3.75 \(\int \frac{x^5 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=60 \[ \frac{a (A b-a B)}{3 b^3 \left (a+b x^3\right )}+\frac{(A b-2 a B) \log \left (a+b x^3\right )}{3 b^3}+\frac{B x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0581286, antiderivative size = 60, 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} \[ \frac{a (A b-a B)}{3 b^3 \left (a+b x^3\right )}+\frac{(A b-2 a B) \log \left (a+b x^3\right )}{3 b^3}+\frac{B x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0350622, size = 50, normalized size = 0.83 \[ \frac{\frac{a (A b-a B)}{a+b x^3}+(A b-2 a B) \log \left (a+b x^3\right )+b B x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

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Maple [A]  time = 0.01, size = 74, normalized size = 1.2 \begin{align*}{\frac{B{x}^{3}}{3\,{b}^{2}}}+{\frac{aA}{3\,{b}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{{a}^{2}B}{3\,{b}^{3} \left ( b{x}^{3}+a \right ) }}+{\frac{\ln \left ( b{x}^{3}+a \right ) A}{3\,{b}^{2}}}-{\frac{2\,\ln \left ( b{x}^{3}+a \right ) Ba}{3\,{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

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

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Maxima [A]  time = 0.970366, size = 81, normalized size = 1.35 \begin{align*} \frac{B x^{3}}{3 \, b^{2}} - \frac{B a^{2} - A a b}{3 \,{\left (b^{4} x^{3} + a b^{3}\right )}} - \frac{{\left (2 \, B a - A b\right )} \log \left (b x^{3} + a\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.66241, size = 165, normalized size = 2.75 \begin{align*} \frac{B b^{2} x^{6} + B a b x^{3} - B a^{2} + A a b -{\left ({\left (2 \, B a b - A b^{2}\right )} x^{3} + 2 \, B a^{2} - A a b\right )} \log \left (b x^{3} + a\right )}{3 \,{\left (b^{4} x^{3} + a b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.20801, size = 56, normalized size = 0.93 \begin{align*} \frac{B x^{3}}{3 b^{2}} - \frac{- A a b + B a^{2}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac{\left (- A b + 2 B a\right ) \log{\left (a + b x^{3} \right )}}{3 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

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

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Giac [A]  time = 1.12053, size = 123, normalized size = 2.05 \begin{align*} \frac{\frac{{\left (b x^{3} + a\right )} B}{b^{2}} + \frac{{\left (2 \, B a - A b\right )} \log \left (\frac{{\left | b x^{3} + a \right |}}{{\left (b x^{3} + a\right )}^{2}{\left | b \right |}}\right )}{b^{2}} - \frac{\frac{B a^{2} b}{b x^{3} + a} - \frac{A a b^{2}}{b x^{3} + a}}{b^{3}}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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