3.2302 \(\int \frac {(a+b \sqrt [3]{x})^2}{x^4} \, dx\)

Optimal. Leaf size=34 \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{4 x^{8/3}}-\frac {3 b^2}{7 x^{7/3}} \]

[Out]

-1/3*a^2/x^3-3/4*a*b/x^(8/3)-3/7*b^2/x^(7/3)

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{4 x^{8/3}}-\frac {3 b^2}{7 x^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^2/x^4,x]

[Out]

-a^2/(3*x^3) - (3*a*b)/(4*x^(8/3)) - (3*b^2)/(7*x^(7/3))

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 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+b \sqrt [3]{x}\right )^2}{x^4} \, dx &=3 \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^{10}} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {a^2}{x^{10}}+\frac {2 a b}{x^9}+\frac {b^2}{x^8}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {a^2}{3 x^3}-\frac {3 a b}{4 x^{8/3}}-\frac {3 b^2}{7 x^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ -\frac {a^2}{3 x^3}-\frac {3 a b}{4 x^{8/3}}-\frac {3 b^2}{7 x^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^2/x^4,x]

[Out]

-1/3*a^2/x^3 - (3*a*b)/(4*x^(8/3)) - (3*b^2)/(7*x^(7/3))

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fricas [A]  time = 0.65, size = 26, normalized size = 0.76 \[ -\frac {36 \, b^{2} x^{\frac {2}{3}} + 63 \, a b x^{\frac {1}{3}} + 28 \, a^{2}}{84 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/84*(36*b^2*x^(2/3) + 63*a*b*x^(1/3) + 28*a^2)/x^3

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giac [A]  time = 0.15, size = 26, normalized size = 0.76 \[ -\frac {36 \, b^{2} x^{\frac {2}{3}} + 63 \, a b x^{\frac {1}{3}} + 28 \, a^{2}}{84 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/84*(36*b^2*x^(2/3) + 63*a*b*x^(1/3) + 28*a^2)/x^3

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maple [A]  time = 0.01, size = 25, normalized size = 0.74 \[ -\frac {3 b^{2}}{7 x^{\frac {7}{3}}}-\frac {3 a b}{4 x^{\frac {8}{3}}}-\frac {a^{2}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2/x^3-3/4*a*b/x^(8/3)-3/7*b^2/x^(7/3)

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maxima [A]  time = 0.92, size = 26, normalized size = 0.76 \[ -\frac {36 \, b^{2} x^{\frac {2}{3}} + 63 \, a b x^{\frac {1}{3}} + 28 \, a^{2}}{84 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/84*(36*b^2*x^(2/3) + 63*a*b*x^(1/3) + 28*a^2)/x^3

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mupad [B]  time = 0.03, size = 26, normalized size = 0.76 \[ -\frac {28\,a^2+36\,b^2\,x^{2/3}+63\,a\,b\,x^{1/3}}{84\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/3))^2/x^4,x)

[Out]

-(28*a^2 + 36*b^2*x^(2/3) + 63*a*b*x^(1/3))/(84*x^3)

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sympy [A]  time = 1.88, size = 32, normalized size = 0.94 \[ - \frac {a^{2}}{3 x^{3}} - \frac {3 a b}{4 x^{\frac {8}{3}}} - \frac {3 b^{2}}{7 x^{\frac {7}{3}}} \]

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/(4*x**(8/3)) - 3*b**2/(7*x**(7/3))

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