\(\int \frac {c+d x^3+e x^6+f x^9}{x^7 (a+b x^3)} \, dx\) [229]

Optimal result

Integrand size = 30, antiderivative size = 95 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=-\frac {c}{6 a x^6}+\frac {b c-a d}{3 a^2 x^3}+\frac {\left (b^2 c-a b d+a^2 e\right ) \log (x)}{a^3}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a+b x^3\right )}{3 a^3 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 95, 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} \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=\frac {b c-a d}{3 a^2 x^3}+\frac {\log (x) \left (a^2 e-a b d+b^2 c\right )}{a^3}-\frac {\log \left (a+b x^3\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^3 b}-\frac {c}{6 a x^6} \]

[In]

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

[Out]

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

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)} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {c}{a x^3}+\frac {-b c+a d}{a^2 x^2}+\frac {b^2 c-a b d+a^2 e}{a^3 x}+\frac {-b^3 c+a b^2 d-a^2 b e+a^3 f}{a^3 (a+b x)}\right ) \, dx,x,x^3\right ) \\ & = -\frac {c}{6 a x^6}+\frac {b c-a d}{3 a^2 x^3}+\frac {\left (b^2 c-a b d+a^2 e\right ) \log (x)}{a^3}-\frac {\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^3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=\frac {-\frac {a \left (a c-2 b c x^3+2 a d x^3\right )}{x^6}+6 \left (b^2 c-a b d+a^2 e\right ) \log (x)+\left (-2 b^2 c+2 a b d-2 a^2 e+\frac {2 a^3 f}{b}\right ) \log \left (a+b x^3\right )}{6 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {c}{6 a \,x^{6}}-\frac {a d -b c}{3 a^{2} x^{3}}+\frac {\left (a^{2} e -a b d +b^{2} c \right ) \ln \left (x \right )}{a^{3}}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{3} b}\)	\(90\)
norman	\(\frac {-\frac {c}{6 a}-\frac {\left (a d -b c \right ) x^{3}}{3 a^{2}}}{x^{6}}+\frac {\left (a^{2} e -a b d +b^{2} c \right ) \ln \left (x \right )}{a^{3}}+\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \ln \left (b \,x^{3}+a \right )}{3 a^{3} b}\)	\(92\)
risch	\(\frac {-\frac {c}{6 a}-\frac {\left (a d -b c \right ) x^{3}}{3 a^{2}}}{x^{6}}+\frac {e \ln \left (x \right )}{a}-\frac {\ln \left (x \right ) b d}{a^{2}}+\frac {\ln \left (x \right ) b^{2} c}{a^{3}}+\frac {\ln \left (-b \,x^{3}-a \right ) f}{3 b}-\frac {\ln \left (-b \,x^{3}-a \right ) e}{3 a}+\frac {b \ln \left (-b \,x^{3}-a \right ) d}{3 a^{2}}-\frac {b^{2} \ln \left (-b \,x^{3}-a \right ) c}{3 a^{3}}\)	\(127\)
parallelrisch	\(\frac {6 \ln \left (x \right ) x^{6} a^{2} b e -6 \ln \left (x \right ) x^{6} a \,b^{2} d +6 \ln \left (x \right ) x^{6} b^{3} c +2 \ln \left (b \,x^{3}+a \right ) x^{6} a^{3} f -2 \ln \left (b \,x^{3}+a \right ) x^{6} a^{2} b e +2 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{2} d -2 \ln \left (b \,x^{3}+a \right ) x^{6} b^{3} c -2 a^{2} b d \,x^{3}+2 a \,b^{2} c \,x^{3}-a^{2} b c}{6 a^{3} x^{6} b}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=-\frac {2 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{6} \log \left (b x^{3} + a\right ) - 6 \, {\left (b^{3} c - a b^{2} d + a^{2} b e\right )} x^{6} \log \left (x\right ) + a^{2} b c - 2 \, {\left (a b^{2} c - a^{2} b d\right )} x^{3}}{6 \, a^{3} b x^{6}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=\frac {{\left (b^{2} c - a b d + a^{2} e\right )} \log \left (x^{3}\right )}{3 \, a^{3}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{3} b} + \frac {2 \, {\left (b c - a d\right )} x^{3} - a c}{6 \, a^{2} x^{6}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.29 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=\frac {{\left (b^{2} c - a b d + a^{2} e\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3} b} - \frac {3 \, b^{2} c x^{6} - 3 \, a b d x^{6} + 3 \, a^{2} e x^{6} - 2 \, a b c x^{3} + 2 \, a^{2} d x^{3} + a^{2} c}{6 \, a^{3} x^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^7 \left (a+b x^3\right )} \, dx=\frac {\ln \left (x\right )\,\left (e\,a^2-d\,a\,b+c\,b^2\right )}{a^3}-\frac {\frac {c}{6\,a}+\frac {x^3\,\left (a\,d-b\,c\right )}{3\,a^2}}{x^6}-\frac {\ln \left (b\,x^3+a\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^3\,b} \]

[In]

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

[Out]

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