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

Optimal. Leaf size=143 \[ \frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{5/3} b^{4/3}}-\frac {\log (a+b x)}{18 a^{5/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{4/3}}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}-\frac {\sqrt [3]{x}}{2 b (a+b x)^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {47, 51, 58, 617, 204, 31} \begin {gather*} \frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{5/3} b^{4/3}}-\frac {\log (a+b x)}{18 a^{5/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{4/3}}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}-\frac {\sqrt [3]{x}}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(1/3)/(a + b*x)^3,x]

[Out]

-x^(1/3)/(2*b*(a + b*x)^2) + x^(1/3)/(6*a*b*(a + b*x)) - ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3)
)]/(3*Sqrt[3]*a^(5/3)*b^(4/3)) + Log[a^(1/3) + b^(1/3)*x^(1/3)]/(6*a^(5/3)*b^(4/3)) - Log[a + b*x]/(18*a^(5/3)
*b^(4/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{x}}{(a+b x)^3} \, dx &=-\frac {\sqrt [3]{x}}{2 b (a+b x)^2}+\frac {\int \frac {1}{x^{2/3} (a+b x)^2} \, dx}{6 b}\\ &=-\frac {\sqrt [3]{x}}{2 b (a+b x)^2}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}+\frac {\int \frac {1}{x^{2/3} (a+b x)} \, dx}{9 a b}\\ &=-\frac {\sqrt [3]{x}}{2 b (a+b x)^2}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}-\frac {\log (a+b x)}{18 a^{5/3} b^{4/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{x}\right )}{6 a^{4/3} b^{5/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{6 a^{5/3} b^{4/3}}\\ &=-\frac {\sqrt [3]{x}}{2 b (a+b x)^2}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{5/3} b^{4/3}}-\frac {\log (a+b x)}{18 a^{5/3} b^{4/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}\right )}{3 a^{5/3} b^{4/3}}\\ &=-\frac {\sqrt [3]{x}}{2 b (a+b x)^2}+\frac {\sqrt [3]{x}}{6 a b (a+b x)}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{5/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{5/3} b^{4/3}}-\frac {\log (a+b x)}{18 a^{5/3} b^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.19 \begin {gather*} \frac {3 x^{4/3} \, _2F_1\left (\frac {4}{3},3;\frac {7}{3};-\frac {b x}{a}\right )}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(1/3)/(a + b*x)^3,x]

[Out]

(3*x^(4/3)*Hypergeometric2F1[4/3, 3, 7/3, -((b*x)/a)])/(4*a^3)

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IntegrateAlgebraic [A]  time = 0.24, size = 163, normalized size = 1.14 \begin {gather*} -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{18 a^{5/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{9 a^{5/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{4/3}}+\frac {b x^{4/3}-2 a \sqrt [3]{x}}{6 a b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*x^(1/3) + b*x^(4/3))/(6*a*b*(a + b*x)^2) - ArcTan[1/Sqrt[3] - (2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))]/(3*
Sqrt[3]*a^(5/3)*b^(4/3)) + Log[a^(1/3) + b^(1/3)*x^(1/3)]/(9*a^(5/3)*b^(4/3)) - Log[a^(2/3) - a^(1/3)*b^(1/3)*
x^(1/3) + b^(2/3)*x^(2/3)]/(18*a^(5/3)*b^(4/3))

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fricas [B]  time = 1.21, size = 501, normalized size = 3.50 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{\frac {2}{3}} - \left (a^{2} b\right )^{\frac {1}{3}} a + \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x^{\frac {1}{3}}}{b x + a}\right ) - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 3 \, {\left (a^{2} b^{2} x - 2 \, a^{3} b\right )} x^{\frac {1}{3}}}{18 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (a^{2} b\right )^{\frac {1}{3}} a - 2 \, \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {2}{3}} + \left (a^{2} b\right )^{\frac {1}{3}} a - \left (a^{2} b\right )^{\frac {2}{3}} x^{\frac {1}{3}}\right ) + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{\frac {1}{3}} + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 3 \, {\left (a^{2} b^{2} x - 2 \, a^{3} b\right )} x^{\frac {1}{3}}}{18 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x - a^2 + 3*sqrt(1/3)*(
2*a*b*x^(2/3) - (a^2*b)^(1/3)*a + (a^2*b)^(2/3)*x^(1/3))*sqrt(-(a^2*b)^(1/3)/b) - 3*(a^2*b)^(1/3)*a*x^(1/3))/(
b*x + a)) - (b^2*x^2 + 2*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/3))
+ 2*(b^2*x^2 + 2*a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(1/3) + (a^2*b)^(2/3)) + 3*(a^2*b^2*x - 2*a^3*b)*x^(1/3
))/(a^3*b^4*x^2 + 2*a^4*b^3*x + a^5*b^2), 1/18*(6*sqrt(1/3)*(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b)*sqrt((a^2*b)^(1/
3)/b)*arctan(-sqrt(1/3)*((a^2*b)^(1/3)*a - 2*(a^2*b)^(2/3)*x^(1/3))*sqrt((a^2*b)^(1/3)/b)/a^2) - (b^2*x^2 + 2*
a*b*x + a^2)*(a^2*b)^(2/3)*log(a*b*x^(2/3) + (a^2*b)^(1/3)*a - (a^2*b)^(2/3)*x^(1/3)) + 2*(b^2*x^2 + 2*a*b*x +
a^2)*(a^2*b)^(2/3)*log(a*b*x^(1/3) + (a^2*b)^(2/3)) + 3*(a^2*b^2*x - 2*a^3*b)*x^(1/3))/(a^3*b^4*x^2 + 2*a^4*b
^3*x + a^5*b^2)]

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giac [A]  time = 1.11, size = 148, normalized size = 1.03 \begin {gather*} -\frac {\left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2} b} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b^{2}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b^{2}} + \frac {b x^{\frac {4}{3}} - 2 \, a x^{\frac {1}{3}}}{6 \, {\left (b x + a\right )}^{2} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(-a/b)^(1/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/(a^2*b) + 1/9*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*
x^(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^2*b^2) + 1/18*(-a*b^2)^(1/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a
/b)^(2/3))/(a^2*b^2) + 1/6*(b*x^(4/3) - 2*a*x^(1/3))/((b*x + a)^2*a*b)

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maple [A]  time = 0.01, size = 132, normalized size = 0.92 \begin {gather*} \frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}+\frac {\ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}-\frac {\ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}+\frac {\frac {x^{\frac {4}{3}}}{6 a}-\frac {x^{\frac {1}{3}}}{3 b}}{\left (b x +a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(1/18/a*x^(4/3)-1/9/b*x^(1/3))/(b*x+a)^2+1/9/b^2/a/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))-1/18/b^2/a/(a/b)^(2/3
)*ln(x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))+1/9/b^2/a/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*
x^(1/3)-1))

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maxima [A]  time = 2.96, size = 152, normalized size = 1.06 \begin {gather*} \frac {b x^{\frac {4}{3}} - 2 \, a x^{\frac {1}{3}}}{6 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {\log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^(4/3) - 2*a*x^(1/3))/(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b) + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/3) -
(a/b)^(1/3))/(a/b)^(1/3))/(a*b^2*(a/b)^(2/3)) - 1/18*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(a*b^2*(
a/b)^(2/3)) + 1/9*log(x^(1/3) + (a/b)^(1/3))/(a*b^2*(a/b)^(2/3))

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mupad [B]  time = 0.24, size = 146, normalized size = 1.02 \begin {gather*} \frac {\frac {x^{4/3}}{6\,a}-\frac {x^{1/3}}{3\,b}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {\ln \left (\frac {b^{2/3}}{a^{2/3}}+\frac {b\,x^{1/3}}{a}\right )}{9\,a^{5/3}\,b^{4/3}}+\frac {\ln \left (\frac {b\,x^{1/3}}{a}+\frac {b^{2/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{5/3}\,b^{4/3}}-\frac {\ln \left (\frac {b\,x^{1/3}}{a}-\frac {b^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{5/3}\,b^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/3)/(a + b*x)^3,x)

[Out]

(x^(4/3)/(6*a) - x^(1/3)/(3*b))/(a^2 + b^2*x^2 + 2*a*b*x) + log(b^(2/3)/a^(2/3) + (b*x^(1/3))/a)/(9*a^(5/3)*b^
(4/3)) + (log((b*x^(1/3))/a + (b^(2/3)*(3^(1/2)*1i - 1))/(2*a^(2/3)))*(3^(1/2)*1i - 1))/(18*a^(5/3)*b^(4/3)) -
(log((b*x^(1/3))/a - (b^(2/3)*(3^(1/2)*1i + 1))/(2*a^(2/3)))*(3^(1/2)*1i + 1))/(18*a^(5/3)*b^(4/3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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