3.1.40 \(\int \frac {1}{(c+d x) (a+b x^3)^{2/3}} \, dx\) [40]

Optimal. Leaf size=332 \[ \frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c \left (a+b x^3\right )^{2/3}}+\frac {d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \left (b c^3-a d^3\right )^{2/3}}-\frac {d \tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} \left (b c^3-a d^3\right )^{2/3}}-\frac {d \log \left (c^3+d^3 x^3\right )}{3 \left (b c^3-a d^3\right )^{2/3}}+\frac {d \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{2 \left (b c^3-a d^3\right )^{2/3}}+\frac {d \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 \left (b c^3-a d^3\right )^{2/3}} \]

[Out]

x*(1+b*x^3/a)^(2/3)*AppellF1(1/3,2/3,1,4/3,-b*x^3/a,-d^3*x^3/c^3)/c/(b*x^3+a)^(2/3)-1/3*d*ln(d^3*x^3+c^3)/(-a*
d^3+b*c^3)^(2/3)+1/2*d*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/(-a*d^3+b*c^3)^(2/3)+1/2*d*ln((-a*d^3+b*c^
3)^(1/3)+d*(b*x^3+a)^(1/3))/(-a*d^3+b*c^3)^(2/3)+1/3*d*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)^(2/3)*3^(1/2)-1/3*d*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)^(2/3)*3^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 332, 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, 441, 440, 503, 455, 60, 631, 210, 31} \begin {gather*} \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c \left (a+b x^3\right )^{2/3}}+\frac {d \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} \left (b c^3-a d^3\right )^{2/3}}-\frac {d \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} \left (b c^3-a d^3\right )^{2/3}}-\frac {d \log \left (c^3+d^3 x^3\right )}{3 \left (b c^3-a d^3\right )^{2/3}}+\frac {d \log \left (\frac {x \sqrt [3]{b c^3-a d^3}}{c}-\sqrt [3]{a+b x^3}\right )}{2 \left (b c^3-a d^3\right )^{2/3}}+\frac {d \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 \left (b c^3-a d^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/(c*(a + b*x^3)^(2/3)) + (
d*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)^(2/3)) - (
d*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)^(2/3)) - (d*Lo
g[c^3 + d^3*x^3])/(3*(b*c^3 - a*d^3)^(2/3)) + (d*Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3)])/(2*(b*c
^3 - a*d^3)^(2/3)) + (d*Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)])/(2*(b*c^3 - a*d^3)^(2/3))

Rule 31

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

Rule 60

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-
Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x
)^(1/3)], x] + Dist[3/(2*b*q^2), 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 440

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

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 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 503

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si
mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/
(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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) \left (a+b x^3\right )^{2/3}} \, dx &=\int \frac {1}{(c+d x) \left (a+b x^3\right )^{2/3}} \, dx\\ \end {align*}

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Mathematica [F]
time = 7.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x) \left (a+b x^3\right )^{2/3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d x +c \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x+c)/(b*x^3+a)^(2/3),x)

[Out]

int(1/(d*x+c)/(b*x^3+a)^(2/3),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)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(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)^(2/3),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}{\left (a + b x^{3}\right )^{\frac {2}{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)**(2/3),x)

[Out]

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

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(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 )}^{2/3}\,\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)^(2/3)*(c + d*x)),x)

[Out]

int(1/((a + b*x^3)^(2/3)*(c + d*x)), x)

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