3.72 \(\int \frac{x^8 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=82 \[ -\frac{a^2 (A b-a B)}{3 b^4 \left (a+b x^3\right )}+\frac{x^3 (A b-2 a B)}{3 b^3}-\frac{a (2 A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}+\frac{B x^6}{6 b^2} \]

[Out]

((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^6)/(6*b^2) - (a^2*(A*b - a*B))/(3*b^4*(a + b*x^3)) - (a*(2*A*b - 3*a*B)*Log
[a + b*x^3])/(3*b^4)

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Rubi [A]  time = 0.0905451, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 77} \[ -\frac{a^2 (A b-a B)}{3 b^4 \left (a+b x^3\right )}+\frac{x^3 (A b-2 a B)}{3 b^3}-\frac{a (2 A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}+\frac{B x^6}{6 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^6)/(6*b^2) - (a^2*(A*b - a*B))/(3*b^4*(a + b*x^3)) - (a*(2*A*b - 3*a*B)*Log
[a + b*x^3])/(3*b^4)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(a+b x)^2} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{A b-2 a B}{b^3}+\frac{B x}{b^2}-\frac{a^2 (-A b+a B)}{b^3 (a+b x)^2}+\frac{a (-2 A b+3 a B)}{b^3 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=\frac{(A b-2 a B) x^3}{3 b^3}+\frac{B x^6}{6 b^2}-\frac{a^2 (A b-a B)}{3 b^4 \left (a+b x^3\right )}-\frac{a (2 A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0654002, size = 72, normalized size = 0.88 \[ \frac{\frac{2 a^2 (a B-A b)}{a+b x^3}+2 b x^3 (A b-2 a B)+2 a (3 a B-2 A b) \log \left (a+b x^3\right )+b^2 B x^6}{6 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

(2*b*(A*b - 2*a*B)*x^3 + b^2*B*x^6 + (2*a^2*(-(A*b) + a*B))/(a + b*x^3) + 2*a*(-2*A*b + 3*a*B)*Log[a + b*x^3])
/(6*b^4)

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Maple [A]  time = 0.016, size = 97, normalized size = 1.2 \begin{align*}{\frac{B{x}^{6}}{6\,{b}^{2}}}+{\frac{A{x}^{3}}{3\,{b}^{2}}}-{\frac{2\,B{x}^{3}a}{3\,{b}^{3}}}-{\frac{{a}^{2}A}{3\,{b}^{3} \left ( b{x}^{3}+a \right ) }}+{\frac{B{a}^{3}}{3\,{b}^{4} \left ( b{x}^{3}+a \right ) }}-{\frac{2\,a\ln \left ( b{x}^{3}+a \right ) A}{3\,{b}^{3}}}+{\frac{{a}^{2}\ln \left ( b{x}^{3}+a \right ) B}{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

1/6*B*x^6/b^2+1/3/b^2*A*x^3-2/3/b^3*B*x^3*a-1/3*a^2/b^3/(b*x^3+a)*A+1/3*a^3/b^4/(b*x^3+a)*B-2/3*a/b^3*ln(b*x^3
+a)*A+a^2/b^4*ln(b*x^3+a)*B

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Maxima [A]  time = 0.961575, size = 111, normalized size = 1.35 \begin{align*} \frac{B a^{3} - A a^{2} b}{3 \,{\left (b^{5} x^{3} + a b^{4}\right )}} + \frac{B b x^{6} - 2 \,{\left (2 \, B a - A b\right )} x^{3}}{6 \, b^{3}} + \frac{{\left (3 \, B a^{2} - 2 \, A a b\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^8*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/3*(B*a^3 - A*a^2*b)/(b^5*x^3 + a*b^4) + 1/6*(B*b*x^6 - 2*(2*B*a - A*b)*x^3)/b^3 + 1/3*(3*B*a^2 - 2*A*a*b)*lo
g(b*x^3 + a)/b^4

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Fricas [A]  time = 1.72699, size = 251, normalized size = 3.06 \begin{align*} \frac{B b^{3} x^{9} -{\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 2 \, B a^{3} - 2 \, A a^{2} b - 2 \,{\left (2 \, B a^{2} b - A a b^{2}\right )} x^{3} + 2 \,{\left (3 \, B a^{3} - 2 \, A a^{2} b +{\left (3 \, B a^{2} b - 2 \, A a b^{2}\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^8*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

1/6*(B*b^3*x^9 - (3*B*a*b^2 - 2*A*b^3)*x^6 + 2*B*a^3 - 2*A*a^2*b - 2*(2*B*a^2*b - A*a*b^2)*x^3 + 2*(3*B*a^3 -
2*A*a^2*b + (3*B*a^2*b - 2*A*a*b^2)*x^3)*log(b*x^3 + a))/(b^5*x^3 + a*b^4)

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Sympy [A]  time = 1.5349, size = 78, normalized size = 0.95 \begin{align*} \frac{B x^{6}}{6 b^{2}} + \frac{a \left (- 2 A b + 3 B a\right ) \log{\left (a + b x^{3} \right )}}{3 b^{4}} + \frac{- A a^{2} b + B a^{3}}{3 a b^{4} + 3 b^{5} x^{3}} - \frac{x^{3} \left (- A b + 2 B a\right )}{3 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

B*x**6/(6*b**2) + a*(-2*A*b + 3*B*a)*log(a + b*x**3)/(3*b**4) + (-A*a**2*b + B*a**3)/(3*a*b**4 + 3*b**5*x**3)
- x**3*(-A*b + 2*B*a)/(3*b**3)

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Giac [A]  time = 1.11033, size = 143, normalized size = 1.74 \begin{align*} \frac{{\left (3 \, B a^{2} - 2 \, A a b\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{4}} + \frac{B b^{2} x^{6} - 4 \, B a b x^{3} + 2 \, A b^{2} x^{3}}{6 \, b^{4}} - \frac{3 \, B a^{2} b x^{3} - 2 \, A a b^{2} x^{3} + 2 \, B a^{3} - A a^{2} b}{3 \,{\left (b x^{3} + a\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

1/3*(3*B*a^2 - 2*A*a*b)*log(abs(b*x^3 + a))/b^4 + 1/6*(B*b^2*x^6 - 4*B*a*b*x^3 + 2*A*b^2*x^3)/b^4 - 1/3*(3*B*a
^2*b*x^3 - 2*A*a*b^2*x^3 + 2*B*a^3 - A*a^2*b)/((b*x^3 + a)*b^4)