3.256 \(\int \frac{c+d x^3+e x^6+f x^9}{x^4 (a+b x^3)^2} \, dx\)

Optimal. Leaf size=109 \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 a^2 b^2 \left (a+b x^3\right )}+\frac{\log \left (a+b x^3\right ) \left (a^3 f-a b^2 d+2 b^3 c\right )}{3 a^3 b^2}-\frac{\log (x) (2 b c-a d)}{a^3}-\frac{c}{3 a^2 x^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14216, antiderivative size = 109, 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} \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 a^2 b^2 \left (a+b x^3\right )}+\frac{\log \left (a+b x^3\right ) \left (a^3 f-a b^2 d+2 b^3 c\right )}{3 a^3 b^2}-\frac{\log (x) (2 b c-a d)}{a^3}-\frac{c}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^4*(a + b*x^3)^2),x]

[Out]

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

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

Mathematica [A]  time = 0.0767274, size = 97, normalized size = 0.89 \[ \frac{\frac{a \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{b^2 \left (a+b x^3\right )}+\frac{\log \left (a+b x^3\right ) \left (a^3 f-a b^2 d+2 b^3 c\right )}{b^2}+3 \log (x) (a d-2 b c)-\frac{a c}{x^3}}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^4*(a + b*x^3)^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.014, size = 132, normalized size = 1.2 \begin{align*}{\frac{f\ln \left ( b{x}^{3}+a \right ) }{3\,{b}^{2}}}-{\frac{d\ln \left ( b{x}^{3}+a \right ) }{3\,{a}^{2}}}+{\frac{2\,bc\ln \left ( b{x}^{3}+a \right ) }{3\,{a}^{3}}}+{\frac{af}{3\,{b}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{e}{3\,b \left ( b{x}^{3}+a \right ) }}+{\frac{d}{3\,a \left ( b{x}^{3}+a \right ) }}-{\frac{bc}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{c}{3\,{x}^{3}{a}^{2}}}+{\frac{d\ln \left ( x \right ) }{{a}^{2}}}-2\,{\frac{bc\ln \left ( x \right ) }{{a}^{3}}} \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)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.954806, size = 157, normalized size = 1.44 \begin{align*} -\frac{a b^{2} c +{\left (2 \, b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{3}}{3 \,{\left (a^{2} b^{3} x^{6} + a^{3} b^{2} x^{3}\right )}} - \frac{{\left (2 \, b c - a d\right )} \log \left (x^{3}\right )}{3 \, a^{3}} + \frac{{\left (2 \, b^{3} c - a b^{2} d + a^{3} f\right )} \log \left (b x^{3} + a\right )}{3 \, a^{3} 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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.49415, size = 340, normalized size = 3.12 \begin{align*} -\frac{a^{2} b^{2} c +{\left (2 \, a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} x^{3} -{\left ({\left (2 \, b^{4} c - a b^{3} d + a^{3} b f\right )} x^{6} +{\left (2 \, a b^{3} c - a^{2} b^{2} d + a^{4} f\right )} x^{3}\right )} \log \left (b x^{3} + a\right ) + 3 \,{\left ({\left (2 \, b^{4} c - a b^{3} d\right )} x^{6} +{\left (2 \, a b^{3} c - a^{2} b^{2} d\right )} x^{3}\right )} \log \left (x\right )}{3 \,{\left (a^{3} b^{3} x^{6} + a^{4} 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)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.05914, size = 177, normalized size = 1.62 \begin{align*} -\frac{{\left (2 \, b c - a d\right )} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac{{\left (2 \, b^{3} c - a b^{2} d + a^{3} f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3} b^{2}} - \frac{a^{2} b f x^{6} + 4 \, b^{3} c x^{3} - 2 \, a b^{2} d x^{3} - a^{3} f x^{3} + 2 \, a^{2} b x^{3} e + 2 \, a b^{2} c}{6 \,{\left (b x^{6} + a x^{3}\right )} a^{2} 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)^2,x, algorithm="giac")

[Out]

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