Integrand size = 38, antiderivative size = 337 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {(b e-2 a h) x}{b^3}+\frac {f x^2}{2 b^2}+\frac {g x^3}{3 b^2}+\frac {h x^4}{4 b^2}+\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 b^3 \left (a+b x^3\right )}-\frac {\left (2 b^{5/3} c-4 a^{2/3} b e-5 a b^{2/3} f+7 a^{5/3} h\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{10/3}}-\frac {\left (b^{2/3} (2 b c-5 a f)+a^{2/3} (4 b e-7 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 \sqrt [3]{a} b^{10/3}}+\frac {\left (b^{2/3} (2 b c-5 a f)+a^{2/3} (4 b e-7 a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 \sqrt [3]{a} b^{10/3}}+\frac {(b d-2 a g) \log \left (a+b x^3\right )}{3 b^3} \] Output:
(-2*a*h+b*e)*x/b^3+1/2*f*x^2/b^2+1/3*g*x^3/b^2+1/4*h*x^4/b^2+1/3*x*(a*(-a* h+b*e)-b*(-a*f+b*c)*x-b*(-a*g+b*d)*x^2)/b^3/(b*x^3+a)-1/9*(2*b^(5/3)*c-4*a ^(2/3)*b*e-5*a*b^(2/3)*f+7*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^(1/3)/b^(10/3)-1/9*(b^(2/3)*(-5*a*f+2*b*c)+a^(2/3) *(-7*a*h+4*b*e))*ln(a^(1/3)+b^(1/3)*x)/a^(1/3)/b^(10/3)+1/18*(b^(2/3)*(-5* a*f+2*b*c)+a^(2/3)*(-7*a*h+4*b*e))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^ 2)/a^(1/3)/b^(10/3)+1/3*(-2*a*g+b*d)*ln(b*x^3+a)/b^3
Time = 0.29 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {36 b^{2/3} (b e-2 a h) x+18 b^{5/3} f x^2+12 b^{5/3} g x^3+9 b^{5/3} h x^4-\frac {12 b^{2/3} \left (b^2 c x^2+a^2 (g+h x)-a b (d+x (e+f x))\right )}{a+b x^3}-\frac {4 \sqrt {3} \left (2 b^2 c-4 a^{2/3} b^{4/3} e-5 a b f+7 a^{5/3} \sqrt [3]{b} h\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {4 \left (-2 b^2 c-4 a^{2/3} b^{4/3} e+5 a b f+7 a^{5/3} \sqrt [3]{b} h\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {2 \left (2 b^2 c+4 a^{2/3} b^{4/3} e-5 a b f-7 a^{5/3} \sqrt [3]{b} h\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}+12 b^{2/3} (b d-2 a g) \log \left (a+b x^3\right )}{36 b^{11/3}} \] Input:
Integrate[(x^4*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x]
Output:
(36*b^(2/3)*(b*e - 2*a*h)*x + 18*b^(5/3)*f*x^2 + 12*b^(5/3)*g*x^3 + 9*b^(5 /3)*h*x^4 - (12*b^(2/3)*(b^2*c*x^2 + a^2*(g + h*x) - a*b*(d + x*(e + f*x)) ))/(a + b*x^3) - (4*Sqrt[3]*(2*b^2*c - 4*a^(2/3)*b^(4/3)*e - 5*a*b*f + 7*a ^(5/3)*b^(1/3)*h)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(1/3) + ( 4*(-2*b^2*c - 4*a^(2/3)*b^(4/3)*e + 5*a*b*f + 7*a^(5/3)*b^(1/3)*h)*Log[a^( 1/3) + b^(1/3)*x])/a^(1/3) + (2*(2*b^2*c + 4*a^(2/3)*b^(4/3)*e - 5*a*b*f - 7*a^(5/3)*b^(1/3)*h)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(1 /3) + 12*b^(2/3)*(b*d - 2*a*g)*Log[a + b*x^3])/(36*b^(11/3))
Time = 1.52 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2367, 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.
| \(\displaystyle \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle \frac {x \left (-b x (b c-a f)-b x^2 (b d-a g)+a (b e-a h)\right )}{3 b^3 \left (a+b x^3\right )}-\frac {\int \frac {-3 a b^2 h x^6-3 a b^2 g x^5-3 a b^2 f x^4-3 a b (b e-a h) x^3-3 a b (b d-a g) x^2-2 a b (b c-a f) x+a^2 (b e-a h)}{b x^3+a}dx}{3 a b^3}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \frac {x \left (-b x (b c-a f)-b x^2 (b d-a g)+a (b e-a h)\right )}{3 b^3 \left (a+b x^3\right )}-\frac {\int \left (-3 a b h x^3-3 a b g x^2-3 a b f x-3 a (b e-2 a h)+\frac {(4 b e-7 a h) a^2-3 b (b d-2 a g) x^2 a-b (2 b c-5 a f) x a}{b x^3+a}\right )dx}{3 a b^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \left (-b x (b c-a f)-b x^2 (b d-a g)+a (b e-a h)\right )}{3 b^3 \left (a+b x^3\right )}-\frac {\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-4 a^{2/3} b e+7 a^{5/3} h-5 a b^{2/3} f+2 b^{5/3} c\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^{2/3} (4 b e-7 a h)+b^{2/3} (2 b c-5 a f)\right )}{6 \sqrt [3]{b}}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^{2/3} (4 b e-7 a h)+b^{2/3} (2 b c-5 a f)\right )}{3 \sqrt [3]{b}}-a (b d-2 a g) \log \left (a+b x^3\right )-3 a x (b e-2 a h)-\frac {3}{2} a b f x^2-a b g x^3-\frac {3}{4} a b h x^4}{3 a b^3}\) |
Input:
Int[(x^4*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x]
Output:
(x*(a*(b*e - a*h) - b*(b*c - a*f)*x - b*(b*d - a*g)*x^2))/(3*b^3*(a + b*x^ 3)) - (-3*a*(b*e - 2*a*h)*x - (3*a*b*f*x^2)/2 - a*b*g*x^3 - (3*a*b*h*x^4)/ 4 + (a^(2/3)*(2*b^(5/3)*c - 4*a^(2/3)*b*e - 5*a*b^(2/3)*f + 7*a^(5/3)*h)*A rcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(1/3)) + (a^( 2/3)*(b^(2/3)*(2*b*c - 5*a*f) + a^(2/3)*(4*b*e - 7*a*h))*Log[a^(1/3) + b^( 1/3)*x])/(3*b^(1/3)) - (a^(2/3)*(b^(2/3)*(2*b*c - 5*a*f) + a^(2/3)*(4*b*e - 7*a*h))*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(1/3)) - a* (b*d - 2*a*g)*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 = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.48
| method | result | size |
| risch | \(\frac {h \,x^{4}}{4 b^{2}}+\frac {g \,x^{3}}{3 b^{2}}+\frac {f \,x^{2}}{2 b^{2}}-\frac {2 a h x}{b^{3}}+\frac {e x}{b^{2}}+\frac {\left (\frac {1}{3} a b f -\frac {1}{3} b^{2} c \right ) x^{2}+\left (-\frac {1}{3} a^{2} h +\frac {1}{3} a b e \right ) x -\frac {a \left (a g -b d \right )}{3}}{b^{3} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (3 b \left (-2 a g +b d \right ) \textit {\_R}^{2}+b \left (-5 a f +2 c b \right ) \textit {\_R} +7 a^{2} h -4 a b e \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{4}}\) | \(162\) |
| default | \(-\frac {-\frac {1}{4} b h \,x^{4}-\frac {1}{3} b g \,x^{3}-\frac {1}{2} b f \,x^{2}+2 a h x -b e x}{b^{3}}+\frac {\frac {\left (\frac {1}{3} a b f -\frac {1}{3} b^{2} c \right ) x^{2}+\left (-\frac {1}{3} a^{2} h +\frac {1}{3} a b e \right ) x -\frac {a \left (a g -b d \right )}{3}}{b \,x^{3}+a}+\frac {\left (7 a^{2} h -4 a b e \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {\left (-5 a b f +2 b^{2} c \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3}+\frac {\left (-6 a b g +3 b^{2} d \right ) \ln \left (b \,x^{3}+a \right )}{9 b}}{b^{3}}\) | \(329\) |
Input:
int(x^4*(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:
1/4*h*x^4/b^2+1/3*g*x^3/b^2+1/2*f*x^2/b^2-2/b^3*a*h*x+e*x/b^2+((1/3*a*b*f- 1/3*b^2*c)*x^2+(-1/3*a^2*h+1/3*a*b*e)*x-1/3*a*(a*g-b*d))/b^3/(b*x^3+a)+1/9 /b^4*sum((3*b*(-2*a*g+b*d)*_R^2+b*(-5*a*f+2*b*c)*_R+7*a^2*h-4*a*b*e)/_R^2* ln(x-_R),_R=RootOf(_Z^3*b+a))
Result contains complex when optimal does not.
Time = 2.31 (sec) , antiderivative size = 16147, normalized size of antiderivative = 47.91 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x^4*(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. \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x**4*(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 = 0.11 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.08 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {a b d - a^{2} g - {\left (b^{2} c - a b f\right )} x^{2} + {\left (a b e - a^{2} h\right )} x}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} + \frac {\sqrt {3} {\left (2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - 5 \, a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4 \, a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} + 7 \, a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{3}} + \frac {3 \, b h x^{4} + 4 \, b g x^{3} + 6 \, b f x^{2} + 12 \, {\left (b e - 2 \, a h\right )} x}{12 \, b^{3}} + \frac {{\left (6 \, b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - 12 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} + 4 \, a b e - 7 \, a^{2} h\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (3 \, b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} - 6 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a b e + 7 \, a^{2} h\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate(x^4*(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*d - a^2*g - (b^2*c - a*b*f)*x^2 + (a*b*e - a^2*h)*x)/(b^4*x^3 + a *b^3) + 1/9*sqrt(3)*(2*b^2*c*(a/b)^(2/3) - 5*a*b*f*(a/b)^(2/3) - 4*a*b*e*( a/b)^(1/3) + 7*a^2*h*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/( a/b)^(1/3))/(a*b^3) + 1/12*(3*b*h*x^4 + 4*b*g*x^3 + 6*b*f*x^2 + 12*(b*e - 2*a*h)*x)/b^3 + 1/18*(6*b^2*d*(a/b)^(2/3) - 12*a*b*g*(a/b)^(2/3) + 2*b^2*c *(a/b)^(1/3) - 5*a*b*f*(a/b)^(1/3) + 4*a*b*e - 7*a^2*h)*log(x^2 - x*(a/b)^ (1/3) + (a/b)^(2/3))/(b^4*(a/b)^(2/3)) + 1/9*(3*b^2*d*(a/b)^(2/3) - 6*a*b* g*(a/b)^(2/3) - 2*b^2*c*(a/b)^(1/3) + 5*a*b*f*(a/b)^(1/3) - 4*a*b*e + 7*a^ 2*h)*log(x + (a/b)^(1/3))/(b^4*(a/b)^(2/3))
Time = 0.13 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.04 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {3} {\left (4 \, a b e - 7 \, a^{2} h + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b c - 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} b^{2}} + \frac {{\left (4 \, a b e - 7 \, a^{2} h - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b c + 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} b^{2}} + \frac {{\left (b d - 2 \, a g\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac {a b d - a^{2} g - {\left (b^{2} c - a b f\right )} x^{2} + {\left (a b e - a^{2} h\right )} x}{3 \, {\left (b x^{3} + a\right )} b^{3}} - \frac {{\left (2 \, b^{6} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a b^{5} f \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a b^{5} e + 7 \, a^{2} b^{4} h\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{7}} + \frac {3 \, b^{6} h x^{4} + 4 \, b^{6} g x^{3} + 6 \, b^{6} f x^{2} + 12 \, b^{6} e x - 24 \, a b^{5} h x}{12 \, b^{8}} \] Input:
integrate(x^4*(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)*(4*a*b*e - 7*a^2*h + 2*(-a*b^2)^(1/3)*b*c - 5*(-a*b^2)^(1/3)*a *f)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)* b^2) + 1/18*(4*a*b*e - 7*a^2*h - 2*(-a*b^2)^(1/3)*b*c + 5*(-a*b^2)^(1/3)*a *f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*b^2) + 1/3*(b *d - 2*a*g)*log(abs(b*x^3 + a))/b^3 + 1/3*(a*b*d - a^2*g - (b^2*c - a*b*f) *x^2 + (a*b*e - a^2*h)*x)/((b*x^3 + a)*b^3) - 1/9*(2*b^6*c*(-a/b)^(1/3) - 5*a*b^5*f*(-a/b)^(1/3) - 4*a*b^5*e + 7*a^2*b^4*h)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^7) + 1/12*(3*b^6*h*x^4 + 4*b^6*g*x^3 + 6*b^6*f*x^2 + 12*b^6*e*x - 24*a*b^5*h*x)/b^8
Time = 0.20 (sec) , antiderivative size = 1241, normalized size of antiderivative = 3.68 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((x^4*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/(a + b*x^3)^2,x)
Output:
symsum(log(root(729*a*b^10*z^3 - 729*a*b^8*d*z^2 + 1458*a^2*b^7*g*z^2 - 21 6*a*b^6*c*e*z - 945*a^3*b^4*f*h*z - 972*a^2*b^5*d*g*z + 540*a^2*b^5*e*f*z + 378*a^2*b^5*c*h*z + 243*a*b^6*d^2*z + 972*a^3*b^4*g^2*z - 630*a^4*b*f*g* h + 72*a*b^4*c*d*e + 360*a^3*b^2*e*f*g + 315*a^3*b^2*d*f*h + 252*a^3*b^2*c *g*h - 180*a^2*b^3*d*e*f - 144*a^2*b^3*c*e*g - 126*a^2*b^3*c*d*h + 588*a^4 *b*e*h^2 - 60*a*b^4*c^2*f - 336*a^3*b^2*e^2*h - 324*a^3*b^2*d*g^2 + 162*a^ 2*b^3*d^2*g + 150*a^2*b^3*c*f^2 - 125*a^3*b^2*f^3 + 64*a^2*b^3*e^3 + 216*a ^4*b*g^3 - 27*a*b^4*d^3 - 343*a^5*h^3 + 8*b^5*c^3, z, k)*((108*a^2*b^3*g - 54*a*b^4*d)/(9*b^4) + (x*(63*a^2*b^3*h - 36*a*b^4*e))/(9*b^4) + 9*root(72 9*a*b^10*z^3 - 729*a*b^8*d*z^2 + 1458*a^2*b^7*g*z^2 - 216*a*b^6*c*e*z - 94 5*a^3*b^4*f*h*z - 972*a^2*b^5*d*g*z + 540*a^2*b^5*e*f*z + 378*a^2*b^5*c*h* z + 243*a*b^6*d^2*z + 972*a^3*b^4*g^2*z - 630*a^4*b*f*g*h + 72*a*b^4*c*d*e + 360*a^3*b^2*e*f*g + 315*a^3*b^2*d*f*h + 252*a^3*b^2*c*g*h - 180*a^2*b^3 *d*e*f - 144*a^2*b^3*c*e*g - 126*a^2*b^3*c*d*h + 588*a^4*b*e*h^2 - 60*a*b^ 4*c^2*f - 336*a^3*b^2*e^2*h - 324*a^3*b^2*d*g^2 + 162*a^2*b^3*d^2*g + 150* a^2*b^3*c*f^2 - 125*a^3*b^2*f^3 + 64*a^2*b^3*e^3 + 216*a^4*b*g^3 - 27*a*b^ 4*d^3 - 343*a^5*h^3 + 8*b^5*c^3, z, k)*a*b^2) + (36*a^3*g^2 + 9*a*b^2*d^2 - 35*a^3*f*h - 8*a*b^2*c*e + 14*a^2*b*c*h - 36*a^2*b*d*g + 20*a^2*b*e*f)/( 9*b^4) + (x*(4*b^3*c^2 + 25*a^2*b*f^2 + 42*a^3*g*h - 20*a*b^2*c*f + 12*a*b ^2*d*e - 21*a^2*b*d*h - 24*a^2*b*e*g))/(9*b^4))*root(729*a*b^10*z^3 - 7...
Time = 0.17 (sec) , antiderivative size = 995, normalized size of antiderivative = 2.95 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int(x^4*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^2,x)
Output:
( - 28*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)* sqrt(3)))*a**2*h + 16*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3 )*x)/(a**(1/3)*sqrt(3)))*a*b*e - 28*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/ 3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b*h*x**3 + 16*b**(1/3)*a**(2/3)*s qrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**2*e*x**3 + 20 *sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b*f - 8*s qrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**2*c + 20*sq rt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**2*f*x**3 - 8 *sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**3*c*x**3 - 14*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a **2*h + 8*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)* x**2)*a*b*e - 14*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b* *(2/3)*x**2)*a*b*h*x**3 + 8*b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3)*a**( 1/3)*x + b**(2/3)*x**2)*b**2*e*x**3 + 28*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a**2*h - 16*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a*b*e + 28*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*a*b*h*x**3 - 16*b**(1/3 )*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*b**2*e*x**3 - 24*b**(2/3)*a**(1/3)*l og(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*g + 12*b**(2/3)*a* *(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*d - 24*b**( 2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*g...