3.5.11 \(\int \frac {1}{x^2 (a+b x)^{2/3}} \, dx\)

Optimal. Leaf size=98 \[ \frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{5/3}}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\sqrt [3]{a+b x}}{a x} \]

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Rubi [A]  time = 0.03, antiderivative size = 98, 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, 57, 617, 204, 31} \begin {gather*} \frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{5/3}}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\sqrt [3]{a+b x}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(2/3)),x]

[Out]

-((a + b*x)^(1/3)/(a*x)) + (2*b*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3)) + (
b*Log[x])/(3*a^(5/3)) - (b*Log[a^(1/3) - (a + b*x)^(1/3)])/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 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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]
&& PosQ[(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^2 (a+b x)^{2/3}} \, dx &=-\frac {\sqrt [3]{a+b x}}{a x}-\frac {(2 b) \int \frac {1}{x (a+b x)^{2/3}} \, dx}{3 a}\\ &=-\frac {\sqrt [3]{a+b x}}{a x}+\frac {b \log (x)}{3 a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{a^{4/3}}\\ &=-\frac {\sqrt [3]{a+b x}}{a x}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{5/3}}-\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 )}{a^{5/3}}\\ &=-\frac {\sqrt [3]{a+b x}}{a x}+\frac {2 b \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{5/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 31, normalized size = 0.32 \begin {gather*} \frac {3 b \sqrt [3]{a+b x} \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};\frac {b x}{a}+1\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(2/3)),x]

[Out]

(3*b*(a + b*x)^(1/3)*Hypergeometric2F1[1/3, 2, 4/3, 1 + (b*x)/a])/a^2

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IntegrateAlgebraic [A]  time = 0.15, size = 128, normalized size = 1.31 \begin {gather*} -\frac {2 b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{3 a^{5/3}}+\frac {b \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{3 a^{5/3}}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} a^{5/3}}-\frac {\sqrt [3]{a+b x}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a + b*x)^(2/3)),x]

[Out]

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

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fricas [B]  time = 0.69, size = 166, normalized size = 1.69 \begin {gather*} \frac {2 \, \sqrt {3} a b x \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, \sqrt {3} \left (-a^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + \left (-a^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{2}}{3 \, a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(2/3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.37, size = 108, normalized size = 1.10 \begin {gather*} \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 {5}{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 {5}{3}}} - \frac {2 \, b^{2} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {1}{3}} b}{a x}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x+a)^(2/3),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^(5/3) + b^2*log((b*x + a)^(2/3)
+ (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2*b^2*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*(b*x + a
)^(1/3)*b/(a*x))/b

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maple [A]  time = 0.01, size = 95, normalized size = 0.97 \begin {gather*} \frac {2 \sqrt {3}\, b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {5}{3}}}-\frac {2 b \ln \left (-a^{\frac {1}{3}}+\left (b x +a \right )^{\frac {1}{3}}\right )}{3 a^{\frac {5}{3}}}+\frac {b \ln \left (a^{\frac {2}{3}}+\left (b x +a \right )^{\frac {1}{3}} a^{\frac {1}{3}}+\left (b x +a \right )^{\frac {2}{3}}\right )}{3 a^{\frac {5}{3}}}-\frac {\left (b x +a \right )^{\frac {1}{3}}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x+a)^(2/3),x)

[Out]

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

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maxima [A]  time = 2.95, size = 106, normalized size = 1.08 \begin {gather*} \frac {2 \, \sqrt {3} b \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 )}{3 \, a^{\frac {5}{3}}} - \frac {{\left (b x + a\right )}^{\frac {1}{3}} b}{{\left (b x + a\right )} a - a^{2}} + \frac {b \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 )}{3 \, a^{\frac {5}{3}}} - \frac {2 \, b \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {5}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 122, normalized size = 1.24 \begin {gather*} -\frac {{\left (a+b\,x\right )}^{1/3}}{a\,x}+\frac {\ln \left (\frac {3\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{a^{2/3}}+\frac {6\,b\,{\left (a+b\,x\right )}^{1/3}}{a}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{3\,a^{5/3}}+\frac {\ln \left (\frac {3\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{a^{2/3}}+\frac {6\,b\,{\left (a+b\,x\right )}^{1/3}}{a}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{3\,a^{5/3}}-\frac {2\,b\,\ln \left ({\left (a+b\,x\right )}^{1/3}-a^{1/3}\right )}{3\,a^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)^(2/3)),x)

[Out]

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

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sympy [C]  time = 2.27, size = 830, normalized size = 8.47

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x+a)**(2/3),x)

[Out]

-2*a**(4/3)*b**(5/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/3)/(9*
a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9*a**2*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma
(4/3)) - 2*a**(4/3)*b**(5/3)*(a/b + x)**(2/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*
gamma(1/3)/(9*a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9*a**2*b**(5/3)*(a/b + x)**(5/3)*exp(2
*I*pi/3)*gamma(4/3)) - 2*a**(4/3)*b**(5/3)*(a/b + x)**(2/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*e
xp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(9*a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9*a**2*b*
*(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3)) + 2*a**(1/3)*b**(8/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*log(1 -
b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/3)/(9*a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9
*a**2*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3)) + 2*a**(1/3)*b**(8/3)*(a/b + x)**(5/3)*log(1 - b**(1
/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(1/3)/(9*a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*
gamma(4/3) - 9*a**2*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3)) + 2*a**(1/3)*b**(8/3)*(a/b + x)**(5/3)
*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(9*a**3*b**(2/3)*(a
/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9*a**2*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3)) + 3*a*b**
2*(a/b + x)*exp(2*I*pi/3)*gamma(1/3)/(9*a**3*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(4/3) - 9*a**2*b**(5
/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3))

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