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

Optimal. Leaf size=111 \[ -\frac {3 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{5/3}}+\frac {b^{2/3} \log (a+b x)}{2 a^{5/3}}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}-\frac {3}{2 a x^{2/3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {3 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{5/3}}+\frac {b^{2/3} \log (a+b x)}{2 a^{5/3}}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}-\frac {3}{2 a x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/3)*(a + b*x)),x]

[Out]

-3/(2*a*x^(2/3)) + (Sqrt[3]*b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/a^(5/3) - (3*b^(2
/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)])/(2*a^(5/3)) + (b^(2/3)*Log[a + b*x])/(2*a^(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 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)} \, dx &=-\frac {3}{2 a x^{2/3}}-\frac {b \int \frac {1}{x^{2/3} (a+b x)} \, dx}{a}\\ &=-\frac {3}{2 a x^{2/3}}+\frac {b^{2/3} \log (a+b x)}{2 a^{5/3}}-\frac {\left (3 \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^{4/3}}-\frac {\left (3 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^{5/3}}\\ &=-\frac {3}{2 a x^{2/3}}-\frac {3 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{5/3}}+\frac {b^{2/3} \log (a+b x)}{2 a^{5/3}}-\frac {\left (3 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^{5/3}}\\ &=-\frac {3}{2 a x^{2/3}}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3}}-\frac {3 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 a^{5/3}}+\frac {b^{2/3} \log (a+b x)}{2 a^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.24 \begin {gather*} -\frac {3 \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};-\frac {b x}{a}\right )}{2 a x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/3)*(a + b*x)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 137, normalized size = 1.23 \begin {gather*} \frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{2 a^{5/3}}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{a^{5/3}}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}-\frac {3}{2 a x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.42, size = 147, normalized size = 1.32 \begin {gather*} \frac {2 \, \sqrt {3} x \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 ) - x \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 ) + 2 \, x \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 \, x^{\frac {1}{3}}}{2 \, a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.14, size = 120, normalized size = 1.08 \begin {gather*} \frac {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 )}{a^{2}} - \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 )}{a^{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 )}{2 \, a^{2}} - \frac {3}{2 \, a 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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 105, normalized size = 0.95 \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 )}{\left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {\ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{\left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\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 )}{2 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {3}{2 a \,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),x)

[Out]

-3/2/a/x^(2/3)-1/a/(a/b)^(2/3)*ln(x^(1/3)+(a/b)^(1/3))+1/2/a/(a/b)^(2/3)*ln(x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^
(2/3))-1/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 = 2.99, size = 112, normalized size = 1.01 \begin {gather*} -\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 )}{a \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 )}{2 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {3}{2 \, a 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 138, normalized size = 1.24 \begin {gather*} \frac {b^{2/3}\,\ln \left (9\,{\left (-a\right )}^{7/3}\,b^{8/3}-9\,a^2\,b^3\,x^{1/3}\right )}{{\left (-a\right )}^{5/3}}-\frac {3}{2\,a\,x^{2/3}}+\frac {b^{2/3}\,\ln \left (9\,{\left (-a\right )}^{7/3}\,b^{8/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-9\,a^2\,b^3\,x^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}}-\frac {b^{2/3}\,\ln \left (9\,{\left (-a\right )}^{7/3}\,b^{8/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+9\,a^2\,b^3\,x^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/3)*(a + b*x)),x)

[Out]

(b^(2/3)*log(9*(-a)^(7/3)*b^(8/3) - 9*a^2*b^3*x^(1/3)))/(-a)^(5/3) - 3/(2*a*x^(2/3)) + (b^(2/3)*log(9*(-a)^(7/
3)*b^(8/3)*((3^(1/2)*1i)/2 - 1/2) - 9*a^2*b^3*x^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(-a)^(5/3) - (b^(2/3)*log(9*(-a
)^(7/3)*b^(8/3)*((3^(1/2)*1i)/2 + 1/2) + 9*a^2*b^3*x^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(-a)^(5/3)

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sympy [A]  time = 34.96, size = 221, normalized size = 1.99 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{3}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {3}{2 a x^{\frac {2}{3}}} & \text {for}\: b = 0 \\- \frac {3}{5 b x^{\frac {5}{3}}} & \text {for}\: a = 0 \\- \frac {3}{2 a x^{\frac {2}{3}}} + \frac {\sqrt [3]{-1} \log {\left (- \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + \sqrt [3]{x} \right )}}{a^{\frac {5}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}} - \frac {\sqrt [3]{-1} \log {\left (4 \left (-1\right )^{\frac {2}{3}} a^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{x} \sqrt [3]{\frac {1}{b}} + 4 x^{\frac {2}{3}} \right )}}{2 a^{\frac {5}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}} - \frac {\sqrt [3]{-1} \sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3}}{3} - \frac {2 \left (-1\right )^{\frac {2}{3}} \sqrt {3} \sqrt [3]{x}}{3 \sqrt [3]{a} \sqrt [3]{\frac {1}{b}}} \right )}}{a^{\frac {5}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(5/3), Eq(a, 0) & Eq(b, 0)), (-3/(2*a*x**(2/3)), Eq(b, 0)), (-3/(5*b*x**(5/3)), Eq(a, 0)), (
-3/(2*a*x**(2/3)) + (-1)**(1/3)*log(-(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + x**(1/3))/(a**(5/3)*(1/b)**(2/3)) - (
-1)**(1/3)*log(4*(-1)**(2/3)*a**(2/3)*(1/b)**(2/3) + 4*(-1)**(1/3)*a**(1/3)*x**(1/3)*(1/b)**(1/3) + 4*x**(2/3)
)/(2*a**(5/3)*(1/b)**(2/3)) - (-1)**(1/3)*sqrt(3)*atan(sqrt(3)/3 - 2*(-1)**(2/3)*sqrt(3)*x**(1/3)/(3*a**(1/3)*
(1/b)**(1/3)))/(a**(5/3)*(1/b)**(2/3)), True))

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