\(\int \frac {x^8}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\) [748]

Optimal result

Integrand size = 24, antiderivative size = 201 \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=-\frac {(b c+a d) \sqrt [3]{a+b x^3}}{b^2 d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3} (b c-a d)^{2/3}}-\frac {c^2 \log \left (c+d x^3\right )}{6 d^{7/3} (b c-a d)^{2/3}}+\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}} \] Output:

-(a*d+b*c)*(b*x^3+a)^(1/3)/b^2/d^2+1/4*(b*x^3+a)^(4/3)/b^2/d-1/3*c^2*arcta 
n(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^(7 
/3)/(-a*d+b*c)^(2/3)-1/6*c^2*ln(d*x^3+c)/d^(7/3)/(-a*d+b*c)^(2/3)+1/2*c^2* 
ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(7/3)/(-a*d+b*c)^(2/3)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.15 \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {-3 \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{a+b x^3} \left (4 b c+3 a d-b d x^3\right )-4 \sqrt {3} b^2 c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+4 b^2 c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 b^2 c^2 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 b^2 d^{7/3} (b c-a d)^{2/3}} \] Input:

Integrate[x^8/((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)*(4*b*c + 3*a*d - b*d*x^3) 
- 4*Sqrt[3]*b^2*c^2*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^ 
(1/3))/Sqrt[3]] + 4*b^2*c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1 
/3)] - 2*b^2*c^2*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)])/(12*b^2*d^(7/3)*(b*c - a*d)^(2/3) 
)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 205, 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.

\(\displaystyle \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^6}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx^3\)

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-\frac {\sqrt {3} c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{7/3} (b c-a d)^{2/3}}-\frac {3 \sqrt [3]{a+b x^3} (a d+b c)}{b^2 d^2}+\frac {3 \left (a+b x^3\right )^{4/3}}{4 b^2 d}-\frac {c^2 \log \left (c+d x^3\right )}{2 d^{7/3} (b c-a d)^{2/3}}+\frac {3 c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3} (b c-a d)^{2/3}}\right )\)

Input:

Int[x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
 

Output:

((-3*(b*c + a*d)*(a + b*x^3)^(1/3))/(b^2*d^2) + (3*(a + b*x^3)^(4/3))/(4*b 
^2*d) - (Sqrt[3]*c^2*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d) 
^(1/3))/Sqrt[3]])/(d^(7/3)*(b*c - a*d)^(2/3)) - (c^2*Log[c + d*x^3])/(2*d^ 
(7/3)*(b*c - a*d)^(2/3)) + (3*c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x 
^3)^(1/3)])/(2*d^(7/3)*(b*c - a*d)^(2/3)))/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 = 1.76 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right ) b^{2} c^{2}}{9}+\frac {4 c^{2} \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) b^{2}}{9}-\frac {4 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right ) b^{2} c^{2}}{9}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d \left (\frac {\left (-d \,x^{3}+4 c \right ) b}{3}+a d \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{4 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{3} b^{2}}\)	\(204\)

Input:

int(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-3/4/((a*d-b*c)/d)^(2/3)*(2/9*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))*b^2*c^2+4/9*c^2*3^(1/2)*arctan(2/3*3^(1/2) 
/((a*d-b*c)/d)^(1/3)*(b*x^3+a)^(1/3)+1/3*3^(1/2))*b^2-4/9*ln((b*x^3+a)^(1/ 
3)-((a*d-b*c)/d)^(1/3))*b^2*c^2+((a*d-b*c)/d)^(2/3)*d*(1/3*(-d*x^3+4*c)*b+ 
a*d)*(b*x^3+a)^(1/3))/d^3/b^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (164) = 328\).

Time = 0.13 (sec) , antiderivative size = 1156, normalized size of antiderivative = 5.75 \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:

integrate(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

[-1/12*(2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*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)) - 4* 
(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(2/3)*b^2*c^2*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)) + 6*sqrt(1/3)* 
(b^3*c^3*d - a*b^2*c^2*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*(4*b^3*c^3*d - 5*a*b^2*c^2*d^2 - 2*a^2*b*c*d^3 + 3*a^3*d^4 
- (b^3*c^2*d^2 - 2*a*b^2*c*d^3 + a^2*b*d^4)*x^3)*(b*x^3 + a)^(1/3))/(b^4*c 
^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5), -1/12*(2*(b^2*c^2*d - 2*a*b*c*d^2 + 
 a^2*d^3)^(2/3)*b^2*c^2*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)) - 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^ 
(2/3)*b^2*c^2*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)) - 12*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*sqrt((b 
^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)^(1/3)/d)*arctan(-sqrt(1/3)*((b^2*c^2*...
 

Sympy [F]

\[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{8}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**8/(b*x**3+a)**(2/3)/(d*x**3+c),x)
 

Output:

Integral(x**8/((a + b*x**3)**(2/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^8/(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
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.53 \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=-\frac {\frac {4 \, b^{2} c^{2} d^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{b c d^{4} - a d^{5}} - \frac {12 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} b^{2} c^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{3} - \sqrt {3} a d^{4}} - \frac {2 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{b c d^{3} - a d^{4}} + \frac {3 \, {\left (4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} d^{3} + 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a d^{3}\right )}}{d^{4}}}{12 \, b^{2}} \] Input:

integrate(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

-1/12*(4*b^2*c^2*d^2*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (- 
(b*c - a*d)/d)^(1/3)))/(b*c*d^4 - a*d^5) - 12*(-b*c*d^2 + a*d^3)^(1/3)*b^2 
*c^2*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^3 - sqrt(3)*a*d^4) - 2*(-b*c*d^2 + a*d 
^3)^(1/3)*b^2*c^2*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^3 - a*d^4) + 3*(4*(b*x^3 + a)^(1 
/3)*b*c*d^2 - (b*x^3 + a)^(4/3)*d^3 + 4*(b*x^3 + a)^(1/3)*a*d^3)/d^4)/b^2
 

Mupad [B] (verification not implemented)

Time = 3.68 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.45 \[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b^2\,d}-\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{1/3}-\frac {\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}+\frac {\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{6\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}{6\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}+\frac {c^2\,\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}-\frac {c^2\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}+\frac {c^2\,\ln \left (3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}-\frac {c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{7/3}\,{\left (a\,d-b\,c\right )}^{2/3}} \] Input:

int(x^8/((a + b*x^3)^(2/3)*(c + d*x^3)),x)
 

Output:

(a + b*x^3)^(4/3)/(4*b^2*d) - ((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^2) 
)*(a + b*x^3)^(1/3) - (log(3*c^2*(a + b*x^3)^(1/3) + ((3^(1/2)*c^2*1i + c^ 
2)*(9*a*d^3 - 9*b*c*d^2))/(6*d^(7/3)*(a*d - b*c)^(2/3)))*(3^(1/2)*c^2*1i + 
 c^2))/(6*d^(7/3)*(a*d - b*c)^(2/3)) + (c^2*log(3*c^2*(a + b*x^3)^(1/3) - 
(c^2*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)*(a*d - b*c)^(2/3))))/(3*d^(7/3)*(a* 
d - b*c)^(2/3)) + (c^2*log(3*c^2*(a + b*x^3)^(1/3) - (c^2*((3^(1/2)*1i)/6 
- 1/6)*(9*a*d^3 - 9*b*c*d^2))/(d^(7/3)*(a*d - b*c)^(2/3)))*((3^(1/2)*1i)/6 
 - 1/6))/(d^(7/3)*(a*d - b*c)^(2/3))
 

Reduce [F]

\[ \int \frac {x^8}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c +\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{3}}d x \] Input:

int(x^8/(b*x^3+a)^(2/3)/(d*x^3+c),x)
 

Output:

int(x**8/((a + b*x**3)**(2/3)*c + (a + b*x**3)**(2/3)*d*x**3),x)