3.7.92 \(\int \frac {x^{2/3}}{(a+b x)^3} \, dx\)

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

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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, 56, 617, 204, 31} \begin {gather*} -\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{4/3} b^{5/3}}+\frac {\log (a+b x)}{18 a^{4/3} b^{5/3}}-\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^{4/3} b^{5/3}}+\frac {x^{2/3}}{3 a b (a+b x)}-\frac {x^{2/3}}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(2/3)/(a + b*x)^3,x]

[Out]

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

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 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(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 {x^{2/3}}{(a+b x)^3} \, dx &=-\frac {x^{2/3}}{2 b (a+b x)^2}+\frac {\int \frac {1}{\sqrt [3]{x} (a+b x)^2} \, dx}{3 b}\\ &=-\frac {x^{2/3}}{2 b (a+b x)^2}+\frac {x^{2/3}}{3 a b (a+b x)}+\frac {\int \frac {1}{\sqrt [3]{x} (a+b x)} \, dx}{9 a b}\\ &=-\frac {x^{2/3}}{2 b (a+b x)^2}+\frac {x^{2/3}}{3 a b (a+b x)}+\frac {\log (a+b x)}{18 a^{4/3} b^{5/3}}+\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 b^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{6 a^{4/3} b^{5/3}}\\ &=-\frac {x^{2/3}}{2 b (a+b x)^2}+\frac {x^{2/3}}{3 a b (a+b x)}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{4/3} b^{5/3}}+\frac {\log (a+b x)}{18 a^{4/3} b^{5/3}}+\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^{4/3} b^{5/3}}\\ &=-\frac {x^{2/3}}{2 b (a+b x)^2}+\frac {x^{2/3}}{3 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^{4/3} b^{5/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{6 a^{4/3} b^{5/3}}+\frac {\log (a+b x)}{18 a^{4/3} b^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.19 \begin {gather*} \frac {3 x^{5/3} \, _2F_1\left (\frac {5}{3},3;\frac {8}{3};-\frac {b x}{a}\right )}{5 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(2/3)/(a + b*x)^3,x]

[Out]

(3*x^(5/3)*Hypergeometric2F1[5/3, 3, 8/3, -((b*x)/a)])/(5*a^3)

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IntegrateAlgebraic [A]  time = 0.24, size = 164, normalized size = 1.15 \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^{4/3} b^{5/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{9 a^{4/3} b^{5/3}}-\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^{4/3} b^{5/3}}+\frac {2 b x^{5/3}-a x^{2/3}}{6 a b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.99, size = 508, normalized size = 3.55 \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 b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} a + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{\frac {2}{3}}\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{\frac {1}{3}}}{b x + a}\right ) + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x^{\frac {1}{3}} - \left (-a b^{2}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (2 \, a b^{3} x - a^{2} b^{2}\right )} x^{\frac {2}{3}}}{18 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\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 b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} b x^{\frac {1}{3}} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x^{\frac {1}{3}} - \left (-a b^{2}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (2 \, a b^{3} x - a^{2} b^{2}\right )} x^{\frac {2}{3}}}{18 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 1.06, size = 149, normalized size = 1.04 \begin {gather*} -\frac {\left (-\frac {a}{b}\right )^{\frac {2}{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 {2}{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^{3}} + \frac {2 \, b x^{\frac {5}{3}} - a x^{\frac {2}{3}}}{6 \, {\left (b x + a\right )}^{2} a b} + \frac {\left (-a b^{2}\right )^{\frac {2}{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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(-a/b)^(2/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/(a^2*b) - 1/9*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*
x^(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^2*b^3) + 1/6*(2*b*x^(5/3) - a*x^(2/3))/((b*x + a)^2*a*b) + 1/18*(-a*b
^2)^(2/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^2*b^3)

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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 {1}{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 {1}{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 {1}{3}} a \,b^{2}}+\frac {\frac {x^{\frac {5}{3}}}{3 a}-\frac {x^{\frac {2}{3}}}{6 b}}{\left (b x +a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(1/9/a*x^(5/3)-1/18/b*x^(2/3))/(b*x+a)^2-1/9/b^2/a/(a/b)^(1/3)*ln(x^(1/3)+(a/b)^(1/3))+1/18/b^2/a/(a/b)^(1/3
)*ln(x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))+1/9/b^2/a*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*
x^(1/3)-1))

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maxima [A]  time = 2.92, size = 153, normalized size = 1.07 \begin {gather*} \frac {2 \, b x^{\frac {5}{3}} - a x^{\frac {2}{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 {1}{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 {1}{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 {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 172, normalized size = 1.20 \begin {gather*} \frac {\frac {x^{5/3}}{3\,a}-\frac {x^{2/3}}{6\,b}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {\ln \left (\frac {1}{9\,a^{5/3}\,{\left (-b\right )}^{4/3}}+\frac {x^{1/3}}{9\,a^2\,b}\right )}{9\,a^{4/3}\,{\left (-b\right )}^{5/3}}+\frac {\ln \left (\frac {x^{1/3}}{9\,a^2\,b}+\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{36\,a^{5/3}\,{\left (-b\right )}^{4/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,{\left (-b\right )}^{5/3}}-\frac {\ln \left (\frac {x^{1/3}}{9\,a^2\,b}+\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{36\,a^{5/3}\,{\left (-b\right )}^{4/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,{\left (-b\right )}^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2/3)/(a + b*x)^3,x)

[Out]

(x^(5/3)/(3*a) - x^(2/3)/(6*b))/(a^2 + b^2*x^2 + 2*a*b*x) + log(1/(9*a^(5/3)*(-b)^(4/3)) + x^(1/3)/(9*a^2*b))/
(9*a^(4/3)*(-b)^(5/3)) + (log(x^(1/3)/(9*a^2*b) + (3^(1/2)*1i - 1)^2/(36*a^(5/3)*(-b)^(4/3)))*(3^(1/2)*1i - 1)
)/(18*a^(4/3)*(-b)^(5/3)) - (log(x^(1/3)/(9*a^2*b) + (3^(1/2)*1i + 1)^2/(36*a^(5/3)*(-b)^(4/3)))*(3^(1/2)*1i +
1))/(18*a^(4/3)*(-b)^(5/3))

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

[Out]

Timed out

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