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

Optimal. Leaf size=263 \[ \frac {c \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{a x^3-b}+\left (a x^3-b\right )^{2/3}\right )}{6 \sqrt [3]{a}}-\frac {d \log \left (-\sqrt [3]{b} \sqrt [3]{a x^3-b}+\left (a x^3-b\right )^{2/3}+b^{2/3}\right )}{6 \sqrt [3]{b}}-\frac {c \log \left (\sqrt [3]{a x^3-b}-\sqrt [3]{a} x\right )}{3 \sqrt [3]{a}}-\frac {c \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{a}}+\frac {x}{\sqrt {3}}}{x}\right )}{\sqrt {3} \sqrt [3]{a}}+\frac {d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{3 \sqrt [3]{b}}+\frac {d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}} \]

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Rubi [A]  time = 0.15, antiderivative size = 163, normalized size of antiderivative = 0.62, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1844, 239, 266, 56, 617, 204, 31} \begin {gather*} -\frac {c \log \left (\sqrt [3]{a x^3-b}-\sqrt [3]{a} x\right )}{2 \sqrt [3]{a}}+\frac {c \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a} x}{\sqrt [3]{a x^3-b}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a}}+\frac {d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{2 \sqrt [3]{b}}+\frac {d \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {d \log (x)}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.18, size = 159, normalized size = 0.60 \begin {gather*} \frac {1}{6} \left (\frac {c \left (\log \left (\frac {a^{2/3} x^2}{\left (a x^3-b\right )^{2/3}}+\frac {\sqrt [3]{a} x}{\sqrt [3]{a x^3-b}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{a} x}{\sqrt [3]{a x^3-b}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a} x}{\sqrt [3]{a x^3-b}}+1}{\sqrt {3}}\right )\right )}{\sqrt [3]{a}}-\frac {3 d \left (a x^3-b\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};1-\frac {a x^3}{b}\right )}{b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 4.62, size = 263, normalized size = 1.00 \begin {gather*} -\frac {c \tan ^{-1}\left (\frac {\frac {x}{\sqrt {3}}+\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{a}}}{x}\right )}{\sqrt {3} \sqrt [3]{a}}+\frac {d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{3 \sqrt [3]{b}}-\frac {c \log \left (-\sqrt [3]{a} x+\sqrt [3]{-b+a x^3}\right )}{3 \sqrt [3]{a}}-\frac {d \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{6 \sqrt [3]{b}}+\frac {c \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{6 \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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((c*x-d)/x/(a*x^3-b)^(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 {c x - d}{{\left (a x^{3} - b\right )}^{\frac {1}{3}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {c x -d}{x \left (a \,x^{3}-b \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 209, normalized size = 0.79 \begin {gather*} -\frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a^{\frac {1}{3}} + \frac {2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {\log \left (a^{\frac {2}{3}} + \frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}} a^{\frac {1}{3}}}{x} + \frac {{\left (a x^{3} - b\right )}^{\frac {2}{3}}}{x^{2}}\right )}{a^{\frac {1}{3}}} + \frac {2 \, \log \left (-a^{\frac {1}{3}} + \frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}}}{x}\right )}{a^{\frac {1}{3}}}\right )} c - \frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} + \frac {\log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} - {\left (a x^{3} - b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}{b^{\frac {1}{3}}}\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.13, size = 165, normalized size = 0.63 \begin {gather*} \frac {d\,\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+b^{1/3}\,d^2\right )}{3\,b^{1/3}}-\frac {\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+\frac {b^{1/3}\,{\left (d-\sqrt {3}\,d\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (d-\sqrt {3}\,d\,1{}\mathrm {i}\right )}{6\,b^{1/3}}-\frac {\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+\frac {b^{1/3}\,{\left (d+\sqrt {3}\,d\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (d+\sqrt {3}\,d\,1{}\mathrm {i}\right )}{6\,b^{1/3}}+\frac {c\,x\,{\left (1-\frac {a\,x^3}{b}\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ \frac {a\,x^3}{b}\right )}{{\left (a\,x^3-b\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.11, size = 80, normalized size = 0.30 \begin {gather*} \frac {c 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 x^{3}}{b}} \right )}}{3 \sqrt [3]{b} \Gamma \left (\frac {4}{3}\right )} + \frac {d \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 e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [3]{a} x \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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