Integrand size = 30, antiderivative size = 297 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=-\frac {c}{8 a^2 x^8}+\frac {2 b c-a d}{5 a^3 x^5}-\frac {3 b^2 c-2 a b d+a^2 e}{2 a^4 x^2}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^4 \left (a+b x^3\right )}+\frac {\left (11 b^3 c-8 a b^2 d+5 a^2 b e-2 a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{14/3} \sqrt [3]{b}}-\frac {\left (11 b^3 c-8 a b^2 d+5 a^2 b e-2 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{14/3} \sqrt [3]{b}}+\frac {\left (11 b^3 c-8 a b^2 d+5 a^2 b e-2 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{14/3} \sqrt [3]{b}} \] Output:
-1/8*c/a^2/x^8+1/5*(-a*d+2*b*c)/a^3/x^5-1/2*(a^2*e-2*a*b*d+3*b^2*c)/a^4/x^ 2-1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*x/a^4/(b*x^3+a)+1/9*(-2*a^3*f+5*a^2*b *e-8*a*b^2*d+11*b^3*c)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3 ^(1/2)/a^(14/3)/b^(1/3)-1/9*(-2*a^3*f+5*a^2*b*e-8*a*b^2*d+11*b^3*c)*ln(a^( 1/3)+b^(1/3)*x)/a^(14/3)/b^(1/3)+1/18*(-2*a^3*f+5*a^2*b*e-8*a*b^2*d+11*b^3 *c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(14/3)/b^(1/3)
Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=\frac {-\frac {45 a^{8/3} c}{x^8}-\frac {72 a^{5/3} (-2 b c+a d)}{x^5}-\frac {180 a^{2/3} \left (3 b^2 c-2 a b d+a^2 e\right )}{x^2}+\frac {120 a^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{a+b x^3}+\frac {40 \sqrt {3} \left (11 b^3 c-8 a b^2 d+5 a^2 b e-2 a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {40 \left (-11 b^3 c+8 a b^2 d-5 a^2 b e+2 a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {20 \left (11 b^3 c-8 a b^2 d+5 a^2 b e-2 a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}}{360 a^{14/3}} \] Input:
Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^9*(a + b*x^3)^2),x]
Output:
((-45*a^(8/3)*c)/x^8 - (72*a^(5/3)*(-2*b*c + a*d))/x^5 - (180*a^(2/3)*(3*b ^2*c - 2*a*b*d + a^2*e))/x^2 + (120*a^(2/3)*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*x)/(a + b*x^3) + (40*Sqrt[3]*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3) + (40*(-11*b ^3*c + 8*a*b^2*d - 5*a^2*b*e + 2*a^3*f)*Log[a^(1/3) + b^(1/3)*x])/b^(1/3) + (20*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*Log[a^(2/3) - a^(1/3)*b ^(1/3)*x + b^(2/3)*x^2])/b^(1/3))/(360*a^(14/3))
Time = 1.16 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2368, 25, 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.
| \(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2368 |
| \(\displaystyle -\frac {\int -\frac {-\frac {2 b^3 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^9}{a^3}+\frac {3 b^3 \left (e a^2-b d a+b^2 c\right ) x^6}{a^2}-3 b^3 \left (\frac {b c}{a}-d\right ) x^3+3 b^3 c}{x^9 \left (b x^3+a\right )}dx}{3 a b^3}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {-\frac {2 b^3 \left (-f a^3+b e a^2-b^2 d a+b^3 c\right ) x^9}{a^3}+\frac {3 b^3 \left (e a^2-b d a+b^2 c\right ) x^6}{a^2}-3 b^3 \left (\frac {b c}{a}-d\right ) x^3+3 b^3 c}{x^9 \left (b x^3+a\right )}dx}{3 a b^3}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle \frac {\int \left (\frac {\left (2 f a^3-5 b e a^2+8 b^2 d a-11 b^3 c\right ) b^3}{a^3 \left (b x^3+a\right )}+\frac {3 \left (e a^2-2 b d a+3 b^2 c\right ) b^3}{a^3 x^3}+\frac {3 (a d-2 b c) b^3}{a^2 x^6}+\frac {3 c b^3}{a x^9}\right )dx}{3 a b^3}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^4 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3 b^3 (2 b c-a d)}{5 a^2 x^5}-\frac {3 b^3 \left (a^2 e-2 a b d+3 b^2 c\right )}{2 a^3 x^2}+\frac {b^{8/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-2 a^3 f+5 a^2 b e-8 a b^2 d+11 b^3 c\right )}{\sqrt {3} a^{11/3}}+\frac {b^{8/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-2 a^3 f+5 a^2 b e-8 a b^2 d+11 b^3 c\right )}{6 a^{11/3}}-\frac {b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (-2 a^3 f+5 a^2 b e-8 a b^2 d+11 b^3 c\right )}{3 a^{11/3}}-\frac {3 b^3 c}{8 a x^8}}{3 a b^3}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^4 \left (a+b x^3\right )}\) |
Input:
Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^9*(a + b*x^3)^2),x]
Output:
-1/3*((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/(a^4*(a + b*x^3)) + ((-3*b^3* c)/(8*a*x^8) + (3*b^3*(2*b*c - a*d))/(5*a^2*x^5) - (3*b^3*(3*b^2*c - 2*a*b *d + a^2*e))/(2*a^3*x^2) + (b^(8/3)*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2* a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(11/3 )) - (b^(8/3)*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*Log[a^(1/3) + b ^(1/3)*x])/(3*a^(11/3)) + (b^(8/3)*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a ^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(11/3)))/(3*a*b ^3)
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]
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {\frac {\left (\frac {1}{3} f \,a^{3}-\frac {1}{3} e \,a^{2} b +\frac {1}{3} d a \,b^{2}-\frac {1}{3} b^{3} c \right ) x}{b \,x^{3}+a}+\frac {\left (2 f \,a^{3}-5 e \,a^{2} b +8 d a \,b^{2}-11 b^{3} c \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}}{a^{4}}-\frac {c}{8 a^{2} x^{8}}-\frac {a d -2 c b}{5 a^{3} x^{5}}-\frac {a^{2} e -2 d a b +3 b^{2} c}{2 a^{4} x^{2}}\) | \(214\) |
| risch | \(\frac {\frac {\left (2 f \,a^{3}-5 e \,a^{2} b +8 d a \,b^{2}-11 b^{3} c \right ) x^{9}}{6 a^{4}}-\frac {\left (5 a^{2} e -8 d a b +11 b^{2} c \right ) x^{6}}{10 a^{3}}-\frac {\left (8 a d -11 c b \right ) x^{3}}{40 a^{2}}-\frac {c}{8 a}}{x^{8} \left (b \,x^{3}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{14} b \,\textit {\_Z}^{3}-8 a^{9} f^{3}+60 a^{8} b e \,f^{2}-96 a^{7} b^{2} d \,f^{2}-150 a^{7} b^{2} e^{2} f +132 a^{6} b^{3} c \,f^{2}+480 a^{6} b^{3} d e f +125 a^{6} b^{3} e^{3}-660 a^{5} b^{4} c e f -384 a^{5} b^{4} d^{2} f -600 a^{5} b^{4} d \,e^{2}+1056 a^{4} b^{5} c d f +825 a^{4} b^{5} c \,e^{2}+960 a^{4} b^{5} d^{2} e -726 a^{3} b^{6} c^{2} f -2640 a^{3} b^{6} c d e -512 a^{3} b^{6} d^{3}+1815 a^{2} b^{7} c^{2} e +2112 a^{2} b^{7} c \,d^{2}-2904 a \,b^{8} c^{2} d +1331 b^{9} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{14} b +24 a^{9} f^{3}-180 a^{8} b e \,f^{2}+288 a^{7} b^{2} d \,f^{2}+450 a^{7} b^{2} e^{2} f -396 a^{6} b^{3} c \,f^{2}-1440 a^{6} b^{3} d e f -375 a^{6} b^{3} e^{3}+1980 a^{5} b^{4} c e f +1152 a^{5} b^{4} d^{2} f +1800 a^{5} b^{4} d \,e^{2}-3168 a^{4} b^{5} c d f -2475 a^{4} b^{5} c \,e^{2}-2880 a^{4} b^{5} d^{2} e +2178 a^{3} b^{6} c^{2} f +7920 a^{3} b^{6} c d e +1536 a^{3} b^{6} d^{3}-5445 a^{2} b^{7} c^{2} e -6336 a^{2} b^{7} c \,d^{2}+8712 a \,b^{8} c^{2} d -3993 b^{9} c^{3}\right ) x +\left (-4 a^{11} f^{2}+20 a^{10} b e f -32 a^{9} b^{2} d f -25 a^{9} b^{2} e^{2}+44 a^{8} b^{3} c f +80 a^{8} b^{3} d e -110 a^{7} b^{4} c e -64 a^{7} b^{4} d^{2}+176 a^{6} b^{5} c d -121 a^{5} b^{6} c^{2}\right ) \textit {\_R} \right )\right )}{9}\) | \(677\) |
Input:
int((f*x^9+e*x^6+d*x^3+c)/x^9/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/a^4*((1/3*f*a^3-1/3*e*a^2*b+1/3*d*a*b^2-1/3*b^3*c)*x/(b*x^3+a)+1/3*(2*a^ 3*f-5*a^2*b*e+8*a*b^2*d-11*b^3*c)*(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))))-1/8*c/a^2/x^8-1/5*(a*d-2*b*c)/a ^3/x^5-1/2*(a^2*e-2*a*b*d+3*b^2*c)/a^4/x^2
Time = 0.10 (sec) , antiderivative size = 959, normalized size of antiderivative = 3.23 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((f*x^9+e*x^6+d*x^3+c)/x^9/(b*x^3+a)^2,x, algorithm="fricas")
Output:
[-1/360*(60*(11*a^2*b^4*c - 8*a^3*b^3*d + 5*a^4*b^2*e - 2*a^5*b*f)*x^9 + 4 5*a^5*b*c + 36*(11*a^3*b^3*c - 8*a^4*b^2*d + 5*a^5*b*e)*x^6 - 9*(11*a^4*b^ 2*c - 8*a^5*b*d)*x^3 + 60*sqrt(1/3)*((11*a*b^5*c - 8*a^2*b^4*d + 5*a^3*b^3 *e - 2*a^4*b^2*f)*x^11 + (11*a^2*b^4*c - 8*a^3*b^3*d + 5*a^4*b^2*e - 2*a^5 *b*f)*x^8)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a ^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^ 2*b)^(1/3)/b))/(b*x^3 + a)) - 20*((11*b^4*c - 8*a*b^3*d + 5*a^2*b^2*e - 2* a^3*b*f)*x^11 + (11*a*b^3*c - 8*a^2*b^2*d + 5*a^3*b*e - 2*a^4*f)*x^8)*(a^2 *b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 40*((11*b^4*c - 8*a*b^3*d + 5*a^2*b^2*e - 2*a^3*b*f)*x^11 + (11*a*b^3*c - 8*a^2*b^2*d + 5*a^3*b*e - 2*a^4*f)*x^8)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^6* b^2*x^11 + a^7*b*x^8), -1/360*(60*(11*a^2*b^4*c - 8*a^3*b^3*d + 5*a^4*b^2* e - 2*a^5*b*f)*x^9 + 45*a^5*b*c + 36*(11*a^3*b^3*c - 8*a^4*b^2*d + 5*a^5*b *e)*x^6 - 9*(11*a^4*b^2*c - 8*a^5*b*d)*x^3 + 120*sqrt(1/3)*((11*a*b^5*c - 8*a^2*b^4*d + 5*a^3*b^3*e - 2*a^4*b^2*f)*x^11 + (11*a^2*b^4*c - 8*a^3*b^3* d + 5*a^4*b^2*e - 2*a^5*b*f)*x^8)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*( 2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a^2) - 20*((11* b^4*c - 8*a*b^3*d + 5*a^2*b^2*e - 2*a^3*b*f)*x^11 + (11*a*b^3*c - 8*a^2*b^ 2*d + 5*a^3*b*e - 2*a^4*f)*x^8)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)* x + (a^2*b)^(1/3)*a) + 40*((11*b^4*c - 8*a*b^3*d + 5*a^2*b^2*e - 2*a^3*...
Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((f*x**9+e*x**6+d*x**3+c)/x**9/(b*x**3+a)**2,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=-\frac {20 \, {\left (11 \, b^{3} c - 8 \, a b^{2} d + 5 \, a^{2} b e - 2 \, a^{3} f\right )} x^{9} + 12 \, {\left (11 \, a b^{2} c - 8 \, a^{2} b d + 5 \, a^{3} e\right )} x^{6} + 15 \, a^{3} c - 3 \, {\left (11 \, a^{2} b c - 8 \, a^{3} d\right )} x^{3}}{120 \, {\left (a^{4} b x^{11} + a^{5} x^{8}\right )}} - \frac {\sqrt {3} {\left (11 \, b^{3} c - 8 \, a b^{2} d + 5 \, a^{2} b e - 2 \, a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (11 \, b^{3} c - 8 \, a b^{2} d + 5 \, a^{2} b e - 2 \, a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{4} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (11 \, b^{3} c - 8 \, a b^{2} d + 5 \, a^{2} b e - 2 \, a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{4} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate((f*x^9+e*x^6+d*x^3+c)/x^9/(b*x^3+a)^2,x, algorithm="maxima")
Output:
-1/120*(20*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*x^9 + 12*(11*a*b^2 *c - 8*a^2*b*d + 5*a^3*e)*x^6 + 15*a^3*c - 3*(11*a^2*b*c - 8*a^3*d)*x^3)/( a^4*b*x^11 + a^5*x^8) - 1/9*sqrt(3)*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2* a^3*f)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^4*b*(a/b)^(2 /3)) + 1/18*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*log(x^2 - x*(a/b) ^(1/3) + (a/b)^(2/3))/(a^4*b*(a/b)^(2/3)) - 1/9*(11*b^3*c - 8*a*b^2*d + 5* a^2*b*e - 2*a^3*f)*log(x + (a/b)^(1/3))/(a^4*b*(a/b)^(2/3))
Time = 0.13 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.15 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=\frac {{\left (11 \, b^{3} c - 8 \, a b^{2} d + 5 \, a^{2} b e - 2 \, a^{3} f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{5}} - \frac {\sqrt {3} {\left (11 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3} c - 8 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2} d + 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b e - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{5} b} - \frac {b^{3} c x - a b^{2} d x + a^{2} b e x - a^{3} f x}{3 \, {\left (b x^{3} + a\right )} a^{4}} - \frac {{\left (11 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3} c - 8 \, \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2} d + 5 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b e - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{5} b} - \frac {60 \, b^{2} c x^{6} - 40 \, a b d x^{6} + 20 \, a^{2} e x^{6} - 16 \, a b c x^{3} + 8 \, a^{2} d x^{3} + 5 \, a^{2} c}{40 \, a^{4} x^{8}} \] Input:
integrate((f*x^9+e*x^6+d*x^3+c)/x^9/(b*x^3+a)^2,x, algorithm="giac")
Output:
1/9*(11*b^3*c - 8*a*b^2*d + 5*a^2*b*e - 2*a^3*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^5 - 1/9*sqrt(3)*(11*(-a*b^2)^(1/3)*b^3*c - 8*(-a*b^2)^(1/ 3)*a*b^2*d + 5*(-a*b^2)^(1/3)*a^2*b*e - 2*(-a*b^2)^(1/3)*a^3*f)*arctan(1/3 *sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^5*b) - 1/3*(b^3*c*x - a*b^2 *d*x + a^2*b*e*x - a^3*f*x)/((b*x^3 + a)*a^4) - 1/18*(11*(-a*b^2)^(1/3)*b^ 3*c - 8*(-a*b^2)^(1/3)*a*b^2*d + 5*(-a*b^2)^(1/3)*a^2*b*e - 2*(-a*b^2)^(1/ 3)*a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^5*b) - 1/40*(60*b^2* c*x^6 - 40*a*b*d*x^6 + 20*a^2*e*x^6 - 16*a*b*c*x^3 + 8*a^2*d*x^3 + 5*a^2*c )/(a^4*x^8)
Time = 6.11 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx=-\frac {\frac {c}{8\,a}+\frac {x^9\,\left (-2\,f\,a^3+5\,e\,a^2\,b-8\,d\,a\,b^2+11\,c\,b^3\right )}{6\,a^4}+\frac {x^3\,\left (8\,a\,d-11\,b\,c\right )}{40\,a^2}+\frac {x^6\,\left (5\,e\,a^2-8\,d\,a\,b+11\,c\,b^2\right )}{10\,a^3}}{b\,x^{11}+a\,x^8}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-2\,f\,a^3+5\,e\,a^2\,b-8\,d\,a\,b^2+11\,c\,b^3\right )}{9\,a^{14/3}\,b^{1/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-2\,f\,a^3+5\,e\,a^2\,b-8\,d\,a\,b^2+11\,c\,b^3\right )}{9\,a^{14/3}\,b^{1/3}}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-2\,f\,a^3+5\,e\,a^2\,b-8\,d\,a\,b^2+11\,c\,b^3\right )}{9\,a^{14/3}\,b^{1/3}} \] Input:
int((c + d*x^3 + e*x^6 + f*x^9)/(x^9*(a + b*x^3)^2),x)
Output:
(log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(1 1*b^3*c - 2*a^3*f - 8*a*b^2*d + 5*a^2*b*e))/(9*a^(14/3)*b^(1/3)) - (log(b^ (1/3)*x + a^(1/3))*(11*b^3*c - 2*a^3*f - 8*a*b^2*d + 5*a^2*b*e))/(9*a^(14/ 3)*b^(1/3)) - (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1 i)/2 - 1/2)*(11*b^3*c - 2*a^3*f - 8*a*b^2*d + 5*a^2*b*e))/(9*a^(14/3)*b^(1 /3)) - (c/(8*a) + (x^9*(11*b^3*c - 2*a^3*f - 8*a*b^2*d + 5*a^2*b*e))/(6*a^ 4) + (x^3*(8*a*d - 11*b*c))/(40*a^2) + (x^6*(11*b^2*c + 5*a^2*e - 8*a*b*d) )/(10*a^3))/(a*x^8 + b*x^11)
Time = 0.16 (sec) , antiderivative size = 819, normalized size of antiderivative = 2.76 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^9 \left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((f*x^9+e*x^6+d*x^3+c)/x^9/(b*x^3+a)^2,x)
Output:
( - 80*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3))) *a**4*f*x**8 + 200*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/ 3)*sqrt(3)))*a**3*b*e*x**8 - 80*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/ 3)*x)/(a**(1/3)*sqrt(3)))*a**3*b*f*x**11 - 320*a**(1/3)*sqrt(3)*atan((a**( 1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*d*x**8 + 200*a**(1/3)*s qrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**2*e*x**1 1 + 440*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)) )*a*b**3*c*x**8 - 320*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a** (1/3)*sqrt(3)))*a*b**3*d*x**11 + 440*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b **(1/3)*x)/(a**(1/3)*sqrt(3)))*b**4*c*x**11 - 40*a**(1/3)*log(a**(2/3) - b **(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*f*x**8 + 100*a**(1/3)*log(a**(2/3 ) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b*e*x**8 - 40*a**(1/3)*log(a **(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b*f*x**11 - 160*a**(1/ 3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*d*x**8 + 100*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2 *e*x**11 + 220*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2 )*a*b**3*c*x**8 - 160*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/ 3)*x**2)*a*b**3*d*x**11 + 220*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**4*c*x**11 + 80*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a** 4*f*x**8 - 200*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a**3*b*e*x**8 + 80*a...