Integrand size = 24, antiderivative size = 188 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3}}+\frac {c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac {c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}} \] Output:
-1/2*c*(b*x^3+a)^(2/3)/d^2+1/5*(b*x^3+a)^(5/3)/b/d-1/3*c*(-a*d+b*c)^(2/3)* 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^(8/3)+1/6*c*(-a*d+b*c)^(2/3)*ln(d*x^3+c)/d^(8/3)-1/2*c*(-a*d+b*c)^(2/3) *ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(8/3)
Time = 0.36 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.21 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {\frac {3 d^{2/3} \left (a+b x^3\right )^{2/3} \left (-5 b c+2 a d+2 b d x^3\right )}{b}-10 \sqrt {3} c (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-10 c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+5 c (b c-a d)^{2/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{30 d^{8/3}} \] Input:
Integrate[(x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x]
Output:
((3*d^(2/3)*(a + b*x^3)^(2/3)*(-5*b*c + 2*a*d + 2*b*d*x^3))/b - 10*Sqrt[3] *c*(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]] - 10*c*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)* (a + b*x^3)^(1/3)] + 5*c*(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)])/(30*d^( 8/3))
Time = 0.54 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {948, 90, 60, 68, 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.
\(\displaystyle \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3 \left (b x^3+a\right )^{2/3}}{d x^3+c}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \int \frac {\left (b x^3+a\right )^{2/3}}{d x^3+c}dx^3}{d}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3}{d}\right )}{d}\right )\) |
\(\Big \downarrow \) 68 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \left (-\frac {3 \int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{-x^6-3}d\left (1-\frac {2 \sqrt [3]{d} \sqrt [3]{b x^3+a}}{\sqrt [3]{b c-a d}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}\right )\) |
Input:
Int[(x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x]
Output:
((3*(a + b*x^3)^(5/3))/(5*b*d) - (c*((3*(a + b*x^3)^(2/3))/(2*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^(2/3)*(b*c - a*d)^(1/3))) + Log[c + d*x^3]/(2*d^(2/3)*(b *c - a*d)^(1/3)) - (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/ (2*d^(2/3)*(b*c - a*d)^(1/3))))/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_))^(1/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) 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 = 1.71 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {\frac {6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} d \left (\left (d \,x^{3}-\frac {5 c}{2}\right ) b +a d \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{5}+b c \left (a d -b c \right ) \left (-2 \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} b \,d^{3}}\) | \(195\) |
Input:
int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/6/((a*d-b*c)/d)^(1/3)*(6/5*((a*d-b*c)/d)^(1/3)*d*((d*x^3-5/2*c)*b+a*d)*( b*x^3+a)^(2/3)+b*c*(a*d-b*c)*(-2*arctan(2/3*3^(1/2)/((a*d-b*c)/d)^(1/3)*(b *x^3+a)^(1/3)+1/3*3^(1/2))*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))))/b/d^3
Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (149) = 298\).
Time = 0.19 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.88 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {10 \, \sqrt {3} b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 5 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} + {\left (b c - a d\right )} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) - 3 \, {\left (2 \, b d x^{3} - 5 \, b c + 2 \, a d\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, b d^{2}} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
Output:
-1/30*(10*sqrt(3)*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan( -1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2) ^(1/3) + sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 5*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*log((b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2 *d^2)/d^2)^(2/3) - (b*x^3 + a)^(2/3)*(b*c - a*d) + (b*c - a*d)*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) - 10*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d ^2)/d^2)^(1/3)*log(-d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^ 3 + a)^(1/3)*(b*c - a*d)) - 3*(2*b*d*x^3 - 5*b*c + 2*a*d)*(b*x^3 + a)^(2/3 ))/(b*d^2)
\[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int \frac {x^{5} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \] Input:
integrate(x**5*(b*x**3+a)**(2/3)/(d*x**3+c),x)
Output:
Integral(x**5*(a + b*x**3)**(2/3)/(c + d*x**3), x)
Exception generated. \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)/(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. 306 vs. \(2 (149) = 298\).
Time = 0.16 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.63 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {{\left (b^{7} c^{2} d^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{6} c d^{4} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{7} c d^{5} - a b^{6} d^{6}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{5} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{4} d^{4}}{10 \, b^{5} d^{5}} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
Output:
-1/3*(b^7*c^2*d^3*(-(b*c - a*d)/d)^(1/3) - a*b^6*c*d^4*(-(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^7*c*d^5 - a*b^6*d^6) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(2/3)*c*ar ctan(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))/d^4 + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c*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 - 1/10*(5*(b*x^3 + a)^(2/3)*b^5*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^4*d^4)/(b^5*d^ 5)
Time = 3.66 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.61 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b\,d}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {a}{2\,b\,d}+\frac {b^2\,c-a\,b\,d}{2\,b^2\,d^2}\right )-\frac {c\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}{d^3}-\frac {c^2\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{16/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}-\frac {c\,\ln \left (\frac {c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}+\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}+\frac {c\,\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}-\frac {c^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}} \] Input:
int((x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x)
Output:
(a + b*x^3)^(5/3)/(5*b*d) - (a + b*x^3)^(2/3)*(a/(2*b*d) + (b^2*c - a*b*d) /(2*b^2*d^2)) - (c*log(((a + b*x^3)^(1/3)*(b^2*c^4 + a^2*c^2*d^2 - 2*a*b*c ^3*d))/d^3 - (c^2*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(9*d^(16/3)))*( a*d - b*c)^(2/3))/(3*d^(8/3)) - (c*log((c^2*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(10/3) + (c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^3)*((3^(1/2 )*1i)/2 - 1/2)*(a*d - b*c)^(2/3))/(3*d^(8/3)) + (c*log((c^2*(a + b*x^3)^(1 /3)*(a*d - b*c)^2)/d^3 - (c^2*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(7/3))/d^ (10/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(2/3))/(3*d^(8/3))
\[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a +2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}+10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b d -10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c}{10 b d} \] Input:
int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
( - 3*(a + b*x**3)**(2/3)*a + 2*(a + b*x**3)**(2/3)*b*x**3 + 10*int(((a + b*x**3)**(2/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b*d - 10* int(((a + b*x**3)**(2/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b **2*c)/(10*b*d)