Integrand size = 19, antiderivative size = 235 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=-\frac {b^2 (2 b c-3 a d) x}{d^3}+\frac {b^3 x^4}{4 d^2}-\frac {(b c-a d)^3 x}{3 c d^3 \left (c+d x^3\right )}-\frac {(b c-a d)^2 (7 b c+2 a d) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} d^{10/3}}+\frac {(b c-a d)^2 (7 b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{10/3}}-\frac {(b c-a d)^2 (7 b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{10/3}} \] Output:
-b^2*(-3*a*d+2*b*c)*x/d^3+1/4*b^3*x^4/d^2-1/3*(-a*d+b*c)^3*x/c/d^3/(d*x^3+ c)-1/9*(-a*d+b*c)^2*(2*a*d+7*b*c)*arctan(1/3*(c^(1/3)-2*d^(1/3)*x)*3^(1/2) /c^(1/3))*3^(1/2)/c^(5/3)/d^(10/3)+1/9*(-a*d+b*c)^2*(2*a*d+7*b*c)*ln(c^(1/ 3)+d^(1/3)*x)/c^(5/3)/d^(10/3)-1/18*(-a*d+b*c)^2*(2*a*d+7*b*c)*ln(c^(2/3)- c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/c^(5/3)/d^(10/3)
Time = 0.13 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=\frac {-36 b^2 \sqrt [3]{d} (2 b c-3 a d) x+9 b^3 d^{4/3} x^4+\frac {12 \sqrt [3]{d} (-b c+a d)^3 x}{c \left (c+d x^3\right )}+\frac {4 \sqrt {3} (b c-a d)^2 (7 b c+2 a d) \arctan \left (\frac {-\sqrt [3]{c}+2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{c^{5/3}}+\frac {4 (b c-a d)^2 (7 b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{5/3}}-\frac {2 (b c-a d)^2 (7 b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{5/3}}}{36 d^{10/3}} \] Input:
Integrate[(a + b*x^3)^3/(c + d*x^3)^2,x]
Output:
(-36*b^2*d^(1/3)*(2*b*c - 3*a*d)*x + 9*b^3*d^(4/3)*x^4 + (12*d^(1/3)*(-(b* c) + a*d)^3*x)/(c*(c + d*x^3)) + (4*Sqrt[3]*(b*c - a*d)^2*(7*b*c + 2*a*d)* ArcTan[(-c^(1/3) + 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/c^(5/3) + (4*(b*c - a* d)^2*(7*b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x])/c^(5/3) - (2*(b*c - a*d)^2* (7*b*c + 2*a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/c^(5/3))/( 36*d^(10/3))
Time = 0.72 (sec) , antiderivative size = 235, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 915 |
| \(\displaystyle \int \left (-\frac {b^2 (2 b c-3 a d)}{d^3}+\frac {3 b d x^3 (b c-a d)^2+(b c-a d)^2 (a d+2 b c)}{d^3 \left (c+d x^3\right )^2}+\frac {b^3 x^3}{d^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {(b c-a d)^2 (2 a d+7 b c) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{3 \sqrt {3} c^{5/3} d^{10/3}}-\frac {b^2 x (2 b c-3 a d)}{d^3}-\frac {(b c-a d)^2 (2 a d+7 b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{10/3}}+\frac {(b c-a d)^2 (2 a d+7 b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{10/3}}-\frac {x (b c-a d)^3}{3 c d^3 \left (c+d x^3\right )}+\frac {b^3 x^4}{4 d^2}\) |
Input:
Int[(a + b*x^3)^3/(c + d*x^3)^2,x]
Output:
-((b^2*(2*b*c - 3*a*d)*x)/d^3) + (b^3*x^4)/(4*d^2) - ((b*c - a*d)^3*x)/(3* c*d^3*(c + d*x^3)) - ((b*c - a*d)^2*(7*b*c + 2*a*d)*ArcTan[(c^(1/3) - 2*d^ (1/3)*x)/(Sqrt[3]*c^(1/3))])/(3*Sqrt[3]*c^(5/3)*d^(10/3)) + ((b*c - a*d)^2 *(7*b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x])/(9*c^(5/3)*d^(10/3)) - ((b*c - a*d)^2*(7*b*c + 2*a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(18 *c^(5/3)*d^(10/3))
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.92 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.65
| method | result | size |
| risch | \(\frac {b^{3} x^{4}}{4 d^{2}}+\frac {3 b^{2} a x}{d^{2}}-\frac {2 b^{3} c x}{d^{3}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{3 c \,d^{3} \left (d \,x^{3}+c \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+c \right )}{\sum }\frac {\left (2 a^{3} d^{3}+3 a^{2} b c \,d^{2}-12 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 d^{4} c}\) | \(153\) |
| default | \(\frac {b^{2} \left (\frac {1}{4} b d \,x^{4}+3 a d x -2 b c x \right )}{d^{3}}+\frac {\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{3 c \left (d \,x^{3}+c \right )}+\frac {\left (2 a^{3} d^{3}+3 a^{2} b c \,d^{2}-12 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right ) \left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right )}{3 c}}{d^{3}}\) | \(215\) |
Input:
int((b*x^3+a)^3/(d*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
1/4*b^3*x^4/d^2+3*b^2/d^2*a*x-2*b^3/d^3*c*x+1/3*(a^3*d^3-3*a^2*b*c*d^2+3*a *b^2*c^2*d-b^3*c^3)/c*x/d^3/(d*x^3+c)+1/9/d^4/c*sum((2*a^3*d^3+3*a^2*b*c*d ^2-12*a*b^2*c^2*d+7*b^3*c^3)/_R^2*ln(x-_R),_R=RootOf(_Z^3*d+c))
Leaf count of result is larger than twice the leaf count of optimal. 494 vs. \(2 (194) = 388\).
Time = 0.10 (sec) , antiderivative size = 1029, normalized size of antiderivative = 4.38 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c)^2,x, algorithm="fricas")
Output:
[1/36*(9*b^3*c^3*d^3*x^7 - 9*(7*b^3*c^4*d^2 - 12*a*b^2*c^3*d^3)*x^4 + 6*sq rt(1/3)*(7*b^3*c^5*d - 12*a*b^2*c^4*d^2 + 3*a^2*b*c^3*d^3 + 2*a^3*c^2*d^4 + (7*b^3*c^4*d^2 - 12*a*b^2*c^3*d^3 + 3*a^2*b*c^2*d^4 + 2*a^3*c*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)) - 2*(7*b^3*c^4 - 12*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 + 2*a^3 *c*d^3 + (7*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + 2*a^3*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) + 4*(7*b^3 *c^4 - 12*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3 + (7*b^3*c^3*d - 12* a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + 2*a^3*d^4)*x^3)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 12*(7*b^3*c^5*d - 12*a*b^2*c^4*d^2 + 3*a^2*b*c^3*d^3 - a^ 3*c^2*d^4)*x)/(c^3*d^5*x^3 + c^4*d^4), 1/36*(9*b^3*c^3*d^3*x^7 - 9*(7*b^3* c^4*d^2 - 12*a*b^2*c^3*d^3)*x^4 + 12*sqrt(1/3)*(7*b^3*c^5*d - 12*a*b^2*c^4 *d^2 + 3*a^2*b*c^3*d^3 + 2*a^3*c^2*d^4 + (7*b^3*c^4*d^2 - 12*a*b^2*c^3*d^3 + 3*a^2*b*c^2*d^4 + 2*a^3*c*d^5)*x^3)*sqrt((c^2*d)^(1/3)/d)*arctan(sqrt(1 /3)*(2*(c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt((c^2*d)^(1/3)/d)/c^2) - 2*( 7*b^3*c^4 - 12*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3 + (7*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + 2*a^3*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) + 4*(7*b^3*c^4 - 12*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3 + (7*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 3*a...
Time = 1.80 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=\frac {b^{3} x^{4}}{4 d^{2}} + x \left (\frac {3 a b^{2}}{d^{2}} - \frac {2 b^{3} c}{d^{3}}\right ) + \frac {x \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{3 c^{2} d^{3} + 3 c d^{4} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} c^{5} d^{10} - 8 a^{9} d^{9} - 36 a^{8} b c d^{8} + 90 a^{7} b^{2} c^{2} d^{7} + 321 a^{6} b^{3} c^{3} d^{6} - 792 a^{5} b^{4} c^{4} d^{5} - 477 a^{4} b^{5} c^{5} d^{4} + 2946 a^{3} b^{6} c^{6} d^{3} - 3465 a^{2} b^{7} c^{7} d^{2} + 1764 a b^{8} c^{8} d - 343 b^{9} c^{9}, \left ( t \mapsto t \log {\left (\frac {9 t c^{2} d^{3}}{2 a^{3} d^{3} + 3 a^{2} b c d^{2} - 12 a b^{2} c^{2} d + 7 b^{3} c^{3}} + x \right )} \right )\right )} \] Input:
integrate((b*x**3+a)**3/(d*x**3+c)**2,x)
Output:
b**3*x**4/(4*d**2) + x*(3*a*b**2/d**2 - 2*b**3*c/d**3) + x*(a**3*d**3 - 3* a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(3*c**2*d**3 + 3*c*d**4*x**3) + RootSum(729*_t**3*c**5*d**10 - 8*a**9*d**9 - 36*a**8*b*c*d**8 + 90*a**7 *b**2*c**2*d**7 + 321*a**6*b**3*c**3*d**6 - 792*a**5*b**4*c**4*d**5 - 477* a**4*b**5*c**5*d**4 + 2946*a**3*b**6*c**6*d**3 - 3465*a**2*b**7*c**7*d**2 + 1764*a*b**8*c**8*d - 343*b**9*c**9, Lambda(_t, _t*log(9*_t*c**2*d**3/(2* a**3*d**3 + 3*a**2*b*c*d**2 - 12*a*b**2*c**2*d + 7*b**3*c**3) + x)))
Time = 0.11 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=-\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x}{3 \, {\left (c d^{4} x^{3} + c^{2} d^{3}\right )}} + \frac {b^{3} d x^{4} - 4 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x}{4 \, d^{3}} + \frac {\sqrt {3} {\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{9 \, c d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{18 \, c d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{9 \, c d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c)^2,x, algorithm="maxima")
Output:
-1/3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x/(c*d^4*x^3 + c^ 2*d^3) + 1/4*(b^3*d*x^4 - 4*(2*b^3*c - 3*a*b^2*d)*x)/d^3 + 1/9*sqrt(3)*(7* b^3*c^3 - 12*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 2*a^3*d^3)*arctan(1/3*sqrt(3)*( 2*x - (c/d)^(1/3))/(c/d)^(1/3))/(c*d^4*(c/d)^(2/3)) - 1/18*(7*b^3*c^3 - 12 *a*b^2*c^2*d + 3*a^2*b*c*d^2 + 2*a^3*d^3)*log(x^2 - x*(c/d)^(1/3) + (c/d)^ (2/3))/(c*d^4*(c/d)^(2/3)) + 1/9*(7*b^3*c^3 - 12*a*b^2*c^2*d + 3*a^2*b*c*d ^2 + 2*a^3*d^3)*log(x + (c/d)^(1/3))/(c*d^4*(c/d)^(2/3))
Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=-\frac {\sqrt {3} {\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-c d^{2}\right )^{\frac {2}{3}} c d^{2}} - \frac {{\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{18 \, \left (-c d^{2}\right )^{\frac {2}{3}} c d^{2}} - \frac {{\left (7 \, b^{3} c^{3} - 12 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{9 \, c^{2} d^{3}} - \frac {b^{3} c^{3} x - 3 \, a b^{2} c^{2} d x + 3 \, a^{2} b c d^{2} x - a^{3} d^{3} x}{3 \, {\left (d x^{3} + c\right )} c d^{3}} + \frac {b^{3} d^{6} x^{4} - 8 \, b^{3} c d^{5} x + 12 \, a b^{2} d^{6} x}{4 \, d^{8}} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c)^2,x, algorithm="giac")
Output:
-1/9*sqrt(3)*(7*b^3*c^3 - 12*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 2*a^3*d^3)*arct an(1/3*sqrt(3)*(2*x + (-c/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2/3)*c*d^2) - 1/18*(7*b^3*c^3 - 12*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 2*a^3*d^3)*log(x^2 + x *(-c/d)^(1/3) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*c*d^2) - 1/9*(7*b^3*c^3 - 12 *a*b^2*c^2*d + 3*a^2*b*c*d^2 + 2*a^3*d^3)*(-c/d)^(1/3)*log(abs(x - (-c/d)^ (1/3)))/(c^2*d^3) - 1/3*(b^3*c^3*x - 3*a*b^2*c^2*d*x + 3*a^2*b*c*d^2*x - a ^3*d^3*x)/((d*x^3 + c)*c*d^3) + 1/4*(b^3*d^6*x^4 - 8*b^3*c*d^5*x + 12*a*b^ 2*d^6*x)/d^8
Time = 0.72 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx=x\,\left (\frac {3\,a\,b^2}{d^2}-\frac {2\,b^3\,c}{d^3}\right )+\frac {b^3\,x^4}{4\,d^2}+\frac {x\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{3\,c\,\left (d^4\,x^3+c\,d^3\right )}+\frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (2\,a\,d+7\,b\,c\right )}{9\,c^{5/3}\,d^{10/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (2\,a\,d+7\,b\,c\right )}{9\,c^{5/3}\,d^{10/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (2\,a\,d+7\,b\,c\right )}{9\,c^{5/3}\,d^{10/3}} \] Input:
int((a + b*x^3)^3/(c + d*x^3)^2,x)
Output:
x*((3*a*b^2)/d^2 - (2*b^3*c)/d^3) + (b^3*x^4)/(4*d^2) + (x*(a^3*d^3 - b^3* c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(3*c*(c*d^3 + d^4*x^3)) + (log(d^(1/ 3)*x + c^(1/3))*(a*d - b*c)^2*(2*a*d + 7*b*c))/(9*c^(5/3)*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)^2*(2*a*d + 7*b*c))/(9*c^(5/3)*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)^2*(2*a*d + 7*b*c ))/(9*c^(5/3)*d^(10/3))
Time = 0.26 (sec) , antiderivative size = 816, normalized size of antiderivative = 3.47 \[ \int \frac {\left (a+b x^3\right )^3}{\left (c+d x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((b*x^3+a)^3/(d*x^3+c)^2,x)
Output:
( - 8*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))* a**3*c*d**3 - 8*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)* sqrt(3)))*a**3*d**4*x**3 - 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**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*d**3*x**3 + 48*c**(1/3)*s qrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a*b**2*c**3*d + 48*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*a*b **2*c**2*d**2*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**4 - 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*x**3 - 4*c**(1/3)*log(c**(2/3) - d* *(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*c*d**3 - 4*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*d**4*x**3 - 6*c**(1/3)*log(c**( 2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**2*b*c**2*d**2 - 6*c**(1/3)* log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**2*b*c*d**3*x**3 + 2 4*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a*b**2*c**3 *d + 24*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a*b** 2*c**2*d**2*x**3 - 14*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/ 3)*x**2)*b**3*c**4 - 14*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**( 2/3)*x**2)*b**3*c**3*d*x**3 + 8*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*a**3*c *d**3 + 8*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*a**3*d**4*x**3 + 12*c**(1...