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

Optimal. Leaf size=50 \[ \frac {(A b-a B) \log \left (a+b x^3\right )}{3 a^2}-\frac {\log (x) (A b-a B)}{a^2}-\frac {A}{3 a x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 50, 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-a B) \log \left (a+b x^3\right )}{3 a^2}-\frac {\log (x) (A b-a B)}{a^2}-\frac {A}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^4*(a + b*x^3)),x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.98 \begin {gather*} \frac {(A b-a B) \log \left (a+b x^3\right )}{3 a^2}+\frac {\log (x) (a B-A b)}{a^2}-\frac {A}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*A/(a*x^3) + ((-(A*b) + a*B)*Log[x])/a^2 + ((A*b - a*B)*Log[a + b*x^3])/(3*a^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 47, normalized size = 0.94 \begin {gather*} -\frac {{\left (B a - A b\right )} x^{3} \log \left (b x^{3} + a\right ) - 3 \, {\left (B a - A b\right )} x^{3} \log \relax (x) + A a}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 69, normalized size = 1.38 \begin {gather*} \frac {{\left (B a - A b\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {{\left (B a b - A b^{2}\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2} b} - \frac {B a x^{3} - A b x^{3} + A a}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 56, normalized size = 1.12 \begin {gather*} -\frac {A b \ln \relax (x )}{a^{2}}+\frac {A b \ln \left (b \,x^{3}+a \right )}{3 a^{2}}+\frac {B \ln \relax (x )}{a}-\frac {B \ln \left (b \,x^{3}+a \right )}{3 a}-\frac {A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 48, normalized size = 0.96 \begin {gather*} -\frac {{\left (B a - A b\right )} \log \left (b x^{3} + a\right )}{3 \, a^{2}} + \frac {{\left (B a - A b\right )} \log \left (x^{3}\right )}{3 \, a^{2}} - \frac {A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.40, size = 46, normalized size = 0.92 \begin {gather*} \frac {\ln \left (b\,x^3+a\right )\,\left (A\,b-B\,a\right )}{3\,a^2}-\frac {A}{3\,a\,x^3}-\frac {\ln \relax (x)\,\left (A\,b-B\,a\right )}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^4*(a + b*x^3)),x)

[Out]

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

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sympy [A]  time = 2.36, size = 41, normalized size = 0.82 \begin {gather*} - \frac {A}{3 a x^{3}} + \frac {\left (- A b + B a\right ) \log {\relax (x )}}{a^{2}} - \frac {\left (- A b + B a\right ) \log {\left (\frac {a}{b} + x^{3} \right )}}{3 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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