3.1.33 \(\int \frac {1}{(c+d x) \sqrt [3]{a+b x^3}} \, dx\) [33]

Optimal. Leaf size=333 \[ -\frac {d x^2 \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{2 c^2 \sqrt [3]{a+b x^3}}+\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b c^3-a d^3}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b c^3-a d^3}}+\frac {\log \left (c^3+d^3 x^3\right )}{3 \sqrt [3]{b c^3-a d^3}}-\frac {\log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b c^3-a d^3}}-\frac {\log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b c^3-a d^3}} \]

[Out]

-1/2*d*x^2*(1+b*x^3/a)^(1/3)*AppellF1(2/3,1/3,1,5/3,-b*x^3/a,-d^3*x^3/c^3)/c^2/(b*x^3+a)^(1/3)+1/3*ln(d^3*x^3+
c^3)/(-a*d^3+b*c^3)^(1/3)-1/2*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/(-a*d^3+b*c^3)^(1/3)-1/2*ln((-a*d^3
+b*c^3)^(1/3)+d*(b*x^3+a)^(1/3))/(-a*d^3+b*c^3)^(1/3)+1/3*arctan(1/3*(1+2*(-a*d^3+b*c^3)^(1/3)*x/c/(b*x^3+a)^(
1/3))*3^(1/2))/(-a*d^3+b*c^3)^(1/3)*3^(1/2)-1/3*arctan(1/3*(1-2*d*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)^(1/3))*3^(1/2
))/(-a*d^3+b*c^3)^(1/3)*3^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {2181, 384, 525, 524, 455, 58, 631, 210, 31} \begin {gather*} -\frac {d x^2 \sqrt [3]{\frac {b x^3}{a}+1} F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{2 c^2 \sqrt [3]{a+b x^3}}+\frac {\text {ArcTan}\left (\frac {\frac {2 x \sqrt [3]{b c^3-a d^3}}{c \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b c^3-a d^3}}-\frac {\text {ArcTan}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b c^3-a d^3}}+\frac {\log \left (c^3+d^3 x^3\right )}{3 \sqrt [3]{b c^3-a d^3}}-\frac {\log \left (\frac {x \sqrt [3]{b c^3-a d^3}}{c}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b c^3-a d^3}}-\frac {\log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b c^3-a d^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d*x^2*(1 + (b*x^3)/a)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/(c^2*(a + b*x^3)
^(1/3)) + ArcTan[(1 + (2*(b*c^3 - a*d^3)^(1/3)*x)/(c*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*(b*c^3 - a*d^3)^(1/
3)) - ArcTan[(1 - (2*d*(a + b*x^3)^(1/3))/(b*c^3 - a*d^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*(b*c^3 - a*d^3)^(1/3)) + Lo
g[c^3 + d^3*x^3]/(3*(b*c^3 - a*d^3)^(1/3)) - Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3)]/(2*(b*c^3 -
a*d^3)^(1/3)) - Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)]/(2*(b*c^3 - a*d^3)^(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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[L
og[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] && NegQ
[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 455

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

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 631

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 2181

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c^3 + d^3*x
^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ
[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {1}{(c+d x) \sqrt [3]{a+b x^3}} \, dx &=\int \frac {1}{(c+d x) \sqrt [3]{a+b x^3}} \, dx\\ \end {align*}

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Mathematica [F]
time = 7.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x) \sqrt [3]{a+b x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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(1/(d*x+c)/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((a + b*x**3)**(1/3)*(c + d*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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