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

Optimal. Leaf size=103 \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 b^4 \left (a+b x^3\right )}+\frac{\log \left (a+b x^3\right ) \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}+\frac{x^3 (b e-2 a f)}{3 b^3}+\frac{f x^6}{6 b^2} \]

[Out]

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

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Rubi [A]  time = 0.14512, antiderivative size = 103, 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 = {1819, 1850} \[ -\frac{a^2 b e+a^3 (-f)-a b^2 d+b^3 c}{3 b^4 \left (a+b x^3\right )}+\frac{\log \left (a+b x^3\right ) \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}+\frac{x^3 (b e-2 a f)}{3 b^3}+\frac{f x^6}{6 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1819

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1)
, Pq, x]*(a + b*x^Simplify[n/(m + 1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && NeQ[m, -1] &&
IGtQ[Simplify[n/(m + 1)], 0] && PolyQ[Pq, x^(m + 1)]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

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

Mathematica [A]  time = 0.0584239, size = 93, normalized size = 0.9 \[ \frac{\frac{2 \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{a+b x^3}+2 \log \left (a+b x^3\right ) \left (3 a^2 f-2 a b e+b^2 d\right )+2 b x^3 (b e-2 a f)+b^2 f x^6}{6 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 142, normalized size = 1.4 \begin{align*}{\frac{f{x}^{6}}{6\,{b}^{2}}}-{\frac{2\,a{x}^{3}f}{3\,{b}^{3}}}+{\frac{{x}^{3}e}{3\,{b}^{2}}}+{\frac{\ln \left ( b{x}^{3}+a \right ){a}^{2}f}{{b}^{4}}}-{\frac{2\,\ln \left ( b{x}^{3}+a \right ) ae}{3\,{b}^{3}}}+{\frac{\ln \left ( b{x}^{3}+a \right ) d}{3\,{b}^{2}}}+{\frac{{a}^{3}f}{3\,{b}^{4} \left ( b{x}^{3}+a \right ) }}-{\frac{{a}^{2}e}{3\,{b}^{3} \left ( b{x}^{3}+a \right ) }}+{\frac{ad}{3\,{b}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{c}{3\,b \left ( b{x}^{3}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.959441, size = 132, normalized size = 1.28 \begin{align*} -\frac{b^{3} c - a b^{2} d + a^{2} b e - a^{3} f}{3 \,{\left (b^{5} x^{3} + a b^{4}\right )}} + \frac{b f x^{6} + 2 \,{\left (b e - 2 \, a f\right )} x^{3}}{6 \, b^{3}} + \frac{{\left (b^{2} d - 2 \, a b e + 3 \, a^{2} f\right )} \log \left (b x^{3} + a\right )}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.13548, size = 305, normalized size = 2.96 \begin{align*} \frac{b^{3} f x^{9} +{\left (2 \, b^{3} e - 3 \, a b^{2} f\right )} x^{6} - 2 \, b^{3} c + 2 \, a b^{2} d - 2 \, a^{2} b e + 2 \, a^{3} f + 2 \,{\left (a b^{2} e - 2 \, a^{2} b f\right )} x^{3} + 2 \,{\left (a b^{2} d - 2 \, a^{2} b e + 3 \, a^{3} f +{\left (b^{3} d - 2 \, a b^{2} e + 3 \, a^{2} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \,{\left (b^{5} x^{3} + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.22015, size = 97, normalized size = 0.94 \begin{align*} \frac{a^{3} f - a^{2} b e + a b^{2} d - b^{3} c}{3 a b^{4} + 3 b^{5} x^{3}} + \frac{f x^{6}}{6 b^{2}} - \frac{x^{3} \left (2 a f - b e\right )}{3 b^{3}} + \frac{\left (3 a^{2} f - 2 a b e + b^{2} d\right ) \log{\left (a + b x^{3} \right )}}{3 b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.05823, size = 278, normalized size = 2.7 \begin{align*} -\frac{1}{6} \, f{\left (\frac{{\left (b x^{3} + a\right )}^{2}{\left (\frac{6 \, a}{b x^{3} + a} - 1\right )}}{b^{4}} + \frac{6 \, a^{2} \log \left (\frac{{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2}{\left | b \right |}}\right )}{b^{4}} - \frac{2 \, a^{3}}{{\left (b x^{3} + a\right )} b^{4}}\right )} + \frac{1}{3} \,{\left (\frac{2 \, a \log \left (\frac{{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2}{\left | b \right |}}\right )}{b^{3}} + \frac{b x^{3} + a}{b^{3}} - \frac{a^{2}}{{\left (b x^{3} + a\right )} b^{3}}\right )} e - \frac{d{\left (\frac{\log \left (\frac{{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2}{\left | b \right |}}\right )}{b} - \frac{a}{{\left (b x^{3} + a\right )} b}\right )}}{3 \, b} - \frac{c}{3 \,{\left (b x^{3} + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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