\(\int \frac {x^2}{(d+e x^4) (a+b x^4+c x^8)} \, dx\) [85]

Optimal result

Integrand size = 27, antiderivative size = 634 \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {\sqrt [4]{c} \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac {\sqrt [4]{c} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac {e^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )}{2 \sqrt {2} \sqrt [4]{d} \left (c d^2-b d e+a e^2\right )}+\frac {e^{5/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )}{2 \sqrt {2} \sqrt [4]{d} \left (c d^2-b d e+a e^2\right )}+\frac {\sqrt [4]{c} \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}+\frac {\sqrt [4]{c} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac {e^{5/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} x}{\sqrt {d}+\sqrt {e} x^2}\right )}{2 \sqrt {2} \sqrt [4]{d} \left (c d^2-b d e+a e^2\right )} \] Output:

-1/4*c^(1/4)*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x/ 
(-b-(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)/(a*e^ 
2-b*d*e+c*d^2)-1/4*c^(1/4)*(e-(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctan(2^(1 
/4)*c^(1/4)*x/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/(-b+(-4*a*c+b^2)^(1/2 
))^(1/4)/(a*e^2-b*d*e+c*d^2)+1/4*e^(5/4)*arctan(-1+2^(1/2)*e^(1/4)*x/d^(1/ 
4))*2^(1/2)/d^(1/4)/(a*e^2-b*d*e+c*d^2)+1/4*e^(5/4)*arctan(1+2^(1/2)*e^(1/ 
4)*x/d^(1/4))*2^(1/2)/d^(1/4)/(a*e^2-b*d*e+c*d^2)+1/4*c^(1/4)*(e+(-b*e+2*c 
*d)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b-(-4*a*c+b^2)^(1/2))^ 
(1/4))*2^(1/4)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)/(a*e^2-b*d*e+c*d^2)+1/4*c^(1/ 
4)*(e-(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b+(-4*a 
*c+b^2)^(1/2))^(1/4))*2^(1/4)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)/(a*e^2-b*d*e+c 
*d^2)-1/4*e^(5/4)*arctanh(2^(1/2)*d^(1/4)*e^(1/4)*x/(d^(1/2)+e^(1/2)*x^2)) 
*2^(1/2)/d^(1/4)/(a*e^2-b*d*e+c*d^2)
 

Mathematica [C] (verified)

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

Time = 0.23 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.36 \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {\sqrt {2} e^{5/4} \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )+\log \left (\sqrt {d}-\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} x+\sqrt {e} x^2\right )-\log \left (\sqrt {d}+\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} x+\sqrt {e} x^2\right )\right )-2 \sqrt [4]{d} \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-c d \log (x-\text {$\#$1})+b e \log (x-\text {$\#$1})+c e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{8 \sqrt [4]{d} \left (c d^2+e (-b d+a e)\right )} \] Input:

Integrate[x^2/((d + e*x^4)*(a + b*x^4 + c*x^8)),x]
 

Output:

(Sqrt[2]*e^(5/4)*(-2*ArcTan[1 - (Sqrt[2]*e^(1/4)*x)/d^(1/4)] + 2*ArcTan[1 
+ (Sqrt[2]*e^(1/4)*x)/d^(1/4)] + Log[Sqrt[d] - Sqrt[2]*d^(1/4)*e^(1/4)*x + 
 Sqrt[e]*x^2] - Log[Sqrt[d] + Sqrt[2]*d^(1/4)*e^(1/4)*x + Sqrt[e]*x^2]) - 
2*d^(1/4)*RootSum[a + b*#1^4 + c*#1^8 & , (-(c*d*Log[x - #1]) + b*e*Log[x 
- #1] + c*e*Log[x - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ])/(8*d^(1/4)*(c*d^2 + e 
*(-(b*d) + a*e)))
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 703, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1836, 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.

\(\displaystyle \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\)

\(\Big \downarrow \) 1836

\(\displaystyle \int \left (\frac {e^2 x^2}{\left (d+e x^4\right ) \left (a e^2-b d e+c d^2\right )}+\frac {x^2 \left (-b e+c d-c e x^4\right )}{\left (a+b x^4+c x^8\right ) \left (a e^2-b d e+c d^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt [4]{c} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right ) \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right )}{2\ 2^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt [4]{c} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right ) \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )}{2\ 2^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b} \left (a e^2-b d e+c d^2\right )}-\frac {e^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )}{2 \sqrt {2} \sqrt [4]{d} \left (a e^2-b d e+c d^2\right )}+\frac {e^{5/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}+1\right )}{2 \sqrt {2} \sqrt [4]{d} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt [4]{c} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right ) \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right )}{2\ 2^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt [4]{c} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right ) \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )}{2\ 2^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b} \left (a e^2-b d e+c d^2\right )}+\frac {e^{5/4} \log \left (-\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} x+\sqrt {d}+\sqrt {e} x^2\right )}{4 \sqrt {2} \sqrt [4]{d} \left (a e^2-b d e+c d^2\right )}-\frac {e^{5/4} \log \left (\sqrt {2} \sqrt [4]{d} \sqrt [4]{e} x+\sqrt {d}+\sqrt {e} x^2\right )}{4 \sqrt {2} \sqrt [4]{d} \left (a e^2-b d e+c d^2\right )}\)

Input:

Int[x^2/((d + e*x^4)*(a + b*x^4 + c*x^8)),x]
 

Output:

-1/2*(c^(1/4)*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(2^(1/4)*c^(1/4 
)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(2^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1 
/4)*(c*d^2 - b*d*e + a*e^2)) - (c^(1/4)*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a* 
c])*ArcTan[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4) 
*(-b + Sqrt[b^2 - 4*a*c])^(1/4)*(c*d^2 - b*d*e + a*e^2)) - (e^(5/4)*ArcTan 
[1 - (Sqrt[2]*e^(1/4)*x)/d^(1/4)])/(2*Sqrt[2]*d^(1/4)*(c*d^2 - b*d*e + a*e 
^2)) + (e^(5/4)*ArcTan[1 + (Sqrt[2]*e^(1/4)*x)/d^(1/4)])/(2*Sqrt[2]*d^(1/4 
)*(c*d^2 - b*d*e + a*e^2)) + (c^(1/4)*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c] 
)*ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)])/(2*2^(3/4)* 
(-b - Sqrt[b^2 - 4*a*c])^(1/4)*(c*d^2 - b*d*e + a*e^2)) + (c^(1/4)*(e - (2 
*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 
- 4*a*c])^(1/4)])/(2*2^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)*(c*d^2 - b*d*e 
 + a*e^2)) + (e^(5/4)*Log[Sqrt[d] - Sqrt[2]*d^(1/4)*e^(1/4)*x + Sqrt[e]*x^ 
2])/(4*Sqrt[2]*d^(1/4)*(c*d^2 - b*d*e + a*e^2)) - (e^(5/4)*Log[Sqrt[d] + S 
qrt[2]*d^(1/4)*e^(1/4)*x + Sqrt[e]*x^2])/(4*Sqrt[2]*d^(1/4)*(c*d^2 - b*d*e 
 + a*e^2))
 

Defintions of rubi rules used

rule 1836

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.))/((a_) + (c_.)*(x_)^ 
(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e 
*x^n)^q/(a + b*x^n + c*x^(2*n))), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] 
 && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[q] && Int 
egerQ[m]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (verified)

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

Time = 0.26 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.31

method	result	size
default	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (c e \,\textit {\_R}^{6}+\left (e b -c d \right ) \textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {e \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d}{e}}}{x^{2}+\left (\frac {d}{e}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d}{e}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d}{e}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d}{e}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {d}{e}\right )^{\frac {1}{4}}}\)	\(195\)
risch	\(\text {Expression too large to display}\)	\(8130\)

Input:

int(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

-1/4/(a*e^2-b*d*e+c*d^2)*sum((c*e*_R^6+(b*e-c*d)*_R^2)/(2*_R^7*c+_R^3*b)*l 
n(x-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))+1/8*e/(a*e^2-b*d*e+c*d^2)/(d/e)^(1/4)* 
2^(1/2)*(ln((x^2-(d/e)^(1/4)*x*2^(1/2)+(d/e)^(1/2))/(x^2+(d/e)^(1/4)*x*2^( 
1/2)+(d/e)^(1/2)))+2*arctan(2^(1/2)/(d/e)^(1/4)*x+1)+2*arctan(2^(1/2)/(d/e 
)^(1/4)*x-1))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:

integrate(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:

integrate(x**2/(e*x**4+d)/(c*x**8+b*x**4+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {x^{2}}{{\left (c x^{8} + b x^{4} + a\right )} {\left (e x^{4} + d\right )}} \,d x } \] Input:

integrate(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="maxima")
 

Output:

-1/8*e^2*(sqrt(2)*log(sqrt(e)*x^2 + sqrt(2)*d^(1/4)*e^(1/4)*x + sqrt(d))/( 
d^(1/4)*e^(3/4)) - sqrt(2)*log(sqrt(e)*x^2 - sqrt(2)*d^(1/4)*e^(1/4)*x + s 
qrt(d))/(d^(1/4)*e^(3/4)) - sqrt(2)*log((2*sqrt(e)*x - sqrt(2)*sqrt(-sqrt( 
d)*sqrt(e)) + sqrt(2)*d^(1/4)*e^(1/4))/(2*sqrt(e)*x + sqrt(2)*sqrt(-sqrt(d 
)*sqrt(e)) + sqrt(2)*d^(1/4)*e^(1/4)))/(sqrt(-sqrt(d)*sqrt(e))*sqrt(e)) - 
sqrt(2)*log((2*sqrt(e)*x - sqrt(2)*sqrt(-sqrt(d)*sqrt(e)) - sqrt(2)*d^(1/4 
)*e^(1/4))/(2*sqrt(e)*x + sqrt(2)*sqrt(-sqrt(d)*sqrt(e)) - sqrt(2)*d^(1/4) 
*e^(1/4)))/(sqrt(-sqrt(d)*sqrt(e))*sqrt(e)))/(c*d^2 - b*d*e + a*e^2) - int 
egrate((c*e*x^6 - (c*d - b*e)*x^2)/(c*x^8 + b*x^4 + a), x)/(c*d^2 - b*d*e 
+ a*e^2)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:

integrate(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 52.16 (sec) , antiderivative size = 325549, normalized size of antiderivative = 513.48 \[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

int(x^2/((d + e*x^4)*(a + b*x^4 + c*x^8)),x)
 

Output:

(atan((((x*(4*a^2*c^9*d*e^10 + 4*a*b*c^9*d^2*e^9) - ((-2*(-((b^9*e^4 + b^5 
*c^4*d^4 + b^4*e^4*(-(4*a*c - b^2)^5)^(1/2) + c^4*d^4*(-(4*a*c - b^2)^5)^( 
1/2) - 8*a*b^3*c^5*d^4 + 16*a^2*b*c^6*d^4 + 80*a^4*b*c^4*e^4 + 128*a^3*c^6 
*d^3*e - 128*a^4*c^5*d*e^3 - 4*b^6*c^3*d^3*e + 61*a^2*b^5*c^2*e^4 - 120*a^ 
3*b^3*c^3*e^4 + a^2*c^2*e^4*(-(4*a*c - b^2)^5)^(1/2) + 6*b^7*c^2*d^2*e^2 - 
 13*a*b^7*c*e^4 - 4*b^8*c*d*e^3 + 240*a^2*b^3*c^4*d^2*e^2 + 6*b^2*c^2*d^2* 
e^2*(-(4*a*c - b^2)^5)^(1/2) - 3*a*b^2*c*e^4*(-(4*a*c - b^2)^5)^(1/2) + 40 
*a*b^4*c^4*d^3*e + 48*a*b^6*c^2*d*e^3 - 4*b*c^3*d^3*e*(-(4*a*c - b^2)^5)^( 
1/2) - 4*b^3*c*d*e^3*(-(4*a*c - b^2)^5)^(1/2) - 66*a*b^5*c^3*d^2*e^2 - 128 
*a^2*b^2*c^5*d^3*e - 200*a^2*b^4*c^3*d*e^3 - 288*a^3*b*c^5*d^2*e^2 + 320*a 
^3*b^2*c^4*d*e^3 - 6*a*c^3*d^2*e^2*(-(4*a*c - b^2)^5)^(1/2) + 8*a*b*c^2*d* 
e^3*(-(4*a*c - b^2)^5)^(1/2))*(a^5*b^8*e^8 + 256*a^5*c^8*d^8 + 256*a^9*c^4 
*e^8 + a*b^8*c^4*d^8 - 16*a^6*b^6*c*e^8 + a*b^12*d^4*e^4 - 4*a^4*b^9*d*e^7 
 - 16*a^2*b^6*c^5*d^8 + 96*a^3*b^4*c^6*d^8 - 256*a^4*b^2*c^7*d^8 + 96*a^7* 
b^4*c^2*e^8 - 256*a^8*b^2*c^3*e^8 - 4*a^2*b^11*d^3*e^5 + 6*a^3*b^10*d^2*e^ 
6 + 1024*a^6*c^7*d^6*e^2 + 1536*a^7*c^6*d^4*e^4 + 1024*a^8*c^5*d^2*e^6 - 9 
2*a^2*b^8*c^3*d^6*e^2 + 52*a^2*b^9*c^2*d^5*e^3 + 512*a^3*b^6*c^4*d^6*e^2 - 
 192*a^3*b^7*c^3*d^5*e^3 - 90*a^3*b^8*c^2*d^4*e^4 - 1152*a^4*b^4*c^5*d^6*e 
^2 - 128*a^4*b^5*c^4*d^5*e^3 + 800*a^4*b^6*c^3*d^4*e^4 - 192*a^4*b^7*c^2*d 
^3*e^5 + 512*a^5*b^2*c^6*d^6*e^2 + 2048*a^5*b^3*c^5*d^5*e^3 - 2240*a^5*...
 

Reduce [F]

\[ \int \frac {x^2}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {x^{2}}{\left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:

int(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x)
 

Output:

int(x^2/(e*x^4+d)/(c*x^8+b*x^4+a),x)