Integrand size = 38, antiderivative size = 306 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {c}{2 a^2 x^2}-\frac {d}{a^2 x}-\frac {x \left (b c-a f+(b d-a g) x+(b e-a h) x^2\right )}{3 a^2 \left (a+b x^3\right )}+\frac {\left (5 b^{4/3} c+4 \sqrt [3]{a} b d-2 a \sqrt [3]{b} f-a^{4/3} g\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} b^{2/3}}+\frac {e \log (x)}{a^2}-\frac {\left (\sqrt [3]{b} (5 b c-2 a f)-\sqrt [3]{a} (4 b d-a g)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3} b^{2/3}}+\frac {\left (\sqrt [3]{b} (5 b c-2 a f)-\sqrt [3]{a} (4 b d-a g)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3} b^{2/3}}-\frac {e \log \left (a+b x^3\right )}{3 a^2} \]
-1/2*c/a^2/x^2-d/a^2/x-1/3*x*(b*c-a*f+(-a*g+b*d)*x+(-a*h+b*e)*x^2)/a^2/(b* x^3+a)+e*ln(x)/a^2-1/9*(b^(1/3)*(-2*a*f+5*b*c)-a^(1/3)*(-a*g+4*b*d))*ln(a^ (1/3)+b^(1/3)*x)/a^(8/3)/b^(2/3)+1/18*(b^(1/3)*(-2*a*f+5*b*c)-a^(1/3)*(-a* g+4*b*d))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(8/3)/b^(2/3)-1/3*e* ln(b*x^3+a)/a^2+1/9*(5*b^(4/3)*c+4*a^(1/3)*b*d-2*a*b^(1/3)*f-a^(4/3)*g)*ar ctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(8/3)/b^(2/3)*3^(1/2)
Time = 0.50 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {\frac {9 a c}{x^2}+\frac {18 a d}{x}+\frac {6 a \left (a^2 h+b^2 x (c+d x)-a b (e+x (f+g x))\right )}{b \left (a+b x^3\right )}+\frac {2 \sqrt {3} \sqrt [3]{a} \left (-5 b^{4/3} c-4 \sqrt [3]{a} b d+2 a \sqrt [3]{b} f+a^{4/3} g\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}-18 a e \log (x)+\frac {2 \sqrt [3]{a} \left (5 b^{4/3} c-4 \sqrt [3]{a} b d-2 a \sqrt [3]{b} f+a^{4/3} g\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}-\frac {\sqrt [3]{a} \left (5 b^{4/3} c-4 \sqrt [3]{a} b d-2 a \sqrt [3]{b} f+a^{4/3} g\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}+6 a e \log \left (a+b x^3\right )}{18 a^3} \]
-1/18*((9*a*c)/x^2 + (18*a*d)/x + (6*a*(a^2*h + b^2*x*(c + d*x) - a*b*(e + x*(f + g*x))))/(b*(a + b*x^3)) + (2*Sqrt[3]*a^(1/3)*(-5*b^(4/3)*c - 4*a^( 1/3)*b*d + 2*a*b^(1/3)*f + a^(4/3)*g)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/S qrt[3]])/b^(2/3) - 18*a*e*Log[x] + (2*a^(1/3)*(5*b^(4/3)*c - 4*a^(1/3)*b*d - 2*a*b^(1/3)*f + a^(4/3)*g)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) - (a^(1/3) *(5*b^(4/3)*c - 4*a^(1/3)*b*d - 2*a*b^(1/3)*f + a^(4/3)*g)*Log[a^(2/3) - a ^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(2/3) + 6*a*e*Log[a + b*x^3])/a^3
Time = 0.81 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2368, 25, 2373, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle -\frac {\int -\frac {-b^2 \left (\frac {b d}{a}-g\right ) x^4-2 b^2 \left (\frac {b c}{a}-f\right ) x^3+3 b^2 e x^2+3 b^2 d x+3 b^2 c}{x^3 \left (b x^3+a\right )}dx}{3 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 a^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-b^2 \left (\frac {b d}{a}-g\right ) x^4-2 b^2 \left (\frac {b c}{a}-f\right ) x^3+3 b^2 e x^2+3 b^2 d x+3 b^2 c}{x^3 \left (b x^3+a\right )}dx}{3 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 a^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \frac {\int \left (\frac {3 e b^2}{a x}+\frac {\left (-3 b e x^2-(4 b d-a g) x-5 b c+2 a f\right ) b^2}{a \left (b x^3+a\right )}+\frac {3 d b^2}{a x^2}+\frac {3 c b^2}{a x^3}\right )dx}{3 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 a^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{4/3} (-g)+4 \sqrt [3]{a} b d-2 a \sqrt [3]{b} f+5 b^{4/3} c\right )}{\sqrt {3} a^{5/3}}+\frac {b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {\sqrt [3]{a} (4 b d-a g)}{\sqrt [3]{b}}-2 a f+5 b c\right )}{6 a^{5/3}}-\frac {b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (5 b c-2 a f)-\sqrt [3]{a} (4 b d-a g)\right )}{3 a^{5/3}}-\frac {3 b^2 c}{2 a x^2}-\frac {3 b^2 d}{a x}-\frac {b^2 e \log \left (a+b x^3\right )}{a}+\frac {3 b^2 e \log (x)}{a}}{3 a b^2}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 a^2 \left (a+b x^3\right )}\) |
-1/3*(x*(b*c - a*f + (b*d - a*g)*x + (b*e - a*h)*x^2))/(a^2*(a + b*x^3)) + ((-3*b^2*c)/(2*a*x^2) - (3*b^2*d)/(a*x) + (b^(4/3)*(5*b^(4/3)*c + 4*a^(1/ 3)*b*d - 2*a*b^(1/3)*f - a^(4/3)*g)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3 ]*a^(1/3))])/(Sqrt[3]*a^(5/3)) + (3*b^2*e*Log[x])/a - (b^(4/3)*(b^(1/3)*(5 *b*c - 2*a*f) - a^(1/3)*(4*b*d - a*g))*Log[a^(1/3) + b^(1/3)*x])/(3*a^(5/3 )) + (b^(5/3)*(5*b*c - 2*a*f - (a^(1/3)*(4*b*d - a*g))/b^(1/3))*Log[a^(2/3 ) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(5/3)) - (b^2*e*Log[a + b*x^3]) /a)/(3*a*b^2)
3.5.19.3.1 Defintions of rubi rules used
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1)*x^m *Pq, a + b*x^n, x], i}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^( Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[x^m*(a + b*x^n)^(p + 1)*ExpandToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x]]] /; F reeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[E xpandIntegrand[(c*x)^m*(Pq/(a + b*x^n)), x], x] /; FreeQ[{a, b, c, m}, x] & & PolyQ[Pq, x] && IntegerQ[n] && !IGtQ[m, 0]
Time = 1.59 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(-\frac {c}{2 a^{2} x^{2}}-\frac {d}{a^{2} x}+\frac {e \ln \left (x \right )}{a^{2}}+\frac {\frac {\left (\frac {a g}{3}-\frac {b d}{3}\right ) x^{2}+\left (\frac {a f}{3}-\frac {b c}{3}\right ) x -\frac {a \left (a h -b e \right )}{3 b}}{b \,x^{3}+a}+\frac {\left (2 a f -5 b c \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {\left (a g -4 b d \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}-\frac {e \ln \left (b \,x^{3}+a \right )}{3}}{a^{2}}\) | \(293\) |
| risch | \(\frac {\frac {\left (a g -4 b d \right ) x^{4}}{3 a^{2}}+\frac {\left (2 a f -5 b c \right ) x^{3}}{6 a^{2}}-\frac {\left (a h -b e \right ) x^{2}}{3 a b}-\frac {x d}{a}-\frac {c}{2 a}}{x^{2} \left (b \,x^{3}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{8} b^{2} \textit {\_Z}^{3}+9 a^{6} b^{2} e \,\textit {\_Z}^{2}+\left (6 a^{5} b f g -15 a^{4} b^{2} c g -24 a^{4} b^{2} d f +27 a^{4} b^{2} e^{2}+60 a^{3} b^{3} c d \right ) \textit {\_Z} +a^{4} g^{3}-12 a^{3} b d \,g^{2}+18 a^{3} b e f g -8 a^{3} b \,f^{3}-45 a^{2} b^{2} c e g +60 a^{2} b^{2} c \,f^{2}+48 a^{2} b^{2} d^{2} g -72 a^{2} b^{2} d e f +27 a^{2} b^{2} e^{3}-150 a \,b^{3} c^{2} f +180 a \,b^{3} c d e -64 a \,b^{3} d^{3}+125 b^{4} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{8} b^{2}-24 \textit {\_R}^{2} a^{6} b^{2} e +\left (-20 a^{5} b f g +50 a^{4} b^{2} c g +80 a^{4} b^{2} d f -36 a^{4} b^{2} e^{2}-200 a^{3} b^{3} c d \right ) \textit {\_R} -3 a^{4} g^{3}+36 a^{3} b d \,g^{2}-36 a^{3} b e f g +24 a^{3} b \,f^{3}+90 a^{2} b^{2} c e g -180 a^{2} b^{2} c \,f^{2}-144 a^{2} b^{2} d^{2} g +144 a^{2} b^{2} d e f +450 a \,b^{3} c^{2} f -360 a \,b^{3} c d e +192 a \,b^{3} d^{3}-375 b^{4} c^{3}\right ) x +\left (a^{7} b g -4 a^{6} b^{2} d \right ) \textit {\_R}^{2}+\left (-6 a^{5} b e g -4 a^{5} b \,f^{2}+20 a^{4} b^{2} c f +24 a^{4} b^{2} d e -25 a^{3} b^{3} c^{2}\right ) \textit {\_R} -27 a^{3} b \,e^{2} g +36 a^{3} b e \,f^{2}-180 a^{2} b^{2} c e f +108 a^{2} b^{2} d \,e^{2}+225 a \,b^{3} c^{2} e \right )\right )}{9}+\frac {e \ln \left (-x \right )}{a^{2}}\) | \(623\) |
-1/2*c/a^2/x^2-d/a^2/x+e*ln(x)/a^2+1/a^2*(((1/3*a*g-1/3*b*d)*x^2+(1/3*a*f- 1/3*b*c)*x-1/3*a*(a*h-b*e)/b)/(b*x^3+a)+1/3*(2*a*f-5*b*c)*(1/3/b/(a/b)^(2/ 3)*ln(x+(a/b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1 /3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))+1/3*(a*g -4*b*d)*(-1/3/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))+1/6/b/(a/b)^(1/3)*ln(x^2-(a/ b)^(1/3)*x+(a/b)^(2/3))+1/3*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a /b)^(1/3)*x-1)))-1/3*e*ln(b*x^3+a))
Result contains complex when optimal does not.
Time = 17.92 (sec) , antiderivative size = 12231, normalized size of antiderivative = 39.97 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {2 \, {\left (4 \, b^{2} d - a b g\right )} x^{4} + 6 \, a b d x + {\left (5 \, b^{2} c - 2 \, a b f\right )} x^{3} + 3 \, a b c - 2 \, {\left (a b e - a^{2} h\right )} x^{2}}{6 \, {\left (a^{2} b^{2} x^{5} + a^{3} b x^{2}\right )}} + \frac {e \log \left (x\right )}{a^{2}} - \frac {\sqrt {3} {\left (4 \, b d \left (\frac {a}{b}\right )^{\frac {2}{3}} - a g \left (\frac {a}{b}\right )^{\frac {2}{3}} + 5 \, b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a f \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3}} - \frac {{\left (6 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} + 4 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - a g \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, b c + 2 \, a f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (3 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + a g \left (\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, b c - 2 \, a f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
-1/6*(2*(4*b^2*d - a*b*g)*x^4 + 6*a*b*d*x + (5*b^2*c - 2*a*b*f)*x^3 + 3*a* b*c - 2*(a*b*e - a^2*h)*x^2)/(a^2*b^2*x^5 + a^3*b*x^2) + e*log(x)/a^2 - 1/ 9*sqrt(3)*(4*b*d*(a/b)^(2/3) - a*g*(a/b)^(2/3) + 5*b*c*(a/b)^(1/3) - 2*a*f *(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/a^3 - 1/ 18*(6*b*e*(a/b)^(2/3) + 4*b*d*(a/b)^(1/3) - a*g*(a/b)^(1/3) - 5*b*c + 2*a* f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*(a/b)^(2/3)) - 1/9*(3*b*e *(a/b)^(2/3) - 4*b*d*(a/b)^(1/3) + a*g*(a/b)^(1/3) + 5*b*c - 2*a*f)*log(x + (a/b)^(1/3))/(a^2*b*(a/b)^(2/3))
Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {e \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2}} + \frac {e \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {\sqrt {3} {\left (5 \, b^{2} c - 2 \, a b f - 4 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d + \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} + \frac {{\left (5 \, b^{2} c - 2 \, a b f + 4 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d - \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} + \frac {{\left (4 \, a^{2} b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} b g \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, a^{2} b^{2} c - 2 \, a^{3} b f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{5} b} - \frac {2 \, {\left (4 \, b^{2} d - a b g\right )} x^{4} + 6 \, a b d x + {\left (5 \, b^{2} c - 2 \, a b f\right )} x^{3} + 3 \, a b c - 2 \, {\left (a b e - a^{2} h\right )} x^{2}}{6 \, {\left (b x^{3} + a\right )} a^{2} b x^{2}} \]
-1/3*e*log(abs(b*x^3 + a))/a^2 + e*log(abs(x))/a^2 + 1/9*sqrt(3)*(5*b^2*c - 2*a*b*f - 4*(-a*b^2)^(1/3)*b*d + (-a*b^2)^(1/3)*a*g)*arctan(1/3*sqrt(3)* (2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)*a^2) + 1/18*(5*b^2*c - 2*a*b*f + 4*(-a*b^2)^(1/3)*b*d - (-a*b^2)^(1/3)*a*g)*log(x^2 + x*(-a/b)^(1 /3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a^2) + 1/9*(4*a^2*b^2*d*(-a/b)^(1/3) - a^3*b*g*(-a/b)^(1/3) + 5*a^2*b^2*c - 2*a^3*b*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^5*b) - 1/6*(2*(4*b^2*d - a*b*g)*x^4 + 6*a*b*d*x + (5*b^2 *c - 2*a*b*f)*x^3 + 3*a*b*c - 2*(a*b*e - a^2*h)*x^2)/((b*x^3 + a)*a^2*b*x^ 2)
Time = 9.83 (sec) , antiderivative size = 1632, normalized size of antiderivative = 5.33 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )^2} \, dx=\text {Too large to display} \]
symsum(log((b^2*e*(25*b^2*c^2 + 4*a^2*f^2 - 3*a^2*e*g - 20*a*b*c*f + 12*a* b*d*e))/(9*a^5) - (root(729*a^8*b^2*z^3 + 729*a^6*b^2*e*z^2 + 54*a^5*b*f*g *z - 216*a^4*b^2*d*f*z - 135*a^4*b^2*c*g*z + 540*a^3*b^3*c*d*z + 243*a^4*b ^2*e^2*z + 18*a^3*b*e*f*g + 180*a*b^3*c*d*e - 72*a^2*b^2*d*e*f - 45*a^2*b^ 2*c*e*g - 12*a^3*b*d*g^2 - 150*a*b^3*c^2*f + 48*a^2*b^2*d^2*g + 60*a^2*b^2 *c*f^2 + 27*a^2*b^2*e^3 - 8*a^3*b*f^3 - 64*a*b^3*d^3 + 125*b^4*c^3 + a^4*g ^3, z, k)*b^2*(25*b^2*c^2 + 4*a^2*f^2 - 9*root(729*a^8*b^2*z^3 + 729*a^6*b ^2*e*z^2 + 54*a^5*b*f*g*z - 216*a^4*b^2*d*f*z - 135*a^4*b^2*c*g*z + 540*a^ 3*b^3*c*d*z + 243*a^4*b^2*e^2*z + 18*a^3*b*e*f*g + 180*a*b^3*c*d*e - 72*a^ 2*b^2*d*e*f - 45*a^2*b^2*c*e*g - 12*a^3*b*d*g^2 - 150*a*b^3*c^2*f + 48*a^2 *b^2*d^2*g + 60*a^2*b^2*c*f^2 + 27*a^2*b^2*e^3 - 8*a^3*b*f^3 - 64*a*b^3*d^ 3 + 125*b^4*c^3 + a^4*g^3, z, k)*a^4*g + 6*a^2*e*g + 36*root(729*a^8*b^2*z ^3 + 729*a^6*b^2*e*z^2 + 54*a^5*b*f*g*z - 216*a^4*b^2*d*f*z - 135*a^4*b^2* c*g*z + 540*a^3*b^3*c*d*z + 243*a^4*b^2*e^2*z + 18*a^3*b*e*f*g + 180*a*b^3 *c*d*e - 72*a^2*b^2*d*e*f - 45*a^2*b^2*c*e*g - 12*a^3*b*d*g^2 - 150*a*b^3* c^2*f + 48*a^2*b^2*d^2*g + 60*a^2*b^2*c*f^2 + 27*a^2*b^2*e^3 - 8*a^3*b*f^3 - 64*a*b^3*d^3 + 125*b^4*c^3 + a^4*g^3, z, k)*a^3*b*d + 36*a*b*e^2*x + 20 0*b^2*c*d*x + 20*a^2*f*g*x + 324*root(729*a^8*b^2*z^3 + 729*a^6*b^2*e*z^2 + 54*a^5*b*f*g*z - 216*a^4*b^2*d*f*z - 135*a^4*b^2*c*g*z + 540*a^3*b^3*c*d *z + 243*a^4*b^2*e^2*z + 18*a^3*b*e*f*g + 180*a*b^3*c*d*e - 72*a^2*b^2*...