3.84 \(\int \frac{A+B x^3}{x^4 (a+b x^3)^2} \, dx\)

Optimal. Leaf size=76 \[ -\frac{A b-a B}{3 a^2 \left (a+b x^3\right )}+\frac{(2 A b-a B) \log \left (a+b x^3\right )}{3 a^3}-\frac{\log (x) (2 A b-a B)}{a^3}-\frac{A}{3 a^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0758433, antiderivative size = 76, 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} \[ -\frac{A b-a B}{3 a^2 \left (a+b x^3\right )}+\frac{(2 A b-a B) \log \left (a+b x^3\right )}{3 a^3}-\frac{\log (x) (2 A b-a B)}{a^3}-\frac{A}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^4*(a + b*x^3)^2),x]

[Out]

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

Mathematica [A]  time = 0.0469824, size = 64, normalized size = 0.84 \[ \frac{\frac{a (a B-A b)}{a+b x^3}+(2 A b-a B) \log \left (a+b x^3\right )+3 \log (x) (a B-2 A b)-\frac{a A}{x^3}}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^4*(a + b*x^3)^2),x]

[Out]

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

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Maple [A]  time = 0.014, size = 87, normalized size = 1.1 \begin{align*} -{\frac{Ab}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{B}{3\,a \left ( b{x}^{3}+a \right ) }}+{\frac{2\,b\ln \left ( b{x}^{3}+a \right ) A}{3\,{a}^{3}}}-{\frac{\ln \left ( b{x}^{3}+a \right ) B}{3\,{a}^{2}}}-{\frac{A}{3\,{a}^{2}{x}^{3}}}-2\,{\frac{A\ln \left ( x \right ) b}{{a}^{3}}}+{\frac{\ln \left ( x \right ) B}{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^4/(b*x^3+a)^2,x)

[Out]

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

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Maxima [A]  time = 0.948898, size = 103, normalized size = 1.36 \begin{align*} \frac{{\left (B a - 2 \, A b\right )} x^{3} - A a}{3 \,{\left (a^{2} b x^{6} + a^{3} x^{3}\right )}} - \frac{{\left (B a - 2 \, A b\right )} \log \left (b x^{3} + a\right )}{3 \, a^{3}} + \frac{{\left (B a - 2 \, A b\right )} \log \left (x^{3}\right )}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.6387, size = 247, normalized size = 3.25 \begin{align*} \frac{{\left (B a^{2} - 2 \, A a b\right )} x^{3} - A a^{2} -{\left ({\left (B a b - 2 \, A b^{2}\right )} x^{6} +{\left (B a^{2} - 2 \, A a b\right )} x^{3}\right )} \log \left (b x^{3} + a\right ) + 3 \,{\left ({\left (B a b - 2 \, A b^{2}\right )} x^{6} +{\left (B a^{2} - 2 \, A a b\right )} x^{3}\right )} \log \left (x\right )}{3 \,{\left (a^{3} b x^{6} + a^{4} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.56255, size = 70, normalized size = 0.92 \begin{align*} \frac{- A a + x^{3} \left (- 2 A b + B a\right )}{3 a^{3} x^{3} + 3 a^{2} b x^{6}} + \frac{\left (- 2 A b + B a\right ) \log{\left (x \right )}}{a^{3}} - \frac{\left (- 2 A b + B a\right ) \log{\left (\frac{a}{b} + x^{3} \right )}}{3 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**4/(b*x**3+a)**2,x)

[Out]

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

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Giac [A]  time = 1.10326, size = 108, normalized size = 1.42 \begin{align*} \frac{{\left (B a - 2 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac{B a x^{3} - 2 \, A b x^{3} - A a}{3 \,{\left (b x^{6} + a x^{3}\right )} a^{2}} - \frac{{\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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