3.697 \(\int \frac {1}{x^{5/3} (a+b x)^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac {10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{11/3}}+\frac {10 b^{2/3} \log (a+b x)}{9 a^{11/3}}+\frac {20 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{11/3}}-\frac {10}{3 a^3 x^{2/3}}+\frac {4}{3 a^2 x^{2/3} (a+b x)}+\frac {1}{2 a x^{2/3} (a+b x)^2} \]

[Out]

-10/3/a^3/x^(2/3)+1/2/a/x^(2/3)/(b*x+a)^2+4/3/a^2/x^(2/3)/(b*x+a)-10/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*x^(1/3))/a^(
11/3)+10/9*b^(2/3)*ln(b*x+a)/a^(11/3)+20/9*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x^(1/3))/a^(1/3)*3^(1/2))/a^(
11/3)*3^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {51, 58, 617, 204, 31} \[ -\frac {10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{11/3}}+\frac {10 b^{2/3} \log (a+b x)}{9 a^{11/3}}+\frac {20 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{11/3}}+\frac {4}{3 a^2 x^{2/3} (a+b x)}-\frac {10}{3 a^3 x^{2/3}}+\frac {1}{2 a x^{2/3} (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/3)*(a + b*x)^3),x]

[Out]

-10/(3*a^3*x^(2/3)) + 1/(2*a*x^(2/3)*(a + b*x)^2) + 4/(3*a^2*x^(2/3)*(a + b*x)) + (20*b^(2/3)*ArcTan[(a^(1/3)
- 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(11/3)) - (10*b^(2/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(3
*a^(11/3)) + (10*b^(2/3)*Log[a + b*x])/(9*a^(11/3))

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 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 {1}{x^{5/3} (a+b x)^3} \, dx &=\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4 \int \frac {1}{x^{5/3} (a+b x)^2} \, dx}{3 a}\\ &=\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4}{3 a^2 x^{2/3} (a+b x)}+\frac {20 \int \frac {1}{x^{5/3} (a+b x)} \, dx}{9 a^2}\\ &=-\frac {10}{3 a^3 x^{2/3}}+\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4}{3 a^2 x^{2/3} (a+b x)}-\frac {(20 b) \int \frac {1}{x^{2/3} (a+b x)} \, dx}{9 a^3}\\ &=-\frac {10}{3 a^3 x^{2/3}}+\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4}{3 a^2 x^{2/3} (a+b x)}+\frac {10 b^{2/3} \log (a+b x)}{9 a^{11/3}}-\frac {\left (10 \sqrt [3]{b}\right ) \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^{10/3}}-\frac {\left (10 b^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{3 a^{11/3}}\\ &=-\frac {10}{3 a^3 x^{2/3}}+\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4}{3 a^2 x^{2/3} (a+b x)}-\frac {10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{11/3}}+\frac {10 b^{2/3} \log (a+b x)}{9 a^{11/3}}-\frac {\left (20 b^{2/3}\right ) \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^{11/3}}\\ &=-\frac {10}{3 a^3 x^{2/3}}+\frac {1}{2 a x^{2/3} (a+b x)^2}+\frac {4}{3 a^2 x^{2/3} (a+b x)}+\frac {20 b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{11/3}}-\frac {10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{11/3}}+\frac {10 b^{2/3} \log (a+b x)}{9 a^{11/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.18 \[ -\frac {3 \, _2F_1\left (-\frac {2}{3},3;\frac {1}{3};-\frac {b x}{a}\right )}{2 a^3 x^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/3)*(a + b*x)^3),x]

[Out]

(-3*Hypergeometric2F1[-2/3, 3, 1/3, -((b*x)/a)])/(2*a^3*x^(2/3))

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fricas [B]  time = 0.49, size = 244, normalized size = 1.61 \[ \frac {40 \, \sqrt {3} {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x^{\frac {1}{3}} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 20 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{\frac {2}{3}} + a b x^{\frac {1}{3}} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 40 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x^{\frac {1}{3}} - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (20 \, b^{2} x^{2} + 32 \, a b x + 9 \, a^{2}\right )} x^{\frac {1}{3}}}{18 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^3,x, algorithm="fricas")

[Out]

1/18*(40*sqrt(3)*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x^(1/3)*(-b^2/a^2)^(2/
3) - sqrt(3)*b)/b) - 20*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(-b^2/a^2)^(1/3)*log(b^2*x^(2/3) + a*b*x^(1/3)*(-b^2/a^2
)^(1/3) + a^2*(-b^2/a^2)^(2/3)) + 40*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*(-b^2/a^2)^(1/3)*log(b*x^(1/3) - a*(-b^2/a^
2)^(1/3)) - 3*(20*b^2*x^2 + 32*a*b*x + 9*a^2)*x^(1/3))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)

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giac [A]  time = 1.08, size = 150, normalized size = 0.99 \[ \frac {20 \, b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{4}} - \frac {20 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4}} - \frac {10 \, \left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{4}} - \frac {20 \, b^{2} x^{2} + 32 \, a b x + 9 \, a^{2}}{6 \, {\left (b x^{\frac {4}{3}} + a x^{\frac {1}{3}}\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^3,x, algorithm="giac")

[Out]

20/9*b*(-a/b)^(1/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/a^4 - 20/9*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^4 - 10/9*(-a*b^2)^(1/3)*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/
3))/a^4 - 1/6*(20*b^2*x^2 + 32*a*b*x + 9*a^2)/((b*x^(4/3) + a*x^(1/3))^2*a^3)

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maple [A]  time = 0.02, size = 139, normalized size = 0.91 \[ -\frac {11 b^{2} x^{\frac {4}{3}}}{6 \left (b x +a \right )^{2} a^{3}}-\frac {7 b \,x^{\frac {1}{3}}}{3 \left (b x +a \right )^{2} a^{2}}-\frac {20 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}-\frac {20 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}+\frac {10 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}-\frac {3}{2 a^{3} x^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/3)/(b*x+a)^3,x)

[Out]

-3/2/a^3/x^(2/3)-11/6/a^3*b^2/(b*x+a)^2*x^(4/3)-7/3/a^2*b/(b*x+a)^2*x^(1/3)-20/9/a^3/(a/b)^(2/3)*ln(x^(1/3)+(a
/b)^(1/3))+10/9/a^3/(a/b)^(2/3)*ln(x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))-20/9/a^3/(a/b)^(2/3)*3^(1/2)*arcta
n(1/3*3^(1/2)*(2/(a/b)^(1/3)*x^(1/3)-1))

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maxima [A]  time = 2.99, size = 154, normalized size = 1.01 \[ -\frac {20 \, b^{2} x^{2} + 32 \, a b x + 9 \, a^{2}}{6 \, {\left (a^{3} b^{2} x^{\frac {8}{3}} + 2 \, a^{4} b x^{\frac {5}{3}} + a^{5} x^{\frac {2}{3}}\right )}} - \frac {20 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {10 \, \log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {20 \, \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/6*(20*b^2*x^2 + 32*a*b*x + 9*a^2)/(a^3*b^2*x^(8/3) + 2*a^4*b*x^(5/3) + a^5*x^(2/3)) - 20/9*sqrt(3)*arctan(1
/3*sqrt(3)*(2*x^(1/3) - (a/b)^(1/3))/(a/b)^(1/3))/(a^3*(a/b)^(2/3)) + 10/9*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) +
(a/b)^(2/3))/(a^3*(a/b)^(2/3)) - 20/9*log(x^(1/3) + (a/b)^(1/3))/(a^3*(a/b)^(2/3))

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mupad [B]  time = 0.17, size = 182, normalized size = 1.20 \[ \frac {20\,b^{2/3}\,\ln \left (540\,{\left (-a\right )}^{19/3}\,b^{8/3}-540\,a^6\,b^3\,x^{1/3}\right )}{9\,{\left (-a\right )}^{11/3}}-\frac {\frac {3}{2\,a}+\frac {10\,b^2\,x^2}{3\,a^3}+\frac {16\,b\,x}{3\,a^2}}{a^2\,x^{2/3}+b^2\,x^{8/3}+2\,a\,b\,x^{5/3}}+\frac {20\,b^{2/3}\,\ln \left (540\,{\left (-a\right )}^{19/3}\,b^{8/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-540\,a^6\,b^3\,x^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,{\left (-a\right )}^{11/3}}-\frac {20\,b^{2/3}\,\ln \left (540\,{\left (-a\right )}^{19/3}\,b^{8/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+540\,a^6\,b^3\,x^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,{\left (-a\right )}^{11/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/3)*(a + b*x)^3),x)

[Out]

(20*b^(2/3)*log(540*(-a)^(19/3)*b^(8/3) - 540*a^6*b^3*x^(1/3)))/(9*(-a)^(11/3)) - (3/(2*a) + (10*b^2*x^2)/(3*a
^3) + (16*b*x)/(3*a^2))/(a^2*x^(2/3) + b^2*x^(8/3) + 2*a*b*x^(5/3)) + (20*b^(2/3)*log(540*(-a)^(19/3)*b^(8/3)*
((3^(1/2)*1i)/2 - 1/2) - 540*a^6*b^3*x^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(9*(-a)^(11/3)) - (20*b^(2/3)*log(540*(-
a)^(19/3)*b^(8/3)*((3^(1/2)*1i)/2 + 1/2) + 540*a^6*b^3*x^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(9*(-a)^(11/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/3)/(b*x+a)**3,x)

[Out]

Timed out

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