\(\int \frac {c+d x^3+e x^6+f x^9}{x^7 (a+b x^3)^2} \, dx\) [257]

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1835, 1634} \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {2 b c-a d}{3 a^3 x^3}-\frac {c}{6 a^2 x^6}-\frac {\log \left (a+b x^3\right ) \left (a^2 e-2 a b d+3 b^2 c\right )}{3 a^4}+\frac {\log (x) \left (a^2 e-2 a b d+3 b^2 c\right )}{a^4}+\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{3 a^3 b \left (a+b x^3\right )} \]

[In]

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

[Out]

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

Rule 1634

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]) &&
GtQ[Expon[Px, x], 2]

Rule 1835

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[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] && Intege
rQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x^3 (a+b x)^2} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {c}{a^2 x^3}+\frac {-2 b c+a d}{a^3 x^2}+\frac {3 b^2 c-2 a b d+a^2 e}{a^4 x}+\frac {-b^3 c+a b^2 d-a^2 b e+a^3 f}{a^3 (a+b x)^2}-\frac {b \left (3 b^2 c-2 a b d+a^2 e\right )}{a^4 (a+b x)}\right ) \, dx,x,x^3\right ) \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=-\frac {\frac {a^2 c}{x^6}+\frac {2 a (-2 b c+a d)}{x^3}+\frac {2 a \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{b \left (a+b x^3\right )}-6 \left (3 b^2 c-2 a b d+a^2 e\right ) \log (x)+2 \left (3 b^2 c-2 a b d+a^2 e\right ) \log \left (a+b x^3\right )}{6 a^4} \]

[In]

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

[Out]

-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

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.60 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {2 \, {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} x^{6} - a^{3} b c + {\left (3 \, a^{2} b^{2} c - 2 \, a^{3} b d\right )} x^{3} - 2 \, {\left ({\left (3 \, b^{4} c - 2 \, a b^{3} d + a^{2} b^{2} e\right )} x^{9} + {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e\right )} x^{6}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left ({\left (3 \, b^{4} c - 2 \, a b^{3} d + a^{2} b^{2} e\right )} x^{9} + {\left (3 \, a b^{3} c - 2 \, a^{2} b^{2} d + a^{3} b e\right )} x^{6}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{2} x^{9} + a^{5} b x^{6}\right )}} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {2 \, {\left (3 \, b^{3} c - 2 \, a b^{2} d + a^{2} b e - a^{3} f\right )} x^{6} - a^{2} b c + {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} x^{3}}{6 \, {\left (a^{3} b^{2} x^{9} + a^{4} b x^{6}\right )}} - \frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} + \frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left (x^{3}\right )}{3 \, a^{4}} \]

[In]

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

[Out]

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)

none

Time = 0.28 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {{\left (3 \, b^{2} c - 2 \, a b d + a^{2} e\right )} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {{\left (3 \, b^{3} c - 2 \, a b^{2} d + a^{2} b e\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac {3 \, b^{4} c x^{3} - 2 \, a b^{3} d x^{3} + a^{2} b^{2} e x^{3} + 4 \, a b^{3} c - 3 \, a^{2} b^{2} d + 2 \, a^{3} b e - a^{4} f}{3 \, {\left (b x^{3} + a\right )} a^{4} b} - \frac {9 \, b^{2} c x^{6} - 6 \, a b d x^{6} + 3 \, a^{2} e x^{6} - 4 \, a b c x^{3} + 2 \, a^{2} d x^{3} + a^{2} c}{6 \, a^{4} x^{6}} \]

[In]

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

[Out]

(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 = 9.75 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (x\right )\,\left (e\,a^2-2\,d\,a\,b+3\,c\,b^2\right )}{a^4}-\frac {\ln \left (b\,x^3+a\right )\,\left (e\,a^2-2\,d\,a\,b+3\,c\,b^2\right )}{3\,a^4}-\frac {\frac {c}{6\,a}+\frac {x^3\,\left (2\,a\,d-3\,b\,c\right )}{6\,a^2}-\frac {x^6\,\left (-f\,a^3+e\,a^2\,b-2\,d\,a\,b^2+3\,c\,b^3\right )}{3\,a^3\,b}}{b\,x^9+a\,x^6} \]

[In]

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

[Out]

(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)