3.282 \(\int \frac{c+d x^3+e x^6+f x^9}{x^4 (a+b x^3)^3} \, dx\)

Optimal. Leaf size=134 \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{6 a^2 b^2 \left (a+b x^3\right )^2}-\frac{a^3 f-a b^2 d+2 b^3 c}{3 a^3 b^2 \left (a+b x^3\right )}+\frac{(3 b c-a d) \log \left (a+b x^3\right )}{3 a^4}-\frac{\log (x) (3 b c-a d)}{a^4}-\frac{c}{3 a^3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.171251, antiderivative size = 134, normalized size of antiderivative = 1., 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} \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{6 a^2 b^2 \left (a+b x^3\right )^2}-\frac{a^3 f-a b^2 d+2 b^3 c}{3 a^3 b^2 \left (a+b x^3\right )}+\frac{(3 b c-a d) \log \left (a+b x^3\right )}{3 a^4}-\frac{\log (x) (3 b c-a d)}{a^4}-\frac{c}{3 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^4*(a + b*x^3)^3),x]

[Out]

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

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]]

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]

Rubi steps

\begin{align*} \int \frac{c+d x^3+e x^6+f x^9}{x^4 \left (a+b x^3\right )^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{c+d x+e x^2+f x^3}{x^2 (a+b x)^3} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{c}{a^3 x^2}+\frac{-3 b c+a d}{a^4 x}+\frac{b^3 c-a b^2 d+a^2 b e-a^3 f}{a^2 b (a+b x)^3}+\frac{2 b^3 c-a b^2 d+a^3 f}{a^3 b (a+b x)^2}-\frac{b (-3 b c+a d)}{a^4 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{c}{3 a^3 x^3}-\frac{b^3 c-a b^2 d+a^2 b e-a^3 f}{6 a^2 b^2 \left (a+b x^3\right )^2}-\frac{2 b^3 c-a b^2 d+a^3 f}{3 a^3 b^2 \left (a+b x^3\right )}-\frac{(3 b c-a d) \log (x)}{a^4}+\frac{(3 b c-a d) \log \left (a+b x^3\right )}{3 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0914598, size = 121, normalized size = 0.9 \[ \frac{\frac{a^2 \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{b^2 \left (a+b x^3\right )^2}-\frac{2 a \left (a^3 f-a b^2 d+2 b^3 c\right )}{b^2 \left (a+b x^3\right )}+2 (3 b c-a d) \log \left (a+b x^3\right )+6 \log (x) (a d-3 b c)-\frac{2 a c}{x^3}}{6 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^4*(a + b*x^3)^3),x]

[Out]

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

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Maple [A]  time = 0.014, size = 163, normalized size = 1.2 \begin{align*}{\frac{af}{6\,{b}^{2} \left ( b{x}^{3}+a \right ) ^{2}}}-{\frac{e}{6\,b \left ( b{x}^{3}+a \right ) ^{2}}}+{\frac{d}{6\,a \left ( b{x}^{3}+a \right ) ^{2}}}-{\frac{bc}{6\,{a}^{2} \left ( b{x}^{3}+a \right ) ^{2}}}-{\frac{d\ln \left ( b{x}^{3}+a \right ) }{3\,{a}^{3}}}+{\frac{bc\ln \left ( b{x}^{3}+a \right ) }{{a}^{4}}}-{\frac{f}{3\,{b}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{d}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{2\,bc}{3\,{a}^{3} \left ( b{x}^{3}+a \right ) }}-{\frac{c}{3\,{a}^{3}{x}^{3}}}+{\frac{d\ln \left ( x \right ) }{{a}^{3}}}-3\,{\frac{bc\ln \left ( x \right ) }{{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^9+e*x^6+d*x^3+c)/x^4/(b*x^3+a)^3,x)

[Out]

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

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Maxima [A]  time = 0.961867, size = 194, normalized size = 1.45 \begin{align*} -\frac{2 \,{\left (3 \, b^{4} c - a b^{3} d + a^{3} b f\right )} x^{6} + 2 \, a^{2} b^{2} c +{\left (9 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}}{6 \,{\left (a^{3} b^{4} x^{9} + 2 \, a^{4} b^{3} x^{6} + a^{5} b^{2} x^{3}\right )}} + \frac{{\left (3 \, b c - a d\right )} \log \left (b x^{3} + a\right )}{3 \, a^{4}} - \frac{{\left (3 \, b c - a d\right )} \log \left (x^{3}\right )}{3 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^4/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.39977, size = 497, normalized size = 3.71 \begin{align*} -\frac{2 \,{\left (3 \, a b^{4} c - a^{2} b^{3} d + a^{4} b f\right )} x^{6} + 2 \, a^{3} b^{2} c +{\left (9 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d + a^{4} b e + a^{5} f\right )} x^{3} - 2 \,{\left ({\left (3 \, b^{5} c - a b^{4} d\right )} x^{9} + 2 \,{\left (3 \, a b^{4} c - a^{2} b^{3} d\right )} x^{6} +{\left (3 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )} \log \left (b x^{3} + a\right ) + 6 \,{\left ({\left (3 \, b^{5} c - a b^{4} d\right )} x^{9} + 2 \,{\left (3 \, a b^{4} c - a^{2} b^{3} d\right )} x^{6} +{\left (3 \, a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}\right )} \log \left (x\right )}{6 \,{\left (a^{4} b^{4} x^{9} + 2 \, a^{5} b^{3} x^{6} + a^{6} b^{2} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^4/(b*x^3+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**9+e*x**6+d*x**3+c)/x**4/(b*x**3+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.09671, size = 234, normalized size = 1.75 \begin{align*} -\frac{{\left (3 \, b c - a d\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac{{\left (3 \, b^{2} c - a b d\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{4} b} + \frac{3 \, b c x^{3} - a d x^{3} - a c}{3 \, a^{4} x^{3}} - \frac{9 \, b^{5} c x^{6} - 3 \, a b^{4} d x^{6} + 22 \, a b^{4} c x^{3} - 8 \, a^{2} b^{3} d x^{3} + 2 \, a^{4} b f x^{3} + 14 \, a^{2} b^{3} c - 6 \, a^{3} b^{2} d + a^{5} f + a^{4} b e}{6 \,{\left (b x^{3} + a\right )}^{2} a^{4} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^4/(b*x^3+a)^3,x, algorithm="giac")

[Out]

-(3*b*c - a*d)*log(abs(x))/a^4 + 1/3*(3*b^2*c - a*b*d)*log(abs(b*x^3 + a))/(a^4*b) + 1/3*(3*b*c*x^3 - a*d*x^3
- a*c)/(a^4*x^3) - 1/6*(9*b^5*c*x^6 - 3*a*b^4*d*x^6 + 22*a*b^4*c*x^3 - 8*a^2*b^3*d*x^3 + 2*a^4*b*f*x^3 + 14*a^
2*b^3*c - 6*a^3*b^2*d + a^5*f + a^4*b*e)/((b*x^3 + a)^2*a^4*b^2)