Integrand size = 23, antiderivative size = 257 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}+\frac {x \left (5 a d+4 a e x-9 b c x^2\right )}{18 a^3 \left (a+b x^3\right )}-\frac {\left (5 \sqrt [3]{b} d+2 \sqrt [3]{a} e\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3} b^{2/3}}+\frac {c \log (x)}{a^3}+\frac {\left (5 \sqrt [3]{b} d-2 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (5 \sqrt [3]{b} d-2 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{2/3}}-\frac {c \log \left (a+b x^3\right )}{3 a^3} \] Output:
1/6*x*(-b*c*x^2+a*e*x+a*d)/a^2/(b*x^3+a)^2+1/18*x*(-9*b*c*x^2+4*a*e*x+5*a* d)/a^3/(b*x^3+a)-1/27*(5*b^(1/3)*d+2*a^(1/3)*e)*arctan(1/3*(a^(1/3)-2*b^(1 /3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(8/3)/b^(2/3)+c*ln(x)/a^3+1/27*(5*b^(1/3 )*d-2*a^(1/3)*e)*ln(a^(1/3)+b^(1/3)*x)/a^(8/3)/b^(2/3)-1/54*(5*b^(1/3)*d-2 *a^(1/3)*e)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(8/3)/b^(2/3)-1/3* c*ln(b*x^3+a)/a^3
Time = 0.15 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\frac {\frac {9 a^2 (c+x (d+e x))}{\left (a+b x^3\right )^2}+\frac {3 a (6 c+x (5 d+4 e x))}{a+b x^3}-\frac {2 \sqrt {3} \sqrt [3]{a} \left (5 \sqrt [3]{b} d+2 \sqrt [3]{a} e\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}+54 c \log (x)+\frac {2 \left (5 \sqrt [3]{a} \sqrt [3]{b} d-2 a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {\left (-5 \sqrt [3]{a} \sqrt [3]{b} d+2 a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}-18 c \log \left (a+b x^3\right )}{54 a^3} \] Input:
Integrate[(c + d*x + e*x^2)/(x*(a + b*x^3)^3),x]
Output:
((9*a^2*(c + x*(d + e*x)))/(a + b*x^3)^2 + (3*a*(6*c + x*(5*d + 4*e*x)))/( a + b*x^3) - (2*Sqrt[3]*a^(1/3)*(5*b^(1/3)*d + 2*a^(1/3)*e)*ArcTan[(1 - (2 *b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(2/3) + 54*c*Log[x] + (2*(5*a^(1/3)*b^(1/ 3)*d - 2*a^(2/3)*e)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) + ((-5*a^(1/3)*b^(1/ 3)*d + 2*a^(2/3)*e)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(2/3 ) - 18*c*Log[a + b*x^3])/(54*a^3)
Time = 1.19 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2368, 25, 2368, 27, 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.
| \(\displaystyle \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle \frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}-\frac {\int -\frac {-\frac {3 b^2 c x^3}{a}+4 b e x^2+5 b d x+6 b c}{x \left (b x^3+a\right )^2}dx}{6 a b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-\frac {3 b^2 c x^3}{a}+4 b e x^2+5 b d x+6 b c}{x \left (b x^3+a\right )^2}dx}{6 a b}+\frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle \frac {\frac {x \left (5 a b d+4 a b e x-9 b^2 c x^2\right )}{3 a^2 \left (a+b x^3\right )}-\frac {\int -\frac {2 \left (2 e x^2 b^2+9 c b^2+5 d x b^2\right )}{x \left (b x^3+a\right )}dx}{3 a b}}{6 a b}+\frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 \int \frac {2 e x^2 b^2+9 c b^2+5 d x b^2}{x \left (b x^3+a\right )}dx}{3 a b}+\frac {x \left (5 a b d+4 a b e x-9 b^2 c x^2\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b}+\frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \frac {\frac {2 \int \left (\frac {9 c b^2}{a x}+\frac {\left (-9 b c x^2+2 a e x+5 a d\right ) b^2}{a \left (b x^3+a\right )}\right )dx}{3 a b}+\frac {x \left (5 a b d+4 a b e x-9 b^2 c x^2\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b}+\frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \left (a d+a e x-b c x^2\right )}{6 a^2 \left (a+b x^3\right )^2}+\frac {\frac {2 \left (-\frac {b^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (2 \sqrt [3]{a} e+5 \sqrt [3]{b} d\right )}{\sqrt {3} a^{2/3}}-\frac {b^{4/3} \left (5 \sqrt [3]{b} d-2 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3}}+\frac {b^{4/3} \left (5 \sqrt [3]{b} d-2 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3}}-\frac {3 b^2 c \log \left (a+b x^3\right )}{a}+\frac {9 b^2 c \log (x)}{a}\right )}{3 a b}+\frac {x \left (5 a b d+4 a b e x-9 b^2 c x^2\right )}{3 a^2 \left (a+b x^3\right )}}{6 a b}\) |
Input:
Int[(c + d*x + e*x^2)/(x*(a + b*x^3)^3),x]
Output:
(x*(a*d + a*e*x - b*c*x^2))/(6*a^2*(a + b*x^3)^2) + ((x*(5*a*b*d + 4*a*b*e *x - 9*b^2*c*x^2))/(3*a^2*(a + b*x^3)) + (2*(-((b^(4/3)*(5*b^(1/3)*d + 2*a ^(1/3)*e)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2 /3))) + (9*b^2*c*Log[x])/a + (b^(4/3)*(5*b^(1/3)*d - 2*a^(1/3)*e)*Log[a^(1 /3) + b^(1/3)*x])/(3*a^(2/3)) - (b^(4/3)*(5*b^(1/3)*d - 2*a^(1/3)*e)*Log[a ^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(2/3)) - (3*b^2*c*Log[a + b*x^3])/a))/(3*a*b))/(6*a*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.05
| method | result | size |
| risch | \(\frac {\frac {2 e b \,x^{5}}{9 a^{2}}+\frac {5 d b \,x^{4}}{18 a^{2}}+\frac {b c \,x^{3}}{3 a^{2}}+\frac {7 e \,x^{2}}{18 a}+\frac {4 d x}{9 a}+\frac {c}{2 a}}{\left (b \,x^{3}+a \right )^{2}}+\frac {c \ln \left (-x \right )}{a^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{9} b^{2} \textit {\_Z}^{3}+27 a^{6} b^{2} c \,\textit {\_Z}^{2}+\left (30 a^{4} b d e +243 a^{3} b^{2} c^{2}\right ) \textit {\_Z} +8 a^{2} e^{3}+270 a b c d e -125 a b \,d^{3}+729 c^{3} b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{8} b^{2}-72 \textit {\_R}^{2} a^{5} b^{2} c +\left (-100 a^{3} b d e -324 a^{2} b^{2} c^{2}\right ) \textit {\_R} -24 a \,e^{3}-540 b c d e +375 b \,d^{3}\right ) x +2 a^{6} b e \,\textit {\_R}^{2}+\left (-36 a^{3} b c e -25 a^{3} b \,d^{2}\right ) \textit {\_R} -486 b \,c^{2} e +675 b c \,d^{2}\right )\right )}{27}\) | \(269\) |
| default | \(\frac {\frac {\frac {2}{9} a b e \,x^{5}+\frac {5}{18} a b d \,x^{4}+\frac {1}{3} a b c \,x^{3}+\frac {7}{18} a^{2} e \,x^{2}+\frac {4}{9} a^{2} d x +\frac {1}{2} a^{2} c}{\left (b \,x^{3}+a \right )^{2}}+\frac {5 a d \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{9}+\frac {2 a e \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9}-\frac {c \ln \left (b \,x^{3}+a \right )}{3}}{a^{3}}+\frac {c \ln \left (x \right )}{a^{3}}\) | \(270\) |
Input:
int((e*x^2+d*x+c)/x/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
Output:
(2/9/a^2*e*b*x^5+5/18/a^2*d*b*x^4+1/3*b/a^2*c*x^3+7/18*e/a*x^2+4/9*d/a*x+1 /2*c/a)/(b*x^3+a)^2+c/a^3*ln(-x)+1/27*sum(_R*ln((-4*_R^3*a^8*b^2-72*_R^2*a ^5*b^2*c+(-100*a^3*b*d*e-324*a^2*b^2*c^2)*_R-24*a*e^3-540*b*c*d*e+375*b*d^ 3)*x+2*a^6*b*e*_R^2+(-36*a^3*b*c*e-25*a^3*b*d^2)*_R-486*b*c^2*e+675*b*c*d^ 2),_R=RootOf(a^9*b^2*_Z^3+27*a^6*b^2*c*_Z^2+(30*a^4*b*d*e+243*a^3*b^2*c^2) *_Z+8*a^2*e^3+270*a*b*c*d*e-125*a*b*d^3+729*c^3*b^2))
Result contains complex when optimal does not.
Time = 0.91 (sec) , antiderivative size = 5229, normalized size of antiderivative = 20.35 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((e*x^2+d*x+c)/x/(b*x^3+a)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x**2+d*x+c)/x/(b*x**3+a)**3,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\frac {4 \, b e x^{5} + 5 \, b d x^{4} + 6 \, b c x^{3} + 7 \, a e x^{2} + 8 \, a d x + 9 \, a c}{18 \, {\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )}} + \frac {c \log \left (x\right )}{a^{3}} + \frac {\sqrt {3} {\left (2 \, a e \left (\frac {a}{b}\right )^{\frac {2}{3}} + 5 \, a d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{4}} - \frac {{\left (18 \, b c \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a e \left (\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, a d\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, a^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (9 \, b c \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a e \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a d\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate((e*x^2+d*x+c)/x/(b*x^3+a)^3,x, algorithm="maxima")
Output:
1/18*(4*b*e*x^5 + 5*b*d*x^4 + 6*b*c*x^3 + 7*a*e*x^2 + 8*a*d*x + 9*a*c)/(a^ 2*b^2*x^6 + 2*a^3*b*x^3 + a^4) + c*log(x)/a^3 + 1/27*sqrt(3)*(2*a*e*(a/b)^ (2/3) + 5*a*d*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1 /3))/a^4 - 1/54*(18*b*c*(a/b)^(2/3) - 2*a*e*(a/b)^(1/3) + 5*a*d)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^3*b*(a/b)^(2/3)) - 1/27*(9*b*c*(a/b)^(2/3 ) + 2*a*e*(a/b)^(1/3) - 5*a*d)*log(x + (a/b)^(1/3))/(a^3*b*(a/b)^(2/3))
Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=-\frac {\sqrt {3} {\left (5 \, b d - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (5 \, b d + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} e\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {c \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3}} + \frac {c \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {4 \, a b e x^{5} + 5 \, a b d x^{4} + 6 \, a b c x^{3} + 7 \, a^{2} e x^{2} + 8 \, a^{2} d x + 9 \, a^{2} c}{18 \, {\left (b x^{3} + a\right )}^{2} a^{3}} - \frac {{\left (2 \, a^{4} b e \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, a^{4} b d\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{7} b} \] Input:
integrate((e*x^2+d*x+c)/x/(b*x^3+a)^3,x, algorithm="giac")
Output:
-1/27*sqrt(3)*(5*b*d - 2*(-a*b^2)^(1/3)*e)*arctan(1/3*sqrt(3)*(2*x + (-a/b )^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)*a^2) - 1/54*(5*b*d + 2*(-a*b^2)^(1/ 3)*e)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a^2) - 1/3* c*log(abs(b*x^3 + a))/a^3 + c*log(abs(x))/a^3 + 1/18*(4*a*b*e*x^5 + 5*a*b* d*x^4 + 6*a*b*c*x^3 + 7*a^2*e*x^2 + 8*a^2*d*x + 9*a^2*c)/((b*x^3 + a)^2*a^ 3) - 1/27*(2*a^4*b*e*(-a/b)^(1/3) + 5*a^4*b*d)*(-a/b)^(1/3)*log(abs(x - (- a/b)^(1/3)))/(a^7*b)
Time = 5.98 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.10 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx=\frac {\frac {c}{2\,a}+\frac {7\,e\,x^2}{18\,a}+\frac {4\,d\,x}{9\,a}+\frac {b\,c\,x^3}{3\,a^2}+\frac {5\,b\,d\,x^4}{18\,a^2}+\frac {2\,b\,e\,x^5}{9\,a^2}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\left (\sum _{k=1}^3\ln \left (\frac {25\,b^2\,c\,d^2-18\,b^2\,c^2\,e}{81\,a^6}-\mathrm {root}\left (19683\,a^9\,b^2\,z^3+19683\,a^6\,b^2\,c\,z^2+810\,a^4\,b\,d\,e\,z+6561\,a^3\,b^2\,c^2\,z+270\,a\,b\,c\,d\,e-125\,a\,b\,d^3+8\,a^2\,e^3+729\,b^2\,c^3,z,k\right )\,\left (\frac {25\,a^3\,b^2\,d^2+36\,c\,e\,a^3\,b^2}{81\,a^6}+\mathrm {root}\left (19683\,a^9\,b^2\,z^3+19683\,a^6\,b^2\,c\,z^2+810\,a^4\,b\,d\,e\,z+6561\,a^3\,b^2\,c^2\,z+270\,a\,b\,c\,d\,e-125\,a\,b\,d^3+8\,a^2\,e^3+729\,b^2\,c^3,z,k\right )\,\left (-\frac {2\,b^2\,e}{3}+\mathrm {root}\left (19683\,a^9\,b^2\,z^3+19683\,a^6\,b^2\,c\,z^2+810\,a^4\,b\,d\,e\,z+6561\,a^3\,b^2\,c^2\,z+270\,a\,b\,c\,d\,e-125\,a\,b\,d^3+8\,a^2\,e^3+729\,b^2\,c^3,z,k\right )\,a^2\,b^3\,x\,36+\frac {24\,b^3\,c\,x}{a}\right )+\frac {x\,\left (900\,d\,e\,a^3\,b^2+2916\,a^2\,b^3\,c^2\right )}{729\,a^6}\right )-\frac {x\,\left (-125\,b^2\,d^3+180\,c\,b^2\,d\,e+8\,a\,b\,e^3\right )}{729\,a^6}\right )\,\mathrm {root}\left (19683\,a^9\,b^2\,z^3+19683\,a^6\,b^2\,c\,z^2+810\,a^4\,b\,d\,e\,z+6561\,a^3\,b^2\,c^2\,z+270\,a\,b\,c\,d\,e-125\,a\,b\,d^3+8\,a^2\,e^3+729\,b^2\,c^3,z,k\right )\right )+\frac {c\,\ln \left (x\right )}{a^3} \] Input:
int((c + d*x + e*x^2)/(x*(a + b*x^3)^3),x)
Output:
(c/(2*a) + (7*e*x^2)/(18*a) + (4*d*x)/(9*a) + (b*c*x^3)/(3*a^2) + (5*b*d*x ^4)/(18*a^2) + (2*b*e*x^5)/(9*a^2))/(a^2 + b^2*x^6 + 2*a*b*x^3) + symsum(l og((25*b^2*c*d^2 - 18*b^2*c^2*e)/(81*a^6) - root(19683*a^9*b^2*z^3 + 19683 *a^6*b^2*c*z^2 + 810*a^4*b*d*e*z + 6561*a^3*b^2*c^2*z + 270*a*b*c*d*e - 12 5*a*b*d^3 + 8*a^2*e^3 + 729*b^2*c^3, z, k)*((25*a^3*b^2*d^2 + 36*a^3*b^2*c *e)/(81*a^6) + root(19683*a^9*b^2*z^3 + 19683*a^6*b^2*c*z^2 + 810*a^4*b*d* e*z + 6561*a^3*b^2*c^2*z + 270*a*b*c*d*e - 125*a*b*d^3 + 8*a^2*e^3 + 729*b ^2*c^3, z, k)*(36*root(19683*a^9*b^2*z^3 + 19683*a^6*b^2*c*z^2 + 810*a^4*b *d*e*z + 6561*a^3*b^2*c^2*z + 270*a*b*c*d*e - 125*a*b*d^3 + 8*a^2*e^3 + 72 9*b^2*c^3, z, k)*a^2*b^3*x - (2*b^2*e)/3 + (24*b^3*c*x)/a) + (x*(2916*a^2* b^3*c^2 + 900*a^3*b^2*d*e))/(729*a^6)) - (x*(8*a*b*e^3 - 125*b^2*d^3 + 180 *b^2*c*d*e))/(729*a^6))*root(19683*a^9*b^2*z^3 + 19683*a^6*b^2*c*z^2 + 810 *a^4*b*d*e*z + 6561*a^3*b^2*c^2*z + 270*a*b*c*d*e - 125*a*b*d^3 + 8*a^2*e^ 3 + 729*b^2*c^3, z, k), k, 1, 3) + (c*log(x))/a^3
Time = 0.17 (sec) , antiderivative size = 783, normalized size of antiderivative = 3.05 \[ \int \frac {c+d x+e x^2}{x \left (a+b x^3\right )^3} \, dx =\text {Too large to display} \] Input:
int((e*x^2+d*x+c)/x/(b*x^3+a)^3,x)
Output:
( - 10*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)* sqrt(3)))*a**2*d - 20*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3 )*x)/(a**(1/3)*sqrt(3)))*a*b*d*x**3 - 10*b**(1/3)*a**(2/3)*sqrt(3)*atan((a **(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**2*d*x**6 - 4*sqrt(3)*atan(( a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*e - 8*sqrt(3)*atan((a**( 1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b*e*x**3 - 4*sqrt(3)*atan((a **(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**2*e*x**6 - 5*b**(1/3)*a** (2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*d - 10*b**( 1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*d*x* *3 - 5*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x** 2)*b**2*d*x**6 + 10*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a**2*d + 20*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a*b*d*x**3 + 10*b**(1/3)*a **(2/3)*log(a**(1/3) + b**(1/3)*x)*b**2*d*x**6 - 18*b**(2/3)*a**(1/3)*log( a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*c - 36*b**(2/3)*a**(1 /3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*c*x**3 - 18*b* *(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**2*c *x**6 - 18*b**(2/3)*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a**2*c - 36*b**(2/ 3)*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a*b*c*x**3 - 18*b**(2/3)*a**(1/3)*l og(a**(1/3) + b**(1/3)*x)*b**2*c*x**6 + 54*b**(2/3)*a**(1/3)*log(x)*a**2*c + 108*b**(2/3)*a**(1/3)*log(x)*a*b*c*x**3 + 54*b**(2/3)*a**(1/3)*log(x...