3.643 \(\int \frac{(a+b x^4)^3}{x^5} \, dx\)

Optimal. Leaf size=40 \[ 3 a^2 b \log (x)-\frac{a^3}{4 x^4}+\frac{3}{4} a b^2 x^4+\frac{b^3 x^8}{8} \]

[Out]

-a^3/(4*x^4) + (3*a*b^2*x^4)/4 + (b^3*x^8)/8 + 3*a^2*b*Log[x]

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Rubi [A]  time = 0.0214073, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ 3 a^2 b \log (x)-\frac{a^3}{4 x^4}+\frac{3}{4} a b^2 x^4+\frac{b^3 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^3/x^5,x]

[Out]

-a^3/(4*x^4) + (3*a*b^2*x^4)/4 + (b^3*x^8)/8 + 3*a^2*b*Log[x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^4\right )^3}{x^5} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{(a+b x)^3}{x^2} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (3 a b^2+\frac{a^3}{x^2}+\frac{3 a^2 b}{x}+b^3 x\right ) \, dx,x,x^4\right )\\ &=-\frac{a^3}{4 x^4}+\frac{3}{4} a b^2 x^4+\frac{b^3 x^8}{8}+3 a^2 b \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0043256, size = 40, normalized size = 1. \[ 3 a^2 b \log (x)-\frac{a^3}{4 x^4}+\frac{3}{4} a b^2 x^4+\frac{b^3 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^3/x^5,x]

[Out]

-a^3/(4*x^4) + (3*a*b^2*x^4)/4 + (b^3*x^8)/8 + 3*a^2*b*Log[x]

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Maple [A]  time = 0.004, size = 35, normalized size = 0.9 \begin{align*} -{\frac{{a}^{3}}{4\,{x}^{4}}}+{\frac{3\,a{b}^{2}{x}^{4}}{4}}+{\frac{{b}^{3}{x}^{8}}{8}}+3\,{a}^{2}b\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^3/x^5,x)

[Out]

-1/4*a^3/x^4+3/4*a*b^2*x^4+1/8*b^3*x^8+3*a^2*b*ln(x)

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Maxima [A]  time = 0.954185, size = 49, normalized size = 1.22 \begin{align*} \frac{1}{8} \, b^{3} x^{8} + \frac{3}{4} \, a b^{2} x^{4} + \frac{3}{4} \, a^{2} b \log \left (x^{4}\right ) - \frac{a^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="maxima")

[Out]

1/8*b^3*x^8 + 3/4*a*b^2*x^4 + 3/4*a^2*b*log(x^4) - 1/4*a^3/x^4

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Fricas [A]  time = 1.4121, size = 86, normalized size = 2.15 \begin{align*} \frac{b^{3} x^{12} + 6 \, a b^{2} x^{8} + 24 \, a^{2} b x^{4} \log \left (x\right ) - 2 \, a^{3}}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="fricas")

[Out]

1/8*(b^3*x^12 + 6*a*b^2*x^8 + 24*a^2*b*x^4*log(x) - 2*a^3)/x^4

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Sympy [A]  time = 0.347013, size = 37, normalized size = 0.92 \begin{align*} - \frac{a^{3}}{4 x^{4}} + 3 a^{2} b \log{\left (x \right )} + \frac{3 a b^{2} x^{4}}{4} + \frac{b^{3} x^{8}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**3/x**5,x)

[Out]

-a**3/(4*x**4) + 3*a**2*b*log(x) + 3*a*b**2*x**4/4 + b**3*x**8/8

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Giac [A]  time = 1.14398, size = 62, normalized size = 1.55 \begin{align*} \frac{1}{8} \, b^{3} x^{8} + \frac{3}{4} \, a b^{2} x^{4} + \frac{3}{4} \, a^{2} b \log \left (x^{4}\right ) - \frac{3 \, a^{2} b x^{4} + a^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="giac")

[Out]

1/8*b^3*x^8 + 3/4*a*b^2*x^4 + 3/4*a^2*b*log(x^4) - 1/4*(3*a^2*b*x^4 + a^3)/x^4