Integrand size = 24, antiderivative size = 211 \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {c (b c-a d) \sqrt [3]{a+b x^3}}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}+\frac {c (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}} \] Output:
c*(-a*d+b*c)*(b*x^3+a)^(1/3)/d^3-1/4*c*(b*x^3+a)^(4/3)/d^2+1/7*(b*x^3+a)^( 7/3)/b/d+1/3*c*(-a*d+b*c)^(4/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(- a*d+b*c)^(1/3))*3^(1/2))*3^(1/2)/d^(10/3)+1/6*c*(-a*d+b*c)^(4/3)*ln(d*x^3+ c)/d^(10/3)-1/2*c*(-a*d+b*c)^(4/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^( 1/3))/d^(10/3)
Time = 0.37 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.23 \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (4 a^2 d^2+a b d \left (-35 c+8 d x^3\right )+b^2 \left (28 c^2-7 c d x^3+4 d^2 x^6\right )\right )}{b}+28 \sqrt {3} c (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-28 c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+14 c (b c-a d)^{4/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{84 d^{10/3}} \] Input:
Integrate[(x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x]
Output:
((3*d^(1/3)*(a + b*x^3)^(1/3)*(4*a^2*d^2 + a*b*d*(-35*c + 8*d*x^3) + b^2*( 28*c^2 - 7*c*d*x^3 + 4*d^2*x^6)))/b + 28*Sqrt[3]*c*(b*c - a*d)^(4/3)*ArcTa n[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 28*c*(b *c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 14*c* (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)])/(84*d^(10/3))
Time = 0.62 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 90, 60, 60, 70, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^3 \left (b x^3+a\right )^{4/3}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \int \frac {\left (b x^3+a\right )^{4/3}}{d x^3+c}dx^3}{d}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \int \frac {\sqrt [3]{b x^3+a}}{d x^3+c}dx^3}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx^3}{d}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 70 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 \sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{-x^6-3}d\left (1-\frac {2 \sqrt [3]{d} \sqrt [3]{b x^3+a}}{\sqrt [3]{b c-a d}}\right )}{\sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 d}-\frac {(b c-a d) \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )}{d}\right )\) |
Input:
Int[(x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x]
Output:
((3*(a + b*x^3)^(7/3))/(7*b*d) - (c*((3*(a + b*x^3)^(4/3))/(4*d) - ((b*c - a*d)*((3*(a + b*x^3)^(1/3))/d - ((b*c - a*d)*(-((Sqrt[3]*ArcTan[(1 - (2*d ^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(d^(1/3)*(b*c - a*d )^(2/3))) - Log[c + d*x^3]/(2*d^(1/3)*(b*c - a*d)^(2/3)) + (3*Log[(b*c - a *d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(1/3)*(b*c - a*d)^(2/3))))/d) )/d))/d)/3
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) , x] + (Simp[3/(2*b*q) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 /3)], x] + Simp[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]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Time = 1.72 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.08
| method | result | size |
| pseudoelliptic | \(\frac {d \left (\left (b \,x^{3}+a \right )^{2} d^{2}-\frac {35 c \left (\frac {b \,x^{3}}{5}+a \right ) b d}{4}+7 b^{2} c^{2}\right ) \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\frac {7 b c \left (a d -b c \right )^{2} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6}}{7 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{4} b}\) | \(228\) |
Input:
int(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/7/((a*d-b*c)/d)^(2/3)*(d*((b*x^3+a)^2*d^2-35/4*c*(1/5*b*x^3+a)*b*d+7*b^2 *c^2)*((a*d-b*c)/d)^(2/3)*(b*x^3+a)^(1/3)+7/6*b*c*(a*d-b*c)^2*(2*arctan(1/ 3*3^(1/2)*(2*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(1/3))/((a*d-b*c)/d)^(1/3))*3^( 1/2)+ln((b*x^3+a)^(2/3)+((a*d-b*c)/d)^(1/3)*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^ (2/3))-2*ln((b*x^3+a)^(1/3)-((a*d-b*c)/d)^(1/3))))/d^4/b
Time = 0.12 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.41 \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {28 \, \sqrt {3} {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 14 \, {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 28 \, {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, b^{2} d^{2} x^{6} + 28 \, b^{2} c^{2} - 35 \, a b c d + 4 \, a^{2} d^{2} - {\left (7 \, b^{2} c d - 8 \, a b d^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{84 \, b d^{3}} \] Input:
integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
Output:
1/84*(28*sqrt(3)*(b^2*c^2 - a*b*c*d)*((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)) + 14*(b^2*c^2 - a*b*c*d)*((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)) - 28*(b^2*c^2 - a*b*c*d)*((b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) + ((b *c - a*d)/d)^(1/3)) + 3*(4*b^2*d^2*x^6 + 28*b^2*c^2 - 35*a*b*c*d + 4*a^2*d ^2 - (7*b^2*c*d - 8*a*b*d^2)*x^3)*(b*x^3 + a)^(1/3))/(b*d^3)
\[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\int \frac {x^{5} \left (a + b x^{3}\right )^{\frac {4}{3}}}{c + d x^{3}}\, dx \] Input:
integrate(x**5*(b*x**3+a)**(4/3)/(d*x**3+c),x)
Output:
Integral(x**5*(a + b*x**3)**(4/3)/(c + d*x**3), x)
Exception generated. \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (171) = 342\).
Time = 0.14 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.65 \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {{\left (b^{10} c^{3} d^{4} - 2 \, a b^{9} c^{2} d^{5} + a^{2} b^{8} c d^{6}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{9} c d^{7} - a b^{8} d^{8}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{8} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{7} c d^{5} - 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{7} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{6} d^{6}}{28 \, b^{7} d^{7}} \] Input:
integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
Output:
1/3*(b^10*c^3*d^4 - 2*a*b^9*c^2*d^5 + a^2*b^8*c*d^6)*(-(b*c - a*d)/d)^(1/3 )*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^9*c*d^7 - a*b^8* d^8) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c^2 - a*c*d)*arctan(1/3*sqr t(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3) )/d^4 - 1/6*(-b*c*d^2 + a*d^3)^(1/3)*(b*c^2 - a*c*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))/d^4 + 1/28*(28*(b*x^3 + a)^(1/3)*b^8*c^2*d^4 - 7*(b*x^3 + a)^(4/3)*b^7*c*d^5 - 28*(b*x^3 + a)^(1/3)*a*b^7*c*d^5 + 4*(b*x^3 + a)^(7/3)*b^6*d^6)/(b^7*d^7)
Time = 3.80 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.65 \[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b\,d}-{\left (b\,x^3+a\right )}^{4/3}\,\left (\frac {a}{4\,b\,d}+\frac {b^2\,c-a\,b\,d}{4\,b^2\,d^2}\right )-\frac {c\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c\,d^2-2\,a\,b\,c^2\,d+b^2\,c^3\right )}{d}-\frac {c\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{10/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}-\frac {c\,\ln \left (\frac {3\,c\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d}-\frac {3\,c\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}+\frac {c\,\ln \left (\frac {3\,c\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d}+\frac {3\,c\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}+\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (b^2\,c-a\,b\,d\right )\,\left (\frac {a}{b\,d}+\frac {b^2\,c-a\,b\,d}{b^2\,d^2}\right )}{b\,d} \] Input:
int((x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x)
Output:
(a + b*x^3)^(7/3)/(7*b*d) - (a + b*x^3)^(4/3)*(a/(4*b*d) + (b^2*c - a*b*d) /(4*b^2*d^2)) - (c*log((3*(a + b*x^3)^(1/3)*(b^2*c^3 + a^2*c*d^2 - 2*a*b*c ^2*d))/d - (c*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(10/3)))*(a*d - b*c)^(4/3))/(3*d^(10/3)) - (c*log((3*c*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/ d - (3*c*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(7/3))/d^(4/3))*((3^(1/2)*1i)/ 2 - 1/2)*(a*d - b*c)^(4/3))/(3*d^(10/3)) + (c*log((3*c*(a + b*x^3)^(1/3)*( a*d - b*c)^2)/d + (3*c*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(4/3))* ((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(4/3))/(3*d^(10/3)) + ((a + b*x^3)^(1/3 )*(b^2*c - a*b*d)*(a/(b*d) + (b^2*c - a*b*d)/(b^2*d^2)))/(b*d)
\[ \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {-24 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d +21 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c +8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b d \,x^{3}-7 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c \,x^{3}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d \,x^{6}+28 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b \,d^{2}-56 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{2} c d +28 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{3} c^{2}}{28 b \,d^{2}} \] Input:
int(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x)
Output:
( - 24*(a + b*x**3)**(1/3)*a**2*d + 21*(a + b*x**3)**(1/3)*a*b*c + 8*(a + b*x**3)**(1/3)*a*b*d*x**3 - 7*(a + b*x**3)**(1/3)*b**2*c*x**3 + 4*(a + b*x **3)**(1/3)*b**2*d*x**6 + 28*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d*x** 3 + b*c*x**3 + b*d*x**6),x)*a**2*b*d**2 - 56*int(((a + b*x**3)**(1/3)*x**5 )/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b**2*c*d + 28*int(((a + b*x* *3)**(1/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b**3*c**2)/(28* b*d**2)