\(\int \frac {c+d x^3+e x^6+f x^9}{x^7 (a+b x^3)^2} \, dx\) [35]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 130 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=-\frac {c}{6 a^2 x^6}+\frac {2 b c-a d}{3 a^3 x^3}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 a^3 b \left (a+b x^3\right )}+\frac {\left (3 b^2 c-2 a b d+a^2 e\right ) \log (x)}{a^4}-\frac {\left (3 b^2 c-2 a b d+a^2 e\right ) \log \left (a+b x^3\right )}{3 a^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=-\frac {\frac {a^2 c}{x^6}+\frac {2 a (-2 b c+a d)}{x^3}+\frac {2 a \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{b \left (a+b x^3\right )}-6 \left (3 b^2 c-2 a b d+a^2 e\right ) \log (x)+2 \left (3 b^2 c-2 a b d+a^2 e\right ) \log \left (a+b x^3\right )}{6 a^4} \] Input:

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^7*(a + b*x^3)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, 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.

\(\displaystyle \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx\)

\(\Big \downarrow \) 2361

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^7*(a + b*x^3)^2),x]
 

Output:

(-1/2*c/(a^2*x^6) + (2*b*c - a*d)/(a^3*x^3) + (b^3*c - a*b^2*d + a^2*b*e - 
 a^3*f)/(a^3*b*(a + b*x^3)) + ((3*b^2*c - 2*a*b*d + a^2*e)*Log[x^3])/a^4 - 
 ((3*b^2*c - 2*a*b*d + a^2*e)*Log[a + b*x^3])/a^4)/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 = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95

method result size
default \(\frac {-\frac {a \left (f \,a^{3}-e \,a^{2} b +d a \,b^{2}-b^{3} c \right )}{b \left (b \,x^{3}+a \right )}+\left (-a^{2} e +2 d a b -3 b^{2} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{4}}-\frac {c}{6 a^{2} x^{6}}-\frac {a d -2 c b}{3 a^{3} x^{3}}+\frac {\left (a^{2} e -2 d a b +3 b^{2} c \right ) \ln \left (x \right )}{a^{4}}\) \(123\)
norman \(\frac {-\frac {c}{6 a}-\frac {\left (2 a d -3 c b \right ) x^{3}}{6 a^{2}}+\frac {\left (f \,a^{3}-e \,a^{2} b +2 d a \,b^{2}-3 b^{3} c \right ) x^{9}}{3 a^{4}}}{x^{6} \left (b \,x^{3}+a \right )}+\frac {\left (a^{2} e -2 d a b +3 b^{2} c \right ) \ln \left (x \right )}{a^{4}}-\frac {\left (a^{2} e -2 d a b +3 b^{2} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{4}}\) \(126\)
risch \(\frac {-\frac {\left (f \,a^{3}-e \,a^{2} b +2 d a \,b^{2}-3 b^{3} c \right ) x^{6}}{3 a^{3} b}-\frac {\left (2 a d -3 c b \right ) x^{3}}{6 a^{2}}-\frac {c}{6 a}}{x^{6} \left (b \,x^{3}+a \right )}+\frac {e \ln \left (x \right )}{a^{2}}-\frac {2 \ln \left (x \right ) d b}{a^{3}}+\frac {3 \ln \left (x \right ) b^{2} c}{a^{4}}-\frac {e \ln \left (b \,x^{3}+a \right )}{3 a^{2}}+\frac {2 \ln \left (b \,x^{3}+a \right ) d b}{3 a^{3}}-\frac {\ln \left (b \,x^{3}+a \right ) b^{2} c}{a^{4}}\) \(149\)
parallelrisch \(\frac {6 \ln \left (x \right ) x^{9} a^{2} b e -12 \ln \left (x \right ) x^{9} a \,b^{2} d +18 \ln \left (x \right ) x^{9} b^{3} c -2 \ln \left (b \,x^{3}+a \right ) x^{9} a^{2} b e +4 \ln \left (b \,x^{3}+a \right ) x^{9} a \,b^{2} d -6 \ln \left (b \,x^{3}+a \right ) x^{9} b^{3} c +2 x^{9} a^{3} f -2 x^{9} a^{2} b e +4 x^{9} a \,b^{2} d -6 b^{3} c \,x^{9}+6 \ln \left (x \right ) x^{6} a^{3} e -12 \ln \left (x \right ) x^{6} a^{2} b d +18 \ln \left (x \right ) x^{6} a \,b^{2} c -2 \ln \left (b \,x^{3}+a \right ) x^{6} a^{3} e +4 \ln \left (b \,x^{3}+a \right ) x^{6} a^{2} b d -6 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{2} c -2 a^{3} d \,x^{3}+3 a^{2} b c \,x^{3}-c \,a^{3}}{6 a^{4} x^{6} \left (b \,x^{3}+a \right )}\) \(258\)

Input:

int((f*x^9+e*x^6+d*x^3+c)/x^7/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.60 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {2 \, {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} x^{6} - a^{3} b c + {\left (3 \, a^{2} b^{2} c - 2 \, a^{3} b d\right )} x^{3} - 2 \, {\left ({\left (3 \, b^{4} c - 2 \, a b^{3} d + a^{2} b^{2} e\right )} x^{9} + {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e\right )} x^{6}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left ({\left (3 \, b^{4} c - 2 \, a b^{3} d + a^{2} b^{2} e\right )} x^{9} + {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e\right )} x^{6}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{2} x^{9} + a^{5} b x^{6}\right )}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^7/(b*x^3+a)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate((f*x**9+e*x**6+d*x**3+c)/x**7/(b*x**3+a)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {2 \, {\left (3 \, b^{3} c - 2 \, a b^{2} d + a^{2} b e - a^{3} f\right )} x^{6} - a^{2} b c + {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} x^{3}}{6 \, {\left (a^{3} b^{2} x^{9} + a^{4} b x^{6}\right )}} - \frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} + \frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left (x^{3}\right )}{3 \, a^{4}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^7/(b*x^3+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {{\left (3 \, b^{3} c - 2 \, a b^{2} d + a^{2} b e\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac {3 \, b^{4} c x^{3} - 2 \, a b^{3} d x^{3} + a^{2} b^{2} e x^{3} + 4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f}{3 \, {\left (b x^{3} + a\right )} a^{4} b} - \frac {9 \, b^{2} c x^{6} - 6 \, a b d x^{6} + 3 \, a^{2} e x^{6} - 4 \, a b c x^{3} + 2 \, a^{2} d x^{3} + a^{2} c}{6 \, a^{4} x^{6}} \] Input:

integrate((f*x^9+e*x^6+d*x^3+c)/x^7/(b*x^3+a)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 6.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (x\right )\,\left (e\,a^2-2\,d\,a\,b+3\,c\,b^2\right )}{a^4}-\frac {\ln \left (b\,x^3+a\right )\,\left (e\,a^2-2\,d\,a\,b+3\,c\,b^2\right )}{3\,a^4}-\frac {\frac {c}{6\,a}+\frac {x^3\,\left (2\,a\,d-3\,b\,c\right )}{6\,a^2}-\frac {x^6\,\left (-f\,a^3+e\,a^2\,b-2\,d\,a\,b^2+3\,c\,b^3\right )}{3\,a^3\,b}}{b\,x^9+a\,x^6} \] Input:

int((c + d*x^3 + e*x^6 + f*x^9)/(x^7*(a + b*x^3)^2),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 453, normalized size of antiderivative = 3.48 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {4 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b d \,x^{6}-6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{2} c \,x^{6}+4 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b d \,x^{6}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{2} c \,x^{6}-12 \,\mathrm {log}\left (x \right ) a^{2} b d \,x^{6}+18 \,\mathrm {log}\left (x \right ) a \,b^{2} c \,x^{6}-2 a^{3} d \,x^{3}-2 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3} e \,x^{6}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3} e \,x^{6}+6 \,\mathrm {log}\left (x \right ) a^{3} e \,x^{6}-2 a^{2} b e \,x^{9}+4 a \,b^{2} d \,x^{9}-a^{3} c +3 a^{2} b c \,x^{3}-6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{3} c \,x^{9}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{3} c \,x^{9}+18 \,\mathrm {log}\left (x \right ) b^{3} c \,x^{9}+2 a^{3} f \,x^{9}-6 b^{3} c \,x^{9}-2 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b e \,x^{9}+4 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{2} d \,x^{9}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b e \,x^{9}+4 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{2} d \,x^{9}+6 \,\mathrm {log}\left (x \right ) a^{2} b e \,x^{9}-12 \,\mathrm {log}\left (x \right ) a \,b^{2} d \,x^{9}}{6 a^{4} x^{6} \left (b \,x^{3}+a \right )} \] Input:

int((f*x^9+e*x^6+d*x^3+c)/x^7/(b*x^3+a)^2,x)
 

Output:

( - 2*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*e*x**6 + 4* 
log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b*d*x**6 - 2*log( 
a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b*e*x**9 - 6*log(a**( 
2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**2*c*x**6 + 4*log(a**(2/3) 
 - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**2*d*x**9 - 6*log(a**(2/3) - b 
**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**3*c*x**9 - 2*log(a**(1/3) + b**(1/3 
)*x)*a**3*e*x**6 + 4*log(a**(1/3) + b**(1/3)*x)*a**2*b*d*x**6 - 2*log(a**( 
1/3) + b**(1/3)*x)*a**2*b*e*x**9 - 6*log(a**(1/3) + b**(1/3)*x)*a*b**2*c*x 
**6 + 4*log(a**(1/3) + b**(1/3)*x)*a*b**2*d*x**9 - 6*log(a**(1/3) + b**(1/ 
3)*x)*b**3*c*x**9 + 6*log(x)*a**3*e*x**6 - 12*log(x)*a**2*b*d*x**6 + 6*log 
(x)*a**2*b*e*x**9 + 18*log(x)*a*b**2*c*x**6 - 12*log(x)*a*b**2*d*x**9 + 18 
*log(x)*b**3*c*x**9 - a**3*c - 2*a**3*d*x**3 + 2*a**3*f*x**9 + 3*a**2*b*c* 
x**3 - 2*a**2*b*e*x**9 + 4*a*b**2*d*x**9 - 6*b**3*c*x**9)/(6*a**4*x**6*(a 
+ b*x**3))