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

Optimal. Leaf size=88 \[ -\frac {a^2 (A b-a B)}{6 b^4 \left (a+b x^3\right )^2}+\frac {a (2 A b-3 a B)}{3 b^4 \left (a+b x^3\right )}+\frac {(A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}+\frac {B x^3}{3 b^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 77} \[ -\frac {a^2 (A b-a B)}{6 b^4 \left (a+b x^3\right )^2}+\frac {a (2 A b-3 a B)}{3 b^4 \left (a+b x^3\right )}+\frac {(A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}+\frac {B x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

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]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 92, normalized size = 1.05 \[ \frac {2 a A b-3 a^2 B}{3 b^4 \left (a+b x^3\right )}+\frac {a^3 B-a^2 A b}{6 b^4 \left (a+b x^3\right )^2}+\frac {(A b-3 a B) \log \left (a+b x^3\right )}{3 b^4}+\frac {B x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 142, normalized size = 1.61 \[ \frac {2 \, B b^{3} x^{9} + 4 \, B a b^{2} x^{6} - 5 \, B a^{3} + 3 \, A a^{2} b - 4 \, {\left (B a^{2} b - A a b^{2}\right )} x^{3} - 2 \, {\left ({\left (3 \, B a b^{2} - A b^{3}\right )} x^{6} + 3 \, B a^{3} - A a^{2} b + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 93, normalized size = 1.06 \[ \frac {B x^{3}}{3 \, b^{3}} - \frac {{\left (3 \, B a - A b\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{4}} + \frac {9 \, B a b^{2} x^{6} - 3 \, A b^{3} x^{6} + 12 \, B a^{2} b x^{3} - 2 \, A a b^{2} x^{3} + 4 \, B a^{3}}{6 \, {\left (b x^{3} + a\right )}^{2} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*x^3/b^3 - 1/3*(3*B*a - A*b)*log(abs(b*x^3 + a))/b^4 + 1/6*(9*B*a*b^2*x^6 - 3*A*b^3*x^6 + 12*B*a^2*b*x^3
- 2*A*a*b^2*x^3 + 4*B*a^3)/((b*x^3 + a)^2*b^4)

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maple [A]  time = 0.05, size = 110, normalized size = 1.25 \[ \frac {B \,x^{3}}{3 b^{3}}-\frac {A \,a^{2}}{6 \left (b \,x^{3}+a \right )^{2} b^{3}}+\frac {B \,a^{3}}{6 \left (b \,x^{3}+a \right )^{2} b^{4}}+\frac {2 A a}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {A \ln \left (b \,x^{3}+a \right )}{3 b^{3}}-\frac {B \,a^{2}}{\left (b \,x^{3}+a \right ) b^{4}}-\frac {B a \ln \left (b \,x^{3}+a \right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 94, normalized size = 1.07 \[ \frac {B x^{3}}{3 \, b^{3}} - \frac {5 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{3}}{6 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} - \frac {{\left (3 \, B a - A b\right )} \log \left (b x^{3} + a\right )}{3 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.40, size = 94, normalized size = 1.07 \[ \frac {B\,x^3}{3\,b^3}-\frac {x^3\,\left (B\,a^2-\frac {2\,A\,a\,b}{3}\right )+\frac {5\,B\,a^3-3\,A\,a^2\,b}{6\,b}}{a^2\,b^3+2\,a\,b^4\,x^3+b^5\,x^6}+\frac {\ln \left (b\,x^3+a\right )\,\left (A\,b-3\,B\,a\right )}{3\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.94, size = 94, normalized size = 1.07 \[ \frac {B x^{3}}{3 b^{3}} + \frac {3 A a^{2} b - 5 B a^{3} + x^{3} \left (4 A a b^{2} - 6 B a^{2} b\right )}{6 a^{2} b^{4} + 12 a b^{5} x^{3} + 6 b^{6} x^{6}} - \frac {\left (- A b + 3 B a\right ) \log {\left (a + b x^{3} \right )}}{3 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x**3/(3*b**3) + (3*A*a**2*b - 5*B*a**3 + x**3*(4*A*a*b**2 - 6*B*a**2*b))/(6*a**2*b**4 + 12*a*b**5*x**3 + 6*b
**6*x**6) - (-A*b + 3*B*a)*log(a + b*x**3)/(3*b**4)

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