\(\int \frac {x^8 (a+b x^3)^{2/3}}{a d-b d x^3} \, dx\) [801]

Optimal result

Integrand size = 28, antiderivative size = 177 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^3 d}-\frac {\left (a+b x^3\right )^{8/3}}{8 b^3 d}-\frac {2^{2/3} a^{8/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^3 d}+\frac {a^{8/3} \log \left (a-b x^3\right )}{3 \sqrt [3]{2} b^3 d}-\frac {a^{8/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.19 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {15 a^2 \left (a+b x^3\right )^{2/3}+6 a b x^3 \left (a+b x^3\right )^{2/3}+3 b^2 x^6 \left (a+b x^3\right )^{2/3}+8\ 2^{2/3} \sqrt {3} a^{8/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+8\ 2^{2/3} a^{8/3} \log \left (-2 \sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}\right )-4\ 2^{2/3} a^{8/3} \log \left (2 a^{2/3}+2^{2/3} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{24 b^3 d} \] Input:

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

Output:

-1/24*(15*a^2*(a + b*x^3)^(2/3) + 6*a*b*x^3*(a + b*x^3)^(2/3) + 3*b^2*x^6* 
(a + b*x^3)^(2/3) + 8*2^(2/3)*Sqrt[3]*a^(8/3)*ArcTan[(1 + (2^(2/3)*(a + b* 
x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 8*2^(2/3)*a^(8/3)*Log[-2*a^(1/3) + 2^(2/3) 
*(a + b*x^3)^(1/3)] - 4*2^(2/3)*a^(8/3)*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^3*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.94, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 = 1.66 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {-4 \,2^{\frac {2}{3}} \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )\right ) a^{\frac {8}{3}}-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (b^{2} x^{6}+2 a b \,x^{3}+5 a^{2}\right )}{24 b^{3} d}\)	\(138\)

Input:

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

Output:

1/24*(-4*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^(8/3)-3*(b*x^3+a)^(2/3) 
*(b^2*x^6+2*a*b*x^3+5*a^2))/b^3/d
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.11 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {8 \cdot 4^{\frac {1}{3}} \sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a^{2} \arctan \left (\frac {4^{\frac {1}{3}} \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 4 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a^{2} \log \left (4^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a - 2 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a\right ) - 8 \cdot 4^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {1}{3}} a^{2} \log \left (-4^{\frac {2}{3}} \left (-a^{2}\right )^{\frac {2}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a\right ) + 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + 5 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{24 \, b^{3} d} \] Input:

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

Output:

-1/24*(8*4^(1/3)*sqrt(3)*(-a^2)^(1/3)*a^2*arctan(1/3*(4^(1/3)*sqrt(3)*(b*x 
^3 + a)^(1/3)*(-a^2)^(1/3) - sqrt(3)*a)/a) + 4*4^(1/3)*(-a^2)^(1/3)*a^2*lo 
g(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) - 8*4^(1/3)*(-a^2)^(1/3)*a^2*log(-4^(2/3)*(-a^2)^(2/3) 
+ 2*(b*x^3 + a)^(1/3)*a) + 3*(b^2*x^6 + 2*a*b*x^3 + 5*a^2)*(b*x^3 + a)^(2/ 
3))/(b^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.88 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {\frac {8 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{d} - \frac {4 \cdot 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{d} + \frac {8 \cdot 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}{d} + \frac {3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {8}{3}} + 4 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2}\right )}}{d}}{24 \, b^{3}} \] Input:

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

Output:

-1/24*(8*sqrt(3)*2^(2/3)*a^(8/3)*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 - 4*2^(2/3)*a^(8/3)*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 + 8*2^(2/3)* 
a^(8/3)*log(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3))/d + 3*((b*x^3 + a)^(8/3) 
 + 4*(b*x^3 + a)^(2/3)*a^2)/d)/b^3
 

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.99 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {\sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{3 \, b^{3} d} + \frac {2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{6 \, b^{3} d} - \frac {2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left ({\left | -2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \right |}\right )}{3 \, b^{3} d} - \frac {{\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{21} d^{7} + 4 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{21} d^{7}}{8 \, b^{24} d^{8}} \] Input:

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

Output:

-1/3*sqrt(3)*2^(2/3)*a^(8/3)*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^3*d) + 1/6*2^(2/3)*a^(8/3)*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^3*d) - 
1/3*2^(2/3)*a^(8/3)*log(abs(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3)))/(b^3*d) 
 - 1/8*((b*x^3 + a)^(8/3)*b^21*d^7 + 4*(b*x^3 + a)^(2/3)*a^2*b^21*d^7)/(b^ 
24*d^8)
 

Mupad [B] (verification not implemented)

Time = 3.98 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.16 \[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^3\,d}-\frac {a^2\,{\left (b\,x^3+a\right )}^{2/3}}{2\,b^3\,d}-\frac {4^{1/3}\,a^{8/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{3\,b^3\,d}-\frac {4^{1/3}\,a^{8/3}\,\ln \left (\frac {4\,a^6\,{\left (b\,x^3+a\right )}^{1/3}}{b^6\,d^2}-\frac {2\,4^{2/3}\,a^{19/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^6\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^3\,d}+\frac {4^{1/3}\,a^{8/3}\,\ln \left (\frac {4\,a^6\,{\left (b\,x^3+a\right )}^{1/3}}{b^6\,d^2}-\frac {18\,4^{2/3}\,a^{19/3}\,{\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2}{b^6\,d^2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^3\,d} \] Input:

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

Output:

(4^(1/3)*a^(8/3)*log((4*a^6*(a + b*x^3)^(1/3))/(b^6*d^2) - (18*4^(2/3)*a^( 
19/3)*((3^(1/2)*1i)/6 + 1/6)^2)/(b^6*d^2))*((3^(1/2)*1i)/6 + 1/6))/(b^3*d) 
 - (a^2*(a + b*x^3)^(2/3))/(2*b^3*d) - (4^(1/3)*a^(8/3)*log((a + b*x^3)^(1 
/3) - 2^(1/3)*a^(1/3)))/(3*b^3*d) - (4^(1/3)*a^(8/3)*log((4*a^6*(a + b*x^3 
)^(1/3))/(b^6*d^2) - (2*4^(2/3)*a^(19/3)*((3^(1/2)*1i)/2 - 1/2)^2)/(b^6*d^ 
2))*((3^(1/2)*1i)/2 - 1/2))/(3*b^3*d) - (a + b*x^3)^(8/3)/(8*b^3*d)
 

Reduce [F]

\[ \int \frac {x^8 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=\frac {3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2}-2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{3}-\left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{6}+16 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{-b^{2} x^{6}+a^{2}}d x \right ) a^{2} b^{2}}{8 b^{3} d} \] Input:

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

Output:

(3*(a + b*x**3)**(2/3)*a**2 - 2*(a + b*x**3)**(2/3)*a*b*x**3 - (a + b*x**3 
)**(2/3)*b**2*x**6 + 16*int(((a + b*x**3)**(2/3)*x**5)/(a**2 - b**2*x**6), 
x)*a**2*b**2)/(8*b**3*d)