Integrand size = 38, antiderivative size = 309 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {f x}{b^2}+\frac {g x^2}{2 b^2}+\frac {h x^3}{3 b^2}-\frac {x \left (b c-a f+(b d-a g) x+(b e-a h) x^2\right )}{3 b^2 \left (a+b x^3\right )}-\frac {\left (b^{4/3} c+2 \sqrt [3]{a} b d-4 a \sqrt [3]{b} f-5 a^{4/3} g\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3} b^{8/3}}+\frac {\left (\sqrt [3]{b} (b c-4 a f)-\sqrt [3]{a} (2 b d-5 a g)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{8/3}}-\frac {\left (b c-4 a f-\frac {\sqrt [3]{a} (2 b d-5 a g)}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{7/3}}+\frac {(b e-2 a h) \log \left (a+b x^3\right )}{3 b^3} \] Output:
f*x/b^2+1/2*g*x^2/b^2+1/3*h*x^3/b^2-1/3*x*(b*c-a*f+(-a*g+b*d)*x+(-a*h+b*e) *x^2)/b^2/(b*x^3+a)-1/9*(b^(4/3)*c+2*a^(1/3)*b*d-4*a*b^(1/3)*f-5*a^(4/3)*g )*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(2/3)/b^(8/3 )+1/9*(b^(1/3)*(-4*a*f+b*c)-a^(1/3)*(-5*a*g+2*b*d))*ln(a^(1/3)+b^(1/3)*x)/ a^(2/3)/b^(8/3)-1/18*(b*c-4*a*f-a^(1/3)*(-5*a*g+2*b*d)/b^(1/3))*ln(a^(2/3) -a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(2/3)/b^(7/3)+1/3*(-2*a*h+b*e)*ln(b*x^3+ a)/b^3
Time = 0.15 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {18 b f x+9 b g x^2+6 b h x^3-\frac {6 \left (a^2 h+b^2 x (c+d x)-a b (e+x (f+g x))\right )}{a+b x^3}+\frac {2 \sqrt {3} \sqrt [3]{b} \left (-b^{4/3} c-2 \sqrt [3]{a} b d+4 a \sqrt [3]{b} f+5 a^{4/3} g\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {2 \sqrt [3]{b} \left (b^{4/3} c-2 \sqrt [3]{a} b d-4 a \sqrt [3]{b} f+5 a^{4/3} g\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{2/3}}-\frac {\sqrt [3]{b} \left (b^{4/3} c-2 \sqrt [3]{a} b d-4 a \sqrt [3]{b} f+5 a^{4/3} g\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{2/3}}+6 (b e-2 a h) \log \left (a+b x^3\right )}{18 b^3} \] Input:
Integrate[(x^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x]
Output:
(18*b*f*x + 9*b*g*x^2 + 6*b*h*x^3 - (6*(a^2*h + b^2*x*(c + d*x) - a*b*(e + x*(f + g*x))))/(a + b*x^3) + (2*Sqrt[3]*b^(1/3)*(-(b^(4/3)*c) - 2*a^(1/3) *b*d + 4*a*b^(1/3)*f + 5*a^(4/3)*g)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqr t[3]])/a^(2/3) + (2*b^(1/3)*(b^(4/3)*c - 2*a^(1/3)*b*d - 4*a*b^(1/3)*f + 5 *a^(4/3)*g)*Log[a^(1/3) + b^(1/3)*x])/a^(2/3) - (b^(1/3)*(b^(4/3)*c - 2*a^ (1/3)*b*d - 4*a*b^(1/3)*f + 5*a^(4/3)*g)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(2/3) + 6*(b*e - 2*a*h)*Log[a + b*x^3])/(18*b^3)
Time = 1.44 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2367, 25, 2426, 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.
| \(\displaystyle \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle -\frac {\int -\frac {3 a b^2 h x^5+3 a b^2 g x^4+3 a b^2 f x^3+3 a b (b e-a h) x^2+2 a b (b d-a g) x+a b (b c-a f)}{b x^3+a}dx}{3 a b^3}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 b^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 a b^2 h x^5+3 a b^2 g x^4+3 a b^2 f x^3+3 a b (b e-a h) x^2+2 a b (b d-a g) x+a b (b c-a f)}{b x^3+a}dx}{3 a b^3}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 b^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \frac {\int \left (3 a b h x^2+3 a b g x+3 a b f+\frac {3 a b (b e-2 a h) x^2+a b (2 b d-5 a g) x+a b (b c-4 a f)}{b x^3+a}\right )dx}{3 a b^3}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 b^2 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\sqrt [3]{a} \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-5 a^{4/3} g+2 \sqrt [3]{a} b d-4 a \sqrt [3]{b} f+b^{4/3} c\right )}{\sqrt {3}}-\frac {1}{6} \sqrt [3]{a} \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (\sqrt [3]{b} (b c-4 a f)-\sqrt [3]{a} (2 b d-5 a g)\right )+\frac {1}{3} \sqrt [3]{a} \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (b c-4 a f)-\sqrt [3]{a} (2 b d-5 a g)\right )+a (b e-2 a h) \log \left (a+b x^3\right )+3 a b f x+\frac {3}{2} a b g x^2+a b h x^3}{3 a b^3}-\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 b^2 \left (a+b x^3\right )}\) |
Input:
Int[(x^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x]
Output:
-1/3*(x*(b*c - a*f + (b*d - a*g)*x + (b*e - a*h)*x^2))/(b^2*(a + b*x^3)) + (3*a*b*f*x + (3*a*b*g*x^2)/2 + a*b*h*x^3 - (a^(1/3)*b^(1/3)*(b^(4/3)*c + 2*a^(1/3)*b*d - 4*a*b^(1/3)*f - 5*a^(4/3)*g)*ArcTan[(a^(1/3) - 2*b^(1/3)*x )/(Sqrt[3]*a^(1/3))])/Sqrt[3] + (a^(1/3)*b^(1/3)*(b^(1/3)*(b*c - 4*a*f) - a^(1/3)*(2*b*d - 5*a*g))*Log[a^(1/3) + b^(1/3)*x])/3 - (a^(1/3)*b^(1/3)*(b ^(1/3)*(b*c - 4*a*f) - a^(1/3)*(2*b*d - 5*a*g))*Log[a^(2/3) - a^(1/3)*b^(1 /3)*x + b^(2/3)*x^2])/6 + a*(b*e - 2*a*h)*Log[a + b*x^3])/(3*a*b^3)
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1) *x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) I nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 ] && LtQ[p, -1] && IGtQ[m, 0]
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.44
| method | result | size |
| risch | \(\frac {h \,x^{3}}{3 b^{2}}+\frac {g \,x^{2}}{2 b^{2}}+\frac {f x}{b^{2}}+\frac {\left (\frac {a g}{3}-\frac {b d}{3}\right ) x^{2}+\left (\frac {a f}{3}-\frac {c b}{3}\right ) x -\frac {a \left (a h -b e \right )}{3 b}}{b^{2} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (3 \left (-2 a h +b e \right ) \textit {\_R}^{2}+\left (-5 a g +2 b d \right ) \textit {\_R} -4 a f +c b \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{3}}\) | \(136\) |
| default | \(\frac {\frac {1}{3} h \,x^{3}+\frac {1}{2} g \,x^{2}+f x}{b^{2}}-\frac {\frac {\left (-\frac {a g}{3}+\frac {b d}{3}\right ) x^{2}+\left (-\frac {a f}{3}+\frac {c b}{3}\right ) x +\frac {a \left (a h -b e \right )}{3 b}}{b \,x^{3}+a}+\frac {\left (4 a f -c b \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {\left (5 a g -2 b d \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}+\frac {\left (6 a h -3 b e \right ) \ln \left (b \,x^{3}+a \right )}{9 b}}{b^{2}}\) | \(301\) |
Input:
int(x^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x,method=_RETURNVERBOS E)
Output:
1/3*h*x^3/b^2+1/2*g*x^2/b^2+f*x/b^2+((1/3*a*g-1/3*b*d)*x^2+(1/3*a*f-1/3*c* b)*x-1/3*a*(a*h-b*e)/b)/b^2/(b*x^3+a)+1/9/b^3*sum((3*(-2*a*h+b*e)*_R^2+(-5 *a*g+2*b*d)*_R-4*a*f+c*b)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Result contains complex when optimal does not.
Time = 2.38 (sec) , antiderivative size = 16285, normalized size of antiderivative = 52.70 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x, algorithm="fr icas")
Output:
Too large to include
Timed out. \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x**3*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**3+a)**2,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {a b e - a^{2} h - {\left (b^{2} d - a b g\right )} x^{2} - {\left (b^{2} c - a b f\right )} x}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} + \frac {2 \, h x^{3} + 3 \, g x^{2} + 6 \, f x}{6 \, b^{2}} + \frac {\sqrt {3} {\left (2 \, b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - 5 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a b f \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{3}} + \frac {{\left (6 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - 12 \, a h \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} - b c + 4 \, a f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (3 \, b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - 6 \, a h \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} + b c - 4 \, a f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate(x^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x, algorithm="ma xima")
Output:
1/3*(a*b*e - a^2*h - (b^2*d - a*b*g)*x^2 - (b^2*c - a*b*f)*x)/(b^4*x^3 + a *b^3) + 1/6*(2*h*x^3 + 3*g*x^2 + 6*f*x)/b^2 + 1/9*sqrt(3)*(2*b^2*d*(a/b)^( 2/3) - 5*a*b*g*(a/b)^(2/3) + b^2*c*(a/b)^(1/3) - 4*a*b*f*(a/b)^(1/3))*arct an(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b^3) + 1/18*(6*b*e*(a/b )^(2/3) - 12*a*h*(a/b)^(2/3) + 2*b*d*(a/b)^(1/3) - 5*a*g*(a/b)^(1/3) - b*c + 4*a*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) + 1/9*( 3*b*e*(a/b)^(2/3) - 6*a*h*(a/b)^(2/3) - 2*b*d*(a/b)^(1/3) + 5*a*g*(a/b)^(1 /3) + b*c - 4*a*f)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(2/3))
Time = 0.13 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {\sqrt {3} {\left (b^{2} c - 4 \, a b f - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d + 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} b^{2}} - \frac {{\left (b^{2} c - 4 \, a b f + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d - 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} b^{2}} + \frac {{\left (b e - 2 \, a h\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac {a b e - a^{2} h - {\left (b^{2} d - a b g\right )} x^{2} - {\left (b^{2} c - a b f\right )} x}{3 \, {\left (b x^{3} + a\right )} b^{3}} - \frac {{\left (2 \, b^{4} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a b^{3} g \left (-\frac {a}{b}\right )^{\frac {1}{3}} + b^{4} c - 4 \, a b^{3} f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{5}} + \frac {2 \, b^{4} h x^{3} + 3 \, b^{4} g x^{2} + 6 \, b^{4} f x}{6 \, b^{6}} \] Input:
integrate(x^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x, algorithm="gi ac")
Output:
-1/9*sqrt(3)*(b^2*c - 4*a*b*f - 2*(-a*b^2)^(1/3)*b*d + 5*(-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)*b ^2) - 1/18*(b^2*c - 4*a*b*f + 2*(-a*b^2)^(1/3)*b*d - 5*(-a*b^2)^(1/3)*a*g) *log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*b^2) + 1/3*(b*e - 2*a*h)*log(abs(b*x^3 + a))/b^3 + 1/3*(a*b*e - a^2*h - (b^2*d - a*b*g)*x^ 2 - (b^2*c - a*b*f)*x)/((b*x^3 + a)*b^3) - 1/9*(2*b^4*d*(-a/b)^(1/3) - 5*a *b^3*g*(-a/b)^(1/3) + b^4*c - 4*a*b^3*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^( 1/3)))/(a*b^5) + 1/6*(2*b^4*h*x^3 + 3*b^4*g*x^2 + 6*b^4*f*x)/b^6
Time = 5.86 (sec) , antiderivative size = 1229, normalized size of antiderivative = 3.98 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((x^3*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x)
Output:
symsum(log((36*a^3*h^2 + 9*a*b^2*e^2 + 2*b^3*c*d - 5*a*b^2*c*g - 8*a*b^2*d *f - 36*a^2*b*e*h + 20*a^2*b*f*g)/(9*b^4) + root(729*a^2*b^9*z^3 + 1458*a^ 3*b^6*h*z^2 - 729*a^2*b^7*e*z^2 + 54*a*b^6*c*d*z - 972*a^3*b^4*e*h*z + 540 *a^3*b^4*f*g*z - 216*a^2*b^5*d*f*z - 135*a^2*b^5*c*g*z + 972*a^4*b^3*h^2*z + 243*a^2*b^5*e^2*z + 360*a^4*b*f*g*h - 18*a*b^4*c*d*e - 180*a^3*b^2*e*f* g - 144*a^3*b^2*d*f*h - 90*a^3*b^2*c*g*h + 72*a^2*b^3*d*e*f + 45*a^2*b^3*c *e*g + 36*a^2*b^3*c*d*h - 324*a^4*b*e*h^2 + 12*a*b^4*c^2*f + 162*a^3*b^2*e ^2*h + 150*a^3*b^2*d*g^2 - 60*a^2*b^3*d^2*g - 48*a^2*b^3*c*f^2 + 64*a^3*b^ 2*f^3 - 27*a^2*b^3*e^3 - 125*a^4*b*g^3 + 8*a*b^4*d^3 + 216*a^5*h^3 - b^5*c ^3, z, k)*((108*a^2*b^3*h - 54*a*b^4*e)/(9*b^4) + (x*(9*b^4*c - 36*a*b^3*f ))/(9*b^3) + 9*root(729*a^2*b^9*z^3 + 1458*a^3*b^6*h*z^2 - 729*a^2*b^7*e*z ^2 + 54*a*b^6*c*d*z - 972*a^3*b^4*e*h*z + 540*a^3*b^4*f*g*z - 216*a^2*b^5* d*f*z - 135*a^2*b^5*c*g*z + 972*a^4*b^3*h^2*z + 243*a^2*b^5*e^2*z + 360*a^ 4*b*f*g*h - 18*a*b^4*c*d*e - 180*a^3*b^2*e*f*g - 144*a^3*b^2*d*f*h - 90*a^ 3*b^2*c*g*h + 72*a^2*b^3*d*e*f + 45*a^2*b^3*c*e*g + 36*a^2*b^3*c*d*h - 324 *a^4*b*e*h^2 + 12*a*b^4*c^2*f + 162*a^3*b^2*e^2*h + 150*a^3*b^2*d*g^2 - 60 *a^2*b^3*d^2*g - 48*a^2*b^3*c*f^2 + 64*a^3*b^2*f^3 - 27*a^2*b^3*e^3 - 125* a^4*b*g^3 + 8*a*b^4*d^3 + 216*a^5*h^3 - b^5*c^3, z, k)*a*b^2) + (x*(4*b^2* d^2 + 25*a^2*g^2 - 3*b^2*c*e - 24*a^2*f*h + 6*a*b*c*h - 20*a*b*d*g + 12*a* b*e*f))/(9*b^3))*root(729*a^2*b^9*z^3 + 1458*a^3*b^6*h*z^2 - 729*a^2*b^...
Time = 0.17 (sec) , antiderivative size = 986, normalized size of antiderivative = 3.19 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int(x^3*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x)
Output:
(8*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt (3)))*a**2*b*f - 2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x )/(a**(1/3)*sqrt(3)))*a*b**2*c + 8*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3 ) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**2*f*x**3 - 2*b**(1/3)*a**(2/3)* sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**3*c*x**3 + 1 0*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b*g - 4* sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*d + 1 0*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*g*x **3 - 4*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**3* d*x**3 + 4*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3) *x**2)*a**2*b*f - b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b **(2/3)*x**2)*a*b**2*c + 4*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1 /3)*x + b**(2/3)*x**2)*a*b**2*f*x**3 - b**(1/3)*a**(2/3)*log(a**(2/3) - b* *(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**3*c*x**3 - 8*b**(1/3)*a**(2/3)*log(a **(1/3) + b**(1/3)*x)*a**2*b*f + 2*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/ 3)*x)*a*b**2*c - 8*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a*b**2*f*x **3 + 2*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*b**3*c*x**3 - 12*b**( 2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*h + 6*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a **2*b*e - 12*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**...