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

Optimal. Leaf size=100 \[ \frac {c \log (x)}{a^2}-\frac {\log \left (a+b x^3\right ) \left (2 a^3 f-a^2 b e+b^3 c\right )}{3 a^2 b^3}+\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{3 a b^3 \left (a+b x^3\right )}+\frac {f x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1821, 1620} \[ \frac {a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 a b^3 \left (a+b x^3\right )}-\frac {\log \left (a+b x^3\right ) \left (-a^2 b e+2 a^3 f+b^3 c\right )}{3 a^2 b^3}+\frac {c \log (x)}{a^2}+\frac {f x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1620

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 1821

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*} \int \frac {c+d x^3+e x^6+f x^9}{x \left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x (a+b x)^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {f}{b^2}+\frac {c}{a^2 x}+\frac {-b^3 c+a b^2 d-a^2 b e+a^3 f}{a b^2 (a+b x)^2}+\frac {-b^3 c+a^2 b e-2 a^3 f}{a^2 b^2 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=\frac {f x^3}{3 b^2}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{3 a b^3 \left (a+b x^3\right )}+\frac {c \log (x)}{a^2}-\frac {\left (b^3 c-a^2 b e+2 a^3 f\right ) \log \left (a+b x^3\right )}{3 a^2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 95, normalized size = 0.95 \[ \frac {\frac {\log \left (a+b x^3\right ) \left (-2 a^3 f+a^2 b e-b^3 c\right )+\frac {a \left (a^3 (-f)+a^2 b \left (e+f x^3\right )+a b^2 \left (f x^6-d\right )+b^3 c\right )}{a+b x^3}}{b^3}+3 c \log (x)}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 145, normalized size = 1.45 \[ \frac {a^{2} b^{2} f x^{6} + a^{3} b f x^{3} + a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f - {\left (a b^{3} c - a^{3} b e + 2 \, a^{4} f + {\left (b^{4} c - a^{2} b^{2} e + 2 \, a^{3} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right ) + 3 \, {\left (b^{4} c x^{3} + a b^{3} c\right )} \log \relax (x)}{3 \, {\left (a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 125, normalized size = 1.25 \[ \frac {f x^{3}}{3 \, b^{2}} + \frac {c \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {{\left (b^{3} c + 2 \, a^{3} f - a^{2} b e\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2} b^{3}} + \frac {b^{4} c x^{3} + 2 \, a^{3} b f x^{3} - a^{2} b^{2} x^{3} e + 2 \, a b^{3} c - a^{2} b^{2} d + a^{4} f}{3 \, {\left (b x^{3} + a\right )} a^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 125, normalized size = 1.25 \[ \frac {f \,x^{3}}{3 b^{2}}-\frac {a^{2} f}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {a e}{3 \left (b \,x^{3}+a \right ) b^{2}}-\frac {2 a f \ln \left (b \,x^{3}+a \right )}{3 b^{3}}+\frac {c}{3 \left (b \,x^{3}+a \right ) a}+\frac {c \ln \relax (x )}{a^{2}}-\frac {c \ln \left (b \,x^{3}+a \right )}{3 a^{2}}-\frac {d}{3 \left (b \,x^{3}+a \right ) b}+\frac {e \ln \left (b \,x^{3}+a \right )}{3 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 100, normalized size = 1.00 \[ \frac {f x^{3}}{3 \, b^{2}} + \frac {b^{3} c - a b^{2} d + a^{2} b e - a^{3} f}{3 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}} + \frac {c \log \left (x^{3}\right )}{3 \, a^{2}} - \frac {{\left (b^{3} c - a^{2} b e + 2 \, a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.03, size = 100, normalized size = 1.00 \[ \frac {f\,x^3}{3\,b^2}+\frac {c\,\ln \relax (x)}{a^2}+\frac {-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3}{3\,a\,b\,\left (b^3\,x^3+a\,b^2\right )}-\frac {\ln \left (b\,x^3+a\right )\,\left (2\,f\,a^3-e\,a^2\,b+c\,b^3\right )}{3\,a^2\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 41.96, size = 95, normalized size = 0.95 \[ \frac {- a^{3} f + a^{2} b e - a b^{2} d + b^{3} c}{3 a^{2} b^{3} + 3 a b^{4} x^{3}} + \frac {f x^{3}}{3 b^{2}} + \frac {c \log {\relax (x )}}{a^{2}} - \frac {\left (2 a^{3} f - a^{2} b e + b^{3} c\right ) \log {\left (\frac {a}{b} + x^{3} \right )}}{3 a^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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