3.22.80 \(\int \frac {-d+c x}{x^4 \sqrt [3]{-b+a x^3}} \, dx\)

Optimal. Leaf size=162 \[ \frac {a d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{9 b^{4/3}}-\frac {a d \log \left (-\sqrt [3]{b} \sqrt [3]{a x^3-b}+\left (a x^3-b\right )^{2/3}+b^{2/3}\right )}{18 b^{4/3}}+\frac {a d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {\left (a x^3-b\right )^{2/3} (3 c x-2 d)}{6 b x^3} \]

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Rubi [A]  time = 0.18, antiderivative size = 142, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1844, 266, 51, 56, 617, 204, 31, 264} \begin {gather*} \frac {a d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{6 b^{4/3}}+\frac {a d \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} b^{4/3}}-\frac {a d \log (x)}{6 b^{4/3}}+\frac {c \left (a x^3-b\right )^{2/3}}{2 b x^2}-\frac {d \left (a x^3-b\right )^{2/3}}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-d + c*x)/(x^4*(-b + a*x^3)^(1/3)),x]

[Out]

-1/3*(d*(-b + a*x^3)^(2/3))/(b*x^3) + (c*(-b + a*x^3)^(2/3))/(2*b*x^2) + (a*d*ArcTan[(b^(1/3) - 2*(-b + a*x^3)
^(1/3))/(Sqrt[3]*b^(1/3))])/(3*Sqrt[3]*b^(4/3)) - (a*d*Log[x])/(6*b^(4/3)) + (a*d*Log[b^(1/3) + (-b + a*x^3)^(
1/3)])/(6*b^(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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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^4 \sqrt [3]{-b+a x^3}} \, dx &=\int \left (-\frac {d}{x^4 \sqrt [3]{-b+a x^3}}+\frac {c}{x^3 \sqrt [3]{-b+a x^3}}\right ) \, dx\\ &=c \int \frac {1}{x^3 \sqrt [3]{-b+a x^3}} \, dx-d \int \frac {1}{x^4 \sqrt [3]{-b+a x^3}} \, dx\\ &=\frac {c \left (-b+a x^3\right )^{2/3}}{2 b x^2}-\frac {1}{3} d \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{-b+a x}} \, dx,x,x^3\right )\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{3 b x^3}+\frac {c \left (-b+a x^3\right )^{2/3}}{2 b x^2}-\frac {(a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{-b+a x}} \, dx,x,x^3\right )}{9 b}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{3 b x^3}+\frac {c \left (-b+a x^3\right )^{2/3}}{2 b x^2}-\frac {a d \log (x)}{6 b^{4/3}}+\frac {(a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{b}+x} \, dx,x,\sqrt [3]{-b+a x^3}\right )}{6 b^{4/3}}-\frac {(a 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 )}{6 b}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{3 b x^3}+\frac {c \left (-b+a x^3\right )^{2/3}}{2 b x^2}-\frac {a d \log (x)}{6 b^{4/3}}+\frac {a d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{6 b^{4/3}}-\frac {(a 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 )}{3 b^{4/3}}\\ &=-\frac {d \left (-b+a x^3\right )^{2/3}}{3 b x^3}+\frac {c \left (-b+a x^3\right )^{2/3}}{2 b x^2}+\frac {a d \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}-\frac {a d \log (x)}{6 b^{4/3}}+\frac {a d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{6 b^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 53, normalized size = 0.33 \begin {gather*} \frac {\left (a x^3-b\right )^{2/3} \left (b c-a d x^2 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};1-\frac {a x^3}{b}\right )\right )}{2 b^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-d + c*x)/(x^4*(-b + a*x^3)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 9.44, size = 162, normalized size = 1.00 \begin {gather*} \frac {(-2 d+3 c x) \left (-b+a x^3\right )^{2/3}}{6 b x^3}+\frac {a d \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {a d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{9 b^{4/3}}-\frac {a d \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{18 b^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-d + c*x)/(x^4*(-b + a*x^3)^(1/3)),x]

[Out]

((-2*d + 3*c*x)*(-b + a*x^3)^(2/3))/(6*b*x^3) + (a*d*ArcTan[1/Sqrt[3] - (2*(-b + a*x^3)^(1/3))/(Sqrt[3]*b^(1/3
))])/(3*Sqrt[3]*b^(4/3)) + (a*d*Log[b^(1/3) + (-b + a*x^3)^(1/3)])/(9*b^(4/3)) - (a*d*Log[b^(2/3) - b^(1/3)*(-
b + a*x^3)^(1/3) + (-b + a*x^3)^(2/3)])/(18*b^(4/3))

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fricas [A]  time = 83.84, size = 352, normalized size = 2.17 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} a b d x^{3} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {2 \, a x^{3} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} b^{\frac {2}{3}} + {\left (a x^{3} - b\right )}^{\frac {1}{3}} b - b^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} - 3 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} b^{\frac {2}{3}} - 3 \, b}{x^{3}}\right ) + 2 \, a b^{\frac {2}{3}} d x^{3} \log \left (\frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}}{x}\right ) - a b^{\frac {2}{3}} d x^{3} \log \left (\frac {{\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}}}{x^{2}}\right ) + 3 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} {\left (3 \, b c x - 2 \, b d\right )}}{18 \, b^{2} x^{3}}, -\frac {6 \, \sqrt {\frac {1}{3}} a b^{\frac {2}{3}} d x^{3} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}}}\right ) - 2 \, a b^{\frac {2}{3}} d x^{3} \log \left (\frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}}{x}\right ) + a b^{\frac {2}{3}} d x^{3} \log \left (\frac {{\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}}}{x^{2}}\right ) - 3 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} {\left (3 \, b c x - 2 \, b d\right )}}{18 \, b^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^4/(a*x^3-b)^(1/3),x, algorithm="fricas")

[Out]

[1/18*(3*sqrt(1/3)*a*b*d*x^3*sqrt(-1/b^(2/3))*log((2*a*x^3 - 3*sqrt(1/3)*(2*(a*x^3 - b)^(2/3)*b^(2/3) + (a*x^3
- b)^(1/3)*b - b^(4/3))*sqrt(-1/b^(2/3)) - 3*(a*x^3 - b)^(1/3)*b^(2/3) - 3*b)/x^3) + 2*a*b^(2/3)*d*x^3*log(((
a*x^3 - b)^(1/3) + b^(1/3))/x) - a*b^(2/3)*d*x^3*log(((a*x^3 - b)^(2/3) - (a*x^3 - b)^(1/3)*b^(1/3) + b^(2/3))
/x^2) + 3*(a*x^3 - b)^(2/3)*(3*b*c*x - 2*b*d))/(b^2*x^3), -1/18*(6*sqrt(1/3)*a*b^(2/3)*d*x^3*arctan(sqrt(1/3)*
(2*(a*x^3 - b)^(1/3) - b^(1/3))/b^(1/3)) - 2*a*b^(2/3)*d*x^3*log(((a*x^3 - b)^(1/3) + b^(1/3))/x) + a*b^(2/3)*
d*x^3*log(((a*x^3 - b)^(2/3) - (a*x^3 - b)^(1/3)*b^(1/3) + b^(2/3))/x^2) - 3*(a*x^3 - b)^(2/3)*(3*b*c*x - 2*b*
d))/(b^2*x^3)]

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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^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^4/(a*x^3-b)^(1/3),x, algorithm="giac")

[Out]

integrate((c*x - d)/((a*x^3 - b)^(1/3)*x^4), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {c x -d}{x^{4} \left (a \,x^{3}-b \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x-d)/x^4/(a*x^3-b)^(1/3),x)

[Out]

int((c*x-d)/x^4/(a*x^3-b)^(1/3),x)

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maxima [A]  time = 0.45, size = 152, normalized size = 0.94 \begin {gather*} -\frac {1}{18} \, {\left (\frac {2 \, \sqrt {3} a \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 {4}{3}}} + \frac {6 \, {\left (a x^{3} - b\right )}^{\frac {2}{3}} a}{{\left (a x^{3} - b\right )} b + b^{2}} + \frac {a \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 {4}{3}}} - \frac {2 \, a \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}{b^{\frac {4}{3}}}\right )} d + \frac {{\left (a x^{3} - b\right )}^{\frac {2}{3}} c}{2 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x^4/(a*x^3-b)^(1/3),x, algorithm="maxima")

[Out]

-1/18*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*(a*x^3 - b)^(1/3) - b^(1/3))/b^(1/3))/b^(4/3) + 6*(a*x^3 - b)^(2/3)*a
/((a*x^3 - b)*b + b^2) + a*log((a*x^3 - b)^(2/3) - (a*x^3 - b)^(1/3)*b^(1/3) + b^(2/3))/b^(4/3) - 2*a*log((a*x
^3 - b)^(1/3) + b^(1/3))/b^(4/3))*d + 1/2*(a*x^3 - b)^(2/3)*c/(b*x^2)

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mupad [B]  time = 1.78, size = 184, normalized size = 1.14 \begin {gather*} \frac {c\,{\left (a\,x^3-b\right )}^{2/3}}{2\,b\,x^2}-\frac {\ln \left (\frac {{\left (a\,d+\sqrt {3}\,a\,d\,1{}\mathrm {i}\right )}^2}{36\,b^{5/3}}+\frac {a^2\,d^2\,{\left (a\,x^3-b\right )}^{1/3}}{9\,b^2}\right )\,\left (a\,d+\sqrt {3}\,a\,d\,1{}\mathrm {i}\right )}{18\,b^{4/3}}-\frac {\ln \left (\frac {{\left (a\,d-\sqrt {3}\,a\,d\,1{}\mathrm {i}\right )}^2}{36\,b^{5/3}}+\frac {a^2\,d^2\,{\left (a\,x^3-b\right )}^{1/3}}{9\,b^2}\right )\,\left (a\,d-\sqrt {3}\,a\,d\,1{}\mathrm {i}\right )}{18\,b^{4/3}}-\frac {d\,{\left (a\,x^3-b\right )}^{2/3}}{3\,b\,x^3}+\frac {a\,d\,\ln \left ({\left (a\,x^3-b\right )}^{1/3}+b^{1/3}\right )}{9\,b^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(d - c*x)/(x^4*(a*x^3 - b)^(1/3)),x)

[Out]

(c*(a*x^3 - b)^(2/3))/(2*b*x^2) - (log((a*d + 3^(1/2)*a*d*1i)^2/(36*b^(5/3)) + (a^2*d^2*(a*x^3 - b)^(1/3))/(9*
b^2))*(a*d + 3^(1/2)*a*d*1i))/(18*b^(4/3)) - (log((a*d - 3^(1/2)*a*d*1i)^2/(36*b^(5/3)) + (a^2*d^2*(a*x^3 - b)
^(1/3))/(9*b^2))*(a*d - 3^(1/2)*a*d*1i))/(18*b^(4/3)) - (d*(a*x^3 - b)^(2/3))/(3*b*x^3) + (a*d*log((a*x^3 - b)
^(1/3) + b^(1/3)))/(9*b^(4/3))

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sympy [C]  time = 2.70, size = 126, normalized size = 0.78 \begin {gather*} c \left (\begin {cases} - \frac {a^{\frac {2}{3}} \left (-1 + \frac {b}{a x^{3}}\right )^{\frac {2}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {2}{3}\right )}{3 b \Gamma \left (\frac {1}{3}\right )} & \text {for}\: \left |{\frac {b}{a x^{3}}}\right | > 1 \\- \frac {a^{\frac {2}{3}} \left (1 - \frac {b}{a x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right )}{3 b \Gamma \left (\frac {1}{3}\right )} & \text {otherwise} \end {cases}\right ) + \frac {d \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [3]{a} x^{4} \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x-d)/x**4/(a*x**3-b)**(1/3),x)

[Out]

c*Piecewise((-a**(2/3)*(-1 + b/(a*x**3))**(2/3)*exp(2*I*pi/3)*gamma(-2/3)/(3*b*gamma(1/3)), Abs(b/(a*x**3)) >
1), (-a**(2/3)*(1 - b/(a*x**3))**(2/3)*gamma(-2/3)/(3*b*gamma(1/3)), True)) + d*gamma(4/3)*hyper((1/3, 4/3), (
7/3,), b*exp_polar(2*I*pi)/(a*x**3))/(3*a**(1/3)*x**4*gamma(7/3))

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