3.2376 \(\int \frac{1}{(a+b \sqrt [3]{x})^3} \, dx\)

Optimal. Leaf size=54 \[ -\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac{6 a}{b^3 \left (a+b \sqrt [3]{x}\right )}+\frac{3 \log \left (a+b \sqrt [3]{x}\right )}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.0299247, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 43} \[ -\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac{6 a}{b^3 \left (a+b \sqrt [3]{x}\right )}+\frac{3 \log \left (a+b \sqrt [3]{x}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^(-3),x]

[Out]

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

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

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])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{(a+b x)^3} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (\frac{a^2}{b^2 (a+b x)^3}-\frac{2 a}{b^2 (a+b x)^2}+\frac{1}{b^2 (a+b x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac{6 a}{b^3 \left (a+b \sqrt [3]{x}\right )}+\frac{3 \log \left (a+b \sqrt [3]{x}\right )}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0364597, size = 45, normalized size = 0.83 \[ \frac{3 \left (\frac{a \left (3 a+4 b \sqrt [3]{x}\right )}{\left (a+b \sqrt [3]{x}\right )^2}+2 \log \left (a+b \sqrt [3]{x}\right )\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^(-3),x]

[Out]

(3*((a*(3*a + 4*b*x^(1/3)))/(a + b*x^(1/3))^2 + 2*Log[a + b*x^(1/3)]))/(2*b^3)

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Maple [B]  time = 0.066, size = 237, normalized size = 4.4 \begin{align*} -{\frac{9\,{a}^{6}}{2\, \left ({b}^{3}x+{a}^{3} \right ) ^{2}{b}^{3}}}+9\,{\frac{{a}^{3}}{{b}^{3} \left ({b}^{3}x+{a}^{3} \right ) }}+{\frac{\ln \left ({b}^{3}x+{a}^{3} \right ) }{{b}^{3}}}-{\frac{13\,{a}^{2}}{2\,b}{x}^{{\frac{2}{3}}} \left ({b}^{2}{x}^{{\frac{2}{3}}}-ab\sqrt [3]{x}+{a}^{2} \right ) ^{-2}}+5\,{\frac{{a}^{3}\sqrt [3]{x}}{{b}^{2} \left ({b}^{2}{x}^{2/3}-ab\sqrt [3]{x}+{a}^{2} \right ) ^{2}}}-3\,{\frac{{a}^{4}}{{b}^{3} \left ({b}^{2}{x}^{2/3}-ab\sqrt [3]{x}+{a}^{2} \right ) ^{2}}}-{\frac{1}{{b}^{3}}\ln \left ({b}^{2}{x}^{{\frac{2}{3}}}-ab\sqrt [3]{x}+{a}^{2} \right ) }+2\,{\frac{\ln \left ( a+b\sqrt [3]{x} \right ) }{{b}^{3}}}-{\frac{{a}^{2}}{{b}^{3}} \left ( a+b\sqrt [3]{x} \right ) ^{-2}}+2\,{\frac{ax}{ \left ({b}^{2}{x}^{2/3}-ab\sqrt [3]{x}+{a}^{2} \right ) ^{2}}}+4\,{\frac{a}{{b}^{3} \left ( a+b\sqrt [3]{x} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/2*a^6/(b^3*x+a^3)^2/b^3+9/b^3*a^3/(b^3*x+a^3)+ln(b^3*x+a^3)/b^3-13/2*a^2/b/(b^2*x^(2/3)-a*b*x^(1/3)+a^2)^2*
x^(2/3)+5*a^3/b^2/(b^2*x^(2/3)-a*b*x^(1/3)+a^2)^2*x^(1/3)-3*a^4/b^3/(b^2*x^(2/3)-a*b*x^(1/3)+a^2)^2-1/b^3*ln(b
^2*x^(2/3)-a*b*x^(1/3)+a^2)+2*ln(a+b*x^(1/3))/b^3-a^2/b^3/(a+b*x^(1/3))^2+2*a/(b^2*x^(2/3)-a*b*x^(1/3)+a^2)^2*
x+4*a/b^3/(a+b*x^(1/3))

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Maxima [A]  time = 0.977779, size = 62, normalized size = 1.15 \begin{align*} \frac{3 \, \log \left (b x^{\frac{1}{3}} + a\right )}{b^{3}} + \frac{6 \, a}{{\left (b x^{\frac{1}{3}} + a\right )} b^{3}} - \frac{3 \, a^{2}}{2 \,{\left (b x^{\frac{1}{3}} + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.44315, size = 243, normalized size = 4.5 \begin{align*} \frac{3 \,{\left (6 \, a^{3} b^{3} x + 3 \, a^{6} + 2 \,{\left (b^{6} x^{2} + 2 \, a^{3} b^{3} x + a^{6}\right )} \log \left (b x^{\frac{1}{3}} + a\right ) +{\left (4 \, a b^{5} x + a^{4} b^{2}\right )} x^{\frac{2}{3}} -{\left (5 \, a^{2} b^{4} x + 2 \, a^{5} b\right )} x^{\frac{1}{3}}\right )}}{2 \,{\left (b^{9} x^{2} + 2 \, a^{3} b^{6} x + a^{6} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(6*a^3*b^3*x + 3*a^6 + 2*(b^6*x^2 + 2*a^3*b^3*x + a^6)*log(b*x^(1/3) + a) + (4*a*b^5*x + a^4*b^2)*x^(2/3)
- (5*a^2*b^4*x + 2*a^5*b)*x^(1/3))/(b^9*x^2 + 2*a^3*b^6*x + a^6*b^3)

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Sympy [A]  time = 0.819298, size = 228, normalized size = 4.22 \begin{align*} \begin{cases} \frac{6 a^{2} \log{\left (\frac{a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac{2}{3}}} + \frac{9 a^{2}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac{2}{3}}} + \frac{12 a b \sqrt [3]{x} \log{\left (\frac{a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac{2}{3}}} + \frac{12 a b \sqrt [3]{x}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac{2}{3}}} + \frac{6 b^{2} x^{\frac{2}{3}} \log{\left (\frac{a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac{2}{3}}} & \text{for}\: b \neq 0 \\\frac{x}{a^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((6*a**2*log(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3)) + 9*a**2/(2*a**2*b**
3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3)) + 12*a*b*x**(1/3)*log(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b**4*x**(1/3
) + 2*b**5*x**(2/3)) + 12*a*b*x**(1/3)/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3)) + 6*b**2*x**(2/3)*l
og(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3)), Ne(b, 0)), (x/a**3, True))

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Giac [A]  time = 1.13608, size = 59, normalized size = 1.09 \begin{align*} \frac{3 \, \log \left ({\left | b x^{\frac{1}{3}} + a \right |}\right )}{b^{3}} + \frac{3 \,{\left (4 \, a x^{\frac{1}{3}} + \frac{3 \, a^{2}}{b}\right )}}{2 \,{\left (b x^{\frac{1}{3}} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(abs(b*x^(1/3) + a))/b^3 + 3/2*(4*a*x^(1/3) + 3*a^2/b)/((b*x^(1/3) + a)^2*b^2)