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

Optimal. Leaf size=125 \[ -\frac{3 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}+\frac{a^{5/3} \log (a+b x)}{2 b^{8/3}}-\frac{\sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{b^{8/3}}-\frac{3 a x^{2/3}}{2 b^2}+\frac{3 x^{5/3}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0700581, antiderivative size = 125, normalized size of antiderivative = 1., 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 = {50, 56, 617, 204, 31} \[ -\frac{3 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}+\frac{a^{5/3} \log (a+b x)}{2 b^{8/3}}-\frac{\sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{b^{8/3}}-\frac{3 a x^{2/3}}{2 b^2}+\frac{3 x^{5/3}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/3)/(a + b*x),x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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]

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 0.0093248, size = 38, normalized size = 0.3 \[ \frac{3 x^{2/3} \left (5 a \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\frac{b x}{a}\right )-5 a+2 b x\right )}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/3)/(a + b*x),x]

[Out]

(3*x^(2/3)*(-5*a + 2*b*x + 5*a*Hypergeometric2F1[2/3, 1, 5/3, -((b*x)/a)]))/(10*b^2)

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Maple [A]  time = 0.007, size = 122, normalized size = 1. \begin{align*}{\frac{3}{5\,b}{x}^{{\frac{5}{3}}}}-{\frac{3\,a}{2\,{b}^{2}}{x}^{{\frac{2}{3}}}}-{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \sqrt [3]{x}+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}}{2\,{b}^{3}}\ln \left ({x}^{{\frac{2}{3}}}-\sqrt [3]{{\frac{a}{b}}}\sqrt [3]{x}+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}\sqrt{3}}{{b}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sqrt [3]{x}{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50858, size = 381, normalized size = 3.05 \begin{align*} \frac{10 \, \sqrt{3} a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} b x^{\frac{1}{3}} \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} + \sqrt{3} a}{3 \, a}\right ) - 5 \, a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (-b x^{\frac{1}{3}} \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} + a x^{\frac{2}{3}} - a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}}\right ) + 10 \, a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (b \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} + a x^{\frac{1}{3}}\right ) + 3 \,{\left (2 \, b x - 5 \, a\right )} x^{\frac{2}{3}}}{10 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.09942, size = 186, normalized size = 1.49 \begin{align*} -\frac{a \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 )}{b^{2}} - \frac{\sqrt{3} \left (-a b^{2}\right )^{\frac{2}{3}} a \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 )}{b^{4}} + \frac{\left (-a b^{2}\right )^{\frac{2}{3}} a \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 \, b^{4}} + \frac{3 \,{\left (2 \, b^{4} x^{\frac{5}{3}} - 5 \, a b^{3} x^{\frac{2}{3}}\right )}}{10 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-a/b)^(2/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/b^2 - sqrt(3)*(-a*b^2)^(2/3)*a*arctan(1/3*sqrt(3)*(2*x^(1/3)
+ (-a/b)^(1/3))/(-a/b)^(1/3))/b^4 + 1/2*(-a*b^2)^(2/3)*a*log(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/b^
4 + 3/10*(2*b^4*x^(5/3) - 5*a*b^3*x^(2/3))/b^5