3.74 \(\int \frac{(a+b x)^3}{x^6} \, dx\)

Optimal. Leaf size=36 \[ \frac{b (a+b x)^4}{20 a^2 x^4}-\frac{(a+b x)^4}{5 a x^5} \]

[Out]

-(a + b*x)^4/(5*a*x^5) + (b*(a + b*x)^4)/(20*a^2*x^4)

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Rubi [A]  time = 0.005332, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {45, 37} \[ \frac{b (a+b x)^4}{20 a^2 x^4}-\frac{(a+b x)^4}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/x^6,x]

[Out]

-(a + b*x)^4/(5*a*x^5) + (b*(a + b*x)^4)/(20*a^2*x^4)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+b x)^3}{x^6} \, dx &=-\frac{(a+b x)^4}{5 a x^5}-\frac{b \int \frac{(a+b x)^3}{x^5} \, dx}{5 a}\\ &=-\frac{(a+b x)^4}{5 a x^5}+\frac{b (a+b x)^4}{20 a^2 x^4}\\ \end{align*}

Mathematica [A]  time = 0.006572, size = 41, normalized size = 1.14 \[ -\frac{3 a^2 b}{4 x^4}-\frac{a^3}{5 x^5}-\frac{a b^2}{x^3}-\frac{b^3}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/x^6,x]

[Out]

-a^3/(5*x^5) - (3*a^2*b)/(4*x^4) - (a*b^2)/x^3 - b^3/(2*x^2)

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Maple [A]  time = 0.004, size = 36, normalized size = 1. \begin{align*} -{\frac{{b}^{2}a}{{x}^{3}}}-{\frac{{a}^{3}}{5\,{x}^{5}}}-{\frac{3\,{a}^{2}b}{4\,{x}^{4}}}-{\frac{{b}^{3}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/x^6,x)

[Out]

-b^2*a/x^3-1/5*a^3/x^5-3/4*a^2*b/x^4-1/2*b^3/x^2

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Maxima [A]  time = 1.07295, size = 47, normalized size = 1.31 \begin{align*} -\frac{10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^6,x, algorithm="maxima")

[Out]

-1/20*(10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)/x^5

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Fricas [A]  time = 1.5316, size = 81, normalized size = 2.25 \begin{align*} -\frac{10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^6,x, algorithm="fricas")

[Out]

-1/20*(10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)/x^5

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Sympy [A]  time = 0.438314, size = 37, normalized size = 1.03 \begin{align*} - \frac{4 a^{3} + 15 a^{2} b x + 20 a b^{2} x^{2} + 10 b^{3} x^{3}}{20 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/x**6,x)

[Out]

-(4*a**3 + 15*a**2*b*x + 20*a*b**2*x**2 + 10*b**3*x**3)/(20*x**5)

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Giac [A]  time = 1.13052, size = 47, normalized size = 1.31 \begin{align*} -\frac{10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^6,x, algorithm="giac")

[Out]

-1/20*(10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)/x^5