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

Optimal. Leaf size=225 \[ -\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a^3 x^3-b^3}}{\sqrt {3} a}+\frac {x}{\sqrt {3}}}{x}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a^3 x^3-b^3}}{\sqrt {3} b}\right )}{\sqrt {3}}-\frac {2}{3} \tanh ^{-1}\left (\frac {-a x-b}{2 \sqrt [3]{a^3 x^3-b^3}-a x+b}\right )-\frac {1}{6} \log \left (-b \sqrt [3]{a^3 x^3-b^3}+\left (a^3 x^3-b^3\right )^{2/3}+b^2\right )+\frac {1}{6} \log \left (a x \sqrt [3]{a^3 x^3-b^3}+\left (a^3 x^3-b^3\right )^{2/3}+a^2 x^2\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 131, normalized size of antiderivative = 0.58, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1844, 239, 266, 56, 617, 204, 31} \begin {gather*} \frac {1}{2} \log \left (\sqrt [3]{a^3 x^3-b^3}+b\right )-\frac {1}{2} \log \left (\sqrt [3]{a^3 x^3-b^3}-a x\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 a x}{\sqrt [3]{a^3 x^3-b^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {b-2 \sqrt [3]{a^3 x^3-b^3}}{\sqrt {3} b}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 239

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + (2*Rt[b, 3]*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [C]  time = 0.18, size = 158, normalized size = 0.70 \begin {gather*} \frac {1}{6} \left (-2 \log \left (1-\frac {a x}{\sqrt [3]{a^3 x^3-b^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 a x}{\sqrt [3]{a^3 x^3-b^3}}+1}{\sqrt {3}}\right )-\frac {3 \left (a^3 x^3-b^3\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};1-\frac {a^3 x^3}{b^3}\right )}{b^2}+\log \left (\frac {a x}{\sqrt [3]{a^3 x^3-b^3}}+\frac {a^2 x^2}{\left (a^3 x^3-b^3\right )^{2/3}}+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3]*ArcTan[(1 + (2*a*x)/(-b^3 + a^3*x^3)^(1/3))/Sqrt[3]] - (3*(-b^3 + a^3*x^3)^(2/3)*Hypergeometric2F1[
2/3, 1, 5/3, 1 - (a^3*x^3)/b^3])/b^2 - 2*Log[1 - (a*x)/(-b^3 + a^3*x^3)^(1/3)] + Log[1 + (a^2*x^2)/(-b^3 + a^3
*x^3)^(2/3) + (a*x)/(-b^3 + a^3*x^3)^(1/3)])/6

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IntegrateAlgebraic [A]  time = 4.56, size = 225, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\frac {x}{\sqrt {3}}+\frac {2 \sqrt [3]{-b^3+a^3 x^3}}{\sqrt {3} a}}{x}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-b^3+a^3 x^3}}{\sqrt {3} b}\right )}{\sqrt {3}}-\frac {2}{3} \tanh ^{-1}\left (\frac {-b-a x}{b-a x+2 \sqrt [3]{-b^3+a^3 x^3}}\right )-\frac {1}{6} \log \left (b^2-b \sqrt [3]{-b^3+a^3 x^3}+\left (-b^3+a^3 x^3\right )^{2/3}\right )+\frac {1}{6} \log \left (a^2 x^2+a x \sqrt [3]{-b^3+a^3 x^3}+\left (-b^3+a^3 x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x - b}{{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {a x -b}{x \left (a^{3} x^{3}-b^{3}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 227, normalized size = 1.01 \begin {gather*} -\frac {1}{6} \, a {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a + \frac {2 \, {\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, a}\right )}{a} - \frac {\log \left (a^{2} + \frac {{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}} a}{x} + \frac {{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{a} + \frac {2 \, \log \left (-a + \frac {{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}}}{x}\right )}{a}\right )} - \frac {1}{6} \, b {\left (\frac {2 \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (b - 2 \, {\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, b}\right )}{b} + \frac {\log \left (b^{2} - {\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}} b + {\left (a^{3} x^{3} - b^{3}\right )}^{\frac {2}{3}}\right )}{b} - \frac {2 \, \log \left (b + {\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}}\right )}{b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.02, size = 166, normalized size = 0.74 \begin {gather*} \frac {\ln \left (b^3+b^2\,{\left (a^3\,x^3-b^3\right )}^{1/3}\right )}{3}+\ln \left (9\,b^3\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+b^2\,{\left (a^3\,x^3-b^3\right )}^{1/3}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (9\,b^3\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2+b^2\,{\left (a^3\,x^3-b^3\right )}^{1/3}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {a\,x\,{\left (1-\frac {a^3\,x^3}{b^3}\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ \frac {a^3\,x^3}{b^3}\right )}{{\left (a^3\,x^3-b^3\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b^3 + b^2*(a^3*x^3 - b^3)^(1/3))/3 + log(9*b^3*((3^(1/2)*1i)/6 - 1/6)^2 + b^2*(a^3*x^3 - b^3)^(1/3))*((3^(
1/2)*1i)/6 - 1/6) - log(9*b^3*((3^(1/2)*1i)/6 + 1/6)^2 + b^2*(a^3*x^3 - b^3)^(1/3))*((3^(1/2)*1i)/6 + 1/6) + (
a*x*(1 - (a^3*x^3)/b^3)^(1/3)*hypergeom([1/3, 1/3], 4/3, (a^3*x^3)/b^3))/(a^3*x^3 - b^3)^(1/3)

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sympy [C]  time = 4.55, size = 80, normalized size = 0.36 \begin {gather*} \frac {a x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {a^{3} x^{3}}{b^{3}}} \right )}}{3 b \Gamma \left (\frac {4}{3}\right )} + \frac {b \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b^{3} e^{2 i \pi }}{a^{3} x^{3}}} \right )}}{3 a x \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x*exp(-I*pi/3)*gamma(1/3)*hyper((1/3, 1/3), (4/3,), a**3*x**3/b**3)/(3*b*gamma(4/3)) + b*gamma(1/3)*hyper((1
/3, 1/3), (4/3,), b**3*exp_polar(2*I*pi)/(a**3*x**3))/(3*a*x*gamma(4/3))

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