3.1.90 \(\int \frac {x^5 (A+B x^3)}{(a+b x^3)^3} \, dx\)

Optimal. Leaf size=66 \[ -\frac {A b-2 a B}{3 b^3 \left (a+b x^3\right )}+\frac {a (A b-a B)}{6 b^3 \left (a+b x^3\right )^2}+\frac {B \log \left (a+b x^3\right )}{3 b^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 66, 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} \begin {gather*} -\frac {A b-2 a B}{3 b^3 \left (a+b x^3\right )}+\frac {a (A b-a B)}{6 b^3 \left (a+b x^3\right )^2}+\frac {B \log \left (a+b x^3\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

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

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^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x (A+B x)}{(a+b x)^3} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {a (-A b+a B)}{b^2 (a+b x)^3}+\frac {A b-2 a B}{b^2 (a+b x)^2}+\frac {B}{b^2 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=\frac {a (A b-a B)}{6 b^3 \left (a+b x^3\right )^2}-\frac {A b-2 a B}{3 b^3 \left (a+b x^3\right )}+\frac {B \log \left (a+b x^3\right )}{3 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 64, normalized size = 0.97 \begin {gather*} \frac {3 a^2 B-a b \left (A-4 B x^3\right )+2 B \left (a+b x^3\right )^2 \log \left (a+b x^3\right )-2 A b^2 x^3}{6 b^3 \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^5*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

IntegrateAlgebraic[(x^5*(A + B*x^3))/(a + b*x^3)^3, x]

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fricas [A]  time = 0.75, size = 89, normalized size = 1.35 \begin {gather*} \frac {2 \, {\left (2 \, B a b - A b^{2}\right )} x^{3} + 3 \, B a^{2} - A a b + 2 \, {\left (B b^{2} x^{6} + 2 \, B a b x^{3} + B a^{2}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 61, normalized size = 0.92 \begin {gather*} \frac {B \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac {2 \, {\left (2 \, B a - A b\right )} x^{3} + \frac {3 \, B a^{2} - A a b}{b}}{6 \, {\left (b x^{3} + a\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 81, normalized size = 1.23 \begin {gather*} \frac {A a}{6 \left (b \,x^{3}+a \right )^{2} b^{2}}-\frac {B \,a^{2}}{6 \left (b \,x^{3}+a \right )^{2} b^{3}}-\frac {A}{3 \left (b \,x^{3}+a \right ) b^{2}}+\frac {2 B a}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {B \ln \left (b \,x^{3}+a \right )}{3 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^3+A)/(b*x^3+a)^3,x)

[Out]

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

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maxima [A]  time = 0.47, size = 72, normalized size = 1.09 \begin {gather*} \frac {2 \, {\left (2 \, B a b - A b^{2}\right )} x^{3} + 3 \, B a^{2} - A a b}{6 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\right )}} + \frac {B \log \left (b x^{3} + a\right )}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.38, size = 70, normalized size = 1.06 \begin {gather*} \frac {\frac {3\,B\,a^2-A\,a\,b}{6\,b^3}-\frac {x^3\,\left (A\,b-2\,B\,a\right )}{3\,b^2}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\frac {B\,\ln \left (b\,x^3+a\right )}{3\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(A + B*x^3))/(a + b*x^3)^3,x)

[Out]

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

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sympy [A]  time = 3.69, size = 70, normalized size = 1.06 \begin {gather*} \frac {B \log {\left (a + b x^{3} \right )}}{3 b^{3}} + \frac {- A a b + 3 B a^{2} + x^{3} \left (- 2 A b^{2} + 4 B a b\right )}{6 a^{2} b^{3} + 12 a b^{4} x^{3} + 6 b^{5} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**3+A)/(b*x**3+a)**3,x)

[Out]

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

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