*b 
*e*f-2*_R*c*d*f)*ln(x-_R),_R=RootOf((a*g^2-b*f*g+c*f^2)*_Z^4+(-2*a*e*g+b*d 
*g+b*e*f-2*c*d*f)*_Z^2+a*e^2-b*d*e+c*d^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8845 vs. \(2 (234) = 468\).

Time = 0.35 (sec) , antiderivative size = 8845, normalized size of antiderivative = 32.76 \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\text {Too large to display} \] Input:

integrate(1/(c*d^2-b*d*e+a*e^2-(2*a*e*g-b*d*g-b*e*f+2*c*d*f)*x^2+(a*g^2-b* 
f*g+c*f^2)*x^4),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\text {Timed out} \] Input:

integrate(1/(c*d**2-b*d*e+a*e**2-(2*a*e*g-b*d*g-b*e*f+2*c*d*f)*x**2+(a*g** 
2-b*f*g+c*f**2)*x**4),x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\int { \frac {1}{{\left (c f^{2} - b f g + a g^{2}\right )} x^{4} + c d^{2} - b d e + a e^{2} - {\left (2 \, c d f - b e f - b d g + 2 \, a e g\right )} x^{2}} \,d x } \] Input:

integrate(1/(c*d^2-b*d*e+a*e^2-(2*a*e*g-b*d*g-b*e*f+2*c*d*f)*x^2+(a*g^2-b* 
f*g+c*f^2)*x^4),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(c*d^2-b*d*e+a*e^2-(2*a*e*g-b*d*g-b*e*f+2*c*d*f)*x^2+(a*g^2-b* 
f*g+c*f^2)*x^4),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{-4,[0,0,2]%%%},[1,0,3,0]%%%}+%%%{%%{poly1[%%%{2,[0,0,1 
]%%%},%%%
 

Mupad [B] (verification not implemented)

Time = 23.57 (sec) , antiderivative size = 25137, normalized size of antiderivative = 93.10 \[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((x*(4*a^3*g^6 + 4*c^3*f^6 - 4*b^3*f^3*g^3 + 12*a*b^2*f^2*g^4 + 12*a* 
c^2*f^4*g^2 + 12*a^2*c*f^2*g^4 + 12*b^2*c*f^4*g^2 - 12*a^2*b*f*g^5 - 12*b* 
c^2*f^5*g - 24*a*b*c*f^3*g^3) + (-(b^3*d*g + b^3*e*f + d*g*(-(4*a*c - b^2) 
^3)^(1/2) - e*f*(-(4*a*c - b^2)^3)^(1/2) + 8*a*c^2*d*f - 2*a*b^2*e*g - 2*b 
^2*c*d*f + 8*a^2*c*e*g - 4*a*b*c*d*g - 4*a*b*c*e*f)/(8*(a*b^4*e^4*f^2 + b^ 
4*c*d^4*g^2 - b^5*d*e^3*f^2 - b^5*d^3*e*g^2 + 16*a^2*c^3*d^4*g^2 + 16*a^3* 
c^2*e^4*f^2 + 16*a^2*c^3*d^2*e^2*f^2 + 16*a^3*c^2*d^2*e^2*g^2 + 2*b^5*d^2* 
e^2*f*g - 8*a*b^2*c^2*d^4*g^2 - 8*a^2*b^2*c*e^4*f^2 + a*b^4*d^2*e^2*g^2 + 
b^4*c*d^2*e^2*f^2 - 16*a^2*b*c^2*d*e^3*f^2 - 16*a^2*b*c^2*d^3*e*g^2 - 2*a* 
b^4*d*e^3*f*g - 2*b^4*c*d^3*e*f*g - 8*a*b^2*c^2*d^2*e^2*f^2 - 8*a^2*b^2*c* 
d^2*e^2*g^2 + 8*a*b^3*c*d*e^3*f^2 + 8*a*b^3*c*d^3*e*g^2 - 32*a^2*c^3*d^3*e 
*f*g - 32*a^3*c^2*d*e^3*f*g + 16*a*b^2*c^2*d^3*e*f*g - 16*a*b^3*c*d^2*e^2* 
f*g + 16*a^2*b^2*c*d*e^3*f*g + 32*a^2*b*c^2*d^2*e^2*f*g)))^(1/2)*(x*(-(b^3 
*d*g + b^3*e*f + d*g*(-(4*a*c - b^2)^3)^(1/2) - e*f*(-(4*a*c - b^2)^3)^(1/ 
2) + 8*a*c^2*d*f - 2*a*b^2*e*g - 2*b^2*c*d*f + 8*a^2*c*e*g - 4*a*b*c*d*g - 
 4*a*b*c*e*f)/(8*(a*b^4*e^4*f^2 + b^4*c*d^4*g^2 - b^5*d*e^3*f^2 - b^5*d^3* 
e*g^2 + 16*a^2*c^3*d^4*g^2 + 16*a^3*c^2*e^4*f^2 + 16*a^2*c^3*d^2*e^2*f^2 + 
 16*a^3*c^2*d^2*e^2*g^2 + 2*b^5*d^2*e^2*f*g - 8*a*b^2*c^2*d^4*g^2 - 8*a^2* 
b^2*c*e^4*f^2 + a*b^4*d^2*e^2*g^2 + b^4*c*d^2*e^2*f^2 - 16*a^2*b*c^2*d*e^3 
*f^2 - 16*a^2*b*c^2*d^3*e*g^2 - 2*a*b^4*d*e^3*f*g - 2*b^4*c*d^3*e*f*g -...
 

Reduce [F]

\[ \int \frac {1}{c d^2-b d e+a e^2-(2 c d f-b e f-b d g+2 a e g) x^2+\left (c f^2-b f g+a g^2\right ) x^4} \, dx=\int \frac {1}{c \,d^{2}-b d e +a \,e^{2}-\left (2 a e g -b d g -b e f +2 c d f \right ) x^{2}+\left (a \,g^{2}-b f g +c \,f^{2}\right ) x^{4}}d x \] Input:

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

Output:

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