Integrand size = 22, antiderivative size = 643 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=-\frac {2 \left (3 b^2 c^2 d+2 a c^3 d-2 a b^3 e-6 a^2 b c e\right ) x}{e^2}-\frac {\left (4 b c^3 d-b^4 e-12 a b^2 c e-6 a^2 c^2 e\right ) x^2}{2 e^2}-\frac {c \left (c^3 d-4 b^3 e-12 a b c e\right ) x^3}{3 e^2}+\frac {c^2 \left (3 b^2+2 a c\right ) x^4}{2 e}+\frac {4 b c^3 x^5}{5 e}+\frac {c^4 x^6}{6 e}-\frac {\left (b \sqrt [3]{d}+a \sqrt [3]{e}\right ) \left (4 c^3 d^2+6 c^2 \left (b d^{5/3} \sqrt [3]{e}-a d^{4/3} e^{2/3}\right )-12 a b c d e-e \left (b^3 d+3 a b^2 d^{2/3} \sqrt [3]{e}-3 a^2 b \sqrt [3]{d} e^{2/3}-a^3 e\right )\right ) \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt {3} d^{2/3} e^{8/3}}+\frac {\left (\sqrt [3]{e} \left (6 b^2 c^2 d^2+4 a c^3 d^2-4 a b^3 d e-12 a^2 b c d e+a^4 e^2\right )+\sqrt [3]{d} \left (b^4 d e+12 a b^2 c d e+6 a^2 c^2 d e-4 b \left (c^3 d^2+a^3 e^2\right )\right )\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} e^{8/3}}-\frac {\left (6 b^2 c^2 d^2+4 a c^3 d^2-4 a b^3 d e-12 a^2 b c d e+a^4 e^2+\frac {\sqrt [3]{d} \left (b^4 d e+12 a b^2 c d e+6 a^2 c^2 d e-4 b \left (c^3 d^2+a^3 e^2\right )\right )}{\sqrt [3]{e}}\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{6 d^{2/3} e^{7/3}}+\frac {\left (c^4 d^2-12 a b c^2 d e+6 a^2 b^2 e^2-4 c e \left (b^3 d-a^3 e\right )\right ) \log \left (d+e x^3\right )}{3 e^3} \] Output:
-2*(-6*a^2*b*c*e-2*a*b^3*e+2*a*c^3*d+3*b^2*c^2*d)*x/e^2-1/2*(-6*a^2*c^2*e- 12*a*b^2*c*e-b^4*e+4*b*c^3*d)*x^2/e^2-1/3*c*(-12*a*b*c*e-4*b^3*e+c^3*d)*x^ 3/e^2+1/2*c^2*(2*a*c+3*b^2)*x^4/e+4/5*b*c^3*x^5/e+1/6*c^4*x^6/e-1/3*(b*d^( 1/3)+a*e^(1/3))*(4*c^3*d^2+6*c^2*(b*d^(5/3)*e^(1/3)-a*d^(4/3)*e^(2/3))-12* a*b*c*d*e-e*(b^3*d+3*a*b^2*d^(2/3)*e^(1/3)-3*a^2*b*d^(1/3)*e^(2/3)-a^3*e)) *arctan(1/3*(d^(1/3)-2*e^(1/3)*x)*3^(1/2)/d^(1/3))*3^(1/2)/d^(2/3)/e^(8/3) +1/3*(e^(1/3)*(a^4*e^2-12*a^2*b*c*d*e-4*a*b^3*d*e+4*a*c^3*d^2+6*b^2*c^2*d^ 2)+d^(1/3)*(b^4*d*e+12*a*b^2*c*d*e+6*a^2*c^2*d*e-4*b*(a^3*e^2+c^3*d^2)))*l n(d^(1/3)+e^(1/3)*x)/d^(2/3)/e^(8/3)-1/6*(6*b^2*c^2*d^2+4*a*c^3*d^2-4*a*b^ 3*d*e-12*a^2*b*c*d*e+a^4*e^2+d^(1/3)*(b^4*d*e+12*a*b^2*c*d*e+6*a^2*c^2*d*e -4*b*(a^3*e^2+c^3*d^2))/e^(1/3))*ln(d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2) /d^(2/3)/e^(7/3)+1/3*(c^4*d^2-12*a*b*c^2*d*e+6*a^2*b^2*e^2-4*c*e*(-a^3*e+b ^3*d))*ln(e*x^3+d)/e^3
Time = 0.23 (sec) , antiderivative size = 678, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=\frac {60 e^{2/3} \left (-3 b^2 c^2 d-2 a c^3 d+2 a b^3 e+6 a^2 b c e\right ) x+15 e^{2/3} \left (-4 b c^3 d+b^4 e+12 a b^2 c e+6 a^2 c^2 e\right ) x^2+10 c e^{2/3} \left (-c^3 d+4 b^3 e+12 a b c e\right ) x^3+15 c^2 \left (3 b^2+2 a c\right ) e^{5/3} x^4+24 b c^3 e^{5/3} x^5+5 c^4 e^{5/3} x^6+\frac {10 \sqrt {3} \left (b \sqrt [3]{d}+a \sqrt [3]{e}\right ) \left (-4 c^3 d^2+c^2 \left (-6 b d^{5/3} \sqrt [3]{e}+6 a d^{4/3} e^{2/3}\right )+12 a b c d e+e \left (b^3 d+3 a b^2 d^{2/3} \sqrt [3]{e}-3 a^2 b \sqrt [3]{d} e^{2/3}-a^3 e\right )\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{d^{2/3}}+\frac {10 \left (4 a c^3 d^2 \sqrt [3]{e}+b^4 d^{4/3} e+6 a^2 c^2 d^{4/3} e-4 a b^3 d e^{4/3}+a^4 e^{7/3}+6 b^2 \left (c^2 d^2 \sqrt [3]{e}+2 a c d^{4/3} e\right )-4 b \left (c^3 d^{7/3}+3 a^2 c d e^{4/3}+a^3 \sqrt [3]{d} e^2\right )\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{d^{2/3}}-\frac {5 \left (4 a c^3 d^2 \sqrt [3]{e}+b^4 d^{4/3} e+6 a^2 c^2 d^{4/3} e-4 a b^3 d e^{4/3}+a^4 e^{7/3}+6 b^2 \left (c^2 d^2 \sqrt [3]{e}+2 a c d^{4/3} e\right )-4 b \left (c^3 d^{7/3}+3 a^2 c d e^{4/3}+a^3 \sqrt [3]{d} e^2\right )\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{d^{2/3}}+\frac {10 \left (c^4 d^2-12 a b c^2 d e+6 a^2 b^2 e^2+4 c e \left (-b^3 d+a^3 e\right )\right ) \log \left (d+e x^3\right )}{\sqrt [3]{e}}}{30 e^{8/3}} \] Input:
Integrate[(a + b*x + c*x^2)^4/(d + e*x^3),x]
Output:
(60*e^(2/3)*(-3*b^2*c^2*d - 2*a*c^3*d + 2*a*b^3*e + 6*a^2*b*c*e)*x + 15*e^ (2/3)*(-4*b*c^3*d + b^4*e + 12*a*b^2*c*e + 6*a^2*c^2*e)*x^2 + 10*c*e^(2/3) *(-(c^3*d) + 4*b^3*e + 12*a*b*c*e)*x^3 + 15*c^2*(3*b^2 + 2*a*c)*e^(5/3)*x^ 4 + 24*b*c^3*e^(5/3)*x^5 + 5*c^4*e^(5/3)*x^6 + (10*Sqrt[3]*(b*d^(1/3) + a* e^(1/3))*(-4*c^3*d^2 + c^2*(-6*b*d^(5/3)*e^(1/3) + 6*a*d^(4/3)*e^(2/3)) + 12*a*b*c*d*e + e*(b^3*d + 3*a*b^2*d^(2/3)*e^(1/3) - 3*a^2*b*d^(1/3)*e^(2/3 ) - a^3*e))*ArcTan[(1 - (2*e^(1/3)*x)/d^(1/3))/Sqrt[3]])/d^(2/3) + (10*(4* a*c^3*d^2*e^(1/3) + b^4*d^(4/3)*e + 6*a^2*c^2*d^(4/3)*e - 4*a*b^3*d*e^(4/3 ) + a^4*e^(7/3) + 6*b^2*(c^2*d^2*e^(1/3) + 2*a*c*d^(4/3)*e) - 4*b*(c^3*d^( 7/3) + 3*a^2*c*d*e^(4/3) + a^3*d^(1/3)*e^2))*Log[d^(1/3) + e^(1/3)*x])/d^( 2/3) - (5*(4*a*c^3*d^2*e^(1/3) + b^4*d^(4/3)*e + 6*a^2*c^2*d^(4/3)*e - 4*a *b^3*d*e^(4/3) + a^4*e^(7/3) + 6*b^2*(c^2*d^2*e^(1/3) + 2*a*c*d^(4/3)*e) - 4*b*(c^3*d^(7/3) + 3*a^2*c*d*e^(4/3) + a^3*d^(1/3)*e^2))*Log[d^(2/3) - d^ (1/3)*e^(1/3)*x + e^(2/3)*x^2])/d^(2/3) + (10*(c^4*d^2 - 12*a*b*c^2*d*e + 6*a^2*b^2*e^2 + 4*c*e*(-(b^3*d) + a^3*e))*Log[d + e*x^3])/e^(1/3))/(30*e^( 8/3))
Time = 2.08 (sec) , antiderivative size = 643, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2426, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \int \left (-\frac {x \left (-6 a^2 c^2 e-12 a b^2 c e+b^4 (-e)+4 b c^3 d\right )}{e^2}-\frac {2 \left (-6 a^2 b c e-2 a b^3 e+2 a c^3 d+3 b^2 c^2 d\right )}{e^2}+\frac {a^4 e^2-12 a^2 b c d e-x \left (-4 b \left (a^3 e^2+c^3 d^2\right )+6 a^2 c^2 d e+12 a b^2 c d e+b^4 d e\right )+x^2 \left (-4 c e \left (b^3 d-a^3 e\right )+6 a^2 b^2 e^2-12 a b c^2 d e+c^4 d^2\right )-4 a b^3 d e+4 a c^3 d^2+6 b^2 c^2 d^2}{e^2 \left (d+e x^3\right )}-\frac {c x^2 \left (-12 a b c e-4 b^3 e+c^3 d\right )}{e^2}+\frac {2 c^2 x^3 \left (2 a c+3 b^2\right )}{e}+\frac {4 b c^3 x^4}{e}+\frac {c^4 x^5}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {x^2 \left (-6 a^2 c^2 e-12 a b^2 c e+b^4 (-e)+4 b c^3 d\right )}{2 e^2}-\frac {2 x \left (-6 a^2 b c e-2 a b^3 e+2 a c^3 d+3 b^2 c^2 d\right )}{e^2}-\frac {\left (a \sqrt [3]{e}+b \sqrt [3]{d}\right ) \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right ) \left (-e \left (a^3 (-e)-3 a^2 b \sqrt [3]{d} e^{2/3}+3 a b^2 d^{2/3} \sqrt [3]{e}+b^3 d\right )+6 c^2 \left (b d^{5/3} \sqrt [3]{e}-a d^{4/3} e^{2/3}\right )-12 a b c d e+4 c^3 d^2\right )}{\sqrt {3} d^{2/3} e^{8/3}}+\frac {\log \left (d+e x^3\right ) \left (-4 c e \left (b^3 d-a^3 e\right )+6 a^2 b^2 e^2-12 a b c^2 d e+c^4 d^2\right )}{3 e^3}-\frac {\log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right ) \left (a^4 e^2-12 a^2 b c d e+\frac {\sqrt [3]{d} \left (-4 b \left (a^3 e^2+c^3 d^2\right )+6 a^2 c^2 d e+12 a b^2 c d e+b^4 d e\right )}{\sqrt [3]{e}}-4 a b^3 d e+4 a c^3 d^2+6 b^2 c^2 d^2\right )}{6 d^{2/3} e^{7/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \left (\sqrt [3]{e} \left (a^4 e^2-12 a^2 b c d e-4 a b^3 d e+4 a c^3 d^2+6 b^2 c^2 d^2\right )+\sqrt [3]{d} \left (-4 b \left (a^3 e^2+c^3 d^2\right )+6 a^2 c^2 d e+12 a b^2 c d e+b^4 d e\right )\right )}{3 d^{2/3} e^{8/3}}-\frac {c x^3 \left (-12 a b c e-4 b^3 e+c^3 d\right )}{3 e^2}+\frac {c^2 x^4 \left (2 a c+3 b^2\right )}{2 e}+\frac {4 b c^3 x^5}{5 e}+\frac {c^4 x^6}{6 e}\) |
Input:
Int[(a + b*x + c*x^2)^4/(d + e*x^3),x]
Output:
(-2*(3*b^2*c^2*d + 2*a*c^3*d - 2*a*b^3*e - 6*a^2*b*c*e)*x)/e^2 - ((4*b*c^3 *d - b^4*e - 12*a*b^2*c*e - 6*a^2*c^2*e)*x^2)/(2*e^2) - (c*(c^3*d - 4*b^3* e - 12*a*b*c*e)*x^3)/(3*e^2) + (c^2*(3*b^2 + 2*a*c)*x^4)/(2*e) + (4*b*c^3* x^5)/(5*e) + (c^4*x^6)/(6*e) - ((b*d^(1/3) + a*e^(1/3))*(4*c^3*d^2 + 6*c^2 *(b*d^(5/3)*e^(1/3) - a*d^(4/3)*e^(2/3)) - 12*a*b*c*d*e - e*(b^3*d + 3*a*b ^2*d^(2/3)*e^(1/3) - 3*a^2*b*d^(1/3)*e^(2/3) - a^3*e))*ArcTan[(d^(1/3) - 2 *e^(1/3)*x)/(Sqrt[3]*d^(1/3))])/(Sqrt[3]*d^(2/3)*e^(8/3)) + ((e^(1/3)*(6*b ^2*c^2*d^2 + 4*a*c^3*d^2 - 4*a*b^3*d*e - 12*a^2*b*c*d*e + a^4*e^2) + d^(1/ 3)*(b^4*d*e + 12*a*b^2*c*d*e + 6*a^2*c^2*d*e - 4*b*(c^3*d^2 + a^3*e^2)))*L og[d^(1/3) + e^(1/3)*x])/(3*d^(2/3)*e^(8/3)) - ((6*b^2*c^2*d^2 + 4*a*c^3*d ^2 - 4*a*b^3*d*e - 12*a^2*b*c*d*e + a^4*e^2 + (d^(1/3)*(b^4*d*e + 12*a*b^2 *c*d*e + 6*a^2*c^2*d*e - 4*b*(c^3*d^2 + a^3*e^2)))/e^(1/3))*Log[d^(2/3) - d^(1/3)*e^(1/3)*x + e^(2/3)*x^2])/(6*d^(2/3)*e^(7/3)) + ((c^4*d^2 - 12*a*b *c^2*d*e + 6*a^2*b^2*e^2 - 4*c*e*(b^3*d - a^3*e))*Log[d + e*x^3])/(3*e^3)
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.54
| method | result | size |
| risch | \(\frac {c^{4} x^{6}}{6 e}+\frac {4 b \,c^{3} x^{5}}{5 e}+\frac {a \,c^{3} x^{4}}{e}+\frac {3 b^{2} c^{2} x^{4}}{2 e}+\frac {4 a b \,c^{2} x^{3}}{e}+\frac {4 b^{3} c \,x^{3}}{3 e}-\frac {c^{4} d \,x^{3}}{3 e^{2}}+\frac {3 a^{2} c^{2} x^{2}}{e}+\frac {6 a \,b^{2} c \,x^{2}}{e}+\frac {b^{4} x^{2}}{2 e}-\frac {2 b \,c^{3} d \,x^{2}}{e^{2}}+\frac {12 a^{2} b c x}{e}+\frac {4 a \,b^{3} x}{e}-\frac {4 a \,c^{3} d x}{e^{2}}-\frac {6 b^{2} c^{2} d x}{e^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{3}+d \right )}{\sum }\frac {\left (\left (4 a^{3} c \,e^{2}+6 a^{2} b^{2} e^{2}-12 a b \,c^{2} d e -4 b^{3} c d e +c^{4} d^{2}\right ) \textit {\_R}^{2}+\left (4 a^{3} b \,e^{2}-6 a^{2} c^{2} d e -12 a \,b^{2} c d e -b^{4} d e +4 d^{2} c^{3} b \right ) \textit {\_R} +a^{4} e^{2}-12 a^{2} b c d e -4 a \,b^{3} d e +4 a \,c^{3} d^{2}+6 b^{2} c^{2} d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 e^{3}}\) | \(350\) |
| default | \(\frac {\frac {1}{6} c^{4} x^{6} e +\frac {4}{5} c^{3} b \,x^{5} e +a \,c^{3} e \,x^{4}+\frac {3}{2} b^{2} c^{2} e \,x^{4}+4 a b \,c^{2} e \,x^{3}+\frac {4}{3} b^{3} c e \,x^{3}-\frac {1}{3} c^{4} d \,x^{3}+3 a^{2} c^{2} e \,x^{2}+6 a \,b^{2} c e \,x^{2}+\frac {1}{2} b^{4} e \,x^{2}-2 b \,c^{3} d \,x^{2}+12 a^{2} b c e x +4 a \,b^{3} e x -4 a \,c^{3} d x -6 b^{2} c^{2} d x}{e^{2}}+\frac {\left (a^{4} e^{2}-12 a^{2} b c d e -4 a \,b^{3} d e +4 a \,c^{3} d^{2}+6 b^{2} c^{2} d^{2}\right ) \left (\frac {\ln \left (x +\left (\frac {d}{e}\right )^{\frac {1}{3}}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{3}} x +\left (\frac {d}{e}\right )^{\frac {2}{3}}\right )}{6 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{e}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}\right )+\left (4 a^{3} b \,e^{2}-6 a^{2} c^{2} d e -12 a \,b^{2} c d e -b^{4} d e +4 d^{2} c^{3} b \right ) \left (-\frac {\ln \left (x +\left (\frac {d}{e}\right )^{\frac {1}{3}}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{3}} x +\left (\frac {d}{e}\right )^{\frac {2}{3}}\right )}{6 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{e}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {1}{3}}}\right )+\frac {\left (4 a^{3} c \,e^{2}+6 a^{2} b^{2} e^{2}-12 a b \,c^{2} d e -4 b^{3} c d e +c^{4} d^{2}\right ) \ln \left (x^{3} e +d \right )}{3 e}}{e^{2}}\) | \(489\) |
Input:
int((c*x^2+b*x+a)^4/(e*x^3+d),x,method=_RETURNVERBOSE)
Output:
1/6*c^4*x^6/e+4/5*b*c^3*x^5/e+1/e*a*c^3*x^4+3/2/e*b^2*c^2*x^4+4/e*a*b*c^2* x^3+4/3/e*b^3*c*x^3-1/3/e^2*c^4*d*x^3+3/e*a^2*c^2*x^2+6/e*a*b^2*c*x^2+1/2/ e*b^4*x^2-2/e^2*b*c^3*d*x^2+12/e*a^2*b*c*x+4/e*a*b^3*x-4/e^2*a*c^3*d*x-6/e ^2*b^2*c^2*d*x+1/3/e^3*sum(((4*a^3*c*e^2+6*a^2*b^2*e^2-12*a*b*c^2*d*e-4*b^ 3*c*d*e+c^4*d^2)*_R^2+(4*a^3*b*e^2-6*a^2*c^2*d*e-12*a*b^2*c*d*e-b^4*d*e+4* b*c^3*d^2)*_R+a^4*e^2-12*a^2*b*c*d*e-4*a*b^3*d*e+4*a*c^3*d^2+6*b^2*c^2*d^2 )/_R^2*ln(x-_R),_R=RootOf(_Z^3*e+d))
Result contains complex when optimal does not.
Time = 77.47 (sec) , antiderivative size = 47284, normalized size of antiderivative = 73.54 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=\text {Too large to display} \] Input:
integrate((c*x^2+b*x+a)^4/(e*x^3+d),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=\text {Timed out} \] Input:
integrate((c*x**2+b*x+a)**4/(e*x**3+d),x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c*x^2+b*x+a)^4/(e*x^3+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(e>0)', see `assume?` for more de tails)Is e
Time = 0.13 (sec) , antiderivative size = 773, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx =\text {Too large to display} \] Input:
integrate((c*x^2+b*x+a)^4/(e*x^3+d),x, algorithm="giac")
Output:
-1/3*sqrt(3)*(6*b^2*c^2*d^2*e + 4*a*c^3*d^2*e - 4*a*b^3*d*e^2 - 12*a^2*b*c *d*e^2 + a^4*e^3 - 4*(-d*e^2)^(1/3)*b*c^3*d^2 + (-d*e^2)^(1/3)*b^4*d*e + 1 2*(-d*e^2)^(1/3)*a*b^2*c*d*e + 6*(-d*e^2)^(1/3)*a^2*c^2*d*e - 4*(-d*e^2)^( 1/3)*a^3*b*e^2)*arctan(1/3*sqrt(3)*(2*x + (-d/e)^(1/3))/(-d/e)^(1/3))/((-d *e^2)^(2/3)*e^2) - 1/6*(6*b^2*c^2*d^2*e + 4*a*c^3*d^2*e - 4*a*b^3*d*e^2 - 12*a^2*b*c*d*e^2 + a^4*e^3 + 4*(-d*e^2)^(1/3)*b*c^3*d^2 - (-d*e^2)^(1/3)*b ^4*d*e - 12*(-d*e^2)^(1/3)*a*b^2*c*d*e - 6*(-d*e^2)^(1/3)*a^2*c^2*d*e + 4* (-d*e^2)^(1/3)*a^3*b*e^2)*log(x^2 + x*(-d/e)^(1/3) + (-d/e)^(2/3))/((-d*e^ 2)^(2/3)*e^2) + 1/3*(c^4*d^2 - 4*b^3*c*d*e - 12*a*b*c^2*d*e + 6*a^2*b^2*e^ 2 + 4*a^3*c*e^2)*log(abs(e*x^3 + d))/e^3 + 1/30*(5*c^4*e^5*x^6 + 24*b*c^3* e^5*x^5 + 45*b^2*c^2*e^5*x^4 + 30*a*c^3*e^5*x^4 - 10*c^4*d*e^4*x^3 + 40*b^ 3*c*e^5*x^3 + 120*a*b*c^2*e^5*x^3 - 60*b*c^3*d*e^4*x^2 + 15*b^4*e^5*x^2 + 180*a*b^2*c*e^5*x^2 + 90*a^2*c^2*e^5*x^2 - 180*b^2*c^2*d*e^4*x - 120*a*c^3 *d*e^4*x + 120*a*b^3*e^5*x + 360*a^2*b*c*e^5*x)/e^6 - 1/3*(4*b*c^3*d^2*e^1 1*(-d/e)^(1/3) - b^4*d*e^12*(-d/e)^(1/3) - 12*a*b^2*c*d*e^12*(-d/e)^(1/3) - 6*a^2*c^2*d*e^12*(-d/e)^(1/3) + 4*a^3*b*e^13*(-d/e)^(1/3) + 6*b^2*c^2*d^ 2*e^11 + 4*a*c^3*d^2*e^11 - 4*a*b^3*d*e^12 - 12*a^2*b*c*d*e^12 + a^4*e^13) *(-d/e)^(1/3)*log(abs(x - (-d/e)^(1/3)))/(d*e^13)
Time = 5.77 (sec) , antiderivative size = 2971, normalized size of antiderivative = 4.62 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx=\text {Too large to display} \] Input:
int((a + b*x + c*x^2)^4/(d + e*x^3),x)
Output:
x^2*((b^4 + 6*a^2*c^2 + 12*a*b^2*c)/(2*e) - (2*b*c^3*d)/e^2) - x^3*((c^4*d )/(3*e^2) - (4*b*c*(3*a*c + b^2))/(3*e)) + symsum(log(root(27*d^2*e^9*z^3 + 324*a*b*c^2*d^3*e^7*z^2 + 108*b^3*c*d^3*e^7*z^2 - 108*a^3*c*d^2*e^8*z^2 - 162*a^2*b^2*d^2*e^8*z^2 - 27*c^4*d^4*e^6*z^2 - 72*a*b*c^6*d^5*e^4*z + 21 6*a^2*b^2*c^4*d^4*e^5*z + 144*a^3*b^3*c^2*d^3*e^6*z - 108*a^5*b^2*c*d^2*e^ 7*z + 108*a^2*b^5*c*d^3*e^6*z - 36*a^4*b*c^3*d^3*e^6*z + 36*a*b^4*c^3*d^4* e^5*z + 144*b^3*c^5*d^5*e^4*z + 90*b^6*c^2*d^4*e^5*z - 144*a^3*c^5*d^4*e^5 *z + 90*a^6*c^2*d^2*e^7*z + 171*a^4*b^4*d^2*e^7*z + 36*a*b^7*d^3*e^6*z + 3 6*a^7*b*d*e^8*z + 9*c^8*d^6*e^3*z + 36*a^7*b^4*c*d^2*e^6 - 36*a^7*b*c^4*d^ 3*e^5 - 36*a^4*b^7*c*d^3*e^5 - 36*a^4*b*c^7*d^5*e^3 - 36*a*b^7*c^4*d^5*e^3 + 36*a*b^4*c^7*d^6*e^2 + 12*a*b^10*c*d^4*e^4 + 108*a^5*b^5*c^2*d^3*e^5 - 108*a^5*b^2*c^5*d^4*e^4 + 108*a^2*b^5*c^5*d^5*e^3 - 96*a^6*b^3*c^3*d^3*e^5 + 96*a^3*b^6*c^3*d^4*e^4 - 96*a^3*b^3*c^6*d^5*e^3 - 54*a^8*b^2*c^2*d^2*e^ 6 - 54*a^2*b^8*c^2*d^4*e^4 - 54*a^2*b^2*c^8*d^6*e^2 - 9*a^4*b^4*c^4*d^4*e^ 4 - 12*a^10*b*c*d*e^7 - 12*a*b*c^10*d^7*e - 6*b^6*c^6*d^6*e^2 + 4*b^9*c^3* d^5*e^3 - 6*a^6*c^6*d^4*e^4 - 4*a^9*c^3*d^2*e^6 - 4*a^3*c^9*d^6*e^2 - 6*a^ 6*b^6*d^2*e^6 + 4*a^3*b^9*d^3*e^5 + 4*b^3*c^9*d^7*e + 4*a^9*b^3*d*e^7 - b^ 12*d^4*e^4 - c^12*d^8 - a^12*e^8, z, k)*((x*(3*a^4*e^5 + 12*a*c^3*d^2*e^3 + 18*b^2*c^2*d^2*e^3 - 12*a*b^3*d*e^4 - 36*a^2*b*c*d*e^4))/e^3 - (6*c^4*d^ 3*e^3 + 36*a^2*b^2*d*e^5 - 24*b^3*c*d^2*e^4 + 24*a^3*c*d*e^5 - 72*a*b*c...
Time = 0.16 (sec) , antiderivative size = 1391, normalized size of antiderivative = 2.16 \[ \int \frac {\left (a+b x+c x^2\right )^4}{d+e x^3} \, dx =\text {Too large to display} \] Input:
int((c*x^2+b*x+a)^4/(e*x^3+d),x)
Output:
( - 10*e**(1/3)*d**(2/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)* sqrt(3)))*a**4*e**3 + 120*e**(1/3)*d**(2/3)*sqrt(3)*atan((d**(1/3) - 2*e** (1/3)*x)/(d**(1/3)*sqrt(3)))*a**2*b*c*d*e**2 + 40*e**(1/3)*d**(2/3)*sqrt(3 )*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*a*b**3*d*e**2 - 40*e* *(1/3)*d**(2/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3))) *a*c**3*d**2*e - 60*e**(1/3)*d**(2/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)* x)/(d**(1/3)*sqrt(3)))*b**2*c**2*d**2*e - 40*sqrt(3)*atan((d**(1/3) - 2*e* *(1/3)*x)/(d**(1/3)*sqrt(3)))*a**3*b*d*e**3 + 60*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*a**2*c**2*d**2*e**2 + 120*sqrt(3)*atan(( d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*a*b**2*c*d**2*e**2 + 10*sqrt( 3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*b**4*d**2*e**2 - 40* sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*b*c**3*d**3*e - 5*e**(1/3)*d**(2/3)*log(d**(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*a **4*e**3 + 60*e**(1/3)*d**(2/3)*log(d**(2/3) - e**(1/3)*d**(1/3)*x + e**(2 /3)*x**2)*a**2*b*c*d*e**2 + 20*e**(1/3)*d**(2/3)*log(d**(2/3) - e**(1/3)*d **(1/3)*x + e**(2/3)*x**2)*a*b**3*d*e**2 - 20*e**(1/3)*d**(2/3)*log(d**(2/ 3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*a*c**3*d**2*e - 30*e**(1/3)*d**( 2/3)*log(d**(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*b**2*c**2*d**2*e + 10*e**(1/3)*d**(2/3)*log(d**(1/3) + e**(1/3)*x)*a**4*e**3 - 120*e**(1/3) *d**(2/3)*log(d**(1/3) + e**(1/3)*x)*a**2*b*c*d*e**2 - 40*e**(1/3)*d**(...