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

Optimal. Leaf size=176 \[ -\frac{d \log \left (c+d x^3\right )}{6 c^{5/3} \sqrt [3]{b c-a d}}+\frac{d \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} c^{5/3} \sqrt [3]{b c-a d}}-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2} \]

[Out]

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

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Rubi [A]  time = 0.226174, antiderivative size = 235, normalized size of antiderivative = 1.34, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {494, 453, 200, 31, 634, 617, 204, 628} \[ \frac{d \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt{3} \sqrt [3]{c}}\right )}{\sqrt{3} c^{5/3} \sqrt [3]{b c-a d}}-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 453

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 31

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-b x^3}{x^3 \left (c-(b c-a d) x^3\right )} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{a}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2}-\frac{d \operatorname{Subst}\left (\int \frac{1}{c+(-b c+a d) x^3} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{c}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2}-\frac{d \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 c^{5/3}}-\frac{d \operatorname{Subst}\left (\int \frac{2 \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 c^{5/3}}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2}+\frac{d \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \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 c^{4/3}}-\frac{d \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 c^{5/3} \sqrt [3]{b c-a d}}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2}+\frac{d \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \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 c^{5/3} \sqrt [3]{b c-a d}}+\frac{d \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 )}{c^{5/3} \sqrt [3]{b c-a d}}\\ &=-\frac{\left (a+b x^3\right )^{2/3}}{2 a c x^2}-\frac{d \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} c^{5/3} \sqrt [3]{b c-a d}}+\frac{d \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{5/3} \sqrt [3]{b c-a d}}-\frac{d \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 c^{5/3} \sqrt [3]{b c-a d}}\\ \end{align*}

Mathematica [C]  time = 0.0572724, size = 124, normalized size = 0.7 \[ \frac{-3 x^3 \left (c+d x^3\right ) (b c-a d) \, _2F_1\left (\frac{4}{3},2;\frac{7}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-4 c \left (a+b x^3\right ) \left (c+3 d x^3\right ) \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{8 c^3 x^2 \left (a+b x^3\right )^{4/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*c*(a + b*x^3)*(c + 3*d*x^3)*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 3*(b*c - a
*d)*x^3*(c + d*x^3)*Hypergeometric2F1[4/3, 2, 7/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(8*c^3*x^2*(a + b*x^3)^
(4/3))

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( d{x}^{3}+c \right ) }{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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