Integrand size = 38, antiderivative size = 258 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\frac {g x}{b}+\frac {h x^2}{2 b}-\frac {\left (b^{4/3} d+\sqrt [3]{a} b e-a \sqrt [3]{b} g-a^{4/3} h\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3}}+\frac {c \log (x)}{a}+\frac {\left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{5/3}}-\frac {\left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} b^{5/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b} \]
g*x/b+1/2*h*x^2/b+c*ln(x)/a+1/3*(b^(1/3)*(-a*g+b*d)-a^(1/3)*(-a*h+b*e))*ln (a^(1/3)+b^(1/3)*x)/a^(2/3)/b^(5/3)-1/6*(b^(1/3)*(-a*g+b*d)-a^(1/3)*(-a*h+ b*e))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(2/3)/b^(5/3)-1/3*(-a*f+ b*c)*ln(b*x^3+a)/a/b-1/3*(b^(4/3)*d+a^(1/3)*b*e-a*b^(1/3)*g-a^(4/3)*h)*arc tan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(2/3)/b^(5/3)*3^(1/2)
Time = 0.18 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\frac {6 a b^{2/3} g x+3 a b^{2/3} h x^2+2 \sqrt {3} \sqrt [3]{a} \left (-b^{4/3} d-\sqrt [3]{a} b e+a \sqrt [3]{b} g+a^{4/3} h\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+6 b^{5/3} c \log (x)+2 \sqrt [3]{a} \left (b^{4/3} d-\sqrt [3]{a} b e-a \sqrt [3]{b} g+a^{4/3} h\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-\sqrt [3]{a} \left (b^{4/3} d-\sqrt [3]{a} b e-a \sqrt [3]{b} g+a^{4/3} h\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-2 b^{2/3} (b c-a f) \log \left (a+b x^3\right )}{6 a b^{5/3}} \]
(6*a*b^(2/3)*g*x + 3*a*b^(2/3)*h*x^2 + 2*Sqrt[3]*a^(1/3)*(-(b^(4/3)*d) - a ^(1/3)*b*e + a*b^(1/3)*g + a^(4/3)*h)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/S qrt[3]] + 6*b^(5/3)*c*Log[x] + 2*a^(1/3)*(b^(4/3)*d - a^(1/3)*b*e - a*b^(1 /3)*g + a^(4/3)*h)*Log[a^(1/3) + b^(1/3)*x] - a^(1/3)*(b^(4/3)*d - a^(1/3) *b*e - a*b^(1/3)*g + a^(4/3)*h)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)* x^2] - 2*b^(2/3)*(b*c - a*f)*Log[a + b*x^3])/(6*a*b^(5/3))
Time = 0.61 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.99, 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.
| \(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \int \left (\frac {-b x^2 (b c-a f)+a (b d-a g)+a x (b e-a h)}{a b \left (a+b x^3\right )}+\frac {c}{a x}+\frac {g}{b}+\frac {h x}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{4/3} (-h)+\sqrt [3]{a} b e-a \sqrt [3]{b} g+b^{4/3} d\right )}{\sqrt {3} a^{2/3} b^{5/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}-a g+b d\right )}{6 a^{2/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right )}{3 a^{2/3} b^{5/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {c \log (x)}{a}+\frac {g x}{b}+\frac {h x^2}{2 b}\) |
(g*x)/b + (h*x^2)/(2*b) - ((b^(4/3)*d + a^(1/3)*b*e - a*b^(1/3)*g - a^(4/3 )*h)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*b ^(5/3)) + (c*Log[x])/a + ((b^(1/3)*(b*d - a*g) - a^(1/3)*(b*e - a*h))*Log[ a^(1/3) + b^(1/3)*x])/(3*a^(2/3)*b^(5/3)) - ((b*d - a*g - (a^(1/3)*(b*e - a*h))/b^(1/3))*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(2/3)* b^(4/3)) - ((b*c - a*f)*Log[a + b*x^3])/(3*a*b)
3.5.8.3.1 Defintions of rubi rules used
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 = 1.55 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(\frac {\frac {1}{2} h \,x^{2}+g x}{b}+\frac {c \ln \left (x \right )}{a}+\frac {\left (-a^{2} g +a b d \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (-a^{2} h +a e b \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (a f b -b^{2} c \right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{b a}\) | \(259\) |
| risch | \(\frac {h \,x^{2}}{2 b}+\frac {g x}{b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{3} b^{2} \textit {\_Z}^{3}+\left (-3 a^{3} b^{2} f +3 a^{2} c \,b^{3}\right ) \textit {\_Z}^{2}+\left (3 a^{4} b g h -3 a^{3} b^{2} d h -3 a^{3} b^{2} e g +3 a^{3} b^{2} f^{2}-6 a^{2} b^{3} c f +3 a^{2} b^{3} d e +3 c^{2} a \,b^{4}\right ) \textit {\_Z} -a^{5} h^{3}+3 a^{4} b e \,h^{2}-3 a^{4} b f g h +a^{4} b \,g^{3}+3 a^{3} b^{2} c g h +3 a^{3} b^{2} d f h -3 a^{3} b^{2} d \,g^{2}-3 a^{3} b^{2} e^{2} h +3 a^{3} b^{2} e f g -a^{3} b^{2} f^{3}-3 a^{2} b^{3} c d h -3 a^{2} b^{3} c e g +3 a^{2} b^{3} c \,f^{2}+3 a^{2} b^{3} d^{2} g -3 a^{2} b^{3} d e f +a^{2} b^{3} e^{3}-3 a \,b^{4} c^{2} f +3 a \,b^{4} c d e -a \,b^{4} d^{3}+b^{5} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{2} b^{2}+\left (11 a^{2} b^{2} f -8 a \,b^{3} c \right ) \textit {\_R}^{2}+\left (-10 a^{3} b g h +10 a^{2} b^{2} d h +10 a^{2} b^{2} e g -10 a^{2} b^{2} f^{2}+14 a \,b^{3} c f -10 a \,b^{3} d e -4 b^{4} c^{2}\right ) \textit {\_R} +3 a^{4} h^{3}-9 a^{3} b e \,h^{2}+9 a^{3} b f g h -3 a^{3} b \,g^{3}-6 a^{2} b^{2} c g h -9 a^{2} b^{2} d f h +9 a^{2} b^{2} d \,g^{2}+9 a^{2} b^{2} e^{2} h -9 a^{2} b^{2} e f g +3 a^{2} b^{2} f^{3}+6 a \,b^{3} c d h +6 a \,b^{3} c e g -6 a \,b^{3} c \,f^{2}-9 a \,b^{3} d^{2} g +9 a \,b^{3} d e f -3 a \,b^{3} e^{3}+3 b^{4} c^{2} f -6 b^{4} c d e +3 b^{4} d^{3}\right ) x +\left (-a^{3} b h +a^{2} e \,b^{2}\right ) \textit {\_R}^{2}+\left (a^{3} b f h -a^{3} b \,g^{2}+2 a^{2} b^{2} c h +2 a^{2} b^{2} d g -a^{2} b^{2} e f -2 a \,b^{3} c e -a \,b^{3} d^{2}\right ) \textit {\_R} -3 a^{2} b^{2} c f h +3 a^{2} b^{2} c \,g^{2}+3 a \,b^{3} c^{2} h -6 a \,b^{3} c d g +3 a \,b^{3} c e f -3 b^{4} c^{2} e +3 b^{4} c \,d^{2}\right )}{3 b}+\frac {c \ln \left (x \right )}{a}\) | \(788\) |
1/b*(1/2*h*x^2+g*x)+c*ln(x)/a+((-a^2*g+a*b*d)*(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^2*h+a*b*e)*(-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*f-b^2*c)*ln(b*x^3+a)/b)/b/a
Result contains complex when optimal does not.
Time = 84.14 (sec) , antiderivative size = 15327, normalized size of antiderivative = 59.41 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.12 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\frac {c \log \left (x\right )}{a} + \frac {h x^{2} + 2 \, g x}{2 \, b} + \frac {\sqrt {3} {\left (a b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - a^{2} h \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} g \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2} b} - \frac {{\left (2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b d - a^{2} g\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b d + a^{2} g\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
c*log(x)/a + 1/2*(h*x^2 + 2*g*x)/b + 1/3*sqrt(3)*(a*b*e*(a/b)^(2/3) - a^2* h*(a/b)^(2/3) + a*b*d*(a/b)^(1/3) - a^2*g*(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^2*c*(a/b)^(2/3) - 2*a* b*f*(a/b)^(2/3) - a*b*e*(a/b)^(1/3) + a^2*h*(a/b)^(1/3) + a*b*d - a^2*g)*l og(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*b^2*(a/b)^(2/3)) - 1/3*(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) - a*b *d + a^2*g)*log(x + (a/b)^(1/3))/(a*b^2*(a/b)^(2/3))
Time = 0.27 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.08 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\frac {c \log \left ({\left | x \right |}\right )}{a} - \frac {\sqrt {3} {\left (b^{2} d - a b g - \left (-a b^{2}\right )^{\frac {1}{3}} b e + \left (-a b^{2}\right )^{\frac {1}{3}} a h\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} - \frac {{\left (b^{2} d - a b g + \left (-a b^{2}\right )^{\frac {1}{3}} b e - \left (-a b^{2}\right )^{\frac {1}{3}} a h\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} - \frac {{\left (b c - a f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a b} + \frac {b h x^{2} + 2 \, b g x}{2 \, b^{2}} - \frac {{\left (a^{2} b^{3} e \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} b^{2} h \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} b^{3} d - a^{3} b^{2} g\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{3} b^{3}} \]
c*log(abs(x))/a - 1/3*sqrt(3)*(b^2*d - a*b*g - (-a*b^2)^(1/3)*b*e + (-a*b^ 2)^(1/3)*a*h)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b ^2)^(2/3)*b) - 1/6*(b^2*d - a*b*g + (-a*b^2)^(1/3)*b*e - (-a*b^2)^(1/3)*a* h)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*b) - 1/3*(b*c - a*f)*log(abs(b*x^3 + a))/(a*b) + 1/2*(b*h*x^2 + 2*b*g*x)/b^2 - 1/3*(a^2* b^3*e*(-a/b)^(1/3) - a^3*b^2*h*(-a/b)^(1/3) + a^2*b^3*d - a^3*b^2*g)*(-a/b )^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b^3)
Time = 9.19 (sec) , antiderivative size = 1731, normalized size of antiderivative = 6.71 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx=\text {Too large to display} \]
symsum(log(b^2*c*d^2 - root(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5 *c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3*e*g*z - 9*a^3*b^3*d*h*z - 18*a^2*b^4* c*f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^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^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5* h^3, z, k)*(a^3*g^2 - root(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5* c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3*e*g*z - 9*a^3*b^3*d*h*z - 18*a^2*b^4*c *f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^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^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h ^3, z, k)*((x*(33*a^2*b^4*f - 24*a*b^5*c))/b^2 + 3*a^2*b^2*e - 3*a^3*b*h - 36*root(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5*c*z^2 + 9*a^4*b^2* g*h*z - 9*a^3*b^3*e*g*z - 9*a^3*b^3*d*h*z - 18*a^2*b^4*c*f*z + 9*a^2*b^4*d *e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^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*...