3.2370 \(\int \frac {1}{(a+b \sqrt [3]{x})^2 x^2} \, dx\)

Optimal. Leaf size=80 \[ \frac {12 b^3 \log \left (a+b \sqrt [3]{x}\right )}{a^5}-\frac {4 b^3 \log (x)}{a^5}-\frac {3 b^3}{a^4 \left (a+b \sqrt [3]{x}\right )}-\frac {9 b^2}{a^4 \sqrt [3]{x}}+\frac {3 b}{a^3 x^{2/3}}-\frac {1}{a^2 x} \]

[Out]

-3*b^3/a^4/(a+b*x^(1/3))-1/a^2/x+3*b/a^3/x^(2/3)-9*b^2/a^4/x^(1/3)+12*b^3*ln(a+b*x^(1/3))/a^5-4*b^3*ln(x)/a^5

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Rubi [A]  time = 0.05, antiderivative size = 80, 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, 44} \[ -\frac {3 b^3}{a^4 \left (a+b \sqrt [3]{x}\right )}-\frac {9 b^2}{a^4 \sqrt [3]{x}}+\frac {12 b^3 \log \left (a+b \sqrt [3]{x}\right )}{a^5}-\frac {4 b^3 \log (x)}{a^5}+\frac {3 b}{a^3 x^{2/3}}-\frac {1}{a^2 x} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^(1/3))^2*x^2),x]

[Out]

(-3*b^3)/(a^4*(a + b*x^(1/3))) - 1/(a^2*x) + (3*b)/(a^3*x^(2/3)) - (9*b^2)/(a^4*x^(1/3)) + (12*b^3*Log[a + b*x
^(1/3)])/a^5 - (4*b^3*Log[x])/a^5

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 {1}{\left (a+b \sqrt [3]{x}\right )^2 x^2} \, dx &=3 \operatorname {Subst}\left (\int \frac {1}{x^4 (a+b x)^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^4}-\frac {2 b}{a^3 x^3}+\frac {3 b^2}{a^4 x^2}-\frac {4 b^3}{a^5 x}+\frac {b^4}{a^4 (a+b x)^2}+\frac {4 b^4}{a^5 (a+b x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 b^3}{a^4 \left (a+b \sqrt [3]{x}\right )}-\frac {1}{a^2 x}+\frac {3 b}{a^3 x^{2/3}}-\frac {9 b^2}{a^4 \sqrt [3]{x}}+\frac {12 b^3 \log \left (a+b \sqrt [3]{x}\right )}{a^5}-\frac {4 b^3 \log (x)}{a^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 77, normalized size = 0.96 \[ \frac {-\frac {a^4-2 a^3 b \sqrt [3]{x}+6 a^2 b^2 x^{2/3}+12 a b^3 x}{a x+b x^{4/3}}+12 b^3 \log \left (a+b \sqrt [3]{x}\right )-4 b^3 \log (x)}{a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x^(1/3))^2*x^2),x]

[Out]

(-((a^4 - 2*a^3*b*x^(1/3) + 6*a^2*b^2*x^(2/3) + 12*a*b^3*x)/(a*x + b*x^(4/3))) + 12*b^3*Log[a + b*x^(1/3)] - 4
*b^3*Log[x])/a^5

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fricas [A]  time = 0.74, size = 122, normalized size = 1.52 \[ -\frac {4 \, a^{3} b^{3} x + a^{6} - 12 \, {\left (b^{6} x^{2} + a^{3} b^{3} x\right )} \log \left (b x^{\frac {1}{3}} + a\right ) + 12 \, {\left (b^{6} x^{2} + a^{3} b^{3} x\right )} \log \left (x^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, a b^{5} x + 3 \, a^{4} b^{2}\right )} x^{\frac {2}{3}} - 3 \, {\left (2 \, a^{2} b^{4} x + a^{5} b\right )} x^{\frac {1}{3}}}{a^{5} b^{3} x^{2} + a^{8} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a^3*b^3*x + a^6 - 12*(b^6*x^2 + a^3*b^3*x)*log(b*x^(1/3) + a) + 12*(b^6*x^2 + a^3*b^3*x)*log(x^(1/3)) + 3*
(4*a*b^5*x + 3*a^4*b^2)*x^(2/3) - 3*(2*a^2*b^4*x + a^5*b)*x^(1/3))/(a^5*b^3*x^2 + a^8*x)

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giac [A]  time = 0.17, size = 77, normalized size = 0.96 \[ \frac {12 \, b^{3} \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{a^{5}} - \frac {4 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{5}} - \frac {12 \, a b^{3} x + 6 \, a^{2} b^{2} x^{\frac {2}{3}} - 2 \, a^{3} b x^{\frac {1}{3}} + a^{4}}{{\left (b x^{\frac {1}{3}} + a\right )} a^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b^3*log(abs(b*x^(1/3) + a))/a^5 - 4*b^3*log(abs(x))/a^5 - (12*a*b^3*x + 6*a^2*b^2*x^(2/3) - 2*a^3*b*x^(1/3)
+ a^4)/((b*x^(1/3) + a)*a^5*x)

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maple [A]  time = 0.01, size = 73, normalized size = 0.91 \[ -\frac {3 b^{3}}{\left (b \,x^{\frac {1}{3}}+a \right ) a^{4}}-\frac {4 b^{3} \ln \relax (x )}{a^{5}}+\frac {12 b^{3} \ln \left (b \,x^{\frac {1}{3}}+a \right )}{a^{5}}-\frac {9 b^{2}}{a^{4} x^{\frac {1}{3}}}+\frac {3 b}{a^{3} x^{\frac {2}{3}}}-\frac {1}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b^3/a^4/(b*x^(1/3)+a)-1/a^2/x+3*b/a^3/x^(2/3)-9*b^2/a^4/x^(1/3)+12*b^3*ln(b*x^(1/3)+a)/a^5-4*b^3*ln(x)/a^5

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maxima [A]  time = 0.57, size = 73, normalized size = 0.91 \[ -\frac {12 \, b^{3} x + 6 \, a b^{2} x^{\frac {2}{3}} - 2 \, a^{2} b x^{\frac {1}{3}} + a^{3}}{a^{4} b x^{\frac {4}{3}} + a^{5} x} + \frac {12 \, b^{3} \log \left (b x^{\frac {1}{3}} + a\right )}{a^{5}} - \frac {4 \, b^{3} \log \relax (x)}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(12*b^3*x + 6*a*b^2*x^(2/3) - 2*a^2*b*x^(1/3) + a^3)/(a^4*b*x^(4/3) + a^5*x) + 12*b^3*log(b*x^(1/3) + a)/a^5
- 4*b^3*log(x)/a^5

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mupad [B]  time = 1.16, size = 67, normalized size = 0.84 \[ \frac {24\,b^3\,\mathrm {atanh}\left (\frac {2\,b\,x^{1/3}}{a}+1\right )}{a^5}-\frac {\frac {1}{a}-\frac {2\,b\,x^{1/3}}{a^2}+\frac {12\,b^3\,x}{a^4}+\frac {6\,b^2\,x^{2/3}}{a^3}}{a\,x+b\,x^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^(1/3))^2),x)

[Out]

(24*b^3*atanh((2*b*x^(1/3))/a + 1))/a^5 - (1/a - (2*b*x^(1/3))/a^2 + (12*b^3*x)/a^4 + (6*b^2*x^(2/3))/a^3)/(a*
x + b*x^(4/3))

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sympy [A]  time = 5.57, size = 272, normalized size = 3.40 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {1}{a^{2} x} & \text {for}\: b = 0 \\- \frac {3}{5 b^{2} x^{\frac {5}{3}}} & \text {for}\: a = 0 \\- \frac {a^{4} x^{\frac {2}{3}}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} + \frac {2 a^{3} b x}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} - \frac {6 a^{2} b^{2} x^{\frac {4}{3}}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} - \frac {4 a b^{3} x^{\frac {5}{3}} \log {\relax (x )}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} + \frac {12 a b^{3} x^{\frac {5}{3}} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} - \frac {12 a b^{3} x^{\frac {5}{3}}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} - \frac {4 b^{4} x^{2} \log {\relax (x )}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} + \frac {12 b^{4} x^{2} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{a^{6} x^{\frac {5}{3}} + a^{5} b x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(5/3), Eq(a, 0) & Eq(b, 0)), (-1/(a**2*x), Eq(b, 0)), (-3/(5*b**2*x**(5/3)), Eq(a, 0)), (-a*
*4*x**(2/3)/(a**6*x**(5/3) + a**5*b*x**2) + 2*a**3*b*x/(a**6*x**(5/3) + a**5*b*x**2) - 6*a**2*b**2*x**(4/3)/(a
**6*x**(5/3) + a**5*b*x**2) - 4*a*b**3*x**(5/3)*log(x)/(a**6*x**(5/3) + a**5*b*x**2) + 12*a*b**3*x**(5/3)*log(
a/b + x**(1/3))/(a**6*x**(5/3) + a**5*b*x**2) - 12*a*b**3*x**(5/3)/(a**6*x**(5/3) + a**5*b*x**2) - 4*b**4*x**2
*log(x)/(a**6*x**(5/3) + a**5*b*x**2) + 12*b**4*x**2*log(a/b + x**(1/3))/(a**6*x**(5/3) + a**5*b*x**2), True))

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