\(\int \frac {(a+b x^3)^3}{(c+d x^3)^3} \, dx\) [28]

Optimal result

Integrand size = 19, antiderivative size = 287 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{6 c d^3 \left (c+d x^3\right )^2}+\frac {(b c-a d)^2 (13 b c+5 a d) x}{18 c^2 d^3 \left (c+d x^3\right )}+\frac {(b c-a d) \left (14 b^2 c^2+8 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{9 \sqrt {3} c^{8/3} d^{10/3}}-\frac {(b c-a d) \left (14 b^2 c^2+8 a b c d+5 a^2 d^2\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{27 c^{8/3} d^{10/3}}+\frac {(b c-a d) \left (14 b^2 c^2+8 a b c d+5 a^2 d^2\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{54 c^{8/3} d^{10/3}} \] Output:

b^3*x/d^3-1/6*(-a*d+b*c)^3*x/c/d^3/(d*x^3+c)^2+1/18*(-a*d+b*c)^2*(5*a*d+13 
*b*c)*x/c^2/d^3/(d*x^3+c)+1/27*(-a*d+b*c)*(5*a^2*d^2+8*a*b*c*d+14*b^2*c^2) 
*arctan(1/3*(c^(1/3)-2*d^(1/3)*x)*3^(1/2)/c^(1/3))*3^(1/2)/c^(8/3)/d^(10/3 
)-1/27*(-a*d+b*c)*(5*a^2*d^2+8*a*b*c*d+14*b^2*c^2)*ln(c^(1/3)+d^(1/3)*x)/c 
^(8/3)/d^(10/3)+1/54*(-a*d+b*c)*(5*a^2*d^2+8*a*b*c*d+14*b^2*c^2)*ln(c^(2/3 
)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/c^(8/3)/d^(10/3)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {54 b^3 \sqrt [3]{d} x+\frac {9 \sqrt [3]{d} (-b c+a d)^3 x}{c \left (c+d x^3\right )^2}+\frac {3 \sqrt [3]{d} (b c-a d)^2 (13 b c+5 a d) x}{c^2 \left (c+d x^3\right )}+\frac {2 \sqrt {3} \left (14 b^3 c^3-6 a b^2 c^2 d-3 a^2 b c d^2-5 a^3 d^3\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{c^{8/3}}-\frac {2 \left (14 b^3 c^3-6 a b^2 c^2 d-3 a^2 b c d^2-5 a^3 d^3\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{8/3}}+\frac {\left (14 b^3 c^3-6 a b^2 c^2 d-3 a^2 b c d^2-5 a^3 d^3\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{8/3}}}{54 d^{10/3}} \] Input:

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

Output:

(54*b^3*d^(1/3)*x + (9*d^(1/3)*(-(b*c) + a*d)^3*x)/(c*(c + d*x^3)^2) + (3* 
d^(1/3)*(b*c - a*d)^2*(13*b*c + 5*a*d)*x)/(c^2*(c + d*x^3)) + (2*Sqrt[3]*( 
14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTan[(1 - (2*d^( 
1/3)*x)/c^(1/3))/Sqrt[3]])/c^(8/3) - (2*(14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^ 
2*b*c*d^2 - 5*a^3*d^3)*Log[c^(1/3) + d^(1/3)*x])/c^(8/3) + ((14*b^3*c^3 - 
6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x 
 + d^(2/3)*x^2])/c^(8/3))/(54*d^(10/3))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {915, 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.

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

\(\Big \downarrow \) 915

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(b c-a d) \left (5 a^2 d^2+8 a b c d+14 b^2 c^2\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{9 \sqrt {3} c^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2+8 a b c d+14 b^2 c^2\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{54 c^{8/3} d^{10/3}}-\frac {(b c-a d) \left (5 a^2 d^2+8 a b c d+14 b^2 c^2\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{27 c^{8/3} d^{10/3}}+\frac {x (b c-a d)^2 (5 a d+13 b c)}{18 c^2 d^3 \left (c+d x^3\right )}-\frac {x (b c-a d)^3}{6 c d^3 \left (c+d x^3\right )^2}+\frac {b^3 x}{d^3}\)

Input:

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

Output:

(b^3*x)/d^3 - ((b*c - a*d)^3*x)/(6*c*d^3*(c + d*x^3)^2) + ((b*c - a*d)^2*( 
13*b*c + 5*a*d)*x)/(18*c^2*d^3*(c + d*x^3)) + ((b*c - a*d)*(14*b^2*c^2 + 8 
*a*b*c*d + 5*a^2*d^2)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/( 
9*Sqrt[3]*c^(8/3)*d^(10/3)) - ((b*c - a*d)*(14*b^2*c^2 + 8*a*b*c*d + 5*a^2 
*d^2)*Log[c^(1/3) + d^(1/3)*x])/(27*c^(8/3)*d^(10/3)) + ((b*c - a*d)*(14*b 
^2*c^2 + 8*a*b*c*d + 5*a^2*d^2)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)* 
x^2])/(54*c^(8/3)*d^(10/3))
 

Defintions of rubi rules used

rule 915

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Int[PolynomialDivide[(a + b*x^n)^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a 
, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILtQ[q, 
0] && GeQ[p, -q]
 

rule 2009

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.92 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {b^{3} x}{d^{3}}+\frac {\frac {d \left (5 a^{3} d^{3}+3 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +13 b^{3} c^{3}\right ) x^{4}}{18 c^{2}}+\frac {\left (4 a^{3} d^{3}-3 a^{2} b c \,d^{2}-6 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right ) x}{9 c}}{d^{3} \left (d \,x^{3}+c \right )^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+c \right )}{\sum }\frac {\left (5 a^{3} d^{3}+3 a^{2} b c \,d^{2}+6 a \,b^{2} c^{2} d -14 b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{27 d^{4} c^{2}}\)	\(179\)
default	\(\frac {b^{3} x}{d^{3}}+\frac {\frac {\frac {d \left (5 a^{3} d^{3}+3 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +13 b^{3} c^{3}\right ) x^{4}}{18 c^{2}}+\frac {\left (4 a^{3} d^{3}-3 a^{2} b c \,d^{2}-6 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right ) x}{9 c}}{\left (d \,x^{3}+c \right )^{2}}+\frac {\left (5 a^{3} d^{3}+3 a^{2} b c \,d^{2}+6 a \,b^{2} c^{2} d -14 b^{3} c^{3}\right ) \left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right )}{9 c^{2}}}{d^{3}}\)	\(247\)

Input:

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

Output:

b^3*x/d^3+(1/18*d*(5*a^3*d^3+3*a^2*b*c*d^2-21*a*b^2*c^2*d+13*b^3*c^3)/c^2* 
x^4+1/9*(4*a^3*d^3-3*a^2*b*c*d^2-6*a*b^2*c^2*d+5*b^3*c^3)/c*x)/d^3/(d*x^3+ 
c)^2+1/27/d^4/c^2*sum((5*a^3*d^3+3*a^2*b*c*d^2+6*a*b^2*c^2*d-14*b^3*c^3)/_ 
R^2*ln(x-_R),_R=RootOf(_Z^3*d+c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (246) = 492\).

Time = 0.12 (sec) , antiderivative size = 1407, normalized size of antiderivative = 4.90 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/54*(54*b^3*c^4*d^3*x^7 + 3*(49*b^3*c^5*d^2 - 21*a*b^2*c^4*d^3 + 3*a^2*b 
*c^3*d^4 + 5*a^3*c^2*d^5)*x^4 - 3*sqrt(1/3)*(14*b^3*c^6*d - 6*a*b^2*c^5*d^ 
2 - 3*a^2*b*c^4*d^3 - 5*a^3*c^3*d^4 + (14*b^3*c^4*d^3 - 6*a*b^2*c^3*d^4 - 
3*a^2*b*c^2*d^5 - 5*a^3*c*d^6)*x^6 + 2*(14*b^3*c^5*d^2 - 6*a*b^2*c^4*d^3 - 
 3*a^2*b*c^3*d^4 - 5*a^3*c^2*d^5)*x^3)*sqrt(-(c^2*d)^(1/3)/d)*log((2*c*d*x 
^3 - 3*(c^2*d)^(1/3)*c*x - c^2 + 3*sqrt(1/3)*(2*c*d*x^2 + (c^2*d)^(2/3)*x 
- (c^2*d)^(1/3)*c)*sqrt(-(c^2*d)^(1/3)/d))/(d*x^3 + c)) + (14*b^3*c^5 - 6* 
a*b^2*c^4*d - 3*a^2*b*c^3*d^2 - 5*a^3*c^2*d^3 + (14*b^3*c^3*d^2 - 6*a*b^2* 
c^2*d^3 - 3*a^2*b*c*d^4 - 5*a^3*d^5)*x^6 + 2*(14*b^3*c^4*d - 6*a*b^2*c^3*d 
^2 - 3*a^2*b*c^2*d^3 - 5*a^3*c*d^4)*x^3)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2* 
d)^(2/3)*x + (c^2*d)^(1/3)*c) - 2*(14*b^3*c^5 - 6*a*b^2*c^4*d - 3*a^2*b*c^ 
3*d^2 - 5*a^3*c^2*d^3 + (14*b^3*c^3*d^2 - 6*a*b^2*c^2*d^3 - 3*a^2*b*c*d^4 
- 5*a^3*d^5)*x^6 + 2*(14*b^3*c^4*d - 6*a*b^2*c^3*d^2 - 3*a^2*b*c^2*d^3 - 5 
*a^3*c*d^4)*x^3)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) + 6*(14*b^3*c^6* 
d - 6*a*b^2*c^5*d^2 - 3*a^2*b*c^4*d^3 + 4*a^3*c^3*d^4)*x)/(c^4*d^6*x^6 + 2 
*c^5*d^5*x^3 + c^6*d^4), 1/54*(54*b^3*c^4*d^3*x^7 + 3*(49*b^3*c^5*d^2 - 21 
*a*b^2*c^4*d^3 + 3*a^2*b*c^3*d^4 + 5*a^3*c^2*d^5)*x^4 - 6*sqrt(1/3)*(14*b^ 
3*c^6*d - 6*a*b^2*c^5*d^2 - 3*a^2*b*c^4*d^3 - 5*a^3*c^3*d^4 + (14*b^3*c^4* 
d^3 - 6*a*b^2*c^3*d^4 - 3*a^2*b*c^2*d^5 - 5*a^3*c*d^6)*x^6 + 2*(14*b^3*c^5 
*d^2 - 6*a*b^2*c^4*d^3 - 3*a^2*b*c^3*d^4 - 5*a^3*c^2*d^5)*x^3)*sqrt((c^...
 

Sympy [A] (verification not implemented)

Time = 4.04 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {b^{3} x}{d^{3}} + \frac {x^{4} \cdot \left (5 a^{3} d^{4} + 3 a^{2} b c d^{3} - 21 a b^{2} c^{2} d^{2} + 13 b^{3} c^{3} d\right ) + x \left (8 a^{3} c d^{3} - 6 a^{2} b c^{2} d^{2} - 12 a b^{2} c^{3} d + 10 b^{3} c^{4}\right )}{18 c^{4} d^{3} + 36 c^{3} d^{4} x^{3} + 18 c^{2} d^{5} x^{6}} + \operatorname {RootSum} {\left (19683 t^{3} c^{8} d^{10} - 125 a^{9} d^{9} - 225 a^{8} b c d^{8} - 585 a^{7} b^{2} c^{2} d^{7} + 483 a^{6} b^{3} c^{3} d^{6} + 558 a^{5} b^{4} c^{4} d^{5} + 2574 a^{4} b^{5} c^{5} d^{4} - 1644 a^{3} b^{6} c^{6} d^{3} - 252 a^{2} b^{7} c^{7} d^{2} - 3528 a b^{8} c^{8} d + 2744 b^{9} c^{9}, \left ( t \mapsto t \log {\left (\frac {27 t c^{3} d^{3}}{5 a^{3} d^{3} + 3 a^{2} b c d^{2} + 6 a b^{2} c^{2} d - 14 b^{3} c^{3}} + x \right )} \right )\right )} \] Input:

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

Output:

b**3*x/d**3 + (x**4*(5*a**3*d**4 + 3*a**2*b*c*d**3 - 21*a*b**2*c**2*d**2 + 
 13*b**3*c**3*d) + x*(8*a**3*c*d**3 - 6*a**2*b*c**2*d**2 - 12*a*b**2*c**3* 
d + 10*b**3*c**4))/(18*c**4*d**3 + 36*c**3*d**4*x**3 + 18*c**2*d**5*x**6) 
+ RootSum(19683*_t**3*c**8*d**10 - 125*a**9*d**9 - 225*a**8*b*c*d**8 - 585 
*a**7*b**2*c**2*d**7 + 483*a**6*b**3*c**3*d**6 + 558*a**5*b**4*c**4*d**5 + 
 2574*a**4*b**5*c**5*d**4 - 1644*a**3*b**6*c**6*d**3 - 252*a**2*b**7*c**7* 
d**2 - 3528*a*b**8*c**8*d + 2744*b**9*c**9, Lambda(_t, _t*log(27*_t*c**3*d 
**3/(5*a**3*d**3 + 3*a**2*b*c*d**2 + 6*a*b**2*c**2*d - 14*b**3*c**3) + x)) 
)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {b^{3} x}{d^{3}} + \frac {{\left (13 \, b^{3} c^{3} d - 21 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + 5 \, a^{3} d^{4}\right )} x^{4} + 2 \, {\left (5 \, b^{3} c^{4} - 6 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 4 \, a^{3} c d^{3}\right )} x}{18 \, {\left (c^{2} d^{5} x^{6} + 2 \, c^{3} d^{4} x^{3} + c^{4} d^{3}\right )}} - \frac {\sqrt {3} {\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{27 \, c^{2} d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{54 \, c^{2} d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{27 \, c^{2} d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \] Input:

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

Output:

b^3*x/d^3 + 1/18*((13*b^3*c^3*d - 21*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + 5*a^3 
*d^4)*x^4 + 2*(5*b^3*c^4 - 6*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + 4*a^3*c*d^3)* 
x)/(c^2*d^5*x^6 + 2*c^3*d^4*x^3 + c^4*d^3) - 1/27*sqrt(3)*(14*b^3*c^3 - 6* 
a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(1/3*sqrt(3)*(2*x - (c/d)^( 
1/3))/(c/d)^(1/3))/(c^2*d^4*(c/d)^(2/3)) + 1/54*(14*b^3*c^3 - 6*a*b^2*c^2* 
d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(x^2 - x*(c/d)^(1/3) + (c/d)^(2/3))/(c^2 
*d^4*(c/d)^(2/3)) - 1/27*(14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a 
^3*d^3)*log(x + (c/d)^(1/3))/(c^2*d^4*(c/d)^(2/3))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {b^{3} x}{d^{3}} + \frac {\sqrt {3} {\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-c d^{2}\right )^{\frac {2}{3}} c^{2} d^{2}} + \frac {{\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{54 \, \left (-c d^{2}\right )^{\frac {2}{3}} c^{2} d^{2}} + \frac {{\left (14 \, b^{3} c^{3} - 6 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{27 \, c^{3} d^{3}} + \frac {13 \, b^{3} c^{3} d x^{4} - 21 \, a b^{2} c^{2} d^{2} x^{4} + 3 \, a^{2} b c d^{3} x^{4} + 5 \, a^{3} d^{4} x^{4} + 10 \, b^{3} c^{4} x - 12 \, a b^{2} c^{3} d x - 6 \, a^{2} b c^{2} d^{2} x + 8 \, a^{3} c d^{3} x}{18 \, {\left (d x^{3} + c\right )}^{2} c^{2} d^{3}} \] Input:

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

Output:

b^3*x/d^3 + 1/27*sqrt(3)*(14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a 
^3*d^3)*arctan(1/3*sqrt(3)*(2*x + (-c/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2 
/3)*c^2*d^2) + 1/54*(14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^ 
3)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*c^2*d^2) + 1/2 
7*(14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*(-c/d)^(1/3)*lo 
g(abs(x - (-c/d)^(1/3)))/(c^3*d^3) + 1/18*(13*b^3*c^3*d*x^4 - 21*a*b^2*c^2 
*d^2*x^4 + 3*a^2*b*c*d^3*x^4 + 5*a^3*d^4*x^4 + 10*b^3*c^4*x - 12*a*b^2*c^3 
*d*x - 6*a^2*b*c^2*d^2*x + 8*a^3*c*d^3*x)/((d*x^3 + c)^2*c^2*d^3)
 

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx=\frac {\frac {x\,\left (4\,a^3\,d^3-3\,a^2\,b\,c\,d^2-6\,a\,b^2\,c^2\,d+5\,b^3\,c^3\right )}{9\,c}+\frac {x^4\,\left (5\,a^3\,d^4+3\,a^2\,b\,c\,d^3-21\,a\,b^2\,c^2\,d^2+13\,b^3\,c^3\,d\right )}{18\,c^2}}{c^2\,d^3+2\,c\,d^4\,x^3+d^5\,x^6}+\frac {b^3\,x}{d^3}+\frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a^2\,d^2+8\,a\,b\,c\,d+14\,b^2\,c^2\right )}{27\,c^{8/3}\,d^{10/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a^2\,d^2+8\,a\,b\,c\,d+14\,b^2\,c^2\right )}{27\,c^{8/3}\,d^{10/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a^2\,d^2+8\,a\,b\,c\,d+14\,b^2\,c^2\right )}{27\,c^{8/3}\,d^{10/3}} \] Input:

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

Output:

((x*(4*a^3*d^3 + 5*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(9*c) + (x^4* 
(5*a^3*d^4 + 13*b^3*c^3*d - 21*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3))/(18*c^2))/( 
c^2*d^3 + d^5*x^6 + 2*c*d^4*x^3) + (b^3*x)/d^3 + (log(d^(1/3)*x + c^(1/3)) 
*(a*d - b*c)*(5*a^2*d^2 + 14*b^2*c^2 + 8*a*b*c*d))/(27*c^(8/3)*d^(10/3)) - 
 (log(3^(1/2)*c^(1/3)*1i - 2*d^(1/3)*x + c^(1/3))*((3^(1/2)*1i)/2 + 1/2)*( 
a*d - b*c)*(5*a^2*d^2 + 14*b^2*c^2 + 8*a*b*c*d))/(27*c^(8/3)*d^(10/3)) + ( 
log(3^(1/2)*c^(1/3)*1i + 2*d^(1/3)*x - c^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(a* 
d - b*c)*(5*a^2*d^2 + 14*b^2*c^2 + 8*a*b*c*d))/(27*c^(8/3)*d^(10/3))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1238, normalized size of antiderivative = 4.31 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 10*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3))) 
*a**3*c**2*d**3 - 20*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**( 
1/3)*sqrt(3)))*a**3*c*d**4*x**3 - 10*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d 
**(1/3)*x)/(c**(1/3)*sqrt(3)))*a**3*d**5*x**6 - 6*c**(1/3)*sqrt(3)*atan((c 
**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a**2*b*c**3*d**2 - 12*c**(1/3) 
*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a**2*b*c**2*d* 
*3*x**3 - 6*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt 
(3)))*a**2*b*c*d**4*x**6 - 12*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3) 
*x)/(c**(1/3)*sqrt(3)))*a*b**2*c**4*d - 24*c**(1/3)*sqrt(3)*atan((c**(1/3) 
 - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a*b**2*c**3*d**2*x**3 - 12*c**(1/3)*s 
qrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a*b**2*c**2*d**3 
*x**6 + 28*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt( 
3)))*b**3*c**5 + 56*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1 
/3)*sqrt(3)))*b**3*c**4*d*x**3 + 28*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d* 
*(1/3)*x)/(c**(1/3)*sqrt(3)))*b**3*c**3*d**2*x**6 - 5*c**(1/3)*log(c**(2/3 
) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*c**2*d**3 - 10*c**(1/3)*log( 
c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*c*d**4*x**3 - 5*c**(1 
/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*d**5*x**6 - 3 
*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**2*b*c**3* 
d**2 - 6*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a...