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

Optimal. Leaf size=129 \[ \frac{5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac{5 a^{2/3} \log (a+b x)}{6 b^{8/3}}+\frac{5 a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{8/3}}-\frac{x^{5/3}}{b (a+b x)}+\frac{5 x^{2/3}}{2 b^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0470636, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {47, 50, 56, 617, 204, 31} \[ \frac{5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac{5 a^{2/3} \log (a+b x)}{6 b^{8/3}}+\frac{5 a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{8/3}}-\frac{x^{5/3}}{b (a+b x)}+\frac{5 x^{2/3}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/3)/(a + b*x)^2,x]

[Out]

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

Rule 47

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

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

Mathematica [C]  time = 0.0048991, size = 27, normalized size = 0.21 \[ \frac{3 x^{8/3} \, _2F_1\left (2,\frac{8}{3};\frac{11}{3};-\frac{b x}{a}\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/3)/(a + b*x)^2,x]

[Out]

(3*x^(8/3)*Hypergeometric2F1[2, 8/3, 11/3, -((b*x)/a)])/(8*a^2)

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 123, normalized size = 1. \begin{align*}{\frac{3}{2\,{b}^{2}}{x}^{{\frac{2}{3}}}}+{\frac{a}{{b}^{2} \left ( bx+a \right ) }{x}^{{\frac{2}{3}}}}+{\frac{5\,a}{3\,{b}^{3}}\ln \left ( \sqrt [3]{x}+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{5\,a}{6\,{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{5\,a\sqrt{3}}{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)^2,x)

[Out]

3/2*x^(2/3)/b^2+1/b^2*a*x^(2/3)/(b*x+a)+5/3/b^3*a/(1/b*a)^(1/3)*ln(x^(1/3)+(1/b*a)^(1/3))-5/6/b^3*a/(1/b*a)^(1
/3)*ln(x^(2/3)-(1/b*a)^(1/3)*x^(1/3)+(1/b*a)^(2/3))-5/3/b^3*a*3^(1/2)/(1/b*a)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b
*a)^(1/3)*x^(1/3)-1))

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.55877, size = 420, normalized size = 3.26 \begin{align*} -\frac{10 \, \sqrt{3}{\left (b x + a\right )} \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 \,{\left (b x + a\right )} \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 \,{\left (b x + a\right )} \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 (3 \, b x + 5 \, a\right )} x^{\frac{2}{3}}}{6 \,{\left (b^{3} x + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.07725, size = 182, normalized size = 1.41 \begin{align*} \frac{5 \, \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 )}{3 \, b^{2}} + \frac{a x^{\frac{2}{3}}}{{\left (b x + a\right )} b^{2}} + \frac{3 \, x^{\frac{2}{3}}}{2 \, b^{2}} + \frac{5 \, \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 )}{3 \, b^{4}} - \frac{5 \, \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 )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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