Integrand size = 24, antiderivative size = 266 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{8/3}}{8 b^3 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^3 d}-\frac {c^3 (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{14/3}}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{14/3}}-\frac {c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}} \] Output:
-1/2*c^3*(b*x^3+a)^(2/3)/d^4+1/5*(a^2*d^2+a*b*c*d+b^2*c^2)*(b*x^3+a)^(5/3) /b^3/d^3-1/8*(2*a*d+b*c)*(b*x^3+a)^(8/3)/b^3/d^2+1/11*(b*x^3+a)^(11/3)/b^3 /d-1/3*c^3*(-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^(14/3)+1/6*c^3*(-a*d+b*c)^(2/3)*ln(d*x^3+c) /d^(14/3)-1/2*c^3*(-a*d+b*c)^(2/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^( 1/3))/d^(14/3)
Time = 0.87 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {\frac {3 d^{2/3} \left (a+b x^3\right )^{2/3} \left (18 a^3 d^3+3 a^2 b d^2 \left (11 c-4 d x^3\right )+2 a b^2 d \left (44 c^2-11 c d x^3+5 d^2 x^6\right )+b^3 \left (-220 c^3+88 c^2 d x^3-55 c d^2 x^6+40 d^3 x^9\right )\right )}{b^3}-440 \sqrt {3} c^3 (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-440 c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+220 c^3 (b c-a d)^{2/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{1320 d^{14/3}} \] Input:
Integrate[(x^11*(a + b*x^3)^(2/3))/(c + d*x^3),x]
Output:
((3*d^(2/3)*(a + b*x^3)^(2/3)*(18*a^3*d^3 + 3*a^2*b*d^2*(11*c - 4*d*x^3) + 2*a*b^2*d*(44*c^2 - 11*c*d*x^3 + 5*d^2*x^6) + b^3*(-220*c^3 + 88*c^2*d*x^ 3 - 55*c*d^2*x^6 + 40*d^3*x^9)))/b^3 - 440*Sqrt[3]*c^3*(b*c - a*d)^(2/3)*A rcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 440 *c^3*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 220*c^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)])/(1320*d^(14/3))
Time = 0.76 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 99, 2009}
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.
| \(\displaystyle \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9 \left (b x^3+a\right )^{2/3}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (-\frac {\left (b x^3+a\right )^{2/3} c^3}{d^3 \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{8/3}}{b^2 d}+\frac {(-b c-2 a d) \left (b x^3+a\right )^{5/3}}{b^2 d^2}+\frac {\left (b^2 c^2+a b d c+a^2 d^2\right ) \left (b x^3+a\right )^{2/3}}{b^2 d^3}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{5 b^3 d^3}-\frac {\sqrt {3} c^3 (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{14/3}}-\frac {3 \left (a+b x^3\right )^{8/3} (2 a d+b c)}{8 b^3 d^2}+\frac {3 \left (a+b x^3\right )^{11/3}}{11 b^3 d}+\frac {c^3 (b c-a d)^{2/3} \log \left (c+d x^3\right )}{2 d^{14/3}}-\frac {3 c^3 (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3}}-\frac {3 c^3 \left (a+b x^3\right )^{2/3}}{2 d^4}\right )\) |
Input:
Int[(x^11*(a + b*x^3)^(2/3))/(c + d*x^3),x]
Output:
((-3*c^3*(a + b*x^3)^(2/3))/(2*d^4) + (3*(b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(5/3))/(5*b^3*d^3) - (3*(b*c + 2*a*d)*(a + b*x^3)^(8/3))/(8*b^3*d ^2) + (3*(a + b*x^3)^(11/3))/(11*b^3*d) - (Sqrt[3]*c^3*(b*c - a*d)^(2/3)*A rcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/d^(1 4/3) + (c^3*(b*c - a*d)^(2/3)*Log[c + d*x^3])/(2*d^(14/3)) - (3*c^3*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(14/3 )))/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]]
Time = 1.78 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.02
| method | result | size |
| pseudoelliptic | \(\frac {\frac {27 \left (\left (\frac {20}{9} b^{2} x^{6}-\frac {5}{3} a b \,x^{3}+a^{2}\right ) \left (b \,x^{3}+a \right ) d^{3}+\frac {11 c \left (b \,x^{3}+a \right ) \left (-\frac {5 b \,x^{3}}{3}+a \right ) b \,d^{2}}{6}+\frac {44 b^{2} c^{2} \left (b \,x^{3}+a \right ) d}{9}-\frac {110 b^{3} c^{3}}{9}\right ) d \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}{110}+b^{3} c^{3} \left (a d -b c \right ) \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} b^{3} d^{5}}\) | \(272\) |
Input:
int(x^11*(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/6*(27/110*((20/9*b^2*x^6-5/3*a*b*x^3+a^2)*(b*x^3+a)*d^3+11/6*c*(b*x^3+a) *(-5/3*b*x^3+a)*b*d^2+44/9*b^2*c^2*(b*x^3+a)*d-110/9*b^3*c^3)*d*(b*x^3+a)^ (2/3)*((a*d-b*c)/d)^(1/3)+b^3*c^3*(a*d-b*c)*(-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))))/((a*d-b*c)/d)^(1/3)/b^3/d^5
Leaf count of result is larger than twice the leaf count of optimal. 455 vs. \(2 (219) = 438\).
Time = 0.22 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.71 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {440 \, \sqrt {3} b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 220 \, b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} + {\left (b c - a d\right )} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 440 \, b^{3} c^{3} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) - 3 \, {\left (40 \, b^{3} d^{3} x^{9} - 5 \, {\left (11 \, b^{3} c d^{2} - 2 \, a b^{2} d^{3}\right )} x^{6} - 220 \, b^{3} c^{3} + 88 \, a b^{2} c^{2} d + 33 \, a^{2} b c d^{2} + 18 \, a^{3} d^{3} + 2 \, {\left (44 \, b^{3} c^{2} d - 11 \, a b^{2} c d^{2} - 6 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{1320 \, b^{3} d^{4}} \] Input:
integrate(x^11*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
Output:
-1/1320*(440*sqrt(3)*b^3*c^3*(-(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)) + 220*b^3*c^3*(-(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)) - 440*b^3*c^3*(-(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*(40*b^3*d^3*x^9 - 5*(11*b^ 3*c*d^2 - 2*a*b^2*d^3)*x^6 - 220*b^3*c^3 + 88*a*b^2*c^2*d + 33*a^2*b*c*d^2 + 18*a^3*d^3 + 2*(44*b^3*c^2*d - 11*a*b^2*c*d^2 - 6*a^2*b*d^3)*x^3)*(b*x^ 3 + a)^(2/3))/(b^3*d^4)
\[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int \frac {x^{11} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \] Input:
integrate(x**11*(b*x**3+a)**(2/3)/(d*x**3+c),x)
Output:
Integral(x**11*(a + b*x**3)**(2/3)/(c + d*x**3), x)
Exception generated. \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^11*(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
Time = 0.15 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.54 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {{\left (b^{37} c^{4} d^{7} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{36} c^{3} d^{8} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{37} c d^{11} - a b^{36} d^{12}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{6}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{6}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{33} c^{3} d^{7} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{32} c^{2} d^{8} + 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{31} c d^{9} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{31} c d^{9} - 40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b^{30} d^{10} + 110 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a b^{30} d^{10} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} b^{30} d^{10}}{440 \, b^{33} d^{11}} \] Input:
integrate(x^11*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
Output:
-1/3*(b^37*c^4*d^7*(-(b*c - a*d)/d)^(1/3) - a*b^36*c^3*d^8*(-(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^37*c*d^11 - a*b^36*d^12) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(2 /3)*c^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))/d^6 + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^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))/d^6 - 1/440*(220*(b*x^3 + a)^(2/3)*b^33*c^3*d^7 - 88*(b*x^3 + a)^(5/ 3)*b^32*c^2*d^8 + 55*(b*x^3 + a)^(8/3)*b^31*c*d^9 - 88*(b*x^3 + a)^(5/3)*a *b^31*c*d^9 - 40*(b*x^3 + a)^(11/3)*b^30*d^10 + 110*(b*x^3 + a)^(8/3)*a*b^ 30*d^10 - 88*(b*x^3 + a)^(5/3)*a^2*b^30*d^10)/(b^33*d^11)
Time = 3.87 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.84 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\left (\frac {3\,a^2}{5\,b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{5\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{5/3}-\left (\frac {3\,a}{8\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{8\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{8/3}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {a^3}{2\,b^3\,d}+\frac {\left (\frac {3\,a^2}{b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{2\,b^3\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{11/3}}{11\,b^3\,d}-\frac {c^3\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^6\,d^2-2\,a\,b\,c^7\,d+b^2\,c^8\right )}{d^7}-\frac {c^6\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{28/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{14/3}}-\frac {c^3\,\ln \left (\frac {c^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{22/3}}+\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^7}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{14/3}}+\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^7}-\frac {c^6\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{7/3}}{4\,d^{22/3}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{d^{14/3}} \] Input:
int((x^11*(a + b*x^3)^(2/3))/(c + d*x^3),x)
Output:
((3*a^2)/(5*b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(5*b^3*d))*(a + b*x^3)^(5/3) - ((3*a)/(8*b^3*d) + (b^4*c - a* b^3*d)/(8*b^6*d^2))*(a + b*x^3)^(8/3) - (a + b*x^3)^(2/3)*(a^3/(2*b^3*d) + (((3*a^2)/(b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(b^3*d))*(b^4*c - a*b^3*d))/(2*b^3*d)) + (a + b*x^3)^(11/3)/( 11*b^3*d) - (c^3*log(((a + b*x^3)^(1/3)*(b^2*c^8 + a^2*c^6*d^2 - 2*a*b*c^7 *d))/d^7 - (c^6*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(9*d^(28/3)))*(a* d - b*c)^(2/3))/(3*d^(14/3)) - (c^3*log((c^6*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(22/3) + (c^6*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^7)*((3^(1/ 2)*1i)/2 - 1/2)*(a*d - b*c)^(2/3))/(3*d^(14/3)) + (c^3*log((c^6*(a + b*x^3 )^(1/3)*(a*d - b*c)^2)/d^7 - (c^6*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(7/3))/(4 *d^(22/3)))*((3^(1/2)*1i)/6 + 1/6)*(a*d - b*c)^(2/3))/d^(14/3)
\[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {18 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{3} d^{2}+33 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} b c d -12 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} b \,d^{2} x^{3}-132 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} c^{2}-22 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} c d \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} d^{2} x^{6}+88 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} c^{2} x^{3}-55 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} c d \,x^{6}+40 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} d^{2} x^{9}+440 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{3} c^{2} d -440 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{4} c^{3}}{440 b^{3} d^{3}} \] Input:
int(x^11*(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
(18*(a + b*x**3)**(2/3)*a**3*d**2 + 33*(a + b*x**3)**(2/3)*a**2*b*c*d - 12 *(a + b*x**3)**(2/3)*a**2*b*d**2*x**3 - 132*(a + b*x**3)**(2/3)*a*b**2*c** 2 - 22*(a + b*x**3)**(2/3)*a*b**2*c*d*x**3 + 10*(a + b*x**3)**(2/3)*a*b**2 *d**2*x**6 + 88*(a + b*x**3)**(2/3)*b**3*c**2*x**3 - 55*(a + b*x**3)**(2/3 )*b**3*c*d*x**6 + 40*(a + b*x**3)**(2/3)*b**3*d**2*x**9 + 440*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**3*c**2*d - 440*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**4*c**3)/(440*b**3*d**3)