Integrand size = 38, antiderivative size = 292 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=-\frac {c}{4 a x^4}-\frac {d}{3 a x^3}-\frac {e}{2 a x^2}+\frac {b c-a f}{a^2 x}-\frac {\left (b^{5/3} c-a^{2/3} b e-a b^{2/3} f+a^{5/3} h\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3} \sqrt [3]{b}}-\frac {(b d-a g) \log (x)}{a^2}-\frac {\left (b^{2/3} (b c-a f)+a^{2/3} (b e-a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3} \sqrt [3]{b}}+\frac {\left (b^{2/3} (b c-a f)+a^{2/3} (b e-a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3} \sqrt [3]{b}}+\frac {(b d-a g) \log \left (a+b x^3\right )}{3 a^2} \] Output:
-1/4*c/a/x^4-1/3*d/a/x^3-1/2*e/a/x^2+(-a*f+b*c)/a^2/x-1/3*(b^(5/3)*c-a^(2/ 3)*b*e-a*b^(2/3)*f+a^(5/3)*h)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^( 1/3))*3^(1/2)/a^(7/3)/b^(1/3)-(-a*g+b*d)*ln(x)/a^2-1/3*(b^(2/3)*(-a*f+b*c) +a^(2/3)*(-a*h+b*e))*ln(a^(1/3)+b^(1/3)*x)/a^(7/3)/b^(1/3)+1/6*(b^(2/3)*(- a*f+b*c)+a^(2/3)*(-a*h+b*e))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^( 7/3)/b^(1/3)+1/3*(-a*g+b*d)*ln(b*x^3+a)/a^2
Time = 0.30 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=\frac {-\frac {3 a^{4/3} c}{x^4}-\frac {4 a^{4/3} d}{x^3}-\frac {6 a^{4/3} e}{x^2}+\frac {12 \sqrt [3]{a} (b c-a f)}{x}-\frac {4 \sqrt {3} \left (b^{5/3} c-a^{2/3} b e-a b^{2/3} f+a^{5/3} h\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+12 \sqrt [3]{a} (-b d+a g) \log (x)+\frac {4 \left (-b^{5/3} c-a^{2/3} b e+a b^{2/3} f+a^{5/3} h\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {2 \left (b^{5/3} c+a^{2/3} b e-a b^{2/3} f-a^{5/3} h\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}+4 \sqrt [3]{a} (b d-a g) \log \left (a+b x^3\right )}{12 a^{7/3}} \] Input:
Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^5*(a + b*x^3)),x]
Output:
((-3*a^(4/3)*c)/x^4 - (4*a^(4/3)*d)/x^3 - (6*a^(4/3)*e)/x^2 + (12*a^(1/3)* (b*c - a*f))/x - (4*Sqrt[3]*(b^(5/3)*c - a^(2/3)*b*e - a*b^(2/3)*f + a^(5/ 3)*h)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3) + 12*a^(1/3)*(- (b*d) + a*g)*Log[x] + (4*(-(b^(5/3)*c) - a^(2/3)*b*e + a*b^(2/3)*f + a^(5/ 3)*h)*Log[a^(1/3) + b^(1/3)*x])/b^(1/3) + (2*(b^(5/3)*c + a^(2/3)*b*e - a* b^(2/3)*f - a^(5/3)*h)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^( 1/3) + 4*a^(1/3)*(b*d - a*g)*Log[a + b*x^3])/(12*a^(7/3))
Time = 1.04 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \int \left (\frac {b x (b c-a f)+b x^2 (b d-a g)-a (b e-a h)}{a^2 \left (a+b x^3\right )}+\frac {a f-b c}{a^2 x^2}+\frac {a g-b d}{a^2 x}+\frac {c}{a x^5}+\frac {d}{a x^4}+\frac {e}{a x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-a^{2/3} b e+a^{5/3} h-a b^{2/3} f+b^{5/3} c\right )}{\sqrt {3} a^{7/3} \sqrt [3]{b}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^{2/3} (b e-a h)+b^{2/3} (b c-a f)\right )}{6 a^{7/3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^{2/3} (b e-a h)+b^{2/3} (b c-a f)\right )}{3 a^{7/3} \sqrt [3]{b}}+\frac {b c-a f}{a^2 x}+\frac {(b d-a g) \log \left (a+b x^3\right )}{3 a^2}-\frac {\log (x) (b d-a g)}{a^2}-\frac {c}{4 a x^4}-\frac {d}{3 a x^3}-\frac {e}{2 a x^2}\) |
Input:
Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^5*(a + b*x^3)),x]
Output:
-1/4*c/(a*x^4) - d/(3*a*x^3) - e/(2*a*x^2) + (b*c - a*f)/(a^2*x) - ((b^(5/ 3)*c - a^(2/3)*b*e - a*b^(2/3)*f + a^(5/3)*h)*ArcTan[(a^(1/3) - 2*b^(1/3)* x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(7/3)*b^(1/3)) - ((b*d - a*g)*Log[x])/a^ 2 - ((b^(2/3)*(b*c - a*f) + a^(2/3)*(b*e - a*h))*Log[a^(1/3) + b^(1/3)*x]) /(3*a^(7/3)*b^(1/3)) + ((b^(2/3)*(b*c - a*f) + a^(2/3)*(b*e - a*h))*Log[a^ (2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(7/3)*b^(1/3)) + ((b*d - a* g)*Log[a + b*x^3])/(3*a^2)
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 = 0.14 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(\frac {\left (a^{2} h -a b e \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (-a b f +b^{2} c \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (-a b g +b^{2} d \right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{a^{2}}-\frac {e}{2 a \,x^{2}}-\frac {c}{4 a \,x^{4}}-\frac {d}{3 a \,x^{3}}-\frac {a f -c b}{a^{2} x}+\frac {\left (a g -b d \right ) \ln \left (x \right )}{a^{2}}\) | \(292\) |
| risch | \(\frac {-\frac {\left (a f -c b \right ) x^{3}}{a^{2}}-\frac {e \,x^{2}}{2 a}-\frac {d x}{3 a}-\frac {c}{4 a}}{x^{4}}+\frac {\ln \left (x \right ) g}{a}-\frac {\ln \left (x \right ) b d}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} b \,\textit {\_Z}^{3}+\left (3 a^{6} b g -3 a^{5} b^{2} d \right ) \textit {\_Z}^{2}+\left (-3 a^{5} b f h +3 a^{5} b \,g^{2}+3 a^{4} b^{2} c h -6 a^{4} b^{2} d g +3 a^{4} b^{2} e f -3 a^{3} b^{3} c e +3 a^{3} b^{3} d^{2}\right ) \textit {\_Z} -a^{5} h^{3}+3 a^{4} b e \,h^{2}-3 a^{4} b f g h +a^{4} b \,g^{3}+3 a^{3} b^{2} c g h +3 a^{3} b^{2} d f h -3 a^{3} b^{2} d \,g^{2}-3 a^{3} b^{2} e^{2} h +3 a^{3} b^{2} e f g -a^{3} b^{2} f^{3}-3 a^{2} b^{3} c d h -3 a^{2} b^{3} c e g +3 a^{2} b^{3} c \,f^{2}+3 a^{2} b^{3} d^{2} g -3 a^{2} b^{3} d e f +a^{2} b^{3} e^{3}-3 a \,b^{4} c^{2} f +3 a \,b^{4} c d e -a \,b^{4} d^{3}+b^{5} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{7} b +\left (-8 a^{6} b g +8 a^{5} b^{2} d \right ) \textit {\_R}^{2}+\left (10 a^{5} b f h -4 a^{5} b \,g^{2}-10 a^{4} b^{2} c h +8 a^{4} b^{2} d g -10 a^{4} b^{2} e f +10 a^{3} b^{3} c e -4 a^{3} b^{3} d^{2}\right ) \textit {\_R} +3 a^{5} h^{3}-9 a^{4} b e \,h^{2}+6 a^{4} b f g h -6 a^{3} b^{2} c g h -6 a^{3} b^{2} d f h +9 a^{3} b^{2} e^{2} h -6 a^{3} b^{2} e f g +3 a^{3} b^{2} f^{3}+6 a^{2} b^{3} c d h +6 a^{2} b^{3} c e g -9 a^{2} b^{3} c \,f^{2}+6 a^{2} b^{3} d e f -3 a^{2} b^{3} e^{3}+9 a \,b^{4} c^{2} f -6 a \,b^{4} c d e -3 b^{5} c^{3}\right ) x +\left (-a^{6} b f +a^{5} b^{2} c \right ) \textit {\_R}^{2}+\left (-a^{6} h^{2}+2 a^{5} b e h +2 a^{5} b f g -2 a^{4} b^{2} c g -2 a^{4} b^{2} d f -a^{4} b^{2} e^{2}+2 a^{3} b^{3} c d \right ) \textit {\_R} +3 a^{5} g \,h^{2}-3 a^{4} b d \,h^{2}-6 a^{4} b e g h +3 a^{4} b f \,g^{2}-3 a^{3} b^{2} c \,g^{2}+6 a^{3} b^{2} d e h -6 a^{3} b^{2} d f g +3 a^{3} b^{2} e^{2} g +6 a^{2} b^{3} c d g +3 a^{2} b^{3} d^{2} f -3 a^{2} b^{3} d \,e^{2}-3 a \,b^{4} c \,d^{2}\right )\right )}{3}\) | \(864\) |
Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^5/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
((a^2*h-a*b*e)*(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)))+(-a*b*f+b^2*c)*(-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*(-a*b*g+b^2*d)/b*ln(b*x ^3+a))/a^2-1/2*e/a/x^2-1/4*c/a/x^4-1/3*d/a/x^3-(a*f-b*c)/a^2/x+(a*g-b*d)/a ^2*ln(x)
Result contains complex when optimal does not.
Time = 64.62 (sec) , antiderivative size = 14892, normalized size of antiderivative = 51.00 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^5/(b*x^3+a),x, algorithm="fric as")
Output:
Too large to include
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**5/(b*x**3+a),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=-\frac {{\left (b d - a g\right )} \log \left (x\right )}{a^{2}} + \frac {\sqrt {3} {\left (b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{3}} + \frac {{\left (2 \, b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b e - a^{2} h\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} - b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b e + a^{2} h\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {6 \, a e x^{2} - 12 \, {\left (b c - a f\right )} x^{3} + 4 \, a d x + 3 \, a c}{12 \, a^{2} x^{4}} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^5/(b*x^3+a),x, algorithm="maxi ma")
Output:
-(b*d - a*g)*log(x)/a^2 + 1/3*sqrt(3)*(b^2*c*(a/b)^(2/3) - a*b*f*(a/b)^(2/ 3) - a*b*e*(a/b)^(1/3) + a^2*h*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b )^(1/3))/(a/b)^(1/3))/a^3 + 1/6*(2*b^2*d*(a/b)^(2/3) - 2*a*b*g*(a/b)^(2/3) + b^2*c*(a/b)^(1/3) - a*b*f*(a/b)^(1/3) + a*b*e - a^2*h)*log(x^2 - x*(a/b )^(1/3) + (a/b)^(2/3))/(a^2*b*(a/b)^(2/3)) + 1/3*(b^2*d*(a/b)^(2/3) - a*b* g*(a/b)^(2/3) - b^2*c*(a/b)^(1/3) + a*b*f*(a/b)^(1/3) - a*b*e + a^2*h)*log (x + (a/b)^(1/3))/(a^2*b*(a/b)^(2/3)) - 1/12*(6*a*e*x^2 - 12*(b*c - a*f)*x ^3 + 4*a*d*x + 3*a*c)/(a^2*x^4)
Time = 0.12 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=\frac {{\left (b d - a g\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2}} - \frac {{\left (b d - a g\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} a b e - \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} h + \left (-a b^{2}\right )^{\frac {2}{3}} b c - \left (-a b^{2}\right )^{\frac {2}{3}} a f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{3} b} - \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} a b e - \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} h - \left (-a b^{2}\right )^{\frac {2}{3}} b c + \left (-a b^{2}\right )^{\frac {2}{3}} a f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{3} b} - \frac {{\left (a^{2} b^{3} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} b^{2} f \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} b^{2} e + a^{4} b h\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{5} b} - \frac {6 \, a e x^{2} - 12 \, {\left (b c - a f\right )} x^{3} + 4 \, a d x + 3 \, a c}{12 \, a^{2} x^{4}} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^5/(b*x^3+a),x, algorithm="giac ")
Output:
1/3*(b*d - a*g)*log(abs(b*x^3 + a))/a^2 - (b*d - a*g)*log(abs(x))/a^2 - 1/ 3*sqrt(3)*((-a*b^2)^(1/3)*a*b*e - (-a*b^2)^(1/3)*a^2*h + (-a*b^2)^(2/3)*b* c - (-a*b^2)^(2/3)*a*f)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/ 3))/(a^3*b) - 1/6*((-a*b^2)^(1/3)*a*b*e - (-a*b^2)^(1/3)*a^2*h - (-a*b^2)^ (2/3)*b*c + (-a*b^2)^(2/3)*a*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/( a^3*b) - 1/3*(a^2*b^3*c*(-a/b)^(1/3) - a^3*b^2*f*(-a/b)^(1/3) - a^3*b^2*e + a^4*b*h)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^5*b) - 1/12*(6*a*e*x ^2 - 12*(b*c - a*f)*x^3 + 4*a*d*x + 3*a*c)/(a^2*x^4)
Time = 6.69 (sec) , antiderivative size = 1853, normalized size of antiderivative = 6.35 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx=\text {Too large to display} \] Input:
int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^5*(a + b*x^3)),x)
Output:
symsum(log(- root(27*a^7*b*z^3 + 27*a^6*b*g*z^2 - 27*a^5*b^2*d*z^2 - 9*a^5 *b*f*h*z - 18*a^4*b^2*d*g*z + 9*a^4*b^2*e*f*z + 9*a^4*b^2*c*h*z - 9*a^3*b^ 3*c*e*z + 9*a^5*b*g^2*z + 9*a^3*b^3*d^2*z - 3*a^4*b*f*g*h + 3*a*b^4*c*d*e + 3*a^3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g - 3*a^2*b^3*c*d*h + 3*a^4*b*e*h^2 - 3*a*b^4*c^2*f - 3*a^3* b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a^2*b^3*c*f^2 - a^3*b^2* f^3 - a*b^4*d^3 - a^5*h^3 + a^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3, z, k)*((a^3 *b^4*e^2 + a^5*b^2*h^2 - 2*a^2*b^5*c*d + 2*a^3*b^4*c*g + 2*a^3*b^4*d*f - 2 *a^4*b^3*e*h - 2*a^4*b^3*f*g)/a^4 - root(27*a^7*b*z^3 + 27*a^6*b*g*z^2 - 2 7*a^5*b^2*d*z^2 - 9*a^5*b*f*h*z - 18*a^4*b^2*d*g*z + 9*a^4*b^2*e*f*z + 9*a ^4*b^2*c*h*z - 9*a^3*b^3*c*e*z + 9*a^5*b*g^2*z + 9*a^3*b^3*d^2*z - 3*a^4*b *f*g*h + 3*a*b^4*c*d*e + 3*a^3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g *h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g - 3*a^2*b^3*c*d*h + 3*a^4*b*e*h^2 - 3*a*b^4*c^2*f - 3*a^3*b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a ^2*b^3*c*f^2 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h^3 + a^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3, z, k)*((3*a^4*b^4*c - 3*a^5*b^3*f)/a^4 + (x*(24*a^5*b^4*d - 24 *a^6*b^3*g))/a^5 - 36*root(27*a^7*b*z^3 + 27*a^6*b*g*z^2 - 27*a^5*b^2*d*z^ 2 - 9*a^5*b*f*h*z - 18*a^4*b^2*d*g*z + 9*a^4*b^2*e*f*z + 9*a^4*b^2*c*h*z - 9*a^3*b^3*c*e*z + 9*a^5*b*g^2*z + 9*a^3*b^3*d^2*z - 3*a^4*b*f*g*h + 3*a*b ^4*c*d*e + 3*a^3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*...
Time = 0.17 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.84 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^5 \left (a+b x^3\right )} \, dx =\text {Too large to display} \] Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^5/(b*x^3+a),x)
Output:
( - 4*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*s qrt(3)))*a*h*x**4 + 4*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3 )*x)/(a**(1/3)*sqrt(3)))*b*e*x**4 + 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)* x)/(a**(1/3)*sqrt(3)))*a*b*f*x**4 - 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)* x)/(a**(1/3)*sqrt(3)))*b**2*c*x**4 - 2*b**(1/3)*a**(2/3)*log(a**(2/3) - b* *(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*h*x**4 + 2*b**(1/3)*a**(2/3)*log(a**( 2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b*e*x**4 + 4*b**(1/3)*a**(2/3) *log(a**(1/3) + b**(1/3)*x)*a*h*x**4 - 4*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*b*e*x**4 - 4*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3 )*x + b**(2/3)*x**2)*a*g*x**4 + 4*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3 )*a**(1/3)*x + b**(2/3)*x**2)*b*d*x**4 - 4*b**(2/3)*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a*g*x**4 + 4*b**(2/3)*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*b* d*x**4 + 12*b**(2/3)*a**(1/3)*log(x)*a*g*x**4 - 12*b**(2/3)*a**(1/3)*log(x )*b*d*x**4 - 3*b**(2/3)*a**(1/3)*a*c - 4*b**(2/3)*a**(1/3)*a*d*x - 6*b**(2 /3)*a**(1/3)*a*e*x**2 - 12*b**(2/3)*a**(1/3)*a*f*x**3 + 12*b**(2/3)*a**(1/ 3)*b*c*x**3 - 2*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*f* x**4 + 2*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**2*c*x**4 + 4*log(a**(1/3) + b**(1/3)*x)*a*b*f*x**4 - 4*log(a**(1/3) + b**(1/3)*x)*b* *2*c*x**4)/(12*b**(2/3)*a**(1/3)*a**2*x**4)