3.2411 \(\int \frac {(a+\frac {b}{\sqrt [3]{x}})^2}{x^4} \, dx\)

Optimal. Leaf size=34 \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{5 x^{10/3}}-\frac {3 b^2}{11 x^{11/3}} \]

[Out]

-3/11*b^2/x^(11/3)-3/5*a*b/x^(10/3)-1/3*a^2/x^3

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {263, 266, 43} \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{5 x^{10/3}}-\frac {3 b^2}{11 x^{11/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^2/x^4,x]

[Out]

(-3*b^2)/(11*x^(11/3)) - (3*a*b)/(5*x^(10/3)) - a^2/(3*x^3)

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])

Rule 263

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

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]]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^2}{x^4} \, dx &=\int \frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{14/3}} \, dx\\ &=3 \operatorname {Subst}\left (\int \frac {(b+a x)^2}{x^{12}} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {b^2}{x^{12}}+\frac {2 a b}{x^{11}}+\frac {a^2}{x^{10}}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 b^2}{11 x^{11/3}}-\frac {3 a b}{5 x^{10/3}}-\frac {a^2}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.00 \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{5 x^{10/3}}-\frac {3 b^2}{11 x^{11/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^2/x^4,x]

[Out]

(-3*b^2)/(11*x^(11/3)) - (3*a*b)/(5*x^(10/3)) - a^2/(3*x^3)

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fricas [A]  time = 0.79, size = 27, normalized size = 0.79 \[ -\frac {55 \, a^{2} x + 99 \, a b x^{\frac {2}{3}} + 45 \, b^{2} x^{\frac {1}{3}}}{165 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^2/x^4,x, algorithm="fricas")

[Out]

-1/165*(55*a^2*x + 99*a*b*x^(2/3) + 45*b^2*x^(1/3))/x^4

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giac [A]  time = 0.20, size = 26, normalized size = 0.76 \[ -\frac {55 \, a^{2} x^{\frac {2}{3}} + 99 \, a b x^{\frac {1}{3}} + 45 \, b^{2}}{165 \, x^{\frac {11}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^2/x^4,x, algorithm="giac")

[Out]

-1/165*(55*a^2*x^(2/3) + 99*a*b*x^(1/3) + 45*b^2)/x^(11/3)

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maple [A]  time = 0.01, size = 25, normalized size = 0.74 \[ -\frac {a^{2}}{3 x^{3}}-\frac {3 a b}{5 x^{\frac {10}{3}}}-\frac {3 b^{2}}{11 x^{\frac {11}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^(1/3))^2/x^4,x)

[Out]

-3/11*b^2/x^(11/3)-3/5*a*b/x^(10/3)-1/3*a^2/x^3

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maxima [B]  time = 0.53, size = 149, normalized size = 4.38 \[ -\frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{11}}{11 \, b^{9}} + \frac {12 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{10} a}{5 \, b^{9}} - \frac {28 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{9} a^{2}}{3 \, b^{9}} + \frac {21 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{8} a^{3}}{b^{9}} - \frac {30 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{7} a^{4}}{b^{9}} + \frac {28 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{6} a^{5}}{b^{9}} - \frac {84 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{5} a^{6}}{5 \, b^{9}} + \frac {6 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{4} a^{7}}{b^{9}} - \frac {{\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} a^{8}}{b^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^(1/3))^2/x^4,x, algorithm="maxima")

[Out]

-3/11*(a + b/x^(1/3))^11/b^9 + 12/5*(a + b/x^(1/3))^10*a/b^9 - 28/3*(a + b/x^(1/3))^9*a^2/b^9 + 21*(a + b/x^(1
/3))^8*a^3/b^9 - 30*(a + b/x^(1/3))^7*a^4/b^9 + 28*(a + b/x^(1/3))^6*a^5/b^9 - 84/5*(a + b/x^(1/3))^5*a^6/b^9
+ 6*(a + b/x^(1/3))^4*a^7/b^9 - (a + b/x^(1/3))^3*a^8/b^9

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mupad [B]  time = 1.13, size = 24, normalized size = 0.71 \[ -\frac {a^2}{3\,x^3}-\frac {3\,b^2}{11\,x^{11/3}}-\frac {3\,a\,b}{5\,x^{10/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^(1/3))^2/x^4,x)

[Out]

- a^2/(3*x^3) - (3*b^2)/(11*x^(11/3)) - (3*a*b)/(5*x^(10/3))

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sympy [A]  time = 2.98, size = 32, normalized size = 0.94 \[ - \frac {a^{2}}{3 x^{3}} - \frac {3 a b}{5 x^{\frac {10}{3}}} - \frac {3 b^{2}}{11 x^{\frac {11}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**(1/3))**2/x**4,x)

[Out]

-a**2/(3*x**3) - 3*a*b/(5*x**(10/3)) - 3*b**2/(11*x**(11/3))

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