Integrand size = 24, antiderivative size = 241 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}}{b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{4/3}}{4 b^3 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3} (b c-a d)^{2/3}}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{10/3} (b c-a d)^{2/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}} \] Output:
(a^2*d^2+a*b*c*d+b^2*c^2)*(b*x^3+a)^(1/3)/b^3/d^3-1/4*(2*a*d+b*c)*(b*x^3+a )^(4/3)/b^3/d^2+1/7*(b*x^3+a)^(7/3)/b^3/d+1/3*c^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)/(-a*d+b*c)^(2/ 3)+1/6*c^3*ln(d*x^3+c)/d^(10/3)/(-a*d+b*c)^(2/3)-1/2*c^3*ln((-a*d+b*c)^(1/ 3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(10/3)/(-a*d+b*c)^(2/3)
Time = 0.53 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.10 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {3 \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{a+b x^3} \left (18 a^2 d^2+3 a b d \left (7 c-2 d x^3\right )+b^2 \left (28 c^2-7 c d x^3+4 d^2 x^6\right )\right )+28 \sqrt {3} b^3 c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-28 b^3 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+14 b^3 c^3 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{84 b^3 d^{10/3} (b c-a d)^{2/3}} \] Input:
Integrate[x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
(3*d^(1/3)*(b*c - a*d)^(2/3)*(a + b*x^3)^(1/3)*(18*a^2*d^2 + 3*a*b*d*(7*c - 2*d*x^3) + b^2*(28*c^2 - 7*c*d*x^3 + 4*d^2*x^6)) + 28*Sqrt[3]*b^3*c^3*Ar cTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 28*b ^3*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 14*b^3*c^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*b^3*d^(10/3)*(b*c - a*d)^(2/3))
Time = 0.70 (sec) , antiderivative size = 246, 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.
| \(\displaystyle \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (-\frac {c^3}{d^3 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{4/3}}{b^2 d}+\frac {(-b c-2 a d) \sqrt [3]{b x^3+a}}{b^2 d^2}+\frac {b^2 c^2+a b d c+a^2 d^2}{b^2 d^3 \left (b x^3+a\right )^{2/3}}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \sqrt [3]{a+b x^3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{b^3 d^3}+\frac {\sqrt {3} c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{10/3} (b c-a d)^{2/3}}-\frac {3 \left (a+b x^3\right )^{4/3} (2 a d+b c)}{4 b^3 d^2}+\frac {3 \left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{2 d^{10/3} (b c-a d)^{2/3}}-\frac {3 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3} (b c-a d)^{2/3}}\right )\) |
Input:
Int[x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
Output:
((3*(b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(1/3))/(b^3*d^3) - (3*(b*c + 2*a*d)*(a + b*x^3)^(4/3))/(4*b^3*d^2) + (3*(a + b*x^3)^(7/3))/(7*b^3*d) + (Sqrt[3]*c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3)) /Sqrt[3]])/(d^(10/3)*(b*c - a*d)^(2/3)) + (c^3*Log[c + d*x^3])/(2*d^(10/3) *(b*c - a*d)^(2/3)) - (3*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^( 1/3)])/(2*d^(10/3)*(b*c - a*d)^(2/3)))/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 = 1.77 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.97
| method | result | size |
| pseudoelliptic | \(\frac {\frac {27 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (\frac {\left (2 d^{2} x^{6}-\frac {7}{2} c d \,x^{3}+14 c^{2}\right ) b^{2}}{9}+\frac {7 a d \left (-\frac {2 d \,x^{3}}{7}+c \right ) b}{6}+a^{2} d^{2}\right ) d \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{7}+b^{3} c^{3} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} b^{3} d^{4}}\) | \(234\) |
Input:
int(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/6/((a*d-b*c)/d)^(2/3)*(27/7*((a*d-b*c)/d)^(2/3)*(1/9*(2*d^2*x^6-7/2*c*d* x^3+14*c^2)*b^2+7/6*a*d*(-2/7*d*x^3+c)*b+a^2*d^2)*d*(b*x^3+a)^(1/3)+b^3*c^ 3*(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))))/b^3/d^4
Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (201) = 402\).
Time = 0.11 (sec) , antiderivative size = 1322, normalized size of antiderivative = 5.49 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
Output:
[1/84*(14*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^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)) - 28*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^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)) - 42*sqr t(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^ 3)^(1/3)/d)*log((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2 - 2*(b^2*c*d - a*b*d^2)*x ^3 - 3*sqrt(1/3)*(2*(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))*sqrt((-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(1/3) /d) - 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))/(d*x^3 + c)) + 3*(28*b^4*c^4*d - 35*a*b^3*c^3*d^2 + 4*a^2*b^2*c^2* d^3 - 15*a^3*b*c*d^4 + 18*a^4*d^5 + 4*(b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b ^2*d^5)*x^6 - (7*b^4*c^3*d^2 - 8*a*b^3*c^2*d^3 - 5*a^2*b^2*c*d^4 + 6*a^3*b *d^5)*x^3)*(b*x^3 + a)^(1/3))/(b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6), 1/84*(14*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^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)) - 28*(-b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)^(2/3)*b^3*c^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)) - 84*...
\[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] 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. \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] 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 = 0.16 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.54 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {b^{24} c^{3} d^{4} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{25} c d^{7} - a b^{24} d^{8}\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{4} - \sqrt {3} a d^{5}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c d^{4} - a d^{5}\right )}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{20} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{19} c d^{5} + 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{19} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{18} d^{6} - 14 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{18} d^{6} + 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b^{18} d^{6}}{28 \, b^{21} d^{7}} \] Input:
integrate(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
Output:
1/3*b^24*c^3*d^4*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^25*c*d^7 - a*b^24*d^8) - (-b*c*d^2 + a*d^3)^(1/3)*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))/(sqrt(3)*b*c*d^4 - sqrt(3)*a*d^5) - 1/6*(-b*c*d^2 + a*d^ 3)^(1/3)*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))/(b*c*d^4 - a*d^5) + 1/28*(28*(b*x^3 + a)^(1/ 3)*b^20*c^2*d^4 - 7*(b*x^3 + a)^(4/3)*b^19*c*d^5 + 28*(b*x^3 + a)^(1/3)*a* b^19*c*d^5 + 4*(b*x^3 + a)^(7/3)*b^18*d^6 - 14*(b*x^3 + a)^(4/3)*a*b^18*d^ 6 + 28*(b*x^3 + a)^(1/3)*a^2*b^18*d^6)/(b^21*d^7)
Time = 3.98 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.37 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\left (\frac {3\,a^2}{b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{1/3}-\left (\frac {3\,a}{4\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{4\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{4/3}+\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b^3\,d}+\frac {\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}+\frac {3\,c^3\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{2\,d^{4/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}-\frac {c^3\,\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {3\,c^3\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{4/3}}\right )}{3\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}-\frac {c^3\,\ln \left (\frac {3\,c^3\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {3\,c^3\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}} \] Input:
int(x^11/((a + b*x^3)^(2/3)*(c + d*x^3)),x)
Output:
((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))*(a + b*x^3)^(1/3) - ((3*a)/(4*b^3*d) + (b^4*c - a*b^3* d)/(4*b^6*d^2))*(a + b*x^3)^(4/3) + (a + b*x^3)^(7/3)/(7*b^3*d) + (log((3* c^3*(a + b*x^3)^(1/3))/d + (3*c^3*(3^(1/2)*1i + 1)*(a*d - b*c)^(1/3))/(2*d ^(4/3)))*(3^(1/2)*c^3*1i + c^3))/(6*d^(10/3)*(a*d - b*c)^(2/3)) - (c^3*log ((3*c^3*(a + b*x^3)^(1/3))/d - (3*c^3*(a*d - b*c)^(1/3))/d^(4/3)))/(3*d^(1 0/3)*(a*d - b*c)^(2/3)) - (c^3*log((3*c^3*(a + b*x^3)^(1/3))/d - (3*c^3*(( 3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(1/3))/d^(4/3))*((3^(1/2)*1i)/2 - 1/2))/( 3*d^(10/3)*(a*d - b*c)^(2/3))
\[ \int \frac {x^{11}}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c +\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{3}}d x \] Input:
int(x^11/(b*x^3+a)^(2/3)/(d*x^3+c),x)
Output:
int(x**11/((a + b*x**3)**(2/3)*c + (a + b*x**3)**(2/3)*d*x**3),x)