3.7.89 \(\int \frac {1}{x^{5/3} (a+b x)^2} \, dx\)

Optimal. Leaf size=128 \[ -\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{8/3}}+\frac {5 b^{2/3} \log (a+b x)}{6 a^{8/3}}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{8/3}}-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 128, 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, 58, 617, 204, 31} \begin {gather*} -\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{8/3}}+\frac {5 b^{2/3} \log (a+b x)}{6 a^{8/3}}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{8/3}}-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/3)*(a + b*x)^2),x]

[Out]

-5/(2*a^2*x^(2/3)) + 1/(a*x^(2/3)*(a + b*x)) + (5*b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3
))])/(Sqrt[3]*a^(8/3)) - (5*b^(2/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(2*a^(8/3)) + (5*b^(2/3)*Log[a + b*x])/(6*
a^(8/3))

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)^2} \, dx &=\frac {1}{a x^{2/3} (a+b x)}+\frac {5 \int \frac {1}{x^{5/3} (a+b x)} \, dx}{3 a}\\ &=-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)}-\frac {(5 b) \int \frac {1}{x^{2/3} (a+b x)} \, dx}{3 a^2}\\ &=-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)}+\frac {5 b^{2/3} \log (a+b x)}{6 a^{8/3}}-\frac {\left (5 \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 )}{2 a^{7/3}}-\frac {\left (5 b^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{2 a^{8/3}}\\ &=-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{8/3}}+\frac {5 b^{2/3} \log (a+b x)}{6 a^{8/3}}-\frac {\left (5 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 )}{a^{8/3}}\\ &=-\frac {5}{2 a^2 x^{2/3}}+\frac {1}{a x^{2/3} (a+b x)}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{8/3}}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{8/3}}+\frac {5 b^{2/3} \log (a+b x)}{6 a^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.21 \begin {gather*} -\frac {3 \, _2F_1\left (-\frac {2}{3},2;\frac {1}{3};-\frac {b x}{a}\right )}{2 a^2 x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/3)*(a + b*x)^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 155, normalized size = 1.21 \begin {gather*} \frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{6 a^{8/3}}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{3 a^{8/3}}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{8/3}}+\frac {-3 a-5 b x}{2 a^2 x^{2/3} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/3)*(a + b*x)^2),x]

[Out]

(-3*a - 5*b*x)/(2*a^2*x^(2/3)*(a + b*x)) + (5*b^(2/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))
])/(Sqrt[3]*a^(8/3)) - (5*b^(2/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(3*a^(8/3)) + (5*b^(2/3)*Log[a^(2/3) - a^(1/
3)*b^(1/3)*x^(1/3) + b^(2/3)*x^(2/3)])/(6*a^(8/3))

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fricas [B]  time = 0.65, size = 189, normalized size = 1.48 \begin {gather*} \frac {10 \, \sqrt {3} {\left (b x^{2} + a 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 ) - 5 \, {\left (b x^{2} + a 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 ) + 10 \, {\left (b x^{2} + a 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 (5 \, b x + 3 \, a\right )} x^{\frac {1}{3}}}{6 \, {\left (a^{2} b x^{2} + a^{3} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.01, size = 137, normalized size = 1.07 \begin {gather*} \frac {5 \, 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 )}{3 \, a^{3}} - \frac {5 \, \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 )}{3 \, a^{3}} - \frac {b x^{\frac {1}{3}}}{{\left (b x + a\right )} a^{2}} - \frac {5 \, \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 )}{6 \, a^{3}} - \frac {3}{2 \, a^{2} x^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 121, normalized size = 0.95 \begin {gather*} -\frac {b \,x^{\frac {1}{3}}}{\left (b x +a \right ) a^{2}}-\frac {5 \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 )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}+\frac {5 \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 )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}-\frac {3}{2 a^{2} x^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/3)/(b*x+a)^2,x)

[Out]

-3/2/a^2/x^(2/3)-1/a^2*b*x^(1/3)/(b*x+a)-5/3/a^2/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))+5/6/a^2/(a/b)^(2/3)*ln(x^
(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))-5/3/a^2/(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 = 2.94, size = 132, normalized size = 1.03 \begin {gather*} -\frac {5 \, b x + 3 \, a}{2 \, {\left (a^{2} b x^{\frac {5}{3}} + a^{3} x^{\frac {2}{3}}\right )}} - \frac {5 \, \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 )}{3 \, a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {5 \, \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 )}{6 \, a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {5 \, \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/3)/(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/2*(5*b*x + 3*a)/(a^2*b*x^(5/3) + a^3*x^(2/3)) - 5/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/3) - (a/b)^(1/3))/(a
/b)^(1/3))/(a^2*(a/b)^(2/3)) + 5/6*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*(a/b)^(2/3)) - 5/3*lo
g(x^(1/3) + (a/b)^(1/3))/(a^2*(a/b)^(2/3))

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mupad [B]  time = 0.17, size = 166, normalized size = 1.30 \begin {gather*} \frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left (15\,{\left (-1\right )}^{1/3}\,a^{13/3}\,b^{8/3}-15\,a^4\,b^3\,x^{1/3}\right )}{3\,a^{8/3}}-\frac {\frac {3}{2\,a}+\frac {5\,b\,x}{2\,a^2}}{a\,x^{2/3}+b\,x^{5/3}}+\frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left (15\,a^4\,b^3\,x^{1/3}-15\,{\left (-1\right )}^{1/3}\,a^{13/3}\,b^{8/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,a^{8/3}}-\frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left (15\,a^4\,b^3\,x^{1/3}+15\,{\left (-1\right )}^{1/3}\,a^{13/3}\,b^{8/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,a^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/3)*(a + b*x)^2),x)

[Out]

(5*(-1)^(1/3)*b^(2/3)*log(15*(-1)^(1/3)*a^(13/3)*b^(8/3) - 15*a^4*b^3*x^(1/3)))/(3*a^(8/3)) - (3/(2*a) + (5*b*
x)/(2*a^2))/(a*x^(2/3) + b*x^(5/3)) + (5*(-1)^(1/3)*b^(2/3)*log(15*a^4*b^3*x^(1/3) - 15*(-1)^(1/3)*a^(13/3)*b^
(8/3)*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/(3*a^(8/3)) - (5*(-1)^(1/3)*b^(2/3)*log(15*a^4*b^3*x^(1/
3) + 15*(-1)^(1/3)*a^(13/3)*b^(8/3)*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(3*a^(8/3))

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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(1/x**(5/3)/(b*x+a)**2,x)

[Out]

Timed out

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