\(\int \frac {x^{11}}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [730]

Optimal result

Integrand size = 24, antiderivative size = 244 \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{5/3}}{5 b^3 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}+\frac {c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3} \sqrt [3]{b c-a d}}-\frac {c^3 \log \left (c+d x^3\right )}{6 d^{11/3} \sqrt [3]{b c-a d}}+\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} \sqrt [3]{b c-a d}} \] Output:

1/2*(a^2*d^2+a*b*c*d+b^2*c^2)*(b*x^3+a)^(2/3)/b^3/d^3-1/5*(2*a*d+b*c)*(b*x 
^3+a)^(5/3)/b^3/d^2+1/8*(b*x^3+a)^(8/3)/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^(11/3)/(-a*d+b*c) 
^(1/3)-1/6*c^3*ln(d*x^3+c)/d^(11/3)/(-a*d+b*c)^(1/3)+1/2*c^3*ln((-a*d+b*c) 
^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(11/3)/(-a*d+b*c)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.08 \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {3 d^{2/3} \sqrt [3]{b c-a d} \left (a+b x^3\right )^{2/3} \left (9 a^2 d^2-6 a b d \left (-2 c+d x^3\right )+b^2 \left (20 c^2-8 c d x^3+5 d^2 x^6\right )\right )+40 \sqrt {3} b^3 c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+40 b^3 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-20 b^3 c^3 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{120 b^3 d^{11/3} \sqrt [3]{b c-a d}} \] Input:

Integrate[x^11/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

(3*d^(2/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(2/3)*(9*a^2*d^2 - 6*a*b*d*(-2*c 
+ d*x^3) + b^2*(20*c^2 - 8*c*d*x^3 + 5*d^2*x^6)) + 40*Sqrt[3]*b^3*c^3*ArcT 
an[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] + 40*b^3 
*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] - 20*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)])/(120*b^3*d^(11/3)*(b*c - a*d)^(1/3))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 248, 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.

\(\displaystyle \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^9}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {1}{3} \int \left (-\frac {c^3}{d^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{5/3}}{b^2 d}+\frac {(-b c-2 a d) \left (b x^3+a\right )^{2/3}}{b^2 d^2}+\frac {b^2 c^2+a b d c+a^2 d^2}{b^2 d^3 \sqrt [3]{b x^3+a}}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^3 d^3}+\frac {\sqrt {3} c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{11/3} \sqrt [3]{b c-a d}}-\frac {3 \left (a+b x^3\right )^{5/3} (2 a d+b c)}{5 b^3 d^2}+\frac {3 \left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {c^3 \log \left (c+d x^3\right )}{2 d^{11/3} \sqrt [3]{b c-a d}}+\frac {3 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} \sqrt [3]{b c-a d}}\right )\)

Input:

Int[x^11/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

((3*(b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b*x^3)^(2/3))/(2*b^3*d^3) - (3*(b*c 
 + 2*a*d)*(a + b*x^3)^(5/3))/(5*b^3*d^2) + (3*(a + b*x^3)^(8/3))/(8*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^(11/3)*(b*c - a*d)^(1/3)) - (c^3*Log[c + d*x^3])/(2*d^(11/ 
3)*(b*c - a*d)^(1/3)) + (3*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3) 
^(1/3)])/(2*d^(11/3)*(b*c - a*d)^(1/3)))/3
 

Defintions of rubi rules used

rule 99

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]))
 

rule 948

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]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(\frac {\frac {27 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} d \left (\frac {\left (5 d^{2} x^{6}-8 c d \,x^{3}+20 c^{2}\right ) b^{2}}{9}+\frac {4 a \left (-\frac {d \,x^{3}}{2}+c \right ) d b}{3}+a^{2} d^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{20}+b^{3} c^{3} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} b^{3} d^{4}}\)	\(234\)

Input:

int(x^11/(b*x^3+a)^(1/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

1/6/((a*d-b*c)/d)^(1/3)*(27/20*((a*d-b*c)/d)^(1/3)*d*(1/9*(5*d^2*x^6-8*c*d 
*x^3+20*c^2)*b^2+4/3*a*(-1/2*d*x^3+c)*d*b+a^2*d^2)*(b*x^3+a)^(2/3)+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
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 873, normalized size of antiderivative = 3.58 \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

integrate(x^11/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

[-1/120*(20*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^3*log((b*x^3 + a)^(2/3)*d^2 - (b 
*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 40* 
(b*c*d^2 - a*d^3)^(2/3)*b^3*c^3*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3 
)^(1/3)) - 60*sqrt(1/3)*(b^4*c^4*d - a*b^3*c^3*d^2)*sqrt(-(b*c*d^2 - a*d^3 
)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 - 3*sqrt(1/3)*(2*( 
b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^ 
2) - (b*c*d^2 - a*d^3)^(1/3)*(b*c - a*d))*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(b 
*c - a*d)) - 3*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c)) - 3 
*(20*b^3*c^3*d^2 - 8*a*b^2*c^2*d^3 - 3*a^2*b*c*d^4 - 9*a^3*d^5 + 5*(b^3*c* 
d^4 - a*b^2*d^5)*x^6 - 2*(4*b^3*c^2*d^3 - a*b^2*c*d^4 - 3*a^2*b*d^5)*x^3)* 
(b*x^3 + a)^(2/3))/(b^4*c*d^5 - a*b^3*d^6), -1/120*(20*(b*c*d^2 - a*d^3)^( 
2/3)*b^3*c^3*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + 
a)^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 40*(b*c*d^2 - a*d^3)^(2/3)*b^3*c^3 
*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) + 120*sqrt(1/3)*(b^4*c 
^4*d - a*b^3*c^3*d^2)*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))*arctan(sqr 
t(1/3)*(2*(b*x^3 + a)^(1/3)*d - (b*c*d^2 - a*d^3)^(1/3))*sqrt((b*c*d^2 - a 
*d^3)^(1/3)/(b*c - a*d))/d) - 3*(20*b^3*c^3*d^2 - 8*a*b^2*c^2*d^3 - 3*a^2* 
b*c*d^4 - 9*a^3*d^5 + 5*(b^3*c*d^4 - a*b^2*d^5)*x^6 - 2*(4*b^3*c^2*d^3 - a 
*b^2*c*d^4 - 3*a^2*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(b^4*c*d^5 - a*b^3*d^6)]
 

Sympy [F]

\[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**11/(b*x**3+a)**(1/3)/(d*x**3+c),x)
 

Output:

Integral(x**11/((a + b*x**3)**(1/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^11/(b*x^3+a)^(1/3)/(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
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.52 \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {b^{27} c^{3} d^{5} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{28} c d^{8} - a b^{27} d^{9}\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{5} - \sqrt {3} a d^{6}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c d^{5} - a d^{6}\right )}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{23} c^{2} d^{5} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{22} c d^{6} + 20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{22} c d^{6} + 5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{21} d^{7} - 16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{21} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{21} d^{7}}{40 \, b^{24} d^{8}} \] Input:

integrate(x^11/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

1/3*b^27*c^3*d^5*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c 
 - a*d)/d)^(1/3)))/(b^28*c*d^8 - a*b^27*d^9) + (-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))/(sqrt(3)*b*c*d^5 - sqrt(3)*a*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))/(b*c*d^5 - a*d^6) + 1/40*(20*(b*x^3 + a)^(2/ 
3)*b^23*c^2*d^5 - 8*(b*x^3 + a)^(5/3)*b^22*c*d^6 + 20*(b*x^3 + a)^(2/3)*a* 
b^22*c*d^6 + 5*(b*x^3 + a)^(8/3)*b^21*d^7 - 16*(b*x^3 + a)^(5/3)*a*b^21*d^ 
7 + 20*(b*x^3 + a)^(2/3)*a^2*b^21*d^7)/(b^24*d^8)
 

Mupad [B] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.39 \[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\left (\frac {3\,a^2}{2\,b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{2\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\left (\frac {3\,a}{5\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{5\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{5/3}+\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^3\,d}-\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}+\frac {b\,c^7-a\,c^6\,d}{d^{16/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}-\frac {c^6\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{16/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {c^3\,\ln \left (\frac {c^6\,{\left (b\,x^3+a\right )}^{1/3}}{d^5}+\frac {c^6\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{16/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \] Input:

int(x^11/((a + b*x^3)^(1/3)*(c + d*x^3)),x)
 

Output:

((3*a^2)/(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))/(2*b^3*d))*(a + b*x^3)^(2/3) - ((3*a)/(5*b^3*d) + (b^4*c - a* 
b^3*d)/(5*b^6*d^2))*(a + b*x^3)^(5/3) + (a + b*x^3)^(8/3)/(8*b^3*d) - (c^3 
*log((c^6*(a + b*x^3)^(1/3))/d^5 + (b*c^7 - a*c^6*d)/(d^(16/3)*(a*d - b*c) 
^(2/3))))/(3*d^(11/3)*(a*d - b*c)^(1/3)) + (log((c^6*(a + b*x^3)^(1/3))/d^ 
5 - (c^6*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(1/3))/(4*d^(16/3)))*(3^(1/2)*c^3* 
1i + c^3))/(6*d^(11/3)*(a*d - b*c)^(1/3)) - (c^3*log((c^6*(a + b*x^3)^(1/3 
))/d^5 + (c^6*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1/3))/d^(16/3))*((3^(1/2 
)*1i)/2 - 1/2))/(3*d^(11/3)*(a*d - b*c)^(1/3))
 

Reduce [F]

\[ \int \frac {x^{11}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} d \,x^{3}}d x \] Input:

int(x^11/(b*x^3+a)^(1/3)/(d*x^3+c),x)
 

Output:

int(x**11/((a + b*x**3)**(1/3)*c + (a + b*x**3)**(1/3)*d*x**3),x)