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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 103, 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 = {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*}

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Mathematica [A]  time = 0.07, size = 93, normalized size = 0.90 \[ \frac {2 \log \left (a+b x^3\right ) \left (3 a^2 f-2 a b e+b^2 d\right )+\frac {2 \left (a^3 f-a^2 b e+a b^2 d-b^3 c\right )}{a+b x^3}+2 b x^3 (b e-2 a f)+b^2 f x^6}{6 b^4} \]

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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fricas [A]  time = 0.85, size = 143, normalized size = 1.39 \[ \frac {b^{3} f x^{9} + {\left (2 \, b^{3} e - 3 \, a b^{2} f\right )} x^{6} - 2 \, b^{3} c + 2 \, a b^{2} d - 2 \, a^{2} b e + 2 \, a^{3} f + 2 \, {\left (a b^{2} e - 2 \, a^{2} b f\right )} x^{3} + 2 \, {\left (a b^{2} d - 2 \, a^{2} b e + 3 \, a^{3} f + {\left (b^{3} d - 2 \, a b^{2} e + 3 \, a^{2} b f\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \]

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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giac [B]  time = 0.19, size = 206, normalized size = 2.00 \[ -\frac {1}{6} \, f {\left (\frac {{\left (b x^{3} + a\right )}^{2} {\left (\frac {6 \, a}{b x^{3} + a} - 1\right )}}{b^{4}} + \frac {6 \, a^{2} \log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} - \frac {2 \, a^{3}}{{\left (b x^{3} + a\right )} b^{4}}\right )} + \frac {1}{3} \, {\left (\frac {2 \, a \log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b x^{3} + a}{b^{3}} - \frac {a^{2}}{{\left (b x^{3} + a\right )} b^{3}}\right )} e - \frac {d {\left (\frac {\log \left (\frac {{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{3} + a\right )} b}\right )}}{3 \, b} - \frac {c}{3 \, {\left (b x^{3} + a\right )} b} \]

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)

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maple [A]  time = 0.07, size = 142, normalized size = 1.38 \[ \frac {f \,x^{6}}{6 b^{2}}-\frac {2 a f \,x^{3}}{3 b^{3}}+\frac {e \,x^{3}}{3 b^{2}}+\frac {a^{3} f}{3 \left (b \,x^{3}+a \right ) b^{4}}-\frac {a^{2} e}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {a^{2} f \ln \left (b \,x^{3}+a \right )}{b^{4}}+\frac {a d}{3 \left (b \,x^{3}+a \right ) b^{2}}-\frac {2 a e \ln \left (b \,x^{3}+a \right )}{3 b^{3}}-\frac {c}{3 \left (b \,x^{3}+a \right ) b}+\frac {d \ln \left (b \,x^{3}+a \right )}{3 b^{2}} \]

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 = 1.35, size = 98, normalized size = 0.95 \[ -\frac {b^{3} c - a b^{2} d + a^{2} b e - a^{3} f}{3 \, {\left (b^{5} x^{3} + a b^{4}\right )}} + \frac {b f x^{6} + 2 \, {\left (b e - 2 \, a f\right )} x^{3}}{6 \, b^{3}} + \frac {{\left (b^{2} d - 2 \, a b e + 3 \, a^{2} f\right )} \log \left (b x^{3} + a\right )}{3 \, b^{4}} \]

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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mupad [B]  time = 0.09, size = 103, normalized size = 1.00 \[ x^3\,\left (\frac {e}{3\,b^2}-\frac {2\,a\,f}{3\,b^3}\right )+\frac {f\,x^6}{6\,b^2}-\frac {-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3}{3\,b\,\left (b^4\,x^3+a\,b^3\right )}+\frac {\ln \left (b\,x^3+a\right )\,\left (3\,f\,a^2-2\,e\,a\,b+d\,b^2\right )}{3\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.61, size = 100, normalized size = 0.97 \[ x^{3} \left (- \frac {2 a f}{3 b^{3}} + \frac {e}{3 b^{2}}\right ) + \frac {a^{3} f - a^{2} b e + a b^{2} d - b^{3} c}{3 a b^{4} + 3 b^{5} x^{3}} + \frac {f x^{6}}{6 b^{2}} + \frac {\left (3 a^{2} f - 2 a b e + b^{2} d\right ) \log {\left (a + b x^{3} \right )}}{3 b^{4}} \]

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]

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

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