Integrand size = 28, antiderivative size = 223 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {a^3 \left (a+b x^3\right )^{2/3}}{2 b^4 d}-\frac {a^2 \left (a+b x^3\right )^{5/3}}{5 b^4 d}+\frac {a \left (a+b x^3\right )^{8/3}}{8 b^4 d}-\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}-\frac {2^{2/3} a^{11/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^4 d}+\frac {a^{11/3} \log \left (a-b x^3\right )}{3 \sqrt [3]{2} b^4 d}-\frac {a^{11/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^4 d} \] Output:
-1/2*a^3*(b*x^3+a)^(2/3)/b^4/d-1/5*a^2*(b*x^3+a)^(5/3)/b^4/d+1/8*a*(b*x^3+ a)^(8/3)/b^4/d-1/11*(b*x^3+a)^(11/3)/b^4/d-1/3*2^(2/3)*a^(11/3)*arctan(1/3 *(a^(1/3)+2^(2/3)*(b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/b^4/d+1/6*a^(1 1/3)*ln(-b*x^3+a)*2^(2/3)/b^4/d-1/2*a^(11/3)*ln(2^(1/3)*a^(1/3)-(b*x^3+a)^ (1/3))*2^(2/3)/b^4/d
Time = 0.34 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {3 \left (a+b x^3\right )^{2/3} \left (293 a^3+98 a^2 b x^3+65 a b^2 x^6+40 b^3 x^9\right )+440\ 2^{2/3} \sqrt {3} a^{11/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+440\ 2^{2/3} a^{11/3} \log \left (-2 \sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}\right )-220\ 2^{2/3} a^{11/3} \log \left (2 a^{2/3}+2^{2/3} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{1320 b^4 d} \] Input:
Integrate[(x^11*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]
Output:
-1/1320*(3*(a + b*x^3)^(2/3)*(293*a^3 + 98*a^2*b*x^3 + 65*a*b^2*x^6 + 40*b ^3*x^9) + 440*2^(2/3)*Sqrt[3]*a^(11/3)*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1 /3))/a^(1/3))/Sqrt[3]] + 440*2^(2/3)*a^(11/3)*Log[-2*a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3)] - 220*2^(2/3)*a^(11/3)*Log[2*a^(2/3) + 2^(2/3)*a^(1/3)*(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(b^4*d)
Time = 0.61 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {948, 27, 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.
| \(\displaystyle \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9 \left (b x^3+a\right )^{2/3}}{d \left (a-b x^3\right )}dx^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {x^9 \left (b x^3+a\right )^{2/3}}{a-b x^3}dx^3}{3 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {\left (b x^3+a\right )^{2/3} a^3}{b^3 \left (a-b x^3\right )}-\frac {\left (b x^3+a\right )^{2/3} a^2}{b^3}+\frac {\left (b x^3+a\right )^{5/3} a}{b^3}-\frac {\left (b x^3+a\right )^{8/3}}{b^3}\right )dx^3}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {2^{2/3} \sqrt {3} a^{11/3} \arctan \left (\frac {2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{b^4}+\frac {a^{11/3} \log \left (a-b x^3\right )}{\sqrt [3]{2} b^4}-\frac {3 a^{11/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^4}-\frac {3 a^3 \left (a+b x^3\right )^{2/3}}{2 b^4}-\frac {3 a^2 \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac {3 a \left (a+b x^3\right )^{8/3}}{8 b^4}-\frac {3 \left (a+b x^3\right )^{11/3}}{11 b^4}}{3 d}\) |
Input:
Int[(x^11*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]
Output:
((-3*a^3*(a + b*x^3)^(2/3))/(2*b^4) - (3*a^2*(a + b*x^3)^(5/3))/(5*b^4) + (3*a*(a + b*x^3)^(8/3))/(8*b^4) - (3*(a + b*x^3)^(11/3))/(11*b^4) - (2^(2/ 3)*Sqrt[3]*a^(11/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3))/(Sqrt[3]* a^(1/3))])/b^4 + (a^(11/3)*Log[a - b*x^3])/(2^(1/3)*b^4) - (3*a^(11/3)*Log [2^(1/3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(1/3)*b^4))/(3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 1.71 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.67
| method | result | size |
| pseudoelliptic | \(\frac {-220 \,2^{\frac {2}{3}} \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )\right ) a^{\frac {11}{3}}-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (40 b^{3} x^{9}+65 a \,b^{2} x^{6}+98 a^{2} b \,x^{3}+293 a^{3}\right )}{1320 b^{4} d}\) | \(150\) |
Input:
int(x^11*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
Output:
1/1320*(-220*2^(2/3)*(2*arctan(1/3*(a^(1/3)+2^(2/3)*(b*x^3+a)^(1/3))*3^(1/ 2)/a^(1/3))*3^(1/2)+2*ln((b*x^3+a)^(1/3)-2^(1/3)*a^(1/3))-ln((b*x^3+a)^(2/ 3)+2^(1/3)*a^(1/3)*(b*x^3+a)^(1/3)+2^(2/3)*a^(2/3)))*a^(11/3)-3*(b*x^3+a)^ (2/3)*(40*b^3*x^9+65*a*b^2*x^6+98*a^2*b*x^3+293*a^3))/b^4/d
Time = 0.13 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.94 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {440 \cdot 4^{\frac {1}{3}} \sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a^{3} \arctan \left (\frac {4^{\frac {1}{3}} \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 220 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a^{3} \log \left (4^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a - 2 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a\right ) - 440 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a^{3} \log \left (-4^{\frac {2}{3}} \left (-a^{2}\right )^{\frac {2}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a\right ) + 3 \, {\left (40 \, b^{3} x^{9} + 65 \, a b^{2} x^{6} + 98 \, a^{2} b x^{3} + 293 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{1320 \, b^{4} d} \] Input:
integrate(x^11*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
Output:
-1/1320*(440*4^(1/3)*sqrt(3)*(-a^2)^(1/3)*a^3*arctan(1/3*(4^(1/3)*sqrt(3)* (b*x^3 + a)^(1/3)*(-a^2)^(1/3) - sqrt(3)*a)/a) + 220*4^(1/3)*(-a^2)^(1/3)* a^3*log(4^(2/3)*(b*x^3 + a)^(1/3)*(-a^2)^(2/3) + 2*(b*x^3 + a)^(2/3)*a - 2 *4^(1/3)*(-a^2)^(1/3)*a) - 440*4^(1/3)*(-a^2)^(1/3)*a^3*log(-4^(2/3)*(-a^2 )^(2/3) + 2*(b*x^3 + a)^(1/3)*a) + 3*(40*b^3*x^9 + 65*a*b^2*x^6 + 98*a^2*b *x^3 + 293*a^3)*(b*x^3 + a)^(2/3))/(b^4*d)
\[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{11} \left (a + b x^{3}\right )^{\frac {2}{3}}}{- a + b x^{3}}\, dx}{d} \] Input:
integrate(x**11*(b*x**3+a)**(2/3)/(-b*d*x**3+a*d),x)
Output:
-Integral(x**11*(a + b*x**3)**(2/3)/(-a + b*x**3), x)/d
Time = 0.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.82 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {\frac {440 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{d} - \frac {220 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{d} + \frac {440 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}{d} + \frac {3 \, {\left (40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} - 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a + 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3}\right )}}{d}}{1320 \, b^{4}} \] Input:
integrate(x^11*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
Output:
-1/1320*(440*sqrt(3)*2^(2/3)*a^(11/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)* a^(1/3) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/d - 220*2^(2/3)*a^(11/3)*log(2^(2/ 3)*a^(2/3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/d + 44 0*2^(2/3)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3))/d + 3*(40*(b* x^3 + a)^(11/3) - 55*(b*x^3 + a)^(8/3)*a + 88*(b*x^3 + a)^(5/3)*a^2 + 220* (b*x^3 + a)^(2/3)*a^3)/d)/b^4
Time = 0.61 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.96 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {\sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{3 \, b^{4} d} + \frac {2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{6 \, b^{4} d} - \frac {2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left ({\left | -2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \right |}\right )}{3 \, b^{4} d} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b^{40} d^{10} - 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a b^{40} d^{10} + 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} b^{40} d^{10} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{40} d^{10}}{440 \, b^{44} d^{11}} \] Input:
integrate(x^11*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="giac")
Output:
-1/3*sqrt(3)*2^(2/3)*a^(11/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/(b^4*d) + 1/6*2^(2/3)*a^(11/3)*log(2^(2/3) *a^(2/3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/(b^4*d) - 1/3*2^(2/3)*a^(11/3)*log(abs(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3)))/(b^4 *d) - 1/440*(40*(b*x^3 + a)^(11/3)*b^40*d^10 - 55*(b*x^3 + a)^(8/3)*a*b^40 *d^10 + 88*(b*x^3 + a)^(5/3)*a^2*b^40*d^10 + 220*(b*x^3 + a)^(2/3)*a^3*b^4 0*d^10)/(b^44*d^11)
Time = 4.16 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=\frac {a\,{\left (b\,x^3+a\right )}^{8/3}}{8\,b^4\,d}-\frac {a^3\,{\left (b\,x^3+a\right )}^{2/3}}{2\,b^4\,d}-\frac {a^2\,{\left (b\,x^3+a\right )}^{5/3}}{5\,b^4\,d}-\frac {{\left (b\,x^3+a\right )}^{11/3}}{11\,b^4\,d}+\frac {4^{1/3}\,{\left (-a\right )}^{11/3}\,\ln \left (4\,a^8\,{\left (b\,x^3+a\right )}^{1/3}+4\,2^{1/3}\,{\left (-a\right )}^{25/3}\right )}{3\,b^4\,d}-\frac {4^{1/3}\,{\left (-a\right )}^{11/3}\,\ln \left (\frac {4\,a^8\,{\left (b\,x^3+a\right )}^{1/3}}{b^8\,d^2}+\frac {2\,4^{2/3}\,{\left (-a\right )}^{25/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^4\,d}+\frac {4^{1/3}\,{\left (-a\right )}^{11/3}\,\ln \left (\frac {4\,a^8\,{\left (b\,x^3+a\right )}^{1/3}}{b^8\,d^2}+\frac {18\,4^{2/3}\,{\left (-a\right )}^{25/3}\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2}{b^8\,d^2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^4\,d} \] Input:
int((x^11*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x)
Output:
(a*(a + b*x^3)^(8/3))/(8*b^4*d) - (a^3*(a + b*x^3)^(2/3))/(2*b^4*d) - (a^2 *(a + b*x^3)^(5/3))/(5*b^4*d) - (a + b*x^3)^(11/3)/(11*b^4*d) + (4^(1/3)*( -a)^(11/3)*log(4*a^8*(a + b*x^3)^(1/3) + 4*2^(1/3)*(-a)^(25/3)))/(3*b^4*d) - (4^(1/3)*(-a)^(11/3)*log((4*a^8*(a + b*x^3)^(1/3))/(b^8*d^2) + (2*4^(2/ 3)*(-a)^(25/3)*((3^(1/2)*1i)/2 + 1/2)^2)/(b^8*d^2))*((3^(1/2)*1i)/2 + 1/2) )/(3*b^4*d) + (4^(1/3)*(-a)^(11/3)*log((4*a^8*(a + b*x^3)^(1/3))/(b^8*d^2) + (18*4^(2/3)*(-a)^(25/3)*((3^(1/2)*1i)/6 - 1/6)^2)/(b^8*d^2))*((3^(1/2)* 1i)/6 - 1/6))/(b^4*d)
\[ \int \frac {x^{11} \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=\frac {147 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{3}-98 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} b \,x^{3}-65 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} x^{6}-40 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} x^{9}+880 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{-b^{2} x^{6}+a^{2}}d x \right ) a^{3} b^{2}}{440 b^{4} d} \] Input:
int(x^11*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x)
Output:
(147*(a + b*x**3)**(2/3)*a**3 - 98*(a + b*x**3)**(2/3)*a**2*b*x**3 - 65*(a + b*x**3)**(2/3)*a*b**2*x**6 - 40*(a + b*x**3)**(2/3)*b**3*x**9 + 880*int (((a + b*x**3)**(2/3)*x**5)/(a**2 - b**2*x**6),x)*a**3*b**2)/(440*b**4*d)