Integrand size = 28, antiderivative size = 271 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {(b e-2 a f) x^2}{2 b^3}+\frac {f x^5}{5 b^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2}{3 a b^3 \left (a+b x^3\right )}-\frac {\left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{11/3}}-\frac {\left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{11/3}}+\frac {\left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{11/3}} \] Output:
1/2*(-2*a*f+b*e)*x^2/b^3+1/5*f*x^5/b^2+1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)* x^2/a/b^3/(b*x^3+a)-1/9*(8*a^3*f-5*a^2*b*e+2*a*b^2*d+b^3*c)*arctan(1/3*(a^ (1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(4/3)/b^(11/3)-1/9*(8*a^3*f- 5*a^2*b*e+2*a*b^2*d+b^3*c)*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)/b^(11/3)+1/18*(8* a^3*f-5*a^2*b*e+2*a*b^2*d+b^3*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2) /a^(4/3)/b^(11/3)
Time = 0.12 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {45 b^{2/3} (b e-2 a f) x^2+18 b^{5/3} f x^5+\frac {30 b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^2}{a \left (a+b x^3\right )}-\frac {10 \sqrt {3} \left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{4/3}}-\frac {10 \left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{4/3}}+\frac {5 \left (b^3 c+2 a b^2 d-5 a^2 b e+8 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{4/3}}}{90 b^{11/3}} \] Input:
Integrate[(x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^2,x]
Output:
(45*b^(2/3)*(b*e - 2*a*f)*x^2 + 18*b^(5/3)*f*x^5 + (30*b^(2/3)*(b^3*c - a* b^2*d + a^2*b*e - a^3*f)*x^2)/(a*(a + b*x^3)) - (10*Sqrt[3]*(b^3*c + 2*a*b ^2*d - 5*a^2*b*e + 8*a^3*f)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a ^(4/3) - (10*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*Log[a^(1/3) + b^(1/ 3)*x])/a^(4/3) + (5*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(4/3))/(90*b^(11/3))
Time = 0.94 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2367, 25, 2028, 1812, 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.
| \(\displaystyle \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle \frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3 \left (a+b x^3\right )}-\frac {\int -\frac {3 a b^3 f x^7+3 a b^2 (b e-a f) x^4+b \left (2 f a^3-2 b e a^2+2 b^2 d a+b^3 c\right ) x}{b x^3+a}dx}{3 a b^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 a b^3 f x^7+3 a b^2 (b e-a f) x^4+b \left (2 f a^3-2 b e a^2+2 b^2 d a+b^3 c\right ) x}{b x^3+a}dx}{3 a b^4}+\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2028 |
| \(\displaystyle \frac {\int \frac {x \left (3 a b^3 f x^6+3 a b^2 (b e-a f) x^3+b \left (2 f a^3-2 b e a^2+2 b^2 d a+b^3 c\right )\right )}{b x^3+a}dx}{3 a b^4}+\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1812 |
| \(\displaystyle \frac {\int \left (3 a b^2 f x^4+3 a b (b e-2 a f) x+\frac {\left (c b^4+2 a d b^3-5 a^2 e b^2+8 a^3 f b\right ) x}{b x^3+a}\right )dx}{3 a b^4}+\frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^2 \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a b^3 \left (a+b x^3\right )}+\frac {-\frac {\sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (8 a^3 f-5 a^2 b e+2 a b^2 d+b^3 c\right )}{\sqrt {3} \sqrt [3]{a}}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (8 a^3 f-5 a^2 b e+2 a b^2 d+b^3 c\right )}{3 \sqrt [3]{a}}+\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (8 a^3 f-5 a^2 b e+2 a b^2 d+b^3 c\right )}{6 \sqrt [3]{a}}+\frac {3}{5} a b^2 f x^5+\frac {3}{2} a b x^2 (b e-2 a f)}{3 a b^4}\) |
Input:
Int[(x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^2,x]
Output:
((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2)/(3*a*b^3*(a + b*x^3)) + ((3*a*b* (b*e - 2*a*f)*x^2)/2 + (3*a*b^2*f*x^5)/5 - (b^(1/3)*(b^3*c + 2*a*b^2*d - 5 *a^2*b*e + 8*a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sq rt[3]*a^(1/3)) - (b^(1/3)*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*Log[a^ (1/3) + b^(1/3)*x])/(3*a^(1/3)) + (b^(1/3)*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(1/3)))/(3 *a*b^4)
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*( (d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IGtQ[p, 0]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ {a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] && !(E qQ[p, 1] && EqQ[u, 1])
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.48
| method | result | size |
| risch | \(\frac {f \,x^{5}}{5 b^{2}}-\frac {a f \,x^{2}}{b^{3}}+\frac {e \,x^{2}}{2 b^{2}}-\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) x^{2}}{3 a \,b^{3} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (8 f \,a^{3}-5 e \,a^{2} b +2 d a \,b^{2}+b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{9 b^{4} a}\) | \(130\) |
| default | \(-\frac {-\frac {b f \,x^{5}}{5}+\frac {\left (2 a f -b e \right ) x^{2}}{2}}{b^{3}}+\frac {-\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) x^{2}}{3 a \left (b \,x^{3}+a \right )}+\frac {\left (8 f \,a^{3}-5 e \,a^{2} b +2 d a \,b^{2}+b^{3} c \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 a}}{b^{3}}\) | \(197\) |
Input:
int(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/5*f*x^5/b^2-1/b^3*a*f*x^2+1/2/b^2*e*x^2-1/3*(a^3*f-a^2*b*e+a*b^2*d-b^3*c )/a*x^2/b^3/(b*x^3+a)+1/9/b^4/a*sum((8*a^3*f-5*a^2*b*e+2*a*b^2*d+b^3*c)/_R *ln(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.10 (sec) , antiderivative size = 874, normalized size of antiderivative = 3.23 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^2,x, algorithm="fricas")
Output:
[1/90*(18*a^2*b^4*f*x^8 + 9*(5*a^2*b^4*e - 8*a^3*b^3*f)*x^5 + 15*(2*a*b^5* c - 2*a^2*b^4*d + 5*a^3*b^3*e - 8*a^4*b^2*f)*x^2 + 15*sqrt(1/3)*(a^2*b^4*c + 2*a^3*b^3*d - 5*a^4*b^2*e + 8*a^5*b*f + (a*b^5*c + 2*a^2*b^4*d - 5*a^3* b^3*e + 8*a^4*b^2*f)*x^3)*sqrt((-a*b^2)^(1/3)/a)*log((2*b^2*x^3 - a*b + 3* sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b^2)^(1/3)*a)*sqrt((-a*b^2)^ (1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) + 5*(a*b^3*c + 2*a^2*b^2*d - 5 *a^3*b*e + 8*a^4*f + (b^4*c + 2*a*b^3*d - 5*a^2*b^2*e + 8*a^3*b*f)*x^3)*(- a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) - 10*(a*b^ 3*c + 2*a^2*b^2*d - 5*a^3*b*e + 8*a^4*f + (b^4*c + 2*a*b^3*d - 5*a^2*b^2*e + 8*a^3*b*f)*x^3)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a^2*b^6*x^3 + a^3*b^5), 1/90*(18*a^2*b^4*f*x^8 + 9*(5*a^2*b^4*e - 8*a^3*b^3*f)*x^5 + 1 5*(2*a*b^5*c - 2*a^2*b^4*d + 5*a^3*b^3*e - 8*a^4*b^2*f)*x^2 + 30*sqrt(1/3) *(a^2*b^4*c + 2*a^3*b^3*d - 5*a^4*b^2*e + 8*a^5*b*f + (a*b^5*c + 2*a^2*b^4 *d - 5*a^3*b^3*e + 8*a^4*b^2*f)*x^3)*sqrt(-(-a*b^2)^(1/3)/a)*arctan(sqrt(1 /3)*(2*b*x + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) + 5*(a*b^3*c + 2*a ^2*b^2*d - 5*a^3*b*e + 8*a^4*f + (b^4*c + 2*a*b^3*d - 5*a^2*b^2*e + 8*a^3* b*f)*x^3)*(-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3) ) - 10*(a*b^3*c + 2*a^2*b^2*d - 5*a^3*b*e + 8*a^4*f + (b^4*c + 2*a*b^3*d - 5*a^2*b^2*e + 8*a^3*b*f)*x^3)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/( a^2*b^6*x^3 + a^3*b^5)]
Time = 10.08 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.70 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=x^{2} \left (- \frac {a f}{b^{3}} + \frac {e}{2 b^{2}}\right ) + \frac {x^{2} \left (- a^{3} f + a^{2} b e - a b^{2} d + b^{3} c\right )}{3 a^{2} b^{3} + 3 a b^{4} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{4} b^{11} + 512 a^{9} f^{3} - 960 a^{8} b e f^{2} + 384 a^{7} b^{2} d f^{2} + 600 a^{7} b^{2} e^{2} f + 192 a^{6} b^{3} c f^{2} - 480 a^{6} b^{3} d e f - 125 a^{6} b^{3} e^{3} - 240 a^{5} b^{4} c e f + 96 a^{5} b^{4} d^{2} f + 150 a^{5} b^{4} d e^{2} + 96 a^{4} b^{5} c d f + 75 a^{4} b^{5} c e^{2} - 60 a^{4} b^{5} d^{2} e + 24 a^{3} b^{6} c^{2} f - 60 a^{3} b^{6} c d e + 8 a^{3} b^{6} d^{3} - 15 a^{2} b^{7} c^{2} e + 12 a^{2} b^{7} c d^{2} + 6 a b^{8} c^{2} d + b^{9} c^{3}, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a^{3} b^{7}}{64 a^{6} f^{2} - 80 a^{5} b e f + 32 a^{4} b^{2} d f + 25 a^{4} b^{2} e^{2} + 16 a^{3} b^{3} c f - 20 a^{3} b^{3} d e - 10 a^{2} b^{4} c e + 4 a^{2} b^{4} d^{2} + 4 a b^{5} c d + b^{6} c^{2}} + x \right )} \right )\right )} + \frac {f x^{5}}{5 b^{2}} \] Input:
integrate(x*(f*x**9+e*x**6+d*x**3+c)/(b*x**3+a)**2,x)
Output:
x**2*(-a*f/b**3 + e/(2*b**2)) + x**2*(-a**3*f + a**2*b*e - a*b**2*d + b**3 *c)/(3*a**2*b**3 + 3*a*b**4*x**3) + RootSum(729*_t**3*a**4*b**11 + 512*a** 9*f**3 - 960*a**8*b*e*f**2 + 384*a**7*b**2*d*f**2 + 600*a**7*b**2*e**2*f + 192*a**6*b**3*c*f**2 - 480*a**6*b**3*d*e*f - 125*a**6*b**3*e**3 - 240*a** 5*b**4*c*e*f + 96*a**5*b**4*d**2*f + 150*a**5*b**4*d*e**2 + 96*a**4*b**5*c *d*f + 75*a**4*b**5*c*e**2 - 60*a**4*b**5*d**2*e + 24*a**3*b**6*c**2*f - 6 0*a**3*b**6*c*d*e + 8*a**3*b**6*d**3 - 15*a**2*b**7*c**2*e + 12*a**2*b**7* c*d**2 + 6*a*b**8*c**2*d + b**9*c**3, Lambda(_t, _t*log(81*_t**2*a**3*b**7 /(64*a**6*f**2 - 80*a**5*b*e*f + 32*a**4*b**2*d*f + 25*a**4*b**2*e**2 + 16 *a**3*b**3*c*f - 20*a**3*b**3*d*e - 10*a**2*b**4*c*e + 4*a**2*b**4*d**2 + 4*a*b**5*c*d + b**6*c**2) + x))) + f*x**5/(5*b**2)
Time = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.96 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{2}}{3 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}} + \frac {2 \, b f x^{5} + 5 \, {\left (b e - 2 \, a f\right )} x^{2}}{10 \, b^{3}} + \frac {\sqrt {3} {\left (b^{3} c + 2 \, a b^{2} d - 5 \, a^{2} b e + 8 \, a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (b^{3} c + 2 \, a b^{2} d - 5 \, a^{2} b e + 8 \, a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (b^{3} c + 2 \, a b^{2} d - 5 \, a^{2} b e + 8 \, a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:
integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^2,x, algorithm="maxima")
Output:
1/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^2/(a*b^4*x^3 + a^2*b^3) + 1/10*( 2*b*f*x^5 + 5*(b*e - 2*a*f)*x^2)/b^3 + 1/9*sqrt(3)*(b^3*c + 2*a*b^2*d - 5* a^2*b*e + 8*a^3*f)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a* b^4*(a/b)^(1/3)) + 1/18*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*b^4*(a/b)^(1/3)) - 1/9*(b^3*c + 2*a*b^2* d - 5*a^2*b*e + 8*a^3*f)*log(x + (a/b)^(1/3))/(a*b^4*(a/b)^(1/3))
Time = 0.14 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.15 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {3} {\left (b^{3} c + 2 \, a b^{2} d - 5 \, a^{2} b e + 8 \, a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{3}} - \frac {{\left (b^{3} c + 2 \, a b^{2} d - 5 \, a^{2} b e + 8 \, a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{3}} - \frac {{\left (b^{3} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, a^{2} b e \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 8 \, a^{3} f \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2} b^{3}} + \frac {b^{3} c x^{2} - a b^{2} d x^{2} + a^{2} b e x^{2} - a^{3} f x^{2}}{3 \, {\left (b x^{3} + a\right )} a b^{3}} + \frac {2 \, b^{8} f x^{5} + 5 \, b^{8} e x^{2} - 10 \, a b^{7} f x^{2}}{10 \, b^{10}} \] Input:
integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^2,x, algorithm="giac")
Output:
1/9*sqrt(3)*(b^3*c + 2*a*b^2*d - 5*a^2*b*e + 8*a^3*f)*arctan(1/3*sqrt(3)*( 2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(1/3)*a*b^3) - 1/18*(b^3*c + 2 *a*b^2*d - 5*a^2*b*e + 8*a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/( (-a*b^2)^(1/3)*a*b^3) - 1/9*(b^3*c*(-a/b)^(1/3) + 2*a*b^2*d*(-a/b)^(1/3) - 5*a^2*b*e*(-a/b)^(1/3) + 8*a^3*f*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - ( -a/b)^(1/3)))/(a^2*b^3) + 1/3*(b^3*c*x^2 - a*b^2*d*x^2 + a^2*b*e*x^2 - a^3 *f*x^2)/((b*x^3 + a)*a*b^3) + 1/10*(2*b^8*f*x^5 + 5*b^8*e*x^2 - 10*a*b^7*f *x^2)/b^10
Time = 6.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx=x^2\,\left (\frac {e}{2\,b^2}-\frac {a\,f}{b^3}\right )+\frac {f\,x^5}{5\,b^2}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (8\,f\,a^3-5\,e\,a^2\,b+2\,d\,a\,b^2+c\,b^3\right )}{9\,a^{4/3}\,b^{11/3}}+\frac {x^2\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a\,\left (b^4\,x^3+a\,b^3\right )}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (8\,f\,a^3-5\,e\,a^2\,b+2\,d\,a\,b^2+c\,b^3\right )}{9\,a^{4/3}\,b^{11/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (8\,f\,a^3-5\,e\,a^2\,b+2\,d\,a\,b^2+c\,b^3\right )}{9\,a^{4/3}\,b^{11/3}} \] Input:
int((x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3)^2,x)
Output:
x^2*(e/(2*b^2) - (a*f)/b^3) + (f*x^5)/(5*b^2) - (log(b^(1/3)*x + a^(1/3))* (b^3*c + 8*a^3*f + 2*a*b^2*d - 5*a^2*b*e))/(9*a^(4/3)*b^(11/3)) + (x^2*(b^ 3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a*(a*b^3 + b^4*x^3)) + (log(3^(1/2)*a ^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(b^3*c + 8*a^3*f + 2*a*b^2*d - 5*a^2*b*e))/(9*a^(4/3)*b^(11/3)) - (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(b^3*c + 8*a^3*f + 2*a*b^2 *d - 5*a^2*b*e))/(9*a^(4/3)*b^(11/3))
Time = 0.16 (sec) , antiderivative size = 728, normalized size of antiderivative = 2.69 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a)^2,x)
Output:
( - 80*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**4*f + 50*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b*e - 80*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b*f*x** 3 - 20*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b** 2*d + 50*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b **2*e*x**3 - 10*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3))) *a*b**3*c - 20*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))* a*b**3*d*x**3 - 10*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3 )))*b**4*c*x**3 - 120*b**(2/3)*a**(1/3)*a**3*f*x**2 + 75*b**(2/3)*a**(1/3) *a**2*b*e*x**2 - 72*b**(2/3)*a**(1/3)*a**2*b*f*x**5 - 30*b**(2/3)*a**(1/3) *a*b**2*d*x**2 + 45*b**(2/3)*a**(1/3)*a*b**2*e*x**5 + 18*b**(2/3)*a**(1/3) *a*b**2*f*x**8 + 30*b**(2/3)*a**(1/3)*b**3*c*x**2 + 40*log(a**(2/3) - b**( 1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*f - 25*log(a**(2/3) - b**(1/3)*a**(1 /3)*x + b**(2/3)*x**2)*a**3*b*e + 40*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b*f*x**3 + 10*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b** (2/3)*x**2)*a**2*b**2*d - 25*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3) *x**2)*a**2*b**2*e*x**3 + 5*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)* x**2)*a*b**3*c + 10*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a* b**3*d*x**3 + 5*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**4*c *x**3 - 80*log(a**(1/3) + b**(1/3)*x)*a**4*f + 50*log(a**(1/3) + b**(1/...