\(\int \frac {1}{x^2 (d+e x^3)^2 (a+b x^3+c x^6)} \, dx\) [64]

Optimal result

Integrand size = 27, antiderivative size = 1378 \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx =\text {Too large to display} \] Output:

-1/a/d^2/x-1/3*e^3*x^2/d^2/(a*e^2-b*d*e+c*d^2)/(e*x^3+d)+1/6*c^(1/3)*(c^2* 
d^2+b^2*e^2-c*e*(a*e+2*b*d)-(2*b^2*c*d*e-4*a*c^2*d*e-b^3*e^2-b*c*(-3*a*e^2 
+c*d^2))/(-4*a*c+b^2)^(1/2))*arctan(1/3*(1-2*2^(1/3)*c^(1/3)*x/(b-(-4*a*c+ 
b^2)^(1/2))^(1/3))*3^(1/2))*2^(1/3)*3^(1/2)/a/(b-(-4*a*c+b^2)^(1/2))^(1/3) 
/(a*e^2-b*d*e+c*d^2)^2+1/6*c^(1/3)*(c^2*d^2+b^2*e^2-c*e*(a*e+2*b*d)+(2*b^2 
*c*d*e-4*a*c^2*d*e-b^3*e^2-b*c*(-3*a*e^2+c*d^2))/(-4*a*c+b^2)^(1/2))*arcta 
n(1/3*(1-2*2^(1/3)*c^(1/3)*x/(b+(-4*a*c+b^2)^(1/2))^(1/3))*3^(1/2))*2^(1/3 
)*3^(1/2)/a/(b+(-4*a*c+b^2)^(1/2))^(1/3)/(a*e^2-b*d*e+c*d^2)^2+1/9*e^(7/3) 
*(10*c*d^2-e*(-4*a*e+7*b*d))*arctan(1/3*(d^(1/3)-2*e^(1/3)*x)*3^(1/2)/d^(1 
/3))*3^(1/2)/d^(7/3)/(a*e^2-b*d*e+c*d^2)^2+1/6*c^(1/3)*(c^2*d^2+b^2*e^2-c* 
e*(a*e+2*b*d)-(2*b^2*c*d*e-4*a*c^2*d*e-b^3*e^2-b*c*(-3*a*e^2+c*d^2))/(-4*a 
*c+b^2)^(1/2))*ln((b-(-4*a*c+b^2)^(1/2))^(1/3)+2^(1/3)*c^(1/3)*x)*2^(1/3)/ 
a/(b-(-4*a*c+b^2)^(1/2))^(1/3)/(a*e^2-b*d*e+c*d^2)^2+1/6*c^(1/3)*(c^2*d^2+ 
b^2*e^2-c*e*(a*e+2*b*d)+(2*b^2*c*d*e-4*a*c^2*d*e-b^3*e^2-b*c*(-3*a*e^2+c*d 
^2))/(-4*a*c+b^2)^(1/2))*ln((b+(-4*a*c+b^2)^(1/2))^(1/3)+2^(1/3)*c^(1/3)*x 
)*2^(1/3)/a/(b+(-4*a*c+b^2)^(1/2))^(1/3)/(a*e^2-b*d*e+c*d^2)^2+1/9*e^(7/3) 
*(10*c*d^2-e*(-4*a*e+7*b*d))*ln(d^(1/3)+e^(1/3)*x)/d^(7/3)/(a*e^2-b*d*e+c* 
d^2)^2-1/12*c^(1/3)*(c^2*d^2+b^2*e^2-c*e*(a*e+2*b*d)-(2*b^2*c*d*e-4*a*c^2* 
d*e-b^3*e^2-b*c*(-3*a*e^2+c*d^2))/(-4*a*c+b^2)^(1/2))*ln((b-(-4*a*c+b^2)^( 
1/2))^(2/3)-2^(1/3)*c^(1/3)*(b-(-4*a*c+b^2)^(1/2))^(1/3)*x+2^(2/3)*c^(2...
 

Mathematica [C] (verified)

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

Time = 0.50 (sec) , antiderivative size = 475, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\frac {-6 a \sqrt [3]{d} e^3 \left (c d^2+e (-b d+a e)\right ) x^3-18 \sqrt [3]{d} \left (c d^2+e (-b d+a e)\right )^2 \left (d+e x^3\right )+2 \sqrt {3} a e^{7/3} \left (10 c d^2+e (-7 b d+4 a e)\right ) x \left (d+e x^3\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )+2 a e^{7/3} \left (10 c d^2+e (-7 b d+4 a e)\right ) x \left (d+e x^3\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )-a e^{7/3} \left (10 c d^2+e (-7 b d+4 a e)\right ) x \left (d+e x^3\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )-6 d^{7/3} x \left (d+e x^3\right ) \text {RootSum}\left [a+b \text {$\#$1}^3+c \text {$\#$1}^6\&,\frac {b c^2 d^2 \log (x-\text {$\#$1})-2 b^2 c d e \log (x-\text {$\#$1})+2 a c^2 d e \log (x-\text {$\#$1})+b^3 e^2 \log (x-\text {$\#$1})-2 a b c e^2 \log (x-\text {$\#$1})+c^3 d^2 \log (x-\text {$\#$1}) \text {$\#$1}^3-2 b c^2 d e \log (x-\text {$\#$1}) \text {$\#$1}^3+b^2 c e^2 \log (x-\text {$\#$1}) \text {$\#$1}^3-a c^2 e^2 \log (x-\text {$\#$1}) \text {$\#$1}^3}{b \text {$\#$1}+2 c \text {$\#$1}^4}\&\right ]}{18 a d^{7/3} \left (c d^2+e (-b d+a e)\right )^2 x \left (d+e x^3\right )} \] Input:

Integrate[1/(x^2*(d + e*x^3)^2*(a + b*x^3 + c*x^6)),x]
 

Output:

(-6*a*d^(1/3)*e^3*(c*d^2 + e*(-(b*d) + a*e))*x^3 - 18*d^(1/3)*(c*d^2 + e*( 
-(b*d) + a*e))^2*(d + e*x^3) + 2*Sqrt[3]*a*e^(7/3)*(10*c*d^2 + e*(-7*b*d + 
 4*a*e))*x*(d + e*x^3)*ArcTan[(1 - (2*e^(1/3)*x)/d^(1/3))/Sqrt[3]] + 2*a*e 
^(7/3)*(10*c*d^2 + e*(-7*b*d + 4*a*e))*x*(d + e*x^3)*Log[d^(1/3) + e^(1/3) 
*x] - a*e^(7/3)*(10*c*d^2 + e*(-7*b*d + 4*a*e))*x*(d + e*x^3)*Log[d^(2/3) 
- d^(1/3)*e^(1/3)*x + e^(2/3)*x^2] - 6*d^(7/3)*x*(d + e*x^3)*RootSum[a + b 
*#1^3 + c*#1^6 & , (b*c^2*d^2*Log[x - #1] - 2*b^2*c*d*e*Log[x - #1] + 2*a* 
c^2*d*e*Log[x - #1] + b^3*e^2*Log[x - #1] - 2*a*b*c*e^2*Log[x - #1] + c^3* 
d^2*Log[x - #1]*#1^3 - 2*b*c^2*d*e*Log[x - #1]*#1^3 + b^2*c*e^2*Log[x - #1 
]*#1^3 - a*c^2*e^2*Log[x - #1]*#1^3)/(b*#1 + 2*c*#1^4) & ])/(18*a*d^(7/3)* 
(c*d^2 + e*(-(b*d) + a*e))^2*x*(d + e*x^3))
 

Rubi [A] (verified)

Time = 3.98 (sec) , antiderivative size = 1545, normalized size of antiderivative = 1.12, 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.

\(\displaystyle \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx\)

\(\Big \downarrow \) 1836

\(\displaystyle \int \left (\frac {x \left (-c x^3 \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )-(c d-b e) \left (2 a c e+b^2 (-e)+b c d\right )\right )}{a \left (a+b x^3+c x^6\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac {e^3 x \left (e (2 b d-a e)-3 c d^2\right )}{d^2 \left (d+e x^3\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {e^3 x}{d \left (d+e x^3\right )^2 \left (a e^2-b d e+c d^2\right )}+\frac {1}{a d^2 x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {x^2 e^3}{3 d^2 \left (c d^2-b e d+a e^2\right ) \left (e x^3+d\right )}+\frac {\left (3 c d^2-e (2 b d-a e)\right ) \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right ) e^{7/3}}{\sqrt {3} d^{7/3} \left (c d^2-b e d+a e^2\right )^2}+\frac {\arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right ) e^{7/3}}{3 \sqrt {3} d^{7/3} \left (c d^2-b e d+a e^2\right )}+\frac {\left (3 c d^2-e (2 b d-a e)\right ) \log \left (\sqrt [3]{e} x+\sqrt [3]{d}\right ) e^{7/3}}{3 d^{7/3} \left (c d^2-b e d+a e^2\right )^2}+\frac {\log \left (\sqrt [3]{e} x+\sqrt [3]{d}\right ) e^{7/3}}{9 d^{7/3} \left (c d^2-b e d+a e^2\right )}-\frac {\left (3 c d^2-e (2 b d-a e)\right ) \log \left (e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}\right ) e^{7/3}}{6 d^{7/3} \left (c d^2-b e d+a e^2\right )^2}-\frac {\log \left (e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}\right ) e^{7/3}}{18 d^{7/3} \left (c d^2-b e d+a e^2\right )}+\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)-\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b-\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a \sqrt [3]{b-\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}+\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)+\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \sqrt [3]{c} x}{\sqrt [3]{b+\sqrt {b^2-4 a c}}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a \sqrt [3]{b+\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}+\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)-\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b-\sqrt {b^2-4 a c}}\right )}{3\ 2^{2/3} a \sqrt [3]{b-\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}+\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)+\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \log \left (\sqrt [3]{2} \sqrt [3]{c} x+\sqrt [3]{b+\sqrt {b^2-4 a c}}\right )}{3\ 2^{2/3} a \sqrt [3]{b+\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}-\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)-\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b-\sqrt {b^2-4 a c}} x+\left (b-\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6\ 2^{2/3} a \sqrt [3]{b-\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}-\frac {\sqrt [3]{c} \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)+\frac {-e^2 b^3+2 c d e b^2-c \left (c d^2-3 a e^2\right ) b-4 a c^2 d e}{\sqrt {b^2-4 a c}}\right ) \log \left (2^{2/3} c^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{c} \sqrt [3]{b+\sqrt {b^2-4 a c}} x+\left (b+\sqrt {b^2-4 a c}\right )^{2/3}\right )}{6\ 2^{2/3} a \sqrt [3]{b+\sqrt {b^2-4 a c}} \left (c d^2-b e d+a e^2\right )^2}-\frac {1}{a d^2 x}\)

Input:

Int[1/(x^2*(d + e*x^3)^2*(a + b*x^3 + c*x^6)),x]
 

Output:

-(1/(a*d^2*x)) - (e^3*x^2)/(3*d^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x^3)) + ( 
c^(1/3)*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a*e) - (2*b^2*c*d*e - 4*a*c^2*d* 
e - b^3*e^2 - b*c*(c*d^2 - 3*a*e^2))/Sqrt[b^2 - 4*a*c])*ArcTan[(1 - (2*2^( 
1/3)*c^(1/3)*x)/(b - Sqrt[b^2 - 4*a*c])^(1/3))/Sqrt[3]])/(2^(2/3)*Sqrt[3]* 
a*(b - Sqrt[b^2 - 4*a*c])^(1/3)*(c*d^2 - b*d*e + a*e^2)^2) + (c^(1/3)*(c^2 
*d^2 + b^2*e^2 - c*e*(2*b*d + a*e) + (2*b^2*c*d*e - 4*a*c^2*d*e - b^3*e^2 
- b*c*(c*d^2 - 3*a*e^2))/Sqrt[b^2 - 4*a*c])*ArcTan[(1 - (2*2^(1/3)*c^(1/3) 
*x)/(b + Sqrt[b^2 - 4*a*c])^(1/3))/Sqrt[3]])/(2^(2/3)*Sqrt[3]*a*(b + Sqrt[ 
b^2 - 4*a*c])^(1/3)*(c*d^2 - b*d*e + a*e^2)^2) + (e^(7/3)*ArcTan[(d^(1/3) 
- 2*e^(1/3)*x)/(Sqrt[3]*d^(1/3))])/(3*Sqrt[3]*d^(7/3)*(c*d^2 - b*d*e + a*e 
^2)) + (e^(7/3)*(3*c*d^2 - e*(2*b*d - a*e))*ArcTan[(d^(1/3) - 2*e^(1/3)*x) 
/(Sqrt[3]*d^(1/3))])/(Sqrt[3]*d^(7/3)*(c*d^2 - b*d*e + a*e^2)^2) + (c^(1/3 
)*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a*e) - (2*b^2*c*d*e - 4*a*c^2*d*e - b^ 
3*e^2 - b*c*(c*d^2 - 3*a*e^2))/Sqrt[b^2 - 4*a*c])*Log[(b - Sqrt[b^2 - 4*a* 
c])^(1/3) + 2^(1/3)*c^(1/3)*x])/(3*2^(2/3)*a*(b - Sqrt[b^2 - 4*a*c])^(1/3) 
*(c*d^2 - b*d*e + a*e^2)^2) + (c^(1/3)*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a 
*e) + (2*b^2*c*d*e - 4*a*c^2*d*e - b^3*e^2 - b*c*(c*d^2 - 3*a*e^2))/Sqrt[b 
^2 - 4*a*c])*Log[(b + Sqrt[b^2 - 4*a*c])^(1/3) + 2^(1/3)*c^(1/3)*x])/(3*2^ 
(2/3)*a*(b + Sqrt[b^2 - 4*a*c])^(1/3)*(c*d^2 - b*d*e + a*e^2)^2) + (e^(7/3 
)*Log[d^(1/3) + e^(1/3)*x])/(9*d^(7/3)*(c*d^2 - b*d*e + a*e^2)) + (e^(7...
 

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 = 0.73 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.23

method	result	size
default	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6} c +\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (c \left (a c \,e^{2}-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \textit {\_R}^{4}+\left (2 a b c \,e^{2}-2 a \,c^{2} d e -b^{3} e^{2}+2 b^{2} c d e -b \,c^{2} d^{2}\right ) \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{5} c +\textit {\_R}^{2} b}}{3 a \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {e^{3} \left (\frac {\left (\frac {1}{3} a \,e^{2}-\frac {1}{3} b d e +\frac {1}{3} c \,d^{2}\right ) x^{2}}{e \,x^{3}+d}+\left (\frac {4}{3} a \,e^{2}-\frac {7}{3} b d e +\frac {10}{3} c \,d^{2}\right ) \left (-\frac {\ln \left (x +\left (\frac {d}{e}\right )^{\frac {1}{3}}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{3}} x +\left (\frac {d}{e}\right )^{\frac {2}{3}}\right )}{6 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{e}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}\right )\right )}{d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {1}{a \,d^{2} x}\)	\(319\)
risch	\(\text {Expression too large to display}\)	\(26685\)

Input:

int(1/x^2/(e*x^3+d)^2/(c*x^6+b*x^3+a),x,method=_RETURNVERBOSE)
 

Output:

1/3/a/(a*e^2-b*d*e+c*d^2)^2*sum((c*(a*c*e^2-b^2*e^2+2*b*c*d*e-c^2*d^2)*_R^ 
4+(2*a*b*c*e^2-2*a*c^2*d*e-b^3*e^2+2*b^2*c*d*e-b*c^2*d^2)*_R)/(2*_R^5*c+_R 
^2*b)*ln(x-_R),_R=RootOf(_Z^6*c+_Z^3*b+a))-e^3/d^2/(a*e^2-b*d*e+c*d^2)^2*( 
(1/3*a*e^2-1/3*b*d*e+1/3*c*d^2)*x^2/(e*x^3+d)+(4/3*a*e^2-7/3*b*d*e+10/3*c* 
d^2)*(-1/3/e/(d/e)^(1/3)*ln(x+(d/e)^(1/3))+1/6/e/(d/e)^(1/3)*ln(x^2-(d/e)^ 
(1/3)*x+(d/e)^(2/3))+1/3*3^(1/2)/e/(d/e)^(1/3)*arctan(1/3*3^(1/2)*(2/(d/e) 
^(1/3)*x-1))))-1/a/d^2/x
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x^2/(e*x^3+d)^2/(c*x^6+b*x^3+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x**2/(e*x**3+d)**2/(c*x**6+b*x**3+a),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x^2/(e*x^3+d)^2/(c*x^6+b*x^3+a),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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

\[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )} {\left (e x^{3} + d\right )}^{2} x^{2}} \,d x } \] Input:

integrate(1/x^2/(e*x^3+d)^2/(c*x^6+b*x^3+a),x, algorithm="giac")
 

Output:

integrate(1/((c*x^6 + b*x^3 + a)*(e*x^3 + d)^2*x^2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 95.39 (sec) , antiderivative size = 79010, normalized size of antiderivative = 57.34 \[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\text {Too large to display} \] Input:

int(1/(x^2*(d + e*x^3)^2*(a + b*x^3 + c*x^6)),x)
 

Output:

symsum(log(root(8453100546*a^14*b^6*c^2*d^11*e^20*z^9 + 8453100546*a^6*b^6 
*c^10*d^27*e^4*z^9 + 218868661440*a^14*b^3*c^5*d^14*e^17*z^9 + 21886866144 
0*a^9*b^3*c^10*d^24*e^7*z^9 + 4115479104*a^16*b^3*c^3*d^10*e^21*z^9 + 4115 
479104*a^7*b^3*c^12*d^28*e^3*z^9 + 1332854028*a^14*b^7*c*d^10*e^21*z^9 + 3 
998562084*a^5*b^9*c^8*d^26*e^5*z^9 - 25394376744*a^9*b^10*c^3*d^17*e^14*z^ 
9 - 25394376744*a^7*b^10*c^5*d^21*e^10*z^9 - 3788111448*a^5*b^10*c^7*d^25* 
e^6*z^9 + 149770702620*a^10*b^6*c^6*d^19*e^12*z^9 + 22448067840*a^16*b*c^5 
*d^12*e^19*z^9 + 22448067840*a^9*b*c^12*d^26*e^5*z^9 + 158258878272*a^9*b^ 
8*c^5*d^19*e^12*z^9 - 7926973956*a^13*b^7*c^2*d^12*e^19*z^9 - 7926973956*a 
^6*b^7*c^9*d^26*e^5*z^9 - 20764462752*a^16*b^2*c^4*d^11*e^20*z^9 - 2076446 
2752*a^8*b^2*c^12*d^27*e^4*z^9 - 3577660812*a^8*b^12*c^2*d^17*e^14*z^9 - 3 
577660812*a^6*b^12*c^4*d^21*e^10*z^9 + 119535961248*a^12*b^5*c^5*d^16*e^15 
*z^9 + 119535961248*a^9*b^5*c^8*d^22*e^9*z^9 + 527529594240*a^13*b^3*c^6*d 
^16*e^15*z^9 + 527529594240*a^10*b^3*c^9*d^22*e^9*z^9 + 50765370084*a^13*b 
^6*c^3*d^13*e^18*z^9 + 50765370084*a^7*b^6*c^9*d^25*e^6*z^9 - 701502120*a^ 
8*b^13*c*d^16*e^15*z^9 - 256749775920*a^14*b^2*c^6*d^15*e^16*z^9 - 2567497 
75920*a^10*b^2*c^10*d^23*e^8*z^9 + 631351908*a^7*b^14*c*d^17*e^14*z^9 + 50 
508152640*a^15*b^3*c^4*d^12*e^19*z^9 + 50508152640*a^8*b^3*c^11*d^26*e^5*z 
^9 - 414523980*a^15*b^6*c*d^9*e^22*z^9 + 408146688*a^18*b*c^3*d^8*e^23*z^9 
 + 114748209279*a^12*b^6*c^4*d^15*e^16*z^9 + 114748209279*a^8*b^6*c^8*d...
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (d+e x^3\right )^2 \left (a+b x^3+c x^6\right )} \, dx=\int \frac {1}{c \,e^{2} x^{14}+b \,e^{2} x^{11}+2 c d e \,x^{11}+a \,e^{2} x^{8}+2 b d e \,x^{8}+c \,d^{2} x^{8}+2 a d e \,x^{5}+b \,d^{2} x^{5}+a \,d^{2} x^{2}}d x \] Input:

int(1/x^2/(e*x^3+d)^2/(c*x^6+b*x^3+a),x)
 

Output:

int(1/(a*d**2*x**2 + 2*a*d*e*x**5 + a*e**2*x**8 + b*d**2*x**5 + 2*b*d*e*x* 
*8 + b*e**2*x**11 + c*d**2*x**8 + 2*c*d*e*x**11 + c*e**2*x**14),x)