3.5.47 \(\int \frac {(a+b x^3)^{4/3}}{x^7 (c+d x^3)} \, dx\)

Optimal. Leaf size=440 \[ \frac {\sqrt [3]{a+b x^3} \left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right )}{9 a c^3}+\frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{18 a^{2/3} c^3}-\frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{2/3} c^3}-\frac {\log (x) \left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right )}{18 a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^3}-\frac {d^{2/3} (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c^3}+\frac {d \sqrt [3]{a+b x^3} (b c-a d)}{c^3}-\frac {\left (a+b x^3\right )^{4/3} (b c-6 a d)}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6} \]

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Rubi [A]  time = 0.62, antiderivative size = 440, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {446, 103, 149, 156, 50, 57, 617, 204, 31, 58} \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right )}{9 a c^3}+\frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{18 a^{2/3} c^3}-\frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{2/3} c^3}-\frac {\log (x) \left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right )}{18 a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^3}-\frac {d^{2/3} (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \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^3}-\frac {\left (a+b x^3\right )^{4/3} (b c-6 a d)}{18 a c^2 x^3}+\frac {d \sqrt [3]{a+b x^3} (b c-a d)}{c^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(4/3)/(x^7*(c + d*x^3)),x]

[Out]

(d*(b*c - a*d)*(a + b*x^3)^(1/3))/c^3 + ((2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*(a + b*x^3)^(1/3))/(9*a*c^3) - (
(b*c - 6*a*d)*(a + b*x^3)^(4/3))/(18*a*c^2*x^3) - (a + b*x^3)^(7/3)/(6*a*c*x^6) - ((2*b^2*c^2 - 12*a*b*c*d + 9
*a^2*d^2)*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(2/3)*c^3) + (d^(2/3)*(b*c -
a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*c^3) - ((2*b^2*c^2
- 12*a*b*c*d + 9*a^2*d^2)*Log[x])/(18*a^(2/3)*c^3) + (d^(2/3)*(b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*c^3) + ((2
*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(18*a^(2/3)*c^3) - (d^(2/3)*(b*c - a*d)^(
4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*c^3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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]
&& PosQ[(b*c - a*d)/b]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[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 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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^{4/3}}{x^7 \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{x^3 (c+d x)} \, dx,x,x^3\right )\\ &=-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^{4/3} \left (\frac {1}{3} (-b c+6 a d)-\frac {b d x}{3}\right )}{x^2 (c+d x)} \, dx,x,x^3\right )}{6 a c}\\ &=-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x} \left (-\frac {2}{9} \left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right )-\frac {2}{9} b d (2 b c-3 a d) x\right )}{x (c+d x)} \, dx,x,x^3\right )}{6 a c^2}\\ &=-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}+\frac {\left (d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{x} \, dx,x,x^3\right )}{27 a c^3}\\ &=\frac {d (b c-a d) \sqrt [3]{a+b x^3}}{c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{9 a c^3}-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\left (d (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )}{27 c^3}\\ &=\frac {d (b c-a d) \sqrt [3]{a+b x^3}}{c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{9 a c^3}-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \log (x)}{18 a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^3}-\frac {\left (d^{2/3} (b c-a d)^{4/3}\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^3}-\frac {\left (\sqrt [3]{d} (b c-a d)^{5/3}\right ) \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^3}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{18 a^{2/3} c^3}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{18 \sqrt [3]{a} c^3}\\ &=\frac {d (b c-a d) \sqrt [3]{a+b x^3}}{c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{9 a c^3}-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \log (x)}{18 a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{18 a^{2/3} c^3}-\frac {d^{2/3} (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^3}-\frac {\left (d^{2/3} (b c-a d)^{4/3}\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^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{9 a^{2/3} c^3}\\ &=\frac {d (b c-a d) \sqrt [3]{a+b x^3}}{c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{9 a c^3}-\frac {(b c-6 a d) \left (a+b x^3\right )^{4/3}}{18 a c^2 x^3}-\frac {\left (a+b x^3\right )^{7/3}}{6 a c x^6}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{9 \sqrt {3} a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \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^3}-\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \log (x)}{18 a^{2/3} c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^3}+\frac {\left (2 b^2 c^2-12 a b c d+9 a^2 d^2\right ) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{18 a^{2/3} c^3}-\frac {d^{2/3} (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 1.25, size = 429, normalized size = 0.98 \begin {gather*} \frac {x^6 \left (4 \left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \left (-\frac {1}{2} a^{4/3} \left (\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )\right )+3 a \sqrt [3]{a+b x^3}+\frac {3}{4} \left (a+b x^3\right )^{4/3}\right )-9 a^2 d^{2/3} \left (3 d^{4/3} \left (a+b x^3\right )^{4/3}-2 (b c-a d) \left (\sqrt [3]{b c-a d} \left (\log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )-2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}-1}{\sqrt {3}}\right )\right )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3}\right )\right )\right )-18 a c^2 \left (a+b x^3\right )^{7/3}-6 c x^3 \left (a+b x^3\right )^{7/3} (b c-6 a d)}{108 a^2 c^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(4/3)/(x^7*(c + d*x^3)),x]

[Out]

(-18*a*c^2*(a + b*x^3)^(7/3) - 6*c*(b*c - 6*a*d)*x^3*(a + b*x^3)^(7/3) + x^6*(4*(2*b^2*c^2 - 12*a*b*c*d + 9*a^
2*d^2)*(3*a*(a + b*x^3)^(1/3) + (3*(a + b*x^3)^(4/3))/4 - (a^(4/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3)
)/a^(1/3))/Sqrt[3]] - 2*Log[a^(1/3) - (a + b*x^3)^(1/3)] + Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^
3)^(2/3)]))/2) - 9*a^2*d^(2/3)*(3*d^(4/3)*(a + b*x^3)^(4/3) - 2*(b*c - a*d)*(6*d^(1/3)*(a + b*x^3)^(1/3) + (b*
c - a*d)^(1/3)*(-2*Sqrt[3]*ArcTan[(-1 + (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 2*Log[(b*c
- a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/
3) + d^(2/3)*(a + b*x^3)^(2/3)])))))/(108*a^2*c^3*x^6)

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IntegrateAlgebraic [A]  time = 1.07, size = 445, normalized size = 1.01 \begin {gather*} \frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{27 a^{2/3} c^3}+\frac {\left (-9 a^2 d^2+12 a b c d-2 b^2 c^2\right ) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 a^{2/3} c^3}-\frac {\left (9 a^2 d^2-12 a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{2/3} c^3}-\frac {d^{2/3} (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 c^3}+\frac {d^{2/3} (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{b c-a d}}\right )}{\sqrt {3} c^3}+\frac {\sqrt [3]{a+b x^3} \left (-3 a c+6 a d x^3-7 b c x^3\right )}{18 c^2 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^3)^(4/3)/(x^7*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(1/3)*(-3*a*c - 7*b*c*x^3 + 6*a*d*x^3))/(18*c^2*x^6) - ((2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*ArcT
an[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(2/3)*c^3) + (d^(2/3)*(b*c - a*d)^(4/3)*Ar
cTan[1/Sqrt[3] - (2*d^(1/3)*(a + b*x^3)^(1/3))/(Sqrt[3]*(b*c - a*d)^(1/3))])/(Sqrt[3]*c^3) + ((2*b^2*c^2 - 12*
a*b*c*d + 9*a^2*d^2)*Log[-a^(1/3) + (a + b*x^3)^(1/3)])/(27*a^(2/3)*c^3) - (d^(2/3)*(b*c - a*d)^(4/3)*Log[(b*c
- a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*c^3) + ((-2*b^2*c^2 + 12*a*b*c*d - 9*a^2*d^2)*Log[a^(2/3) + a^(
1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*a^(2/3)*c^3) + (d^(2/3)*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(2/
3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(6*c^3)

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fricas [A]  time = 2.02, size = 503, normalized size = 1.14 \begin {gather*} \frac {18 \, \sqrt {3} {\left (a^{2} b c - a^{3} d\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} x^{6} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \sqrt {3} {\left (b c d - a d^{2}\right )}}{3 \, {\left (b c d - a d^{2}\right )}}\right ) - 2 \, \sqrt {3} {\left (2 \, a b^{2} c^{2} - 12 \, a^{2} b c d + 9 \, a^{3} d^{2}\right )} {\left (a^{2}\right )}^{\frac {1}{6}} x^{6} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) - {\left (2 \, b^{2} c^{2} - 12 \, a b c d + 9 \, a^{2} d^{2}\right )} {\left (a^{2}\right )}^{\frac {2}{3}} x^{6} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (2 \, b^{2} c^{2} - 12 \, a b c d + 9 \, a^{2} d^{2}\right )} {\left (a^{2}\right )}^{\frac {2}{3}} x^{6} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) + 9 \, {\left (a^{2} b c - a^{3} d\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} x^{6} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 18 \, {\left (a^{2} b c - a^{3} d\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} x^{6} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, a^{3} c^{2} + {\left (7 \, a^{2} b c^{2} - 6 \, a^{3} c d\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, a^{2} c^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^7/(d*x^3+c),x, algorithm="fricas")

[Out]

1/54*(18*sqrt(3)*(a^2*b*c - a^3*d)*(b*c*d^2 - a*d^3)^(1/3)*x^6*arctan(-1/3*(2*sqrt(3)*(b*c*d^2 - a*d^3)^(2/3)*
(b*x^3 + a)^(1/3) - sqrt(3)*(b*c*d - a*d^2))/(b*c*d - a*d^2)) - 2*sqrt(3)*(2*a*b^2*c^2 - 12*a^2*b*c*d + 9*a^3*
d^2)*(a^2)^(1/6)*x^6*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)*a + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(a^2)^(2/3))/
a^2) - (2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*(a^2)^(2/3)*x^6*log((b*x^3 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^3 +
a)^(1/3)*(a^2)^(2/3)) + 2*(2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*(a^2)^(2/3)*x^6*log((b*x^3 + a)^(1/3)*a - (a^2
)^(2/3)) + 9*(a^2*b*c - a^3*d)*(b*c*d^2 - a*d^3)^(1/3)*x^6*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)
*(b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 18*(a^2*b*c - a^3*d)*(b*c*d^2 - a*d^3)^(1/3)*x^6*log((b*x^3
+ a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) - 3*(3*a^3*c^2 + (7*a^2*b*c^2 - 6*a^3*c*d)*x^3)*(b*x^3 + a)^(1/3))/(a^
2*c^3*x^6)

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giac [A]  time = 0.86, size = 481, normalized size = 1.09 \begin {gather*} \frac {{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{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 )}{3 \, {\left (b c^{4} - a c^{3} d\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \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 )}{3 \, c^{3}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \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 )}{6 \, c^{3}} - \frac {\sqrt {3} {\left (2 \, b^{2} c^{2} - 12 \, a b c d + 9 \, a^{2} d^{2}\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 )}{27 \, a^{\frac {2}{3}} c^{3}} - \frac {{\left (2 \, b^{2} c^{2} - 12 \, a b c d + 9 \, a^{2} d^{2}\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 )}{54 \, a^{\frac {2}{3}} c^{3}} + \frac {{\left (2 \, b^{2} c^{2} - 12 \, a b c d + 9 \, a^{2} d^{2}\right )} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{27 \, a^{\frac {2}{3}} c^{3}} - \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2} c - 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{2} c - 6 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b d + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b d}{18 \, b^{2} c^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^7/(d*x^3+c),x, algorithm="giac")

[Out]

1/3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1
/3)))/(b*c^4 - a*c^3*d) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*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))/c^3 - 1/6*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*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))/c^3 - 1/27*sqrt(3)*(2*b^2*c
^2 - 12*a*b*c*d + 9*a^2*d^2)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/(a^(2/3)*c^3) - 1/54*
(2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/(a^(2/3)*c^3
) + 1/27*(2*b^2*c^2 - 12*a*b*c*d + 9*a^2*d^2)*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/(a^(2/3)*c^3) - 1/18*(7*(b
*x^3 + a)^(4/3)*b^2*c - 4*(b*x^3 + a)^(1/3)*a*b^2*c - 6*(b*x^3 + a)^(4/3)*a*b*d + 6*(b*x^3 + a)^(1/3)*a^2*b*d)
/(b^2*c^2*x^6)

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maple [F]  time = 0.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{\left (d \,x^{3}+c \right ) x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)/x^7/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(4/3)/x^7/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^7/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^7), x)

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mupad [B]  time = 13.25, size = 2841, normalized size = 6.46

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(4/3)/(x^7*(c + d*x^3)),x)

[Out]

log(((((18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) + 9*a*b^4*c^4*d^3*(2*a^2*d^2 + b^2*c^2 -
3*a*b*c*d)*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(1/3))*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(
a^2*c^9))^(2/3))/729 + (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 - 3258*a^2*b^4*c^4*d^2 + 6939*a^3*b^3*c^3*d^3 - 7182*
a^4*b^2*c^2*d^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*c^4))*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*
c^9))^(1/3))/27 - (b^4*d^5*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(1458*a^6*d^6 + 170*b^6*c^6 + 6561*a^2*b^4*c^4*d^2
- 12420*a^3*b^3*c^3*d^3 + 12798*a^4*b^2*c^2*d^4 - 1764*a*b^5*c^5*d - 6804*a^5*b*c*d^5))/(243*c^8))*((729*a^6*d
^6 + 8*b^6*c^6 + 972*a^2*b^4*c^4*d^2 - 3024*a^3*b^3*c^3*d^3 + 4374*a^4*b^2*c^2*d^4 - 144*a*b^5*c^5*d - 2916*a^
5*b*c*d^5)/(19683*a^2*c^9))^(1/3) + log(((((18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) + 81*
a*b^4*c^4*d^3*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))*(-(d^2*(a*d - b*c)^4)/c^9)^(
2/3))/9 + (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 - 3258*a^2*b^4*c^4*d^2 + 6939*a^3*b^3*c^3*d^3 - 7182*a^4*b^2*c^2*d
^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*c^4))*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))/3 - (b^4*d^5*(a + b*x^3)
^(1/3)*(a*d - b*c)^2*(1458*a^6*d^6 + 170*b^6*c^6 + 6561*a^2*b^4*c^4*d^2 - 12420*a^3*b^3*c^3*d^3 + 12798*a^4*b^
2*c^2*d^4 - 1764*a*b^5*c^5*d - 6804*a^5*b*c*d^5))/(243*c^8))*(-(a^4*d^6 + b^4*c^4*d^2 - 4*a*b^3*c^3*d^3 + 6*a^
2*b^2*c^2*d^4 - 4*a^3*b*c*d^5)/(27*c^9))^(1/3) + (((2*a*b^2*c - 3*a^2*b*d)*(a + b*x^3)^(1/3))/(9*c^2) + (b*(a
+ b*x^3)^(4/3)*(6*a*d - 7*b*c))/(18*c^2))/((a + b*x^3)^2 - 2*a*(a + b*x^3) + a^2) + log((((3^(1/2)*1i)/2 - 1/2
)*((((3^(1/2)*1i)/2 + 1/2)*(18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) + 81*a*b^4*c^4*d^3*((
3^(1/2)*1i)/2 - 1/2)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))*(-(d^2*(a*d - b*c)^4)
/c^9)^(2/3))/9 - (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 - 3258*a^2*b^4*c^4*d^2 + 6939*a^3*b^3*c^3*d^3 - 7182*a^4*b^
2*c^2*d^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*c^4))*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))/3 + (b^4*d^5*(a +
b*x^3)^(1/3)*(a*d - b*c)^2*(1458*a^6*d^6 + 170*b^6*c^6 + 6561*a^2*b^4*c^4*d^2 - 12420*a^3*b^3*c^3*d^3 + 12798
*a^4*b^2*c^2*d^4 - 1764*a*b^5*c^5*d - 6804*a^5*b*c*d^5))/(243*c^8))*((3^(1/2)*1i)/2 - 1/2)*(-(a^4*d^6 + b^4*c^
4*d^2 - 4*a*b^3*c^3*d^3 + 6*a^2*b^2*c^2*d^4 - 4*a^3*b*c*d^5)/(27*c^9))^(1/3) - log((((3^(1/2)*1i)/2 + 1/2)*(((
(3^(1/2)*1i)/2 - 1/2)*(18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) - 81*a*b^4*c^4*d^3*((3^(1/
2)*1i)/2 + 1/2)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))*(-(d^2*(a*d - b*c)^4)/c^9)
^(2/3))/9 + (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 - 3258*a^2*b^4*c^4*d^2 + 6939*a^3*b^3*c^3*d^3 - 7182*a^4*b^2*c^2
*d^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*c^4))*(-(d^2*(a*d - b*c)^4)/c^9)^(1/3))/3 + (b^4*d^5*(a + b*x^
3)^(1/3)*(a*d - b*c)^2*(1458*a^6*d^6 + 170*b^6*c^6 + 6561*a^2*b^4*c^4*d^2 - 12420*a^3*b^3*c^3*d^3 + 12798*a^4*
b^2*c^2*d^4 - 1764*a*b^5*c^5*d - 6804*a^5*b*c*d^5))/(243*c^8))*((3^(1/2)*1i)/2 + 1/2)*(-(a^4*d^6 + b^4*c^4*d^2
- 4*a*b^3*c^3*d^3 + 6*a^2*b^2*c^2*d^4 - 4*a^3*b*c*d^5)/(27*c^9))^(1/3) + log((((3^(1/2)*1i)/2 - 1/2)*((((3^(1
/2)*1i)/2 + 1/2)*(18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) + 9*a*b^4*c^4*d^3*((3^(1/2)*1i)
/2 - 1/2)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(1/3))*((9*a^2*
d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(2/3))/729 - (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 - 3258*a^2*b^4*c^4*d
^2 + 6939*a^3*b^3*c^3*d^3 - 7182*a^4*b^2*c^2*d^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*c^4))*((9*a^2*d^2
+ 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(1/3))/27 + (b^4*d^5*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(1458*a^6*d^6 + 17
0*b^6*c^6 + 6561*a^2*b^4*c^4*d^2 - 12420*a^3*b^3*c^3*d^3 + 12798*a^4*b^2*c^2*d^4 - 1764*a*b^5*c^5*d - 6804*a^5
*b*c*d^5))/(243*c^8))*((3^(1/2)*1i)/2 - 1/2)*((729*a^6*d^6 + 8*b^6*c^6 + 972*a^2*b^4*c^4*d^2 - 3024*a^3*b^3*c^
3*d^3 + 4374*a^4*b^2*c^2*d^4 - 144*a*b^5*c^5*d - 2916*a^5*b*c*d^5)/(19683*a^2*c^9))^(1/3) - log((((3^(1/2)*1i)
/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(18*b^5*c^2*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(6*a*d - b*c) - 9*a*b^4*c^4
*d^3*((3^(1/2)*1i)/2 + 1/2)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9
))^(1/3))*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(2/3))/729 + (b^5*d^4*(8*b^6*c^6 - 729*a^6*d^6 -
3258*a^2*b^4*c^4*d^2 + 6939*a^3*b^3*c^3*d^3 - 7182*a^4*b^2*c^2*d^4 + 577*a*b^5*c^5*d + 3645*a^5*b*c*d^5))/(81*
c^4))*((9*a^2*d^2 + 2*b^2*c^2 - 12*a*b*c*d)^3/(a^2*c^9))^(1/3))/27 + (b^4*d^5*(a + b*x^3)^(1/3)*(a*d - b*c)^2*
(1458*a^6*d^6 + 170*b^6*c^6 + 6561*a^2*b^4*c^4*d^2 - 12420*a^3*b^3*c^3*d^3 + 12798*a^4*b^2*c^2*d^4 - 1764*a*b^
5*c^5*d - 6804*a^5*b*c*d^5))/(243*c^8))*((3^(1/2)*1i)/2 + 1/2)*((729*a^6*d^6 + 8*b^6*c^6 + 972*a^2*b^4*c^4*d^2
- 3024*a^3*b^3*c^3*d^3 + 4374*a^4*b^2*c^2*d^4 - 144*a*b^5*c^5*d - 2916*a^5*b*c*d^5)/(19683*a^2*c^9))^(1/3)

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sympy [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((b*x**3+a)**(4/3)/x**7/(d*x**3+c),x)

[Out]

Timed out

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