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

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

[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))

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Rubi [A]  time = 0.0606935, antiderivative size = 152, normalized size of antiderivative = 1., 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 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 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{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*}

Mathematica [C]  time = 0.0057024, 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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Maple [A]  time = 0.014, size = 139, normalized size = 0.9 \begin{align*} -{\frac{3}{2\,{a}^{3}}{x}^{-{\frac{2}{3}}}}-{\frac{11\,{b}^{2}}{6\,{a}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{4}{3}}}}-{\frac{7\,b}{3\,{a}^{2} \left ( bx+a \right ) ^{2}}\sqrt [3]{x}}-{\frac{20}{9\,{a}^{3}}\ln \left ( \sqrt [3]{x}+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{10}{9\,{a}^{3}}\ln \left ({x}^{{\frac{2}{3}}}-\sqrt [3]{{\frac{a}{b}}}\sqrt [3]{x}+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{20\,\sqrt{3}}{9\,{a}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sqrt [3]{x}{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

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/(1/b*a)^(2/3)*ln(x^(1/3)+
(1/b*a)^(1/3))+10/9/a^3/(1/b*a)^(2/3)*ln(x^(2/3)-(1/b*a)^(1/3)*x^(1/3)+(1/b*a)^(2/3))-20/9/a^3/(1/b*a)^(2/3)*3
^(1/2)*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(1/x^(5/3)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.56814, size = 570, normalized size = 3.75 \begin{align*} \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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.07723, size = 203, normalized size = 1.34 \begin{align*} \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}} \end{align*}

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)