3.25.26 \(\int \frac {1}{(a+\frac {b}{\sqrt [3]{x}}) x^3} \, dx\) [2426]

Optimal. Leaf size=93 \[ -\frac {3}{5 b x^{5/3}}+\frac {3 a}{4 b^2 x^{4/3}}-\frac {a^2}{b^3 x}+\frac {3 a^3}{2 b^4 x^{2/3}}-\frac {3 a^4}{b^5 \sqrt [3]{x}}+\frac {3 a^5 \log \left (b+a \sqrt [3]{x}\right )}{b^6}-\frac {a^5 \log (x)}{b^6} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {269, 272, 46} \begin {gather*} \frac {3 a^5 \log \left (a \sqrt [3]{x}+b\right )}{b^6}-\frac {a^5 \log (x)}{b^6}-\frac {3 a^4}{b^5 \sqrt [3]{x}}+\frac {3 a^3}{2 b^4 x^{2/3}}-\frac {a^2}{b^3 x}+\frac {3 a}{4 b^2 x^{4/3}}-\frac {3}{5 b x^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^(1/3))*x^3),x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 272

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+\frac {b}{\sqrt [3]{x}}\right ) x^3} \, dx &=\int \frac {1}{\left (b+a \sqrt [3]{x}\right ) x^{8/3}} \, dx\\ &=3 \text {Subst}\left (\int \frac {1}{x^6 (b+a x)} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {Subst}\left (\int \left (\frac {1}{b x^6}-\frac {a}{b^2 x^5}+\frac {a^2}{b^3 x^4}-\frac {a^3}{b^4 x^3}+\frac {a^4}{b^5 x^2}-\frac {a^5}{b^6 x}+\frac {a^6}{b^6 (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3}{5 b x^{5/3}}+\frac {3 a}{4 b^2 x^{4/3}}-\frac {a^2}{b^3 x}+\frac {3 a^3}{2 b^4 x^{2/3}}-\frac {3 a^4}{b^5 \sqrt [3]{x}}+\frac {3 a^5 \log \left (b+a \sqrt [3]{x}\right )}{b^6}-\frac {a^5 \log (x)}{b^6}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 84, normalized size = 0.90 \begin {gather*} \frac {\frac {b \left (-12 b^4+15 a b^3 \sqrt [3]{x}-20 a^2 b^2 x^{2/3}+30 a^3 b x-60 a^4 x^{4/3}\right )}{x^{5/3}}+60 a^5 \log \left (b+a \sqrt [3]{x}\right )-20 a^5 \log (x)}{20 b^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^(1/3))*x^3),x]

[Out]

((b*(-12*b^4 + 15*a*b^3*x^(1/3) - 20*a^2*b^2*x^(2/3) + 30*a^3*b*x - 60*a^4*x^(4/3)))/x^(5/3) + 60*a^5*Log[b +
a*x^(1/3)] - 20*a^5*Log[x])/(20*b^6)

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Maple [A]
time = 0.20, size = 78, normalized size = 0.84

method	result	size
derivativedivides	\(-\frac {3}{5 b \,x^{\frac {5}{3}}}+\frac {3 a}{4 b^{2} x^{\frac {4}{3}}}-\frac {a^{2}}{b^{3} x}+\frac {3 a^{3}}{2 b^{4} x^{\frac {2}{3}}}-\frac {3 a^{4}}{b^{5} x^{\frac {1}{3}}}+\frac {3 a^{5} \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{6}}-\frac {a^{5} \ln \left (x \right )}{b^{6}}\)	\(78\)
default	\(-\frac {3}{5 b \,x^{\frac {5}{3}}}+\frac {3 a}{4 b^{2} x^{\frac {4}{3}}}-\frac {a^{2}}{b^{3} x}+\frac {3 a^{3}}{2 b^{4} x^{\frac {2}{3}}}-\frac {3 a^{4}}{b^{5} x^{\frac {1}{3}}}+\frac {3 a^{5} \ln \left (b +a \,x^{\frac {1}{3}}\right )}{b^{6}}-\frac {a^{5} \ln \left (x \right )}{b^{6}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 95, normalized size = 1.02 \begin {gather*} \frac {3 \, a^{5} \log \left (a + \frac {b}{x^{\frac {1}{3}}}\right )}{b^{6}} - \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{5}}{5 \, b^{6}} + \frac {15 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{4} a}{4 \, b^{6}} - \frac {10 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} a^{2}}{b^{6}} + \frac {15 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} a^{3}}{b^{6}} - \frac {15 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} a^{4}}{b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^5*log(a + b/x^(1/3))/b^6 - 3/5*(a + b/x^(1/3))^5/b^6 + 15/4*(a + b/x^(1/3))^4*a/b^6 - 10*(a + b/x^(1/3))^3
*a^2/b^6 + 15*(a + b/x^(1/3))^2*a^3/b^6 - 15*(a + b/x^(1/3))*a^4/b^6

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Fricas [A]
time = 0.42, size = 85, normalized size = 0.91 \begin {gather*} \frac {60 \, a^{5} x^{2} \log \left (a x^{\frac {1}{3}} + b\right ) - 60 \, a^{5} x^{2} \log \left (x^{\frac {1}{3}}\right ) - 20 \, a^{2} b^{3} x - 15 \, {\left (4 \, a^{4} b x - a b^{4}\right )} x^{\frac {2}{3}} + 6 \, {\left (5 \, a^{3} b^{2} x - 2 \, b^{5}\right )} x^{\frac {1}{3}}}{20 \, b^{6} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(60*a^5*x^2*log(a*x^(1/3) + b) - 60*a^5*x^2*log(x^(1/3)) - 20*a^2*b^3*x - 15*(4*a^4*b*x - a*b^4)*x^(2/3)
+ 6*(5*a^3*b^2*x - 2*b^5)*x^(1/3))/(b^6*x^2)

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Sympy [A]
time = 0.94, size = 116, normalized size = 1.25 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {1}{2 a x^{2}} & \text {for}\: b = 0 \\- \frac {3}{5 b x^{\frac {5}{3}}} & \text {for}\: a = 0 \\- \frac {a^{5} \log {\left (x \right )}}{b^{6}} + \frac {3 a^{5} \log {\left (\sqrt [3]{x} + \frac {b}{a} \right )}}{b^{6}} - \frac {3 a^{4}}{b^{5} \sqrt [3]{x}} + \frac {3 a^{3}}{2 b^{4} x^{\frac {2}{3}}} - \frac {a^{2}}{b^{3} x} + \frac {3 a}{4 b^{2} x^{\frac {4}{3}}} - \frac {3}{5 b x^{\frac {5}{3}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 81, normalized size = 0.87 \begin {gather*} \frac {3 \, a^{5} \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{b^{6}} - \frac {a^{5} \log \left ({\left | x \right |}\right )}{b^{6}} - \frac {60 \, a^{4} b x^{\frac {4}{3}} - 30 \, a^{3} b^{2} x + 20 \, a^{2} b^{3} x^{\frac {2}{3}} - 15 \, a b^{4} x^{\frac {1}{3}} + 12 \, b^{5}}{20 \, b^{6} x^{\frac {5}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^5*log(abs(a*x^(1/3) + b))/b^6 - a^5*log(abs(x))/b^6 - 1/20*(60*a^4*b*x^(4/3) - 30*a^3*b^2*x + 20*a^2*b^3*x
^(2/3) - 15*a*b^4*x^(1/3) + 12*b^5)/(b^6*x^(5/3))

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Mupad [B]
time = 1.18, size = 71, normalized size = 0.76 \begin {gather*} \frac {6\,a^5\,\mathrm {atanh}\left (\frac {2\,a\,x^{1/3}}{b}+1\right )}{b^6}-\frac {\frac {3}{5\,b}-\frac {3\,a\,x^{1/3}}{4\,b^2}-\frac {3\,a^3\,x}{2\,b^4}+\frac {a^2\,x^{2/3}}{b^3}+\frac {3\,a^4\,x^{4/3}}{b^5}}{x^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b/x^(1/3))),x)

[Out]

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

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