\(\int \frac {c+d x^3+e x^6+f x^9}{x^{16} (a+b x^3)} \, dx\) [10]

Optimal result

Integrand size = 30, antiderivative size = 205 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=-\frac {c}{15 a x^{15}}+\frac {b c-a d}{12 a^2 x^{12}}-\frac {b^2 c-a b d+a^2 e}{9 a^3 x^9}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{6 a^4 x^6}-\frac {b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )}{3 a^5 x^3}-\frac {b^2 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log (x)}{a^6}+\frac {b^2 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^6} \] Output:

-1/15*c/a/x^15+1/12*(-a*d+b*c)/a^2/x^12-1/9*(a^2*e-a*b*d+b^2*c)/a^3/x^9+1/ 
6*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)/a^4/x^6-1/3*b*(-a^3*f+a^2*b*e-a*b^2*d+b^3 
*c)/a^5/x^3-b^2*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(x)/a^6+1/3*b^2*(-a^3*f+a 
^2*b*e-a*b^2*d+b^3*c)*ln(b*x^3+a)/a^6
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=-\frac {\frac {a \left (60 b^4 c x^{12}-30 a b^3 x^9 \left (c+2 d x^3\right )+10 a^2 b^2 x^6 \left (2 c+3 d x^3+6 e x^6\right )-5 a^3 b x^3 \left (3 c+4 d x^3+6 e x^6+12 f x^9\right )+a^4 \left (12 c+15 d x^3+20 e x^6+30 f x^9\right )\right )}{x^{15}}+180 b^2 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log (x)-60 b^2 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{180 a^6} \] Input:

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^16*(a + b*x^3)),x]
 

Output:

-1/180*((a*(60*b^4*c*x^12 - 30*a*b^3*x^9*(c + 2*d*x^3) + 10*a^2*b^2*x^6*(2 
*c + 3*d*x^3 + 6*e*x^6) - 5*a^3*b*x^3*(3*c + 4*d*x^3 + 6*e*x^6 + 12*f*x^9) 
 + a^4*(12*c + 15*d*x^3 + 20*e*x^6 + 30*f*x^9)))/x^15 + 180*b^2*(b^3*c - a 
*b^2*d + a^2*b*e - a^3*f)*Log[x] - 60*b^2*(b^3*c - a*b^2*d + a^2*b*e - a^3 
*f)*Log[a + b*x^3])/a^6
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2361, 2123, 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.

\(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx\)

\(\Big \downarrow \) 2361

\(\displaystyle \frac {1}{3} \int \frac {f x^9+e x^6+d x^3+c}{x^{18} \left (b x^3+a\right )}dx^3\)

\(\Big \downarrow \) 2123

\(\displaystyle \frac {1}{3} \int \left (-\frac {\left (f a^3-b e a^2+b^2 d a-b^3 c\right ) b^3}{a^6 \left (b x^3+a\right )}+\frac {\left (f a^3-b e a^2+b^2 d a-b^3 c\right ) b^2}{a^6 x^3}-\frac {\left (f a^3-b e a^2+b^2 d a-b^3 c\right ) b}{a^5 x^6}+\frac {f a^3-b e a^2+b^2 d a-b^3 c}{a^4 x^9}+\frac {e a^2-b d a+b^2 c}{a^3 x^{12}}+\frac {a d-b c}{a^2 x^{15}}+\frac {c}{a x^{18}}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {b c-a d}{4 a^2 x^{12}}-\frac {a^2 e-a b d+b^2 c}{3 a^3 x^9}-\frac {b^2 \log \left (x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^6}+\frac {b^2 \log \left (a+b x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^6}-\frac {b \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{a^5 x^3}+\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{2 a^4 x^6}-\frac {c}{5 a x^{15}}\right )\)

Input:

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^16*(a + b*x^3)),x]
 

Output:

(-1/5*c/(a*x^15) + (b*c - a*d)/(4*a^2*x^12) - (b^2*c - a*b*d + a^2*e)/(3*a 
^3*x^9) + (b^3*c - a*b^2*d + a^2*b*e - a^3*f)/(2*a^4*x^6) - (b*(b^3*c - a* 
b^2*d + a^2*b*e - a^3*f))/(a^5*x^3) - (b^2*(b^3*c - a*b^2*d + a^2*b*e - a^ 
3*f)*Log[x^3])/a^6 + (b^2*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a + b*x^ 
3])/a^6)/3
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

rule 2361

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Simp[1/n 
  Subst[Int[x^(Simplify[(m + 1)/n] - 1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x 
], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && IntegerQ[S 
implify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {b^{2} \left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{6}}-\frac {a d -c b}{12 a^{2} x^{12}}-\frac {c}{15 a \,x^{15}}-\frac {a^{2} e -d a b +b^{2} c}{9 a^{3} x^{9}}-\frac {f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c}{6 a^{4} x^{6}}+\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) b^{2} \ln \left (x \right )}{a^{6}}+\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) b}{3 a^{5} x^{3}}\)	\(193\)
norman	\(\frac {-\frac {c}{15 a}-\frac {\left (a d -c b \right ) x^{3}}{12 a^{2}}-\frac {\left (a^{2} e -d a b +b^{2} c \right ) x^{6}}{9 a^{3}}-\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) x^{9}}{6 a^{4}}+\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) b \,x^{12}}{3 a^{5}}}{x^{15}}+\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) b^{2} \ln \left (x \right )}{a^{6}}-\frac {b^{2} \left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{6}}\)	\(195\)
risch	\(\frac {-\frac {c}{15 a}-\frac {\left (a d -c b \right ) x^{3}}{12 a^{2}}-\frac {\left (a^{2} e -d a b +b^{2} c \right ) x^{6}}{9 a^{3}}-\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) x^{9}}{6 a^{4}}+\frac {\left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right ) b \,x^{12}}{3 a^{5}}}{x^{15}}+\frac {b^{2} \ln \left (x \right ) f}{a^{3}}-\frac {b^{3} \ln \left (x \right ) e}{a^{4}}+\frac {b^{4} \ln \left (x \right ) d}{a^{5}}-\frac {b^{5} \ln \left (x \right ) c}{a^{6}}-\frac {b^{2} \ln \left (b \,x^{3}+a \right ) f}{3 a^{3}}+\frac {b^{3} \ln \left (b \,x^{3}+a \right ) e}{3 a^{4}}-\frac {b^{4} \ln \left (b \,x^{3}+a \right ) d}{3 a^{5}}+\frac {b^{5} \ln \left (b \,x^{3}+a \right ) c}{3 a^{6}}\)	\(230\)
parallelrisch	\(\frac {180 \ln \left (x \right ) x^{15} a^{3} b^{2} f -180 \ln \left (x \right ) x^{15} a^{2} b^{3} e +180 \ln \left (x \right ) x^{15} a \,b^{4} d -180 \ln \left (x \right ) x^{15} b^{5} c -60 \ln \left (b \,x^{3}+a \right ) x^{15} a^{3} b^{2} f +60 \ln \left (b \,x^{3}+a \right ) x^{15} a^{2} b^{3} e -60 \ln \left (b \,x^{3}+a \right ) x^{15} a \,b^{4} d +60 \ln \left (b \,x^{3}+a \right ) x^{15} b^{5} c +60 a^{4} b f \,x^{12}-60 a^{3} b^{2} e \,x^{12}+60 a^{2} b^{3} d \,x^{12}-60 a \,b^{4} c \,x^{12}-30 a^{5} f \,x^{9}+30 a^{4} b e \,x^{9}-30 a^{3} b^{2} d \,x^{9}+30 a^{2} b^{3} c \,x^{9}-20 a^{5} e \,x^{6}+20 a^{4} b d \,x^{6}-20 a^{3} b^{2} c \,x^{6}-15 a^{5} d \,x^{3}+15 a^{4} b c \,x^{3}-12 c \,a^{5}}{180 a^{6} x^{15}}\)	\(279\)

Input:

int((f*x^9+e*x^6+d*x^3+c)/x^16/(b*x^3+a),x,method=_RETURNVERBOSE)
 

Output:

-1/3*b^2*(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^6*ln(b*x^3+a)-1/12/a^2*(a*d-b*c)/ 
x^12-1/15*c/a/x^15-1/9*(a^2*e-a*b*d+b^2*c)/a^3/x^9-1/6*(a^3*f-a^2*b*e+a*b^ 
2*d-b^3*c)/a^4/x^6+(a^3*f-a^2*b*e+a*b^2*d-b^3*c)/a^6*b^2*ln(x)+1/3*(a^3*f- 
a^2*b*e+a*b^2*d-b^3*c)/a^5*b/x^3
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=\frac {60 \, {\left (b^{5} c - a b^{4} d + a^{2} b^{3} e - a^{3} b^{2} f\right )} x^{15} \log \left (b x^{3} + a\right ) - 180 \, {\left (b^{5} c - a b^{4} d + a^{2} b^{3} e - a^{3} b^{2} f\right )} x^{15} \log \left (x\right ) - 60 \, {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{12} + 30 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d + a^{4} b e - a^{5} f\right )} x^{9} - 20 \, {\left (a^{3} b^{2} c - a^{4} b d + a^{5} e\right )} x^{6} - 12 \, a^{5} c + 15 \, {\left (a^{4} b c - a^{5} d\right )} x^{3}}{180 \, a^{6} x^{15}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^16/(b*x^3+a),x, algorithm="fricas")
 

Output:

1/180*(60*(b^5*c - a*b^4*d + a^2*b^3*e - a^3*b^2*f)*x^15*log(b*x^3 + a) - 
180*(b^5*c - a*b^4*d + a^2*b^3*e - a^3*b^2*f)*x^15*log(x) - 60*(a*b^4*c - 
a^2*b^3*d + a^3*b^2*e - a^4*b*f)*x^12 + 30*(a^2*b^3*c - a^3*b^2*d + a^4*b* 
e - a^5*f)*x^9 - 20*(a^3*b^2*c - a^4*b*d + a^5*e)*x^6 - 12*a^5*c + 15*(a^4 
*b*c - a^5*d)*x^3)/(a^6*x^15)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((f*x**9+e*x**6+d*x**3+c)/x**16/(b*x**3+a),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=\frac {{\left (b^{5} c - a b^{4} d + a^{2} b^{3} e - a^{3} b^{2} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{6}} - \frac {{\left (b^{5} c - a b^{4} d + a^{2} b^{3} e - a^{3} b^{2} f\right )} \log \left (x^{3}\right )}{3 \, a^{6}} - \frac {60 \, {\left (b^{4} c - a b^{3} d + a^{2} b^{2} e - a^{3} b f\right )} x^{12} - 30 \, {\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} x^{9} + 20 \, {\left (a^{2} b^{2} c - a^{3} b d + a^{4} e\right )} x^{6} + 12 \, a^{4} c - 15 \, {\left (a^{3} b c - a^{4} d\right )} x^{3}}{180 \, a^{5} x^{15}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^16/(b*x^3+a),x, algorithm="maxima")
 

Output:

1/3*(b^5*c - a*b^4*d + a^2*b^3*e - a^3*b^2*f)*log(b*x^3 + a)/a^6 - 1/3*(b^ 
5*c - a*b^4*d + a^2*b^3*e - a^3*b^2*f)*log(x^3)/a^6 - 1/180*(60*(b^4*c - a 
*b^3*d + a^2*b^2*e - a^3*b*f)*x^12 - 30*(a*b^3*c - a^2*b^2*d + a^3*b*e - a 
^4*f)*x^9 + 20*(a^2*b^2*c - a^3*b*d + a^4*e)*x^6 + 12*a^4*c - 15*(a^3*b*c 
- a^4*d)*x^3)/(a^5*x^15)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.37 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=-\frac {{\left (b^{5} c - a b^{4} d + a^{2} b^{3} e - a^{3} b^{2} f\right )} \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {{\left (b^{6} c - a b^{5} d + a^{2} b^{4} e - a^{3} b^{3} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{6} b} + \frac {137 \, b^{5} c x^{15} - 137 \, a b^{4} d x^{15} + 137 \, a^{2} b^{3} e x^{15} - 137 \, a^{3} b^{2} f x^{15} - 60 \, a b^{4} c x^{12} + 60 \, a^{2} b^{3} d x^{12} - 60 \, a^{3} b^{2} e x^{12} + 60 \, a^{4} b f x^{12} + 30 \, a^{2} b^{3} c x^{9} - 30 \, a^{3} b^{2} d x^{9} + 30 \, a^{4} b e x^{9} - 30 \, a^{5} f x^{9} - 20 \, a^{3} b^{2} c x^{6} + 20 \, a^{4} b d x^{6} - 20 \, a^{5} e x^{6} + 15 \, a^{4} b c x^{3} - 15 \, a^{5} d x^{3} - 12 \, a^{5} c}{180 \, a^{6} x^{15}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^16/(b*x^3+a),x, algorithm="giac")
 

Output:

-(b^5*c - a*b^4*d + a^2*b^3*e - a^3*b^2*f)*log(abs(x))/a^6 + 1/3*(b^6*c - 
a*b^5*d + a^2*b^4*e - a^3*b^3*f)*log(abs(b*x^3 + a))/(a^6*b) + 1/180*(137* 
b^5*c*x^15 - 137*a*b^4*d*x^15 + 137*a^2*b^3*e*x^15 - 137*a^3*b^2*f*x^15 - 
60*a*b^4*c*x^12 + 60*a^2*b^3*d*x^12 - 60*a^3*b^2*e*x^12 + 60*a^4*b*f*x^12 
+ 30*a^2*b^3*c*x^9 - 30*a^3*b^2*d*x^9 + 30*a^4*b*e*x^9 - 30*a^5*f*x^9 - 20 
*a^3*b^2*c*x^6 + 20*a^4*b*d*x^6 - 20*a^5*e*x^6 + 15*a^4*b*c*x^3 - 15*a^5*d 
*x^3 - 12*a^5*c)/(a^6*x^15)
 

Mupad [B] (verification not implemented)

Time = 6.66 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=\frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3\,b^2+e\,a^2\,b^3-d\,a\,b^4+c\,b^5\right )}{3\,a^6}-\frac {\frac {c}{15\,a}-\frac {x^9\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{6\,a^4}+\frac {x^3\,\left (a\,d-b\,c\right )}{12\,a^2}+\frac {x^6\,\left (e\,a^2-d\,a\,b+c\,b^2\right )}{9\,a^3}+\frac {b\,x^{12}\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^5}}{x^{15}}-\frac {\ln \left (x\right )\,\left (-f\,a^3\,b^2+e\,a^2\,b^3-d\,a\,b^4+c\,b^5\right )}{a^6} \] Input:

int((c + d*x^3 + e*x^6 + f*x^9)/(x^16*(a + b*x^3)),x)
 

Output:

(log(a + b*x^3)*(b^5*c + a^2*b^3*e - a^3*b^2*f - a*b^4*d))/(3*a^6) - (c/(1 
5*a) - (x^9*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(6*a^4) + (x^3*(a*d - b*c 
))/(12*a^2) + (x^6*(b^2*c + a^2*e - a*b*d))/(9*a^3) + (b*x^12*(b^3*c - a^3 
*f - a*b^2*d + a^2*b*e))/(3*a^5))/x^15 - (log(x)*(b^5*c + a^2*b^3*e - a^3* 
b^2*f - a*b^4*d))/a^6
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.01 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{16} \left (a+b x^3\right )} \, dx=\frac {60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{5} c \,x^{15}+60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{5} c \,x^{15}-180 \,\mathrm {log}\left (x \right ) b^{5} c \,x^{15}+15 a^{4} b c \,x^{3}+20 a^{4} b d \,x^{6}+30 a^{4} b e \,x^{9}+60 a^{4} b f \,x^{12}-20 a^{3} b^{2} c \,x^{6}-30 a^{3} b^{2} d \,x^{9}-60 a^{3} b^{2} e \,x^{12}+30 a^{2} b^{3} c \,x^{9}+60 a^{2} b^{3} d \,x^{12}-60 a \,b^{4} c \,x^{12}-12 a^{5} c -15 a^{5} d \,x^{3}-20 a^{5} e \,x^{6}-30 a^{5} f \,x^{9}-60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3} b^{2} f \,x^{15}+60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b^{3} e \,x^{15}-60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{4} d \,x^{15}-60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3} b^{2} f \,x^{15}+60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b^{3} e \,x^{15}-60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{4} d \,x^{15}+180 \,\mathrm {log}\left (x \right ) a^{3} b^{2} f \,x^{15}-180 \,\mathrm {log}\left (x \right ) a^{2} b^{3} e \,x^{15}+180 \,\mathrm {log}\left (x \right ) a \,b^{4} d \,x^{15}}{180 a^{6} x^{15}} \] Input:

int((f*x^9+e*x^6+d*x^3+c)/x^16/(b*x^3+a),x)
 

Output:

( - 60*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b**2*f*x** 
15 + 60*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**3*e*x* 
*15 - 60*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**4*d*x**1 
5 + 60*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**5*c*x**15 - 
60*log(a**(1/3) + b**(1/3)*x)*a**3*b**2*f*x**15 + 60*log(a**(1/3) + b**(1/ 
3)*x)*a**2*b**3*e*x**15 - 60*log(a**(1/3) + b**(1/3)*x)*a*b**4*d*x**15 + 6 
0*log(a**(1/3) + b**(1/3)*x)*b**5*c*x**15 + 180*log(x)*a**3*b**2*f*x**15 - 
 180*log(x)*a**2*b**3*e*x**15 + 180*log(x)*a*b**4*d*x**15 - 180*log(x)*b** 
5*c*x**15 - 12*a**5*c - 15*a**5*d*x**3 - 20*a**5*e*x**6 - 30*a**5*f*x**9 + 
 15*a**4*b*c*x**3 + 20*a**4*b*d*x**6 + 30*a**4*b*e*x**9 + 60*a**4*b*f*x**1 
2 - 20*a**3*b**2*c*x**6 - 30*a**3*b**2*d*x**9 - 60*a**3*b**2*e*x**12 + 30* 
a**2*b**3*c*x**9 + 60*a**2*b**3*d*x**12 - 60*a*b**4*c*x**12)/(180*a**6*x** 
15)