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

Optimal. Leaf size=399 \[ \frac {d \left (a+b x^3\right )^{4/3}}{4 c^2}+\frac {\left (a+b x^3\right )^{4/3} (4 b c-3 a d)}{12 a c^2}+\frac {\sqrt [3]{a+b x^3} (4 b c-3 a d)}{3 c^2}-\frac {\sqrt [3]{a+b x^3} (b c-a d)}{c^2}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^2 \sqrt [3]{d}}+\frac {\sqrt [3]{a} (4 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 c^2}+\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 \sqrt [3]{d}}-\frac {\sqrt [3]{a} (4 b c-3 a d) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} c^2}-\frac {(b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c^2 \sqrt [3]{d}}-\frac {\sqrt [3]{a} \log (x) (4 b c-3 a d)}{6 c^2}-\frac {\left (a+b x^3\right )^{7/3}}{3 a c x^3} \]

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Rubi [A]  time = 0.48, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {446, 103, 156, 50, 57, 617, 204, 31, 58} \begin {gather*} \frac {d \left (a+b x^3\right )^{4/3}}{4 c^2}+\frac {\left (a+b x^3\right )^{4/3} (4 b c-3 a d)}{12 a c^2}+\frac {\sqrt [3]{a+b x^3} (4 b c-3 a d)}{3 c^2}-\frac {\sqrt [3]{a+b x^3} (b c-a d)}{c^2}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^2 \sqrt [3]{d}}+\frac {\sqrt [3]{a} (4 b c-3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 c^2}+\frac {(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^2 \sqrt [3]{d}}-\frac {\sqrt [3]{a} (4 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} c^2}-\frac {(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^2 \sqrt [3]{d}}-\frac {\sqrt [3]{a} \log (x) (4 b c-3 a d)}{6 c^2}-\frac {\left (a+b x^3\right )^{7/3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.55, size = 389, normalized size = 0.97 \begin {gather*} \frac {\frac {(4 b c-3 a d) \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 )}{3 c}+\frac {a \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 )}{4 c \sqrt [3]{d}}-\frac {\left (a+b x^3\right )^{7/3}}{x^3}}{3 a c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*x^3)^(7/3)/x^3) + ((4*b*c - 3*a*d)*(3*a*(a + b*x^3)^(1/3) + (3*(a + b*x^3)^(4/3))/4 - (a^(4/3)*(2*Sq
rt[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))/(3*c) + (a*(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)]))))/(4*c*d^(1/3)))/(3*a*c)

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IntegrateAlgebraic [A]  time = 0.77, size = 398, normalized size = 1.00 \begin {gather*} \frac {\left (4 \sqrt [3]{a} b c-3 a^{4/3} d\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{9 c^2}+\frac {\left (3 a^{4/3} d-4 \sqrt [3]{a} b c\right ) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 c^2}+\frac {\left (3 a^{4/3} d-4 \sqrt [3]{a} b c\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} c^2}-\frac {(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^2 \sqrt [3]{d}}+\frac {(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^2 \sqrt [3]{d}}-\frac {(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^2 \sqrt [3]{d}}-\frac {a \sqrt [3]{a+b x^3}}{3 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 383, normalized size = 0.96 \begin {gather*} \frac {6 \, \sqrt {3} {\left (b c - a d\right )} x^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 2 \, \sqrt {3} {\left (4 \, b c - 3 \, a d\right )} \left (-a\right )^{\frac {1}{3}} x^{3} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + {\left (4 \, b c - 3 \, a d\right )} \left (-a\right )^{\frac {1}{3}} x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + \left (-a\right )^{\frac {2}{3}}\right ) + 3 \, {\left (b c - a d\right )} x^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{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 ) - 2 \, {\left (4 \, b c - 3 \, a d\right )} \left (-a\right )^{\frac {1}{3}} x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-a\right )^{\frac {1}{3}}\right ) - 6 \, {\left (b c - a d\right )} x^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a c}{18 \, c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.86, size = 394, normalized size = 0.99 \begin {gather*} -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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^{3} - a c^{2} d\right )}} - \frac {\sqrt {3} {\left (4 \, a^{\frac {1}{3}} b c - 3 \, a^{\frac {4}{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 )}{9 \, c^{2}} - \frac {{\left (4 \, a^{\frac {1}{3}} b c - 3 \, a^{\frac {4}{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 )}{18 \, c^{2}} + \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^{2} d} + \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^{2} d} + \frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{9 \, a^{\frac {2}{3}} c^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{3 \, c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)
))/(b*c^3 - a*c^2*d) - 1/9*sqrt(3)*(4*a^(1/3)*b*c - 3*a^(4/3)*d)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(
1/3))/a^(1/3))/c^2 - 1/18*(4*a^(1/3)*b*c - 3*a^(4/3)*d)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^
(2/3))/c^2 + 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^2*d) + 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^2*d) + 1/9*(4*a*b*c - 3*a^2*d)*lo
g(abs((b*x^3 + a)^(1/3) - a^(1/3)))/(a^(2/3)*c^2) - 1/3*(b*x^3 + a)^(1/3)*a/(c*x^3)

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maple [F]  time = 0.65, 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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x^3+a)^(4/3)/x^4/(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^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 10.88, size = 2047, normalized size = 5.13

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c^2*(-(a*(3*a*d - 4*b*c)^3)/c^6)^(1/3) + 3*a*d*(a + b*x^3)^(1/3) - 4*b*c*(a + b*x^3)^(1/3))*(-(27*a^4*d^3
- 64*a*b^3*c^3 + 144*a^2*b^2*c^2*d - 108*a^3*b*c*d^2)/(729*c^6))^(1/3) + log(((((81*a*b^4*c^4*d^3*(2*a^2*d^2 +
b^2*c^2 - 3*a*b*c*d)*((a*d - b*c)^4/(c^6*d))^(1/3) - 108*a*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2)*((a*d
- b*c)^4/(c^6*d))^(2/3))/9 + (a*b^5*d^2*(27*a^5*d^5 - 27*b^5*c^5 - 341*a^2*b^3*c^3*d^2 + 332*a^3*b^2*c^2*d^3
+ 162*a*b^4*c^4*d - 153*a^4*b*c*d^4))/(3*c))*((a*d - b*c)^4/(c^6*d))^(1/3))/3 - (a*b^4*d^2*(a + b*x^3)^(1/3)*(
a*d - b*c)^2*(54*a^5*d^5 - 36*b^5*c^5 - 388*a^2*b^3*c^3*d^2 + 450*a^3*b^2*c^2*d^3 + 171*a*b^4*c^4*d - 252*a^4*
b*c*d^4))/(9*c^4))*((a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3)/(27*c^6*d))^(1/3)
+ log((((3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*(108*a*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2 - 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)*((a*d - b*c)^4/(c^6*d))^(1/3))*((a*d -
b*c)^4/(c^6*d))^(2/3))/9 + (a*b^5*d^2*(27*a^5*d^5 - 27*b^5*c^5 - 341*a^2*b^3*c^3*d^2 + 332*a^3*b^2*c^2*d^3 +
162*a*b^4*c^4*d - 153*a^4*b*c*d^4))/(3*c))*((a*d - b*c)^4/(c^6*d))^(1/3))/3 - (a*b^4*d^2*(a + b*x^3)^(1/3)*(a*
d - b*c)^2*(54*a^5*d^5 - 36*b^5*c^5 - 388*a^2*b^3*c^3*d^2 + 450*a^3*b^2*c^2*d^3 + 171*a*b^4*c^4*d - 252*a^4*b*
c*d^4))/(9*c^4))*((3^(1/2)*1i)/2 - 1/2)*((a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^
3)/(27*c^6*d))^(1/3) - log((a*b^4*d^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(54*a^5*d^5 - 36*b^5*c^5 - 388*a^2*b^3*c
^3*d^2 + 450*a^3*b^2*c^2*d^3 + 171*a*b^4*c^4*d - 252*a^4*b*c*d^4))/(9*c^4) - (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/
2)*1i)/2 - 1/2)*(108*a*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2 + 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)*((a*d - b*c)^4/(c^6*d))^(1/3))*((a*d - b*c)^4/(c^6*d))^(2/3))/9 - (a*b^5*d^2*
(27*a^5*d^5 - 27*b^5*c^5 - 341*a^2*b^3*c^3*d^2 + 332*a^3*b^2*c^2*d^3 + 162*a*b^4*c^4*d - 153*a^4*b*c*d^4))/(3*
c))*((a*d - b*c)^4/(c^6*d))^(1/3))/3)*((3^(1/2)*1i)/2 + 1/2)*((a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3
*c^3*d - 4*a^3*b*c*d^3)/(27*c^6*d))^(1/3) + log((a*b^4*d^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(54*a^5*d^5 - 36*b^
5*c^5 - 388*a^2*b^3*c^3*d^2 + 450*a^3*b^2*c^2*d^3 + 171*a*b^4*c^4*d - 252*a^4*b*c*d^4))/(9*c^4) - (((3^(1/2)*1
i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*(108*a*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2 - 27*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)*(-(a*(3*a*d - 4*b*c)^3)/c^6)^(1/3))*(-(a*(3*a*d - 4*b*c)
^3)/c^6)^(2/3))/81 + (a*b^5*d^2*(27*a^5*d^5 - 27*b^5*c^5 - 341*a^2*b^3*c^3*d^2 + 332*a^3*b^2*c^2*d^3 + 162*a*b
^4*c^4*d - 153*a^4*b*c*d^4))/(3*c))*(-(a*(3*a*d - 4*b*c)^3)/c^6)^(1/3))/9)*((3^(1/2)*1i)/2 - 1/2)*(-(27*a^4*d^
3 - 64*a*b^3*c^3 + 144*a^2*b^2*c^2*d - 108*a^3*b*c*d^2)/(729*c^6))^(1/3) - log((a*b^4*d^2*(a + b*x^3)^(1/3)*(a
*d - b*c)^2*(54*a^5*d^5 - 36*b^5*c^5 - 388*a^2*b^3*c^3*d^2 + 450*a^3*b^2*c^2*d^3 + 171*a*b^4*c^4*d - 252*a^4*b
*c*d^4))/(9*c^4) - (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(108*a*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(a*d
- b*c)^2 + 27*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)*(-(a*(3*a*d - 4*b*c)^3)/c
^6)^(1/3))*(-(a*(3*a*d - 4*b*c)^3)/c^6)^(2/3))/81 - (a*b^5*d^2*(27*a^5*d^5 - 27*b^5*c^5 - 341*a^2*b^3*c^3*d^2
+ 332*a^3*b^2*c^2*d^3 + 162*a*b^4*c^4*d - 153*a^4*b*c*d^4))/(3*c))*(-(a*(3*a*d - 4*b*c)^3)/c^6)^(1/3))/9)*((3^
(1/2)*1i)/2 + 1/2)*(-(27*a^4*d^3 - 64*a*b^3*c^3 + 144*a^2*b^2*c^2*d - 108*a^3*b*c*d^2)/(729*c^6))^(1/3) - (a*(
a + b*x^3)^(1/3))/(3*c*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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