3.2.60 \(\int \frac {\sqrt [3]{(a+b x)^3}}{x^2} \, dx\) [160]

Optimal. Leaf size=138 \[ \left (\frac {b}{6 a^3}-\frac {1}{2 a x^2}\right ) (a+b x)^{5/3}-\frac {b^2 \sqrt [3]{(a+b x)^2}}{6 a^2}-\frac {b^2 \left (\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a+b x}}{2 \sqrt [3]{a}+\sqrt [3]{a+b x}}\right )+\frac {3}{2} \log \left (\frac {-\sqrt [3]{a}+\sqrt [3]{a+b x}}{\sqrt [3]{x}}\right )\right )}{9 a \sqrt [3]{a^2}} \]

[Out]

(-1/2/a/x^2+1/6*b/a^3)*(b*x+a)^(5/3)-1/6*b^2/a^2*((b*x+a)^2)^(1/3)-1/9*b^2/a/(a^2)^(1/3)*(3/2*ln(((b*x+a)^(1/3
)-a^(1/3))/x^(1/3))+3^(1/2)*arctan(3^(1/2)*(b*x+a)^(1/3)/((b*x+a)^(1/3)+2*a^(1/3))))

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 0.34, 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 = {1973, 45} \begin {gather*} \frac {b \log (x) \sqrt [3]{(a+b x)^3}}{a+b x}-\frac {a \sqrt [3]{(a+b x)^3}}{x (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3)^(1/3)/x^2,x]

[Out]

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

Rule 45

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 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=\frac {\sqrt [3]{(a+b x)^3} \int \frac {1+\frac {b x}{a}}{x^2} \, dx}{1+\frac {b x}{a}}\\ &=\frac {\sqrt [3]{(a+b x)^3} \int \left (\frac {1}{x^2}+\frac {b}{a x}\right ) \, dx}{1+\frac {b x}{a}}\\ &=-\frac {a \sqrt [3]{(a+b x)^3}}{x (a+b x)}+\frac {b \sqrt [3]{(a+b x)^3} \log (x)}{a+b x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 31, normalized size = 0.22 \begin {gather*} \frac {\sqrt [3]{(a+b x)^3} (-a+b x \log (x))}{x (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3)^(1/3)/x^2,x]

[Out]

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

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Maple [A]
time = 0.12, size = 44, normalized size = 0.32

method	result	size
risch	\(-\frac {\left (\left (b x +a \right )^{3}\right )^{\frac {1}{3}} a}{\left (b x +a \right ) x}+\frac {\left (\left (b x +a \right )^{3}\right )^{\frac {1}{3}} b \ln \left (x \right )}{b x +a}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 11, normalized size = 0.08 \begin {gather*} b \log \left (x\right ) - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(x) - a/x

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Fricas [A]
time = 0.56, size = 13, normalized size = 0.09 \begin {gather*} \frac {b x \log \left (x\right ) - a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*log(x) - a)/x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{\left (a + b x\right )^{3}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 12, normalized size = 0.09 \begin {gather*} b \log \left ({\left | x \right |}\right ) - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(x)) - a/x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left ({\left (a+b\,x\right )}^3\right )}^{1/3}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^3)^(1/3)/x^2,x)

[Out]

int(((a + b*x)^3)^(1/3)/x^2, x)

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