\(\int \frac {(a+b x^3)^{5/3}}{(c+d x^3)^2} \, dx\) [151]

Optimal result

Integrand size = 21, antiderivative size = 301 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx=-\frac {(b c-a d) x \left (a+b x^3\right )^{2/3}}{3 c d \left (c+d x^3\right )}+\frac {b^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}-\frac {(b c-a d)^{2/3} (3 b c+2 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} d^2}-\frac {(b c-a d)^{2/3} (3 b c+2 a d) \log \left (c+d x^3\right )}{18 c^{5/3} d^2}+\frac {(b c-a d)^{2/3} (3 b c+2 a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} d^2}-\frac {b^{5/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 d^2} \] Output:

-1/3*(-a*d+b*c)*x*(b*x^3+a)^(2/3)/c/d/(d*x^3+c)+1/3*b^(5/3)*arctan(1/3*(1+ 
2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/d^2-1/9*(-a*d+b*c)^(2/3)*(2* 
a*d+3*b*c)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^( 
1/2))*3^(1/2)/c^(5/3)/d^2-1/18*(-a*d+b*c)^(2/3)*(2*a*d+3*b*c)*ln(d*x^3+c)/ 
c^(5/3)/d^2+1/6*(-a*d+b*c)^(2/3)*(2*a*d+3*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/ 
3)-(b*x^3+a)^(1/3))/c^(5/3)/d^2-1/2*b^(5/3)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3)) 
/d^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 9.56 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx=\frac {-\frac {12 d (b c-a d) x \left (a+b x^3\right )^{2/3}}{c \left (c+d x^3\right )}+12 \sqrt {3} b^{5/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+\frac {2 i \left (3 i+\sqrt {3}\right ) \left (3 b^2 c^2-a b c d-2 a^2 d^2\right ) \text {arctanh}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d} x}}{\sqrt {3}}\right )}{c^{5/3} \sqrt [3]{b c-a d}}-12 b^{5/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-\frac {2 i \left (-i+\sqrt {3}\right ) \left (3 b^2 c^2-a b c d-2 a^2 d^2\right ) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{5/3} \sqrt [3]{b c-a d}}+6 b^{5/3} \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\frac {\left (1+i \sqrt {3}\right ) \left (3 b^2 c^2-a b c d-2 a^2 d^2\right ) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{5/3} \sqrt [3]{b c-a d}}}{36 d^2} \] Input:

Integrate[(a + b*x^3)^(5/3)/(c + d*x^3)^2,x]
 

Output:

((-12*d*(b*c - a*d)*x*(a + b*x^3)^(2/3))/(c*(c + d*x^3)) + 12*Sqrt[3]*b^(5 
/3)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] + ((2*I) 
*(3*I + Sqrt[3])*(3*b^2*c^2 - a*b*c*d - 2*a^2*d^2)*ArcTanh[(I + ((-I + Sqr 
t[3])*c^(1/3)*(a + b*x^3)^(1/3))/((b*c - a*d)^(1/3)*x))/Sqrt[3]])/(c^(5/3) 
*(b*c - a*d)^(1/3)) - 12*b^(5/3)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] - ( 
(2*I)*(-I + Sqrt[3])*(3*b^2*c^2 - a*b*c*d - 2*a^2*d^2)*Log[2*(b*c - a*d)^( 
1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(c^(5/3)*(b*c - a*d)^ 
(1/3)) + 6*b^(5/3)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b* 
x^3)^(2/3)] + ((1 + I*Sqrt[3])*(3*b^2*c^2 - a*b*c*d - 2*a^2*d^2)*Log[2*(b* 
c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x 
^3)^(1/3) + I*(I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(c^(5/3)*(b*c - a* 
d)^(1/3)))/(36*d^2)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {930, 1026, 769, 901}

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.

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

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

\(\displaystyle \frac {\frac {3 b^2 c \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{d}-\frac {(b c-a d) (2 a d+3 b c) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{d}}{3 c d}-\frac {x \left (a+b x^3\right )^{2/3} (b c-a d)}{3 c d \left (c+d x^3\right )}\)

Input:

Int[(a + b*x^3)^(5/3)/(c + d*x^3)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 930

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 
1))), x] - Simp[1/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q 
- 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( 
p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
 

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.38

method	result	size
pseudoelliptic	\(-\frac {-\frac {3 b^{\frac {5}{3}} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) c^{2} \left (d \,x^{3}+c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{2}-2 \left (a d -b c \right ) \left (a d +\frac {3 b c}{2}\right ) \left (d \,x^{3}+c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )+3 b^{\frac {5}{3}} \sqrt {3}\, \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c^{2} \left (d \,x^{3}+c \right ) \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right )+3 b^{\frac {5}{3}} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) c^{2} \left (d \,x^{3}+c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (a d -b c \right ) \left (-3 d \left (b \,x^{3}+a \right )^{\frac {2}{3}} x c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (a d +\frac {3 b c}{2}\right ) \left (d \,x^{3}+c \right ) \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right )\right )}{9 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} d^{2} c^{2} \left (d \,x^{3}+c \right )}\)	\(416\)

Input:

int((b*x^3+a)^(5/3)/(d*x^3+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/9*(-3/2*b^(5/3)*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/ 
3))/x^2)*c^2*(d*x^3+c)*((a*d-b*c)/c)^(1/3)-2*(a*d-b*c)*(a*d+3/2*b*c)*(d*x^ 
3+c)*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)+3*b^(5/3)*3^(1/2)*((a*d 
-b*c)/c)^(1/3)*c^2*(d*x^3+c)*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)/b^(1/3) 
+x)/x)+3*b^(5/3)*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*c^2*(d*x^3+c)*((a*d-b* 
c)/c)^(1/3)+(a*d-b*c)*(-3*d*(b*x^3+a)^(2/3)*x*c*((a*d-b*c)/c)^(1/3)+(a*d+3 
/2*b*c)*(d*x^3+c)*(-2*arctan(1/3*3^(1/2)*(-2/((a*d-b*c)/c)^(1/3)*(b*x^3+a) 
^(1/3)+x)/x)*3^(1/2)+ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^ 
3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))))/((a*d-b*c)/c)^(1/3)/d^2/c^2/(d*x^3+c 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 631 vs. \(2 (248) = 496\).

Time = 0.38 (sec) , antiderivative size = 631, normalized size of antiderivative = 2.10 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="fricas")
 

Output:

-1/18*(2*sqrt(3)*((3*b*c*d + 2*a*d^2)*x^3 + 3*b*c^2 + 2*a*c*d)*((b^2*c^2 - 
 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arctan(-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sq 
rt(3)*(b*x^3 + a)^(1/3)*c*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3))/((b 
*c - a*d)*x)) + 6*sqrt(3)*(b*c*d*x^3 + b*c^2)*(-b^2)^(1/3)*arctan(-1/3*(sq 
rt(3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^2)^(1/3))/(b*x)) + 6*(b*x^3 + 
a)^(2/3)*(b*c*d - a*d^2)*x - 2*((3*b*c*d + 2*a*d^2)*x^3 + 3*b*c^2 + 2*a*c* 
d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log((c*x*((b^2*c^2 - 2*a*b* 
c*d + a^2*d^2)/c^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d))/x) - 6*(b*c*d*x 
^3 + b*c^2)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + 
3*(b*c*d*x^3 + b*c^2)*(-b^2)^(1/3)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^ 
(1/3)*(-b^2)^(2/3)*x - (b*x^3 + a)^(2/3)*b)/x^2) + ((3*b*c*d + 2*a*d^2)*x^ 
3 + 3*b*c^2 + 2*a*c*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log(-(( 
b*c - a*d)*x^2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3) + (b*x^3 + a)^( 
1/3)*c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(2/3) + (b*x^3 + a)^(2/3)*( 
b*c - a*d))/x^2))/(c*d^3*x^3 + c^2*d^2)
 

Sympy [F]

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

integrate((b*x**3+a)**(5/3)/(d*x**3+c)**2,x)
 

Output:

Integral((a + b*x**3)**(5/3)/(c + d*x**3)**2, x)
 

Maxima [F]

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

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^2, x)
 

Giac [F]

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

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{5/3}}{{\left (d\,x^3+c\right )}^2} \,d x \] Input:

int((a + b*x^3)^(5/3)/(c + d*x^3)^2,x)
 

Output:

int((a + b*x^3)^(5/3)/(c + d*x^3)^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx =\text {Too large to display} \] Input:

int((b*x^3+a)^(5/3)/(d*x^3+c)^2,x)
                                                                                    
                                                                                    
 

Output:

( - (a + b*x**3)**(2/3)*a*b*x + 4*int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 
 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + 
a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 
- 3*b**2*c*d**2*x**9),x)*a**4*c*d**2 + 4*int((a + b*x**3)**(2/3)/(2*a**2*c 
**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d* 
x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2* 
d*x**6 - 3*b**2*c*d**2*x**9),x)*a**4*d**3*x**3 - 10*int((a + b*x**3)**(2/3 
)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b*c**3 - 4* 
a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c**3*x**3 - 6 
*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**3*b*c**2*d - 10*int((a + b*x 
**3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d**3*x**6 - 3*a*b 
*c**3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x**9 - 3*b**2*c** 
3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**3*b*c*d**2*x**3 + 
6*int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 2*a**2*d** 
3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a*b*d**3*x** 
9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9),x)*a**2*b* 
*2*c**3 + 6*int((a + b*x**3)**(2/3)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**3 + 
2*a**2*d**3*x**6 - 3*a*b*c**3 - 4*a*b*c**2*d*x**3 + a*b*c*d**2*x**6 + 2*a* 
b*d**3*x**9 - 3*b**2*c**3*x**3 - 6*b**2*c**2*d*x**6 - 3*b**2*c*d**2*x**9), 
x)*a**2*b**2*c**2*d*x**3 + 2*int(((a + b*x**3)**(2/3)*x**3)/(c**2 + 2*c...