3.5.4 \(\int \frac {1}{x^2 \sqrt [3]{-a+b x}} \, dx\)

Optimal. Leaf size=103 \[ \frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{2 a^{4/3}}-\frac {b \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3}}+\frac {(b x-a)^{2/3}}{a x} \]

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Rubi [A]  time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {51, 56, 617, 204, 31} \begin {gather*} \frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{2 a^{4/3}}-\frac {b \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3}}+\frac {(b x-a)^{2/3}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(-a + b*x)^(1/3)),x]

[Out]

(-a + b*x)^(2/3)/(a*x) - (b*ArcTan[(a^(1/3) - 2*(-a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)) + (b*L
og[x])/(6*a^(4/3)) - (b*Log[a^(1/3) + (-a + b*x)^(1/3)])/(2*a^(4/3))

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 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 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 \sqrt [3]{-a+b x}} \, dx &=\frac {(-a+b x)^{2/3}}{a x}+\frac {b \int \frac {1}{x \sqrt [3]{-a+b x}} \, dx}{3 a}\\ &=\frac {(-a+b x)^{2/3}}{a x}+\frac {b \log (x)}{6 a^{4/3}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+x} \, dx,x,\sqrt [3]{-a+b x}\right )}{2 a^{4/3}}+\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 )}{2 a}\\ &=\frac {(-a+b x)^{2/3}}{a x}+\frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{a}+\sqrt [3]{-a+b x}\right )}{2 a^{4/3}}+\frac {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^{4/3}}\\ &=\frac {(-a+b x)^{2/3}}{a x}-\frac {b \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{-a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{4/3}}+\frac {b \log (x)}{6 a^{4/3}}-\frac {b \log \left (\sqrt [3]{a}+\sqrt [3]{-a+b x}\right )}{2 a^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.35 \begin {gather*} \frac {3 b (b x-a)^{2/3} \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};1-\frac {b x}{a}\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(-a + b*x)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 136, normalized size = 1.32 \begin {gather*} -\frac {b \log \left (\sqrt [3]{b x-a}+\sqrt [3]{a}\right )}{3 a^{4/3}}+\frac {b \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b x-a}+(b x-a)^{2/3}\right )}{6 a^{4/3}}-\frac {b \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b x-a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3}}+\frac {(b x-a)^{2/3}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-a + b*x)^(2/3)/(a*x) - (b*ArcTan[1/Sqrt[3] - (2*(-a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)) - (b
*Log[a^(1/3) + (-a + b*x)^(1/3)])/(3*a^(4/3)) + (b*Log[a^(2/3) - a^(1/3)*(-a + b*x)^(1/3) + (-a + b*x)^(2/3)])
/(6*a^(4/3))

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fricas [A]  time = 1.06, size = 328, normalized size = 3.18 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} a b x \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b x + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x - a\right )}^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} a + \left (-a\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a\right )^{\frac {1}{3}}}{a}} - 3 \, {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} - 3 \, a}{x}\right ) + \left (-a\right )^{\frac {2}{3}} b x \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a\right )^{\frac {2}{3}} b x \log \left ({\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}}\right ) + 6 \, {\left (b x - a\right )}^{\frac {2}{3}} a}{6 \, a^{2} x}, \frac {6 \, \sqrt {\frac {1}{3}} a b x \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}} \arctan \left (\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x - a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a\right )^{\frac {1}{3}}}{a}}\right ) + \left (-a\right )^{\frac {2}{3}} b x \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a\right )^{\frac {2}{3}} b x \log \left ({\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}}\right ) + 6 \, {\left (b x - a\right )}^{\frac {2}{3}} a}{6 \, a^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.43, size = 144, normalized size = 1.40 \begin {gather*} \frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x - a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right )}}{3 \, \left (-a\right )^{\frac {1}{3}}}\right )}{\left (-a\right )^{\frac {1}{3}} a} - \frac {b^{2} \log \left ({\left (b x - a\right )}^{\frac {2}{3}} + {\left (b x - a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right )}{\left (-a\right )^{\frac {1}{3}} a} - \frac {2 \, \left (-a\right )^{\frac {2}{3}} b^{2} \log \left ({\left | {\left (b x - a\right )}^{\frac {1}{3}} - \left (-a\right )^{\frac {1}{3}} \right |}\right )}{a^{2}} + \frac {6 \, {\left (b x - a\right )}^{\frac {2}{3}} b}{a x}}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 103, normalized size = 1.00 \begin {gather*} \frac {\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 {4}{3}}}-\frac {b \ln \left (a^{\frac {1}{3}}+\left (b x -a \right )^{\frac {1}{3}}\right )}{3 a^{\frac {4}{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 )}{6 a^{\frac {4}{3}}}+\frac {\left (b x -a \right )^{\frac {2}{3}}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x-a)^(2/3)/a/x-1/3*b*ln(a^(1/3)+(b*x-a)^(1/3))/a^(4/3)+1/6*b/a^(4/3)*ln(a^(2/3)-(b*x-a)^(1/3)*a^(1/3)+(b*x-
a)^(2/3))+1/3*b/a^(4/3)*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 = 2.99, size = 116, normalized size = 1.13 \begin {gather*} \frac {\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 {4}{3}}} + \frac {{\left (b x - a\right )}^{\frac {2}{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 )}{6 \, a^{\frac {4}{3}}} - \frac {b \log \left ({\left (b x - a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}{3 \, a^{\frac {4}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 133, normalized size = 1.29 \begin {gather*} \frac {{\left (b\,x-a\right )}^{2/3}}{a\,x}-\frac {b\,\ln \left ({\left (b\,x-a\right )}^{1/3}+a^{1/3}\right )}{3\,a^{4/3}}+\frac {\ln \left (\frac {{\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{4\,a^{5/3}}+\frac {b^2\,{\left (b\,x-a\right )}^{1/3}}{a^2}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{4/3}}+\frac {\ln \left (\frac {{\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2}{4\,a^{5/3}}+\frac {b^2\,{\left (b\,x-a\right )}^{1/3}}{a^2}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x - a)^(1/3)),x)

[Out]

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

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sympy [C]  time = 2.24, size = 838, normalized size = 8.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**(5/3)*b**(7/3)*(-a/b + x)**(4/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(2/
3)/(9*a**3*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/
3)*gamma(5/3)) - 2*a**(5/3)*b**(7/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_po
lar(I*pi)/a**(1/3))*gamma(2/3)/(9*a**3*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3)*(
-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3)) - 2*a**(5/3)*b**(7/3)*(-a/b + x)**(4/3)*exp(-2*I*pi/3)*log(1 - b**(
1/3)*(-a/b + x)**(1/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(2/3)/(9*a**3*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/
3)*gamma(5/3) + 9*a**2*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3)) - 2*a**(2/3)*b**(10/3)*(-a/b + x)*
*(7/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(2/3)/(9*a**3*b**(4/3)*(-a/b + x)**
(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3)) - 2*a**(2/3)*b**(
10/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_polar(I*pi)/a**(1/3))*gamma(2/3)/
(9*a**3*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*
gamma(5/3)) - 2*a**(2/3)*b**(10/3)*(-a/b + x)**(7/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(-a/b + x)**(1/3)*exp_pol
ar(5*I*pi/3)/a**(1/3))*gamma(2/3)/(9*a**3*b**(4/3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3
)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3)) + 6*a*b**3*(-a/b + x)**2*exp(2*I*pi/3)*gamma(2/3)/(9*a**3*b**(4/
3)*(-a/b + x)**(4/3)*exp(2*I*pi/3)*gamma(5/3) + 9*a**2*b**(7/3)*(-a/b + x)**(7/3)*exp(2*I*pi/3)*gamma(5/3))

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