3.5.76 \(\int \frac {x}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=149 \[ \frac {\log \left (c+d x^3\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}} \]

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Rubi [A]  time = 0.14, antiderivative size = 208, normalized size of antiderivative = 1.40, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {494, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {\log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(c^(1/3) + (2*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3))/(Sqrt[3]*c^(1/3))]/(Sqrt[3]*c^(1/3)*(b*c - a*d)
^(2/3))) - Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)]/(3*c^(1/3)*(b*c - a*d)^(2/3)) + Log[c^(2/3)
+ ((b*c - a*d)^(2/3)*x^2)/(a + b*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)]/(6*c^(1/3)*(b*c
- a*d)^(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 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 292

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

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 83, normalized size = 0.56 \begin {gather*} \frac {x^2 \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{2 c \left (a+b x^3\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 1.96, size = 319, normalized size = 2.14 \begin {gather*} \frac {i \left (\sqrt {3}+i\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\left (1-i \sqrt {3}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\sqrt {-1-i \sqrt {3}} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt {6} \sqrt [3]{c} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-1 - I*Sqrt[3]]*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^
(1/3) - Sqrt[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(Sqrt[6]*c^(1/3)*(b*c - a*d)^(2/3)) + ((1 - I*Sqrt[3])*Log[2*(b*c
- a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*c^(1/3)*(b*c - a*d)^(2/3)) + ((I/12)*(I + Sqr
t[3])*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (I +
Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(c^(1/3)*(b*c - a*d)^(2/3))

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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(x/(b*x^3+a)^(2/3)/(d*x^3+c),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 {x}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x/((a + b*x**3)**(2/3)*(c + d*x**3)), x)

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