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

Optimal. Leaf size=94 \[ -\frac{(a+b x)^{2/3}}{x}-\frac{b \log (x)}{3 \sqrt [3]{a}}+\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}}+\frac{2 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} \sqrt [3]{a}} \]

[Out]

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

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Rubi [A]  time = 0.0332543, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {47, 55, 617, 204, 31} \[ -\frac{(a+b x)^{2/3}}{x}-\frac{b \log (x)}{3 \sqrt [3]{a}}+\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}}+\frac{2 b \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(2/3)/x^2,x]

[Out]

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

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 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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] && PosQ
[(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{(a+b x)^{2/3}}{x^2} \, dx &=-\frac{(a+b x)^{2/3}}{x}+\frac{1}{3} (2 b) \int \frac{1}{x \sqrt [3]{a+b x}} \, dx\\ &=-\frac{(a+b x)^{2/3}}{x}-\frac{b \log (x)}{3 \sqrt [3]{a}}+b \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}}\\ &=-\frac{(a+b x)^{2/3}}{x}-\frac{b \log (x)}{3 \sqrt [3]{a}}+\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}\right )}{\sqrt [3]{a}}\\ &=-\frac{(a+b x)^{2/3}}{x}+\frac{2 b \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{a}}-\frac{b \log (x)}{3 \sqrt [3]{a}}+\frac{b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}}\\ \end{align*}

Mathematica [C]  time = 0.0140435, size = 33, normalized size = 0.35 \[ \frac{3 b (a+b x)^{5/3} \, _2F_1\left (\frac{5}{3},2;\frac{8}{3};\frac{b x}{a}+1\right )}{5 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(2/3)/x^2,x]

[Out]

(3*b*(a + b*x)^(5/3)*Hypergeometric2F1[5/3, 2, 8/3, 1 + (b*x)/a])/(5*a^2)

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Maple [A]  time = 0.01, size = 92, normalized size = 1. \begin{align*} -{\frac{1}{x} \left ( bx+a \right ) ^{{\frac{2}{3}}}}+{\frac{2\,b}{3}\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}-{\frac{b}{3}\ln \left ( \left ( bx+a \right ) ^{{\frac{2}{3}}}+\sqrt [3]{a}\sqrt [3]{bx+a}+{a}^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{a}}}}+{\frac{2\,b\sqrt{3}}{3}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt [3]{bx+a}}{\sqrt [3]{a}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x+a)^(2/3)/x+2/3*b/a^(1/3)*ln((b*x+a)^(1/3)-a^(1/3))-1/3*b/a^(1/3)*ln((b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+
a^(2/3))+2/3*b*3^(1/2)/a^(1/3)*arctan(1/3*3^(1/2)*(2/a^(1/3)*(b*x+a)^(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((b*x+a)^(2/3)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91553, size = 760, normalized size = 8.09 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}} a b x \sqrt{-\frac{1}{a^{\frac{2}{3}}}} \log \left (\frac{2 \, b x + 3 \, \sqrt{\frac{1}{3}}{\left (2 \,{\left (b x + a\right )}^{\frac{2}{3}} a^{\frac{2}{3}} -{\left (b x + a\right )}^{\frac{1}{3}} a - a^{\frac{4}{3}}\right )} \sqrt{-\frac{1}{a^{\frac{2}{3}}}} - 3 \,{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{2}{3}} + 3 \, a}{x}\right ) - a^{\frac{2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + 2 \, a^{\frac{2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}}\right ) - 3 \,{\left (b x + a\right )}^{\frac{2}{3}} a}{3 \, a x}, \frac{6 \, \sqrt{\frac{1}{3}} a^{\frac{2}{3}} b x \arctan \left (\frac{\sqrt{\frac{1}{3}}{\left (2 \,{\left (b x + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{a^{\frac{1}{3}}}\right ) - a^{\frac{2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + 2 \, a^{\frac{2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}}\right ) - 3 \,{\left (b x + a\right )}^{\frac{2}{3}} a}{3 \, a x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*sqrt(1/3)*a*b*x*sqrt(-1/a^(2/3))*log((2*b*x + 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*a^(2/3) - (b*x + a)^(1/3)
*a - a^(4/3))*sqrt(-1/a^(2/3)) - 3*(b*x + a)^(1/3)*a^(2/3) + 3*a)/x) - a^(2/3)*b*x*log((b*x + a)^(2/3) + (b*x
+ a)^(1/3)*a^(1/3) + a^(2/3)) + 2*a^(2/3)*b*x*log((b*x + a)^(1/3) - a^(1/3)) - 3*(b*x + a)^(2/3)*a)/(a*x), 1/3
*(6*sqrt(1/3)*a^(2/3)*b*x*arctan(sqrt(1/3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3)) - a^(2/3)*b*x*log((b*x + a)^
(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3)) + 2*a^(2/3)*b*x*log((b*x + a)^(1/3) - a^(1/3)) - 3*(b*x + a)^(2/3)*
a)/(a*x)]

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Sympy [C]  time = 3.17599, size = 643, normalized size = 6.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(2/3)/x**2,x)

[Out]

10*a**(8/3)*b*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma
(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) + 10*a**(8/3)*b*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)
**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I
*pi/3)*gamma(8/3)) + 10*a**(8/3)*b*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(5/3)/
(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) - 10*a**(5/3)*b**2*(a/b + x)*e
xp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*
b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) - 10*a**(5/3)*b**2*(a/b + x)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**
(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*p
i/3)*gamma(8/3)) - 10*a**(5/3)*b**2*(a/b + x)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*
gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) + 15*a**2*b**(5/3)*
(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*
gamma(8/3))

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Giac [A]  time = 2.16197, size = 143, normalized size = 1.52 \begin{align*} \frac{\frac{2 \, \sqrt{3} b^{2} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right )}{a^{\frac{1}{3}}} - \frac{b^{2} \log \left ({\left (b x + a\right )}^{\frac{2}{3}} +{\left (b x + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right )}{a^{\frac{1}{3}}} + \frac{2 \, b^{2} \log \left ({\left |{\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{1}{3}}} - \frac{3 \,{\left (b x + a\right )}^{\frac{2}{3}} b}{x}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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