Integrand size = 38, antiderivative size = 258 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=-\frac {c}{2 a x^2}-\frac {d}{a x}+\frac {\left (b^{4/3} c+\sqrt [3]{a} b d-a \sqrt [3]{b} f-a^{4/3} g\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} b^{2/3}}+\frac {e \log (x)}{a}-\frac {\left (\sqrt [3]{b} (b c-a f)-\sqrt [3]{a} (b d-a g)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3} b^{2/3}}+\frac {\left (b c-a f-\frac {\sqrt [3]{a} (b d-a g)}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3} \sqrt [3]{b}}-\frac {(b e-a h) \log \left (a+b x^3\right )}{3 a b} \] Output:
-1/2*c/a/x^2-d/a/x+1/3*(b^(4/3)*c+a^(1/3)*b*d-a*b^(1/3)*f-a^(4/3)*g)*arcta n(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(5/3)/b^(2/3)+e*ln( x)/a-1/3*(b^(1/3)*(-a*f+b*c)-a^(1/3)*(-a*g+b*d))*ln(a^(1/3)+b^(1/3)*x)/a^( 5/3)/b^(2/3)+1/6*(b*c-a*f-a^(1/3)*(-a*g+b*d)/b^(1/3))*ln(a^(2/3)-a^(1/3)*b ^(1/3)*x+b^(2/3)*x^2)/a^(5/3)/b^(1/3)-1/3*(-a*h+b*e)*ln(b*x^3+a)/a/b
Time = 0.20 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\frac {-\frac {3 a^{2/3} c}{x^2}-\frac {6 a^{2/3} d}{x}+\frac {2 \sqrt {3} \left (b^{4/3} c+\sqrt [3]{a} b d-a \sqrt [3]{b} f-a^{4/3} g\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}+6 a^{2/3} e \log (x)-\frac {2 \left (b^{4/3} c-\sqrt [3]{a} b d-a \sqrt [3]{b} f+a^{4/3} g\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {\left (b^{4/3} c-\sqrt [3]{a} b d-a \sqrt [3]{b} f+a^{4/3} g\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}+\frac {2 a^{2/3} (-b e+a h) \log \left (a+b x^3\right )}{b}}{6 a^{5/3}} \] Input:
Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)),x]
Output:
((-3*a^(2/3)*c)/x^2 - (6*a^(2/3)*d)/x + (2*Sqrt[3]*(b^(4/3)*c + a^(1/3)*b* d - a*b^(1/3)*f - a^(4/3)*g)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/ b^(2/3) + 6*a^(2/3)*e*Log[x] - (2*(b^(4/3)*c - a^(1/3)*b*d - a*b^(1/3)*f + a^(4/3)*g)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) + ((b^(4/3)*c - a^(1/3)*b*d - 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) + (2*a^(2/3)*(-(b*e) + a*h)*Log[a + b*x^3])/b)/(6*a^(5/3))
Time = 0.97 (sec) , antiderivative size = 258, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \int \left (\frac {-x (b d-a g)-\left (x^2 (b e-a h)\right )+a f-b c}{a \left (a+b x^3\right )}+\frac {c}{a x^3}+\frac {d}{a x^2}+\frac {e}{a x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{4/3} (-g)+\sqrt [3]{a} b d-a \sqrt [3]{b} f+b^{4/3} c\right )}{\sqrt {3} a^{5/3} b^{2/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {\sqrt [3]{a} (b d-a g)}{\sqrt [3]{b}}-a f+b c\right )}{6 a^{5/3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (b c-a f)-\sqrt [3]{a} (b d-a g)\right )}{3 a^{5/3} b^{2/3}}-\frac {(b e-a h) \log \left (a+b x^3\right )}{3 a b}-\frac {c}{2 a x^2}-\frac {d}{a x}+\frac {e \log (x)}{a}\) |
Input:
Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)),x]
Output:
-1/2*c/(a*x^2) - d/(a*x) + ((b^(4/3)*c + a^(1/3)*b*d - 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)* b^(2/3)) + (e*Log[x])/a - ((b^(1/3)*(b*c - a*f) - a^(1/3)*(b*d - a*g))*Log [a^(1/3) + b^(1/3)*x])/(3*a^(5/3)*b^(2/3)) + ((b*c - a*f - (a^(1/3)*(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^(1/3)) - ((b*e - a*h)*Log[a + b*x^3])/(3*a*b)
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 = 251, normalized size of antiderivative = 0.97
| method | result | size |
| default | \(\frac {\left (a f -c b \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (a g -b d \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (a h -b e \right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{a}-\frac {c}{2 a \,x^{2}}-\frac {d}{a x}+\frac {e \ln \left (x \right )}{a}\) | \(251\) |
| risch | \(\frac {-\frac {d x}{a}-\frac {c}{2 a}}{x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} b^{3} \textit {\_Z}^{3}+\left (-3 a^{5} b^{2} h +3 a^{4} b^{3} e \right ) \textit {\_Z}^{2}+\left (3 a^{5} b \,h^{2}-6 a^{4} b^{2} e h +3 a^{4} b^{2} f g -3 a^{3} b^{3} c g -3 a^{3} b^{3} d f +3 a^{3} b^{3} e^{2}+3 a^{2} b^{4} c d \right ) \textit {\_Z} -a^{5} h^{3}+3 a^{4} b e \,h^{2}-3 a^{4} b f g h +a^{4} b \,g^{3}+3 a^{3} b^{2} c g h +3 a^{3} b^{2} d f h -3 a^{3} b^{2} d \,g^{2}-3 a^{3} b^{2} e^{2} h +3 a^{3} b^{2} e f g -a^{3} b^{2} f^{3}-3 a^{2} b^{3} c d h -3 a^{2} b^{3} c e g +3 a^{2} b^{3} c \,f^{2}+3 a^{2} b^{3} d^{2} g -3 a^{2} b^{3} d e f +a^{2} b^{3} e^{3}-3 a \,b^{4} c^{2} f +3 a \,b^{4} c d e -a \,b^{4} d^{3}+b^{5} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{5} b^{3}+\left (11 a^{5} b^{2} h -8 a^{4} b^{3} e \right ) \textit {\_R}^{2}+\left (-10 a^{5} b \,h^{2}+14 a^{4} b^{2} e h -10 a^{4} b^{2} f g +10 a^{3} b^{3} c g +10 a^{3} b^{3} d f -4 a^{3} b^{3} e^{2}-10 a^{2} b^{4} c d \right ) \textit {\_R} +3 a^{5} h^{3}-6 a^{4} b e \,h^{2}+9 a^{4} b f g h -3 a^{4} b \,g^{3}-9 a^{3} b^{2} c g h -9 a^{3} b^{2} d f h +9 a^{3} b^{2} d \,g^{2}+3 a^{3} b^{2} e^{2} h -6 a^{3} b^{2} e f g +3 a^{3} b^{2} f^{3}+9 a^{2} b^{3} c d h +6 a^{2} b^{3} c e g -9 a^{2} b^{3} c \,f^{2}-9 a^{2} b^{3} d^{2} g +6 a^{2} b^{3} d e f +9 a \,b^{4} c^{2} f -6 a \,b^{4} c d e +3 a \,b^{4} d^{3}-3 b^{5} c^{3}\right ) x +\left (a^{5} b^{2} g -a^{4} b^{3} d \right ) \textit {\_R}^{2}+\left (-a^{5} b g h +a^{4} b^{2} d h -2 a^{4} b^{2} e g -a^{4} b^{2} f^{2}+2 a^{3} b^{3} c f +2 a^{3} b^{3} d e -a^{2} b^{4} c^{2}\right ) \textit {\_R} +3 a^{4} b e g h -3 a^{3} b^{2} d e h -3 a^{3} b^{2} e^{2} g +3 a^{3} b^{2} e \,f^{2}-6 a^{2} b^{3} c e f +3 a^{2} b^{3} d \,e^{2}+3 a \,b^{4} c^{2} e \right )\right )}{3}+\frac {e \ln \left (x \right )}{a}\) | \(827\) |
Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
((a*f-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)))+(a*g-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)*arc tan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))+1/3*(a*h-b*e)/b*ln(b*x^3+a))/a-1/2*c /a/x^2-d/a/x+e*ln(x)/a
Result contains complex when optimal does not.
Time = 33.45 (sec) , antiderivative size = 15424, normalized size of antiderivative = 59.78 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a),x, algorithm="fric as")
Output:
Too large to include
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**3/(b*x**3+a),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\frac {e \log \left (x\right )}{a} - \frac {\sqrt {3} {\left (b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b f \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2} b} - \frac {{\left (2 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a h \left (\frac {a}{b}\right )^{\frac {2}{3}} + b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - a g \left (\frac {a}{b}\right )^{\frac {1}{3}} - b c + a f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - a h \left (\frac {a}{b}\right )^{\frac {2}{3}} - b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + a g \left (\frac {a}{b}\right )^{\frac {1}{3}} + b c - a f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, d x + c}{2 \, a x^{2}} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a),x, algorithm="maxi ma")
Output:
e*log(x)/a - 1/3*sqrt(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))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b )^(1/3))/(a^2*b) - 1/6*(2*b*e*(a/b)^(2/3) - 2*a*h*(a/b)^(2/3) + b*d*(a/b)^ (1/3) - a*g*(a/b)^(1/3) - b*c + a*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3) )/(a*b*(a/b)^(2/3)) - 1/3*(b*e*(a/b)^(2/3) - a*h*(a/b)^(2/3) - b*d*(a/b)^( 1/3) + a*g*(a/b)^(1/3) + b*c - a*f)*log(x + (a/b)^(1/3))/(a*b*(a/b)^(2/3)) - 1/2*(2*d*x + c)/(a*x^2)
Time = 0.13 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\frac {e \log \left ({\left | x \right |}\right )}{a} + \frac {\sqrt {3} {\left (b^{2} c - a b f - \left (-a b^{2}\right )^{\frac {1}{3}} b d + \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {2}{3}} a} + \frac {{\left (b^{2} c - a b f + \left (-a b^{2}\right )^{\frac {1}{3}} b d - \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {2}{3}} a} - \frac {{\left (b e - a h\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a b} + \frac {{\left (a b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b g \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a b^{2} c - a^{2} b f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{3} b} - \frac {2 \, d x + c}{2 \, a x^{2}} \] Input:
integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a),x, algorithm="giac ")
Output:
e*log(abs(x))/a + 1/3*sqrt(3)*(b^2*c - a*b*f - (-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) + 1/6*(b^2*c - a*b*f + (-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) - 1/3*(b*e - a*h)*log(abs(b*x^3 + a))/(a*b) + 1/3*(a*b^2*d*(-a/b)^(1/3) - a^2*b*g*(-a /b)^(1/3) + a*b^2*c - a^2*b*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^ 3*b) - 1/2*(2*d*x + c)/(a*x^2)
Time = 6.22 (sec) , antiderivative size = 6948, normalized size of antiderivative = 26.93 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx=\text {Too large to display} \] Input:
int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x^3*(a + b*x^3)),x)
Output:
symsum(log(-(b^5*c^3*x - a^5*h^3*x - a^2*b^3*d*e^2 + 36*root(27*a^5*b^3*z^ 3 - 27*a^5*b^2*h*z^2 + 27*a^4*b^3*e*z^2 - 18*a^4*b^2*e*h*z + 9*a^4*b^2*f*g *z - 9*a^3*b^3*d*f*z - 9*a^3*b^3*c*g*z + 9*a^2*b^4*c*d*z + 9*a^5*b*h^2*z + 9*a^3*b^3*e^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^5*b^3*x - a^3*b^2*e*f^2 + a^3*b^2*e^2*g - a^3*b^2*f^3*x - a*b^4*c^2*e - a*b^4*d^3*x + a^4*b*g^3*x + root(27*a^5*b^3*z^3 - 27*a^5*b^2*h*z^2 + 27*a^4*b^3*e*z^2 - 18*a^4*b^2*e*h *z + 9*a^4*b^2*f*g*z - 9*a^3*b^3*d*f*z - 9*a^3*b^3*c*g*z + 9*a^2*b^4*c*d*z + 9*a^5*b*h^2*z + 9*a^3*b^3*e^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^2*b^4*c^2 + 3*root(27*a^5*b^3*z^3 - 27*a^5*b^2*h*z^2 + 27*a^4*b^3*e*z^2 - 18*a^4*b^2 *e*h*z + 9*a^4*b^2*f*g*z - 9*a^3*b^3*d*f*z - 9*a^3*b^3*c*g*z + 9*a^2*b^4*c *d*z + 9*a^5*b*h^2*z + 9*a^3*b^3*e^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 - ...
Time = 0.17 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.91 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x^3 \left (a+b x^3\right )} \, dx =\text {Too large to display} \] Input:
int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^3/(b*x^3+a),x)
Output:
( - 2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*s qrt(3)))*a*b*f*x**2 + 2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1 /3)*x)/(a**(1/3)*sqrt(3)))*b**2*c*x**2 - 2*sqrt(3)*atan((a**(1/3) - 2*b**( 1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b*g*x**2 + 2*sqrt(3)*atan((a**(1/3) - 2*b **(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**2*d*x**2 - b**(1/3)*a**(2/3)*log(a**(2 /3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*f*x**2 + b**(1/3)*a**(2/3)* log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**2*c*x**2 + 2*b**(1/ 3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a*b*f*x**2 - 2*b**(1/3)*a**(2/3)*lo g(a**(1/3) + b**(1/3)*x)*b**2*c*x**2 + 2*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*h*x**2 - 2*b**(2/3)*a**(1/3)*log (a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*e*x**2 + 2*b**(2/3)*a **(1/3)*log(a**(1/3) + b**(1/3)*x)*a**2*h*x**2 - 2*b**(2/3)*a**(1/3)*log(a **(1/3) + b**(1/3)*x)*a*b*e*x**2 + 6*b**(2/3)*a**(1/3)*log(x)*a*b*e*x**2 - 3*b**(2/3)*a**(1/3)*a*b*c - 6*b**(2/3)*a**(1/3)*a*b*d*x + log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b*g*x**2 - log(a**(2/3) - b**(1/ 3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**2*d*x**2 - 2*log(a**(1/3) + b**(1/3)*x )*a**2*b*g*x**2 + 2*log(a**(1/3) + b**(1/3)*x)*a*b**2*d*x**2)/(6*b**(2/3)* a**(1/3)*a**2*b*x**2)