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

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

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 347, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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 )^{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 = 0.16, size = 202, normalized size = 0.58 \begin {gather*} \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} \end {gather*}

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)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.71, size = 384, normalized size = 1.11 \begin {gather*} \frac {(3 a d-2 b c) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 \sqrt [3]{a} c^2}-\frac {\sqrt [3]{d} (b c-a d)^{2/3} \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}+\frac {(2 b c-3 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{9 \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 )}{3 c^2}-\frac {(3 a d-2 b c) \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}{\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}-\frac {\left (a+b x^3\right )^{2/3}}{3 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 1030, normalized size = 2.97

result too large to display

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)]

________________________________________________________________________________________

giac [A]  time = 0.81, size = 400, normalized size = 1.15 \begin {gather*} \frac {{\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 )}{3 \, {\left (b c^{3} - a c^{2} d\right )}} - \frac {{\left (2 \, b c - 3 \, a 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 \, a^{\frac {1}{3}} c^{2}} + \frac {\sqrt {3} {\left (2 \, a^{\frac {2}{3}} b c - 3 \, a^{\frac {5}{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 \, a c^{2}} + \frac {{\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 )}{9 \, a^{\frac {2}{3}} c^{2}} + \frac {\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 )}{3 \, c^{2} d} - \frac {{\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 )}{6 \, c^{2} d} - \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{3 \, c x^{3}} \end {gather*}

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]

1/3*(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*c^3 - a*c^2*d) - 1/18*(2*b*c - 3*a*d)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(
1/3)*a^(1/3) + a^(2/3))/(a^(1/3)*c^2) + 1/9*sqrt(3)*(2*a^(2/3)*b*c - 3*a^(5/3)*d)*arctan(1/3*sqrt(3)*(2*(b*x^3
+ a)^(1/3) + a^(1/3))/a^(1/3))/(a*c^2) + 1/9*(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)*c^2) + 1/3*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))/(c^2*d) - 1/6*(-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))/(c^2*d) - 1/3*(b*x^3 + a)^(2/3)/(c*x^3)

________________________________________________________________________________________

maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{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)^(2/3)/x^4/(d*x^3+c),x)

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

mupad [B]  time = 10.57, size = 1908, normalized size = 5.50

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(- ((((6*b^4*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(9*a^2*d^2 + 2*b^2*c^2 - 6*a*b*c*d) - 27*a*b^4*c^4*d^3*(2*
a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*((d*(a*d - b*c)^2)/c^6)^(2/3))*((d*(a*d - b*c)^2)/c^6)^(1/3))/3 - (a*b^5*d^4*(2
7*a^3*d^3 - 19*b^3*c^3 + 64*a*b^2*c^2*d - 72*a^2*b*c*d^2))/(3*c))*((d*(a*d - b*c)^2)/c^6)^(2/3))/9 - (b^5*d^4*
(a + b*x^3)^(1/3)*(4*a*d - 3*b*c)*(3*a^2*d^2 + 2*b^2*c^2 - 5*a*b*c*d)^2)/(27*c^5))*((a^2*d^3 + b^2*c^2*d - 2*a
*b*c*d^2)/(27*c^6))^(1/3) + log(- ((((6*b^4*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(9*a^2*d^2 + 2*b^2*c^2 - 6*a*b
*c*d) - 3*a*b^4*c^4*d^3*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(3*a*d - 2*b*c)^3/(a*c^6))^(2/3))*(-(3*a*d - 2*b*c
)^3/(a*c^6))^(1/3))/9 - (a*b^5*d^4*(27*a^3*d^3 - 19*b^3*c^3 + 64*a*b^2*c^2*d - 72*a^2*b*c*d^2))/(3*c))*(-(3*a*
d - 2*b*c)^3/(a*c^6))^(2/3))/81 - (b^5*d^4*(a + b*x^3)^(1/3)*(4*a*d - 3*b*c)*(3*a^2*d^2 + 2*b^2*c^2 - 5*a*b*c*
d)^2)/(27*c^5))*(-(27*a^3*d^3 - 8*b^3*c^3 + 36*a*b^2*c^2*d - 54*a^2*b*c*d^2)/(729*a*c^6))^(1/3) - log((((3^(1/
2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*(6*b^4*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(9*a^2*d^2 + 2*b^2*c^2 - 6
*a*b*c*d) - 3*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)*(-(3*a*d - 2*b*c)^3/(a*c^
6))^(2/3))*(-(3*a*d - 2*b*c)^3/(a*c^6))^(1/3))/9 + (a*b^5*d^4*(27*a^3*d^3 - 19*b^3*c^3 + 64*a*b^2*c^2*d - 72*a
^2*b*c*d^2))/(3*c))*(-(3*a*d - 2*b*c)^3/(a*c^6))^(2/3))/81 - (b^5*d^4*(a + b*x^3)^(1/3)*(4*a*d - 3*b*c)*(3*a^2
*d^2 + 2*b^2*c^2 - 5*a*b*c*d)^2)/(27*c^5))*((3^(1/2)*1i)/2 + 1/2)*(-(27*a^3*d^3 - 8*b^3*c^3 + 36*a*b^2*c^2*d -
54*a^2*b*c*d^2)/(729*a*c^6))^(1/3) + log((((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(6*b^4*d^3*(a + b*x
^3)^(1/3)*(a*d - b*c)^2*(9*a^2*d^2 + 2*b^2*c^2 - 6*a*b*c*d) + 3*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)*(-(3*a*d - 2*b*c)^3/(a*c^6))^(2/3))*(-(3*a*d - 2*b*c)^3/(a*c^6))^(1/3))/9 - (a*b^5*d^
4*(27*a^3*d^3 - 19*b^3*c^3 + 64*a*b^2*c^2*d - 72*a^2*b*c*d^2))/(3*c))*(-(3*a*d - 2*b*c)^3/(a*c^6))^(2/3))/81 -
(b^5*d^4*(a + b*x^3)^(1/3)*(4*a*d - 3*b*c)*(3*a^2*d^2 + 2*b^2*c^2 - 5*a*b*c*d)^2)/(27*c^5))*((3^(1/2)*1i)/2 -
1/2)*(-(27*a^3*d^3 - 8*b^3*c^3 + 36*a*b^2*c^2*d - 54*a^2*b*c*d^2)/(729*a*c^6))^(1/3) - (a + b*x^3)^(2/3)/(3*c
*x^3) - log((((3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*(6*b^4*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(9*a^2
*d^2 + 2*b^2*c^2 - 6*a*b*c*d) - 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)*((d*
(a*d - b*c)^2)/c^6)^(2/3))*((d*(a*d - b*c)^2)/c^6)^(1/3))/3 + (a*b^5*d^4*(27*a^3*d^3 - 19*b^3*c^3 + 64*a*b^2*c
^2*d - 72*a^2*b*c*d^2))/(3*c))*((d*(a*d - b*c)^2)/c^6)^(2/3))/9 - (b^5*d^4*(a + b*x^3)^(1/3)*(4*a*d - 3*b*c)*(
3*a^2*d^2 + 2*b^2*c^2 - 5*a*b*c*d)^2)/(27*c^5))*((3^(1/2)*1i)/2 + 1/2)*((a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2)/(2
7*c^6))^(1/3) + log((((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(6*b^4*d^3*(a + b*x^3)^(1/3)*(a*d - b*c)^
2*(9*a^2*d^2 + 2*b^2*c^2 - 6*a*b*c*d) + 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)*((d*(a*d - b*c)^2)/c^6)^(2/3))*((d*(a*d - b*c)^2)/c^6)^(1/3))/3 - (a*b^5*d^4*(27*a^3*d^3 - 19*b^3*c^3 + 64
*a*b^2*c^2*d - 72*a^2*b*c*d^2))/(3*c))*((d*(a*d - b*c)^2)/c^6)^(2/3))/9 - (b^5*d^4*(a + b*x^3)^(1/3)*(4*a*d -
3*b*c)*(3*a^2*d^2 + 2*b^2*c^2 - 5*a*b*c*d)^2)/(27*c^5))*((3^(1/2)*1i)/2 - 1/2)*((a^2*d^3 + b^2*c^2*d - 2*a*b*c
*d^2)/(27*c^6))^(1/3)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{3}\right )^{\frac {2}{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)**(2/3)/x**4/(d*x**3+c),x)

[Out]

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

________________________________________________________________________________________