3.682 \(\int \frac{(a+b x^3)^{2/3}}{x^4 (c+d x^3)} \, dx\)

Optimal. Leaf size=347 \[ \frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{\left (a+b x^3\right )^{2/3} (2 b c-3 a d)}{6 a c^2}-\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^2}+\frac{(2 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2}+\frac{(2 b c-3 a d) \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} c^2}-\frac{\log (x) (2 b c-3 a d)}{6 \sqrt [3]{a} c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3} \]

[Out]

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

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Rubi [A]  time = 0.389123, antiderivative size = 347, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {446, 103, 156, 50, 55, 617, 204, 31, 56} \[ \frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{\left (a+b x^3\right )^{2/3} (2 b c-3 a d)}{6 a c^2}-\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^2}+\frac{(2 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2}+\frac{(2 b c-3 a d) \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} c^2}-\frac{\log (x) (2 b c-3 a d)}{6 \sqrt [3]{a} c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/(x^4*(c + d*x^3)),x]

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 55

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] && PosQ
[(b*c - a*d)/b]

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 31

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^{2/3}}{x^4 \left (c+d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{x^2 (c+d x)} \, dx,x,x^3\right )\\ &=-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{2/3} \left (\frac{1}{3} (-2 b c+3 a d)-\frac{2 b d x}{3}\right )}{x (c+d x)} \, dx,x,x^3\right )}{3 a c}\\ &=-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}+\frac{d^2 \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{c+d x} \, dx,x,x^3\right )}{3 c^2}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{x} \, dx,x,x^3\right )}{9 a c^2}\\ &=\frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{(2 b c-3 a d) \left (a+b x^3\right )^{2/3}}{6 a c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [3]{a+b x}} \, dx,x,x^3\right )}{9 c^2}-\frac{(d (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 c^2}\\ &=\frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{(2 b c-3 a d) \left (a+b x^3\right )^{2/3}}{6 a c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}-\frac{(2 b c-3 a d) \log (x)}{6 \sqrt [3]{a} c^2}-\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^2}+\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 c^2}-\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{a} c^2}+\frac{\left (\sqrt [3]{d} (b c-a d)^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 c^2}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{\frac{(b c-a d)^{2/3}}{d^{2/3}}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 c^2}\\ &=\frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{(2 b c-3 a d) \left (a+b x^3\right )^{2/3}}{6 a c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}-\frac{(2 b c-3 a d) \log (x)}{6 \sqrt [3]{a} c^2}-\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^2}+\frac{(2 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2}-\frac{(2 b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{3 \sqrt [3]{a} c^2}-\frac{\left (\sqrt [3]{d} (b c-a d)^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{c^2}\\ &=\frac{d \left (a+b x^3\right )^{2/3}}{2 c^2}+\frac{(2 b c-3 a d) \left (a+b x^3\right )^{2/3}}{6 a c^2}-\frac{\left (a+b x^3\right )^{5/3}}{3 a c x^3}+\frac{(2 b c-3 a d) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt{3}}\right )}{\sqrt{3} c^2}-\frac{(2 b c-3 a d) \log (x)}{6 \sqrt [3]{a} c^2}-\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^2}+\frac{(2 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 \sqrt [3]{a} c^2}+\frac{\sqrt [3]{d} (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2}\\ \end{align*}

Mathematica [C]  time = 0.0989132, size = 202, normalized size = 0.58 \[ \frac{-9 \sqrt [3]{a} d x^3 \left (a+b x^3\right )^{2/3} \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{d \left (b x^3+a\right )}{a d-b c}\right )+2 \sqrt{3} x^3 (2 b c-3 a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )-6 \sqrt [3]{a} c \left (a+b x^3\right )^{2/3}+6 b c x^3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )-9 a d x^3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+9 a d x^3 \log (x)-6 b c x^3 \log (x)}{18 \sqrt [3]{a} c^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/(x^4*(c + d*x^3)),x]

[Out]

(-6*a^(1/3)*c*(a + b*x^3)^(2/3) + 2*Sqrt[3]*(2*b*c - 3*a*d)*x^3*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqr
t[3]] - 9*a^(1/3)*d*x^3*(a + b*x^3)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c) + a*d)] - 6*b
*c*x^3*Log[x] + 9*a*d*x^3*Log[x] + 6*b*c*x^3*Log[a^(1/3) - (a + b*x^3)^(1/3)] - 9*a*d*x^3*Log[a^(1/3) - (a + b
*x^3)^(1/3)])/(18*a^(1/3)*c^2*x^3)

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \left ( d{x}^{3}+c \right ) } \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(2/3)/x^4/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(2/3)/x^4/(d*x^3+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{{\left (d x^{3} + c\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^4/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(2/3)/((d*x^3 + c)*x^4), x)

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Fricas [A]  time = 2.37889, size = 2473, normalized size = 7.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^4/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{\frac{2}{3}}}{x^{4} \left (c + d x^{3}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(2/3)/x**4/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(2/3)/(x**4*(c + d*x**3)), x)

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Giac [A]  time = 2.67912, size = 585, normalized size = 1.69 \begin{align*} \frac{1}{18} \,{\left (\frac{6 \,{\left (b c d \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} - a d^{2} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} \right |}\right )}{b^{3} c^{3} - a b^{2} c^{2} d} + \frac{2 \,{\left (2 \, a^{\frac{1}{3}} b c - 3 \, a^{\frac{4}{3}} d\right )} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{2}{3}} b^{2} c^{2}} + \frac{6 \, \sqrt{3}{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}}}\right )}{b^{2} c^{2} d} - \frac{3 \,{\left (-b c d^{2} + a d^{3}\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \left (-\frac{b c - a d}{d}\right )^{\frac{1}{3}} + \left (-\frac{b c - a d}{d}\right )^{\frac{2}{3}}\right )}{b^{2} c^{2} d} + \frac{2 \, \sqrt{3}{\left (2 \, a^{\frac{5}{3}} b c - 3 \, a^{\frac{8}{3}} d\right )} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right )}{a^{2} b^{2} c^{2}} - \frac{{\left (2 \, a^{\frac{5}{3}} b c - 3 \, a^{\frac{8}{3}} d\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right )}{a^{2} b^{2} c^{2}} - \frac{6 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}}}{b^{2} c x^{3}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^4/(d*x^3+c),x, algorithm="giac")

[Out]

1/18*(6*(b*c*d*(-(b*c - a*d)/d)^(1/3) - a*d^2*(-(b*c - a*d)/d)^(1/3))*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 +
a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^3*c^3 - a*b^2*c^2*d) + 2*(2*a^(1/3)*b*c - 3*a^(4/3)*d)*log(abs((b*x^3 +
a)^(1/3) - a^(1/3)))/(a^(2/3)*b^2*c^2) + 6*sqrt(3)*(-b*c*d^2 + a*d^3)^(2/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)
^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/(b^2*c^2*d) - 3*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 +
a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d) + 2*sqrt(3)*(2*a^(5
/3)*b*c - 3*a^(8/3)*d)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/(a^2*b^2*c^2) - (2*a^(5/3)*
b*c - 3*a^(8/3)*d)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/(a^2*b^2*c^2) - 6*(b*x^3 + a)^
(2/3)/(b^2*c*x^3))*b^2