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

Optimal. Leaf size=140 \[ -\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{7/3} b^{2/3}}+\frac {\log (a+b x)}{9 a^{7/3} b^{2/3}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} b^{2/3}}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}+\frac {x^{2/3}}{2 a (a+b x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {51, 56, 617, 204, 31} \begin {gather*} -\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{7/3} b^{2/3}}+\frac {\log (a+b x)}{9 a^{7/3} b^{2/3}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} b^{2/3}}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}+\frac {x^{2/3}}{2 a (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(1/3)*(a + b*x)^3),x]

[Out]

x^(2/3)/(2*a*(a + b*x)^2) + (2*x^(2/3))/(3*a^2*(a + b*x)) - (2*ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a
^(1/3))])/(3*Sqrt[3]*a^(7/3)*b^(2/3)) - Log[a^(1/3) + b^(1/3)*x^(1/3)]/(3*a^(7/3)*b^(2/3)) + Log[a + b*x]/(9*a
^(7/3)*b^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 {1}{\sqrt [3]{x} (a+b x)^3} \, dx &=\frac {x^{2/3}}{2 a (a+b x)^2}+\frac {2 \int \frac {1}{\sqrt [3]{x} (a+b x)^2} \, dx}{3 a}\\ &=\frac {x^{2/3}}{2 a (a+b x)^2}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}+\frac {2 \int \frac {1}{\sqrt [3]{x} (a+b x)} \, dx}{9 a^2}\\ &=\frac {x^{2/3}}{2 a (a+b x)^2}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}+\frac {\log (a+b x)}{9 a^{7/3} b^{2/3}}+\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 )}{3 a^2 b}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{3 a^{7/3} b^{2/3}}\\ &=\frac {x^{2/3}}{2 a (a+b x)^2}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{7/3} b^{2/3}}+\frac {\log (a+b x)}{9 a^{7/3} b^{2/3}}+\frac {2 \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^{7/3} b^{2/3}}\\ &=\frac {x^{2/3}}{2 a (a+b x)^2}+\frac {2 x^{2/3}}{3 a^2 (a+b x)}-\frac {2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{7/3} b^{2/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{7/3} b^{2/3}}+\frac {\log (a+b x)}{9 a^{7/3} b^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.00, size = 27, normalized size = 0.19 \begin {gather*} \frac {3 x^{2/3} \, _2F_1\left (\frac {2}{3},3;\frac {5}{3};-\frac {b x}{a}\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(1/3)*(a + b*x)^3),x]

[Out]

(3*x^(2/3)*Hypergeometric2F1[2/3, 3, 5/3, -((b*x)/a)])/(2*a^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.16, size = 157, normalized size = 1.12 \begin {gather*} \frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{9 a^{7/3} b^{2/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{9 a^{7/3} b^{2/3}}-\frac {2 \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^{7/3} b^{2/3}}+\frac {x^{2/3} (7 a+4 b x)}{6 a^2 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(2/3)*(7*a + 4*b*x))/(6*a^2*(a + b*x)^2) - (2*ArcTan[1/Sqrt[3] - (2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/(3
*Sqrt[3]*a^(7/3)*b^(2/3)) - (2*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(9*a^(7/3)*b^(2/3)) + Log[a^(2/3) - a^(1/3)*b^(
1/3)*x^(1/3) + b^(2/3)*x^(2/3)]/(9*a^(7/3)*b^(2/3))

________________________________________________________________________________________

fricas [B]  time = 0.89, size = 510, normalized size = 3.64 \begin {gather*} \left [\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}} \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 ) + 2 \, {\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 ) - 4 \, {\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 (4 \, a b^{3} x + 7 \, a^{2} b^{2}\right )} x^{\frac {2}{3}}}{18 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}, \frac {12 \, \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 ) + 2 \, {\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 ) - 4 \, {\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 (4 \, a b^{3} x + 7 \, a^{2} b^{2}\right )} x^{\frac {2}{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(1/x^(1/3)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[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)*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)) + 2*(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)) - 4*(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*(4*a*b^3*x + 7*a^2*b^2
)*x^(2/3))/(a^3*b^4*x^2 + 2*a^4*b^3*x + a^5*b^2), 1/18*(12*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) + 2*(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)) - 4*(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*(4*a*b^3*x + 7*a^2*b^2)*x^(2/3))/(a^3*b^4*x^2 + 2*a
^4*b^3*x + a^5*b^2)]

________________________________________________________________________________________

giac [A]  time = 1.06, size = 143, normalized size = 1.02 \begin {gather*} -\frac {2 \, \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^{3}} - \frac {2 \, \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^{3} b^{2}} + \frac {4 \, b x^{\frac {5}{3}} + 7 \, a x^{\frac {2}{3}}}{6 \, {\left (b x + a\right )}^{2} a^{2}} + \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 )}{9 \, a^{3} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(-a/b)^(2/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/a^3 - 2/9*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^3*b^2) + 1/6*(4*b*x^(5/3) + 7*a*x^(2/3))/((b*x + a)^2*a^2) + 1/9*(-a*b^2)
^(2/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b^2)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 136, normalized size = 0.97 \begin {gather*} \frac {x^{\frac {2}{3}}}{2 \left (b x +a \right )^{2} a}+\frac {2 x^{\frac {2}{3}}}{3 \left (b x +a \right ) a^{2}}+\frac {2 \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^{2} b}-\frac {2 \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^{2} b}+\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 )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.96, size = 151, normalized size = 1.08 \begin {gather*} \frac {4 \, b x^{\frac {5}{3}} + 7 \, a x^{\frac {2}{3}}}{6 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} + \frac {2 \, \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^{2} b \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 )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {2 \, \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*b*x^(5/3) + 7*a*x^(2/3))/(a^2*b^2*x^2 + 2*a^3*b*x + a^4) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/3) -
(a/b)^(1/3))/(a/b)^(1/3))/(a^2*b*(a/b)^(1/3)) + 1/9*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*(a
/b)^(1/3)) - 2/9*log(x^(1/3) + (a/b)^(1/3))/(a^2*b*(a/b)^(1/3))

________________________________________________________________________________________

mupad [B]  time = 0.19, size = 167, normalized size = 1.19 \begin {gather*} \frac {\frac {7\,x^{2/3}}{6\,a}+\frac {2\,b\,x^{5/3}}{3\,a^2}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {2\,\ln \left (\frac {4\,b\,x^{1/3}}{9\,a^4}-\frac {4\,b^{2/3}}{9\,{\left (-a\right )}^{11/3}}\right )}{9\,{\left (-a\right )}^{7/3}\,b^{2/3}}+\frac {\ln \left (\frac {4\,b\,x^{1/3}}{9\,a^4}-\frac {b^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,{\left (-a\right )}^{11/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{7/3}\,b^{2/3}}-\frac {\ln \left (\frac {4\,b\,x^{1/3}}{9\,a^4}-\frac {b^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,{\left (-a\right )}^{11/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{7/3}\,b^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/3)*(a + b*x)^3),x)

[Out]

((7*x^(2/3))/(6*a) + (2*b*x^(5/3))/(3*a^2))/(a^2 + b^2*x^2 + 2*a*b*x) + (2*log((4*b*x^(1/3))/(9*a^4) - (4*b^(2
/3))/(9*(-a)^(11/3))))/(9*(-a)^(7/3)*b^(2/3)) + (log((4*b*x^(1/3))/(9*a^4) - (b^(2/3)*(3^(1/2)*1i - 1)^2)/(9*(
-a)^(11/3)))*(3^(1/2)*1i - 1))/(9*(-a)^(7/3)*b^(2/3)) - (log((4*b*x^(1/3))/(9*a^4) - (b^(2/3)*(3^(1/2)*1i + 1)
^2)/(9*(-a)^(11/3)))*(3^(1/2)*1i + 1))/(9*(-a)^(7/3)*b^(2/3))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________