3.2299 \(\int \frac {(a+b \sqrt [3]{x})^2}{x} \, dx\)

Optimal. Leaf size=28 \[ a^2 \log (x)+6 a b \sqrt [3]{x}+\frac {3}{2} b^2 x^{2/3} \]

[Out]

6*a*b*x^(1/3)+3/2*b^2*x^(2/3)+a^2*ln(x)

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Rubi [A]  time = 0.01, antiderivative size = 28, 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} \[ a^2 \log (x)+6 a b \sqrt [3]{x}+\frac {3}{2} b^2 x^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^2/x,x]

[Out]

6*a*b*x^(1/3) + (3*b^2*x^(2/3))/2 + a^2*Log[x]

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} \, dx &=3 \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (2 a b+\frac {a^2}{x}+b^2 x\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=6 a b \sqrt [3]{x}+\frac {3}{2} b^2 x^{2/3}+a^2 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ a^2 \log (x)+6 a b \sqrt [3]{x}+\frac {3}{2} b^2 x^{2/3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^2/x,x]

[Out]

6*a*b*x^(1/3) + (3*b^2*x^(2/3))/2 + a^2*Log[x]

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fricas [A]  time = 0.64, size = 25, normalized size = 0.89 \[ 3 \, a^{2} \log \left (x^{\frac {1}{3}}\right ) + \frac {3}{2} \, b^{2} x^{\frac {2}{3}} + 6 \, a b x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^2*log(x^(1/3)) + 3/2*b^2*x^(2/3) + 6*a*b*x^(1/3)

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giac [A]  time = 0.17, size = 23, normalized size = 0.82 \[ a^{2} \log \left ({\left | x \right |}\right ) + \frac {3}{2} \, b^{2} x^{\frac {2}{3}} + 6 \, a b x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(abs(x)) + 3/2*b^2*x^(2/3) + 6*a*b*x^(1/3)

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maple [A]  time = 0.00, size = 23, normalized size = 0.82 \[ a^{2} \ln \relax (x )+\frac {3 b^{2} x^{\frac {2}{3}}}{2}+6 a b \,x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*a*b*x^(1/3)+3/2*b^2*x^(2/3)+a^2*ln(x)

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maxima [A]  time = 0.89, size = 22, normalized size = 0.79 \[ a^{2} \log \relax (x) + \frac {3}{2} \, b^{2} x^{\frac {2}{3}} + 6 \, a b x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) + 3/2*b^2*x^(2/3) + 6*a*b*x^(1/3)

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mupad [B]  time = 1.11, size = 25, normalized size = 0.89 \[ 3\,a^2\,\ln \left (x^{1/3}\right )+\frac {3\,b^2\,x^{2/3}}{2}+6\,a\,b\,x^{1/3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/3))^2/x,x)

[Out]

3*a^2*log(x^(1/3)) + (3*b^2*x^(2/3))/2 + 6*a*b*x^(1/3)

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sympy [A]  time = 0.24, size = 27, normalized size = 0.96 \[ a^{2} \log {\relax (x )} + 6 a b \sqrt [3]{x} + \frac {3 b^{2} x^{\frac {2}{3}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) + 6*a*b*x**(1/3) + 3*b**2*x**(2/3)/2

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