Integrand size = 19, antiderivative size = 320 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=\frac {d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) x}{b^5}+\frac {d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) x^4}{4 b^4}+\frac {d^4 (5 b c-2 a d) x^7}{7 b^3}+\frac {d^5 x^{10}}{10 b^2}+\frac {(b c-a d)^5 x}{3 a b^5 \left (a+b x^3\right )}-\frac {(b c-a d)^4 (2 b c+13 a d) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{16/3}}+\frac {(b c-a d)^4 (2 b c+13 a d) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{16/3}}-\frac {(b c-a d)^4 (2 b c+13 a d) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{16/3}} \] Output:
d^2*(-4*a^3*d^3+15*a^2*b*c*d^2-20*a*b^2*c^2*d+10*b^3*c^3)*x/b^5+1/4*d^3*(3 *a^2*d^2-10*a*b*c*d+10*b^2*c^2)*x^4/b^4+1/7*d^4*(-2*a*d+5*b*c)*x^7/b^3+1/1 0*d^5*x^10/b^2+1/3*(-a*d+b*c)^5*x/a/b^5/(b*x^3+a)-1/9*(-a*d+b*c)^4*(13*a*d +2*b*c)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(5/3)/ b^(16/3)+1/9*(-a*d+b*c)^4*(13*a*d+2*b*c)*ln(a^(1/3)+b^(1/3)*x)/a^(5/3)/b^( 16/3)-1/18*(-a*d+b*c)^4*(13*a*d+2*b*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3 )*x^2)/a^(5/3)/b^(16/3)
Time = 0.21 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=\frac {1260 \sqrt [3]{b} d^2 \left (10 b^3 c^3-20 a b^2 c^2 d+15 a^2 b c d^2-4 a^3 d^3\right ) x+315 b^{4/3} d^3 \left (10 b^2 c^2-10 a b c d+3 a^2 d^2\right ) x^4+180 b^{7/3} d^4 (5 b c-2 a d) x^7+126 b^{10/3} d^5 x^{10}+\frac {420 \sqrt [3]{b} (b c-a d)^5 x}{a \left (a+b x^3\right )}+\frac {140 \sqrt {3} (b c-a d)^4 (2 b c+13 a d) \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3}}+\frac {140 (b c-a d)^4 (2 b c+13 a d) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{5/3}}-\frac {70 (b c-a d)^4 (2 b c+13 a d) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{5/3}}}{1260 b^{16/3}} \] Input:
Integrate[(c + d*x^3)^5/(a + b*x^3)^2,x]
Output:
(1260*b^(1/3)*d^2*(10*b^3*c^3 - 20*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 4*a^3*d^ 3)*x + 315*b^(4/3)*d^3*(10*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^4 + 180*b^( 7/3)*d^4*(5*b*c - 2*a*d)*x^7 + 126*b^(10/3)*d^5*x^10 + (420*b^(1/3)*(b*c - a*d)^5*x)/(a*(a + b*x^3)) + (140*Sqrt[3]*(b*c - a*d)^4*(2*b*c + 13*a*d)*A rcTan[(-a^(1/3) + 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/a^(5/3) + (140*(b*c - a *d)^4*(2*b*c + 13*a*d)*Log[a^(1/3) + b^(1/3)*x])/a^(5/3) - (70*(b*c - a*d) ^4*(2*b*c + 13*a*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(5/3 ))/(1260*b^(16/3))
Time = 0.88 (sec) , antiderivative size = 320, 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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 915 |
| \(\displaystyle \int \left (\frac {d^3 x^3 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right )}{b^4}+\frac {d^2 \left (-4 a^3 d^3+15 a^2 b c d^2-20 a b^2 c^2 d+10 b^3 c^3\right )}{b^5}+\frac {5 b d x^3 (b c-a d)^4+(4 a d+b c) (b c-a d)^4}{b^5 \left (a+b x^3\right )^2}+\frac {d^4 x^6 (5 b c-2 a d)}{b^3}+\frac {d^5 x^9}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) (b c-a d)^4 (13 a d+2 b c)}{3 \sqrt {3} a^{5/3} b^{16/3}}-\frac {(b c-a d)^4 (13 a d+2 b c) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{16/3}}+\frac {(b c-a d)^4 (13 a d+2 b c) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{16/3}}+\frac {d^3 x^4 \left (3 a^2 d^2-10 a b c d+10 b^2 c^2\right )}{4 b^4}+\frac {d^2 x \left (-4 a^3 d^3+15 a^2 b c d^2-20 a b^2 c^2 d+10 b^3 c^3\right )}{b^5}+\frac {x (b c-a d)^5}{3 a b^5 \left (a+b x^3\right )}+\frac {d^4 x^7 (5 b c-2 a d)}{7 b^3}+\frac {d^5 x^{10}}{10 b^2}\) |
Input:
Int[(c + d*x^3)^5/(a + b*x^3)^2,x]
Output:
(d^2*(10*b^3*c^3 - 20*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 4*a^3*d^3)*x)/b^5 + ( d^3*(10*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^4)/(4*b^4) + (d^4*(5*b*c - 2*a *d)*x^7)/(7*b^3) + (d^5*x^10)/(10*b^2) + ((b*c - a*d)^5*x)/(3*a*b^5*(a + b *x^3)) - ((b*c - a*d)^4*(2*b*c + 13*a*d)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(S qrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(16/3)) + ((b*c - a*d)^4*(2*b*c + 1 3*a*d)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(5/3)*b^(16/3)) - ((b*c - a*d)^4*(2* b*c + 13*a*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(5/3)* b^(16/3))
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 = 0.91 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.95
| method | result | size |
| risch | \(\frac {d^{5} x^{10}}{10 b^{2}}-\frac {2 d^{5} a \,x^{7}}{7 b^{3}}+\frac {5 d^{4} c \,x^{7}}{7 b^{2}}+\frac {3 d^{5} a^{2} x^{4}}{4 b^{4}}-\frac {5 d^{4} a c \,x^{4}}{2 b^{3}}+\frac {5 d^{3} c^{2} x^{4}}{2 b^{2}}-\frac {4 d^{5} a^{3} x}{b^{5}}+\frac {15 d^{4} a^{2} c x}{b^{4}}-\frac {20 d^{3} a \,c^{2} x}{b^{3}}+\frac {10 d^{2} c^{3} x}{b^{2}}-\frac {\left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -c^{5} b^{5}\right ) x}{3 a \,b^{5} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (13 a^{5} d^{5}-50 a^{4} b c \,d^{4}+70 a^{3} b^{2} c^{2} d^{3}-40 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d +2 c^{5} b^{5}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{6} a}\) | \(304\) |
| default | \(-\frac {d^{2} \left (-\frac {1}{10} d^{3} x^{10} b^{3}+\frac {2}{7} a \,b^{2} d^{3} x^{7}-\frac {5}{7} b^{3} c \,d^{2} x^{7}-\frac {3}{4} a^{2} b \,d^{3} x^{4}+\frac {5}{2} a \,b^{2} c \,d^{2} x^{4}-\frac {5}{2} b^{3} c^{2} d \,x^{4}+4 a^{3} d^{3} x -15 a^{2} b c \,d^{2} x +20 a \,b^{2} c^{2} d x -10 b^{3} c^{3} x \right )}{b^{5}}+\frac {-\frac {\left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -c^{5} b^{5}\right ) x}{3 a \left (b \,x^{3}+a \right )}+\frac {\left (13 a^{5} d^{5}-50 a^{4} b c \,d^{4}+70 a^{3} b^{2} c^{2} d^{3}-40 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d +2 c^{5} b^{5}\right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3 a}}{b^{5}}\) | \(367\) |
Input:
int((d*x^3+c)^5/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/10*d^5*x^10/b^2-2/7*d^5/b^3*a*x^7+5/7*d^4/b^2*c*x^7+3/4*d^5/b^4*a^2*x^4- 5/2*d^4/b^3*a*c*x^4+5/2*d^3/b^2*c^2*x^4-4*d^5/b^5*a^3*x+15*d^4/b^4*a^2*c*x -20*d^3/b^3*a*c^2*x+10*d^2/b^2*c^3*x-1/3*(a^5*d^5-5*a^4*b*c*d^4+10*a^3*b^2 *c^2*d^3-10*a^2*b^3*c^3*d^2+5*a*b^4*c^4*d-b^5*c^5)/a*x/b^5/(b*x^3+a)+1/9/b ^6/a*sum((13*a^5*d^5-50*a^4*b*c*d^4+70*a^3*b^2*c^2*d^3-40*a^2*b^3*c^3*d^2+ 5*a*b^4*c^4*d+2*b^5*c^5)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 789 vs. \(2 (275) = 550\).
Time = 0.11 (sec) , antiderivative size = 1619, normalized size of antiderivative = 5.06 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((d*x^3+c)^5/(b*x^3+a)^2,x, algorithm="fricas")
Output:
[1/1260*(126*a^3*b^5*d^5*x^13 + 18*(50*a^3*b^5*c*d^4 - 13*a^4*b^4*d^5)*x^1 0 + 45*(70*a^3*b^5*c^2*d^3 - 50*a^4*b^4*c*d^4 + 13*a^5*b^3*d^5)*x^7 + 315* (40*a^3*b^5*c^3*d^2 - 70*a^4*b^4*c^2*d^3 + 50*a^5*b^3*c*d^4 - 13*a^6*b^2*d ^5)*x^4 + 210*sqrt(1/3)*(2*a^2*b^6*c^5 + 5*a^3*b^5*c^4*d - 40*a^4*b^4*c^3* d^2 + 70*a^5*b^3*c^2*d^3 - 50*a^6*b^2*c*d^4 + 13*a^7*b*d^5 + (2*a*b^7*c^5 + 5*a^2*b^6*c^4*d - 40*a^3*b^5*c^3*d^2 + 70*a^4*b^4*c^2*d^3 - 50*a^5*b^3*c *d^4 + 13*a^6*b^2*d^5)*x^3)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2 *b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^( 1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - 70*(2*a*b^5*c^5 + 5*a^2*b^4 *c^4*d - 40*a^3*b^3*c^3*d^2 + 70*a^4*b^2*c^2*d^3 - 50*a^5*b*c*d^4 + 13*a^6 *d^5 + (2*b^6*c^5 + 5*a*b^5*c^4*d - 40*a^2*b^4*c^3*d^2 + 70*a^3*b^3*c^2*d^ 3 - 50*a^4*b^2*c*d^4 + 13*a^5*b*d^5)*x^3)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2 *b)^(2/3)*x + (a^2*b)^(1/3)*a) + 140*(2*a*b^5*c^5 + 5*a^2*b^4*c^4*d - 40*a ^3*b^3*c^3*d^2 + 70*a^4*b^2*c^2*d^3 - 50*a^5*b*c*d^4 + 13*a^6*d^5 + (2*b^6 *c^5 + 5*a*b^5*c^4*d - 40*a^2*b^4*c^3*d^2 + 70*a^3*b^3*c^2*d^3 - 50*a^4*b^ 2*c*d^4 + 13*a^5*b*d^5)*x^3)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)) + 42 0*(a^2*b^6*c^5 - 5*a^3*b^5*c^4*d + 40*a^4*b^4*c^3*d^2 - 70*a^5*b^3*c^2*d^3 + 50*a^6*b^2*c*d^4 - 13*a^7*b*d^5)*x)/(a^3*b^7*x^3 + a^4*b^6), 1/1260*(12 6*a^3*b^5*d^5*x^13 + 18*(50*a^3*b^5*c*d^4 - 13*a^4*b^4*d^5)*x^10 + 45*(70* a^3*b^5*c^2*d^3 - 50*a^4*b^4*c*d^4 + 13*a^5*b^3*d^5)*x^7 + 315*(40*a^3*...
Time = 5.54 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.71 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=x^{7} \left (- \frac {2 a d^{5}}{7 b^{3}} + \frac {5 c d^{4}}{7 b^{2}}\right ) + x^{4} \cdot \left (\frac {3 a^{2} d^{5}}{4 b^{4}} - \frac {5 a c d^{4}}{2 b^{3}} + \frac {5 c^{2} d^{3}}{2 b^{2}}\right ) + x \left (- \frac {4 a^{3} d^{5}}{b^{5}} + \frac {15 a^{2} c d^{4}}{b^{4}} - \frac {20 a c^{2} d^{3}}{b^{3}} + \frac {10 c^{3} d^{2}}{b^{2}}\right ) + \frac {x \left (- a^{5} d^{5} + 5 a^{4} b c d^{4} - 10 a^{3} b^{2} c^{2} d^{3} + 10 a^{2} b^{3} c^{3} d^{2} - 5 a b^{4} c^{4} d + b^{5} c^{5}\right )}{3 a^{2} b^{5} + 3 a b^{6} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{5} b^{16} - 2197 a^{15} d^{15} + 25350 a^{14} b c d^{14} - 132990 a^{13} b^{2} c^{2} d^{13} + 418280 a^{12} b^{3} c^{3} d^{12} - 874635 a^{11} b^{4} c^{4} d^{11} + 1271886 a^{10} b^{5} c^{5} d^{10} - 1302400 a^{9} b^{6} c^{6} d^{9} + 922680 a^{8} b^{7} c^{7} d^{8} - 422235 a^{7} b^{8} c^{8} d^{7} + 97570 a^{6} b^{9} c^{9} d^{6} + 7194 a^{5} b^{10} c^{10} d^{5} - 10200 a^{4} b^{11} c^{11} d^{4} + 1435 a^{3} b^{12} c^{12} d^{3} + 330 a^{2} b^{13} c^{13} d^{2} - 60 a b^{14} c^{14} d - 8 b^{15} c^{15}, \left ( t \mapsto t \log {\left (\frac {9 t a^{2} b^{5}}{13 a^{5} d^{5} - 50 a^{4} b c d^{4} + 70 a^{3} b^{2} c^{2} d^{3} - 40 a^{2} b^{3} c^{3} d^{2} + 5 a b^{4} c^{4} d + 2 b^{5} c^{5}} + x \right )} \right )\right )} + \frac {d^{5} x^{10}}{10 b^{2}} \] Input:
integrate((d*x**3+c)**5/(b*x**3+a)**2,x)
Output:
x**7*(-2*a*d**5/(7*b**3) + 5*c*d**4/(7*b**2)) + x**4*(3*a**2*d**5/(4*b**4) - 5*a*c*d**4/(2*b**3) + 5*c**2*d**3/(2*b**2)) + x*(-4*a**3*d**5/b**5 + 15 *a**2*c*d**4/b**4 - 20*a*c**2*d**3/b**3 + 10*c**3*d**2/b**2) + x*(-a**5*d* *5 + 5*a**4*b*c*d**4 - 10*a**3*b**2*c**2*d**3 + 10*a**2*b**3*c**3*d**2 - 5 *a*b**4*c**4*d + b**5*c**5)/(3*a**2*b**5 + 3*a*b**6*x**3) + RootSum(729*_t **3*a**5*b**16 - 2197*a**15*d**15 + 25350*a**14*b*c*d**14 - 132990*a**13*b **2*c**2*d**13 + 418280*a**12*b**3*c**3*d**12 - 874635*a**11*b**4*c**4*d** 11 + 1271886*a**10*b**5*c**5*d**10 - 1302400*a**9*b**6*c**6*d**9 + 922680* a**8*b**7*c**7*d**8 - 422235*a**7*b**8*c**8*d**7 + 97570*a**6*b**9*c**9*d* *6 + 7194*a**5*b**10*c**10*d**5 - 10200*a**4*b**11*c**11*d**4 + 1435*a**3* b**12*c**12*d**3 + 330*a**2*b**13*c**13*d**2 - 60*a*b**14*c**14*d - 8*b**1 5*c**15, Lambda(_t, _t*log(9*_t*a**2*b**5/(13*a**5*d**5 - 50*a**4*b*c*d**4 + 70*a**3*b**2*c**2*d**3 - 40*a**2*b**3*c**3*d**2 + 5*a*b**4*c**4*d + 2*b **5*c**5) + x))) + d**5*x**10/(10*b**2)
Time = 0.12 (sec) , antiderivative size = 509, normalized size of antiderivative = 1.59 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=\frac {{\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} x}{3 \, {\left (a b^{6} x^{3} + a^{2} b^{5}\right )}} + \frac {14 \, b^{3} d^{5} x^{10} + 20 \, {\left (5 \, b^{3} c d^{4} - 2 \, a b^{2} d^{5}\right )} x^{7} + 35 \, {\left (10 \, b^{3} c^{2} d^{3} - 10 \, a b^{2} c d^{4} + 3 \, a^{2} b d^{5}\right )} x^{4} + 140 \, {\left (10 \, b^{3} c^{3} d^{2} - 20 \, a b^{2} c^{2} d^{3} + 15 \, a^{2} b c d^{4} - 4 \, a^{3} d^{5}\right )} x}{140 \, b^{5}} + \frac {\sqrt {3} {\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate((d*x^3+c)^5/(b*x^3+a)^2,x, algorithm="maxima")
Output:
1/3*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5 *a^4*b*c*d^4 - a^5*d^5)*x/(a*b^6*x^3 + a^2*b^5) + 1/140*(14*b^3*d^5*x^10 + 20*(5*b^3*c*d^4 - 2*a*b^2*d^5)*x^7 + 35*(10*b^3*c^2*d^3 - 10*a*b^2*c*d^4 + 3*a^2*b*d^5)*x^4 + 140*(10*b^3*c^3*d^2 - 20*a*b^2*c^2*d^3 + 15*a^2*b*c*d ^4 - 4*a^3*d^5)*x)/b^5 + 1/9*sqrt(3)*(2*b^5*c^5 + 5*a*b^4*c^4*d - 40*a^2*b ^3*c^3*d^2 + 70*a^3*b^2*c^2*d^3 - 50*a^4*b*c*d^4 + 13*a^5*d^5)*arctan(1/3* sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b^6*(a/b)^(2/3)) - 1/18*(2*b^5 *c^5 + 5*a*b^4*c^4*d - 40*a^2*b^3*c^3*d^2 + 70*a^3*b^2*c^2*d^3 - 50*a^4*b* c*d^4 + 13*a^5*d^5)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*b^6*(a/b)^(2 /3)) + 1/9*(2*b^5*c^5 + 5*a*b^4*c^4*d - 40*a^2*b^3*c^3*d^2 + 70*a^3*b^2*c^ 2*d^3 - 50*a^4*b*c*d^4 + 13*a^5*d^5)*log(x + (a/b)^(1/3))/(a*b^6*(a/b)^(2/ 3))
Time = 0.13 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.65 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=-\frac {\sqrt {3} {\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b^{4}} - \frac {{\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b^{4}} - \frac {{\left (2 \, b^{5} c^{5} + 5 \, a b^{4} c^{4} d - 40 \, a^{2} b^{3} c^{3} d^{2} + 70 \, a^{3} b^{2} c^{2} d^{3} - 50 \, a^{4} b c d^{4} + 13 \, a^{5} d^{5}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2} b^{5}} + \frac {b^{5} c^{5} x - 5 \, a b^{4} c^{4} d x + 10 \, a^{2} b^{3} c^{3} d^{2} x - 10 \, a^{3} b^{2} c^{2} d^{3} x + 5 \, a^{4} b c d^{4} x - a^{5} d^{5} x}{3 \, {\left (b x^{3} + a\right )} a b^{5}} + \frac {14 \, b^{18} d^{5} x^{10} + 100 \, b^{18} c d^{4} x^{7} - 40 \, a b^{17} d^{5} x^{7} + 350 \, b^{18} c^{2} d^{3} x^{4} - 350 \, a b^{17} c d^{4} x^{4} + 105 \, a^{2} b^{16} d^{5} x^{4} + 1400 \, b^{18} c^{3} d^{2} x - 2800 \, a b^{17} c^{2} d^{3} x + 2100 \, a^{2} b^{16} c d^{4} x - 560 \, a^{3} b^{15} d^{5} x}{140 \, b^{20}} \] Input:
integrate((d*x^3+c)^5/(b*x^3+a)^2,x, algorithm="giac")
Output:
-1/9*sqrt(3)*(2*b^5*c^5 + 5*a*b^4*c^4*d - 40*a^2*b^3*c^3*d^2 + 70*a^3*b^2* c^2*d^3 - 50*a^4*b*c*d^4 + 13*a^5*d^5)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1 /3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)*a*b^4) - 1/18*(2*b^5*c^5 + 5*a*b^4*c^4* d - 40*a^2*b^3*c^3*d^2 + 70*a^3*b^2*c^2*d^3 - 50*a^4*b*c*d^4 + 13*a^5*d^5) *log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a*b^4) - 1/9*(2* b^5*c^5 + 5*a*b^4*c^4*d - 40*a^2*b^3*c^3*d^2 + 70*a^3*b^2*c^2*d^3 - 50*a^4 *b*c*d^4 + 13*a^5*d^5)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^2*b^5) + 1/3*(b^5*c^5*x - 5*a*b^4*c^4*d*x + 10*a^2*b^3*c^3*d^2*x - 10*a^3*b^2*c^2* d^3*x + 5*a^4*b*c*d^4*x - a^5*d^5*x)/((b*x^3 + a)*a*b^5) + 1/140*(14*b^18* d^5*x^10 + 100*b^18*c*d^4*x^7 - 40*a*b^17*d^5*x^7 + 350*b^18*c^2*d^3*x^4 - 350*a*b^17*c*d^4*x^4 + 105*a^2*b^16*d^5*x^4 + 1400*b^18*c^3*d^2*x - 2800* a*b^17*c^2*d^3*x + 2100*a^2*b^16*c*d^4*x - 560*a^3*b^15*d^5*x)/b^20
Time = 0.43 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.30 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx=x\,\left (\frac {10\,c^3\,d^2}{b^2}-\frac {2\,a\,\left (\frac {2\,a\,\left (\frac {2\,a\,d^5}{b^3}-\frac {5\,c\,d^4}{b^2}\right )}{b}-\frac {a^2\,d^5}{b^4}+\frac {10\,c^2\,d^3}{b^2}\right )}{b}+\frac {a^2\,\left (\frac {2\,a\,d^5}{b^3}-\frac {5\,c\,d^4}{b^2}\right )}{b^2}\right )-x^7\,\left (\frac {2\,a\,d^5}{7\,b^3}-\frac {5\,c\,d^4}{7\,b^2}\right )+x^4\,\left (\frac {a\,\left (\frac {2\,a\,d^5}{b^3}-\frac {5\,c\,d^4}{b^2}\right )}{2\,b}-\frac {a^2\,d^5}{4\,b^4}+\frac {5\,c^2\,d^3}{2\,b^2}\right )+\frac {d^5\,x^{10}}{10\,b^2}-\frac {x\,\left (a^5\,d^5-5\,a^4\,b\,c\,d^4+10\,a^3\,b^2\,c^2\,d^3-10\,a^2\,b^3\,c^3\,d^2+5\,a\,b^4\,c^4\,d-b^5\,c^5\right )}{3\,a\,\left (b^6\,x^3+a\,b^5\right )}+\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,{\left (a\,d-b\,c\right )}^4\,\left (13\,a\,d+2\,b\,c\right )}{9\,a^{5/3}\,b^{16/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^4\,\left (13\,a\,d+2\,b\,c\right )}{9\,a^{5/3}\,b^{16/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^4\,\left (13\,a\,d+2\,b\,c\right )}{9\,a^{5/3}\,b^{16/3}} \] Input:
int((c + d*x^3)^5/(a + b*x^3)^2,x)
Output:
x*((10*c^3*d^2)/b^2 - (2*a*((2*a*((2*a*d^5)/b^3 - (5*c*d^4)/b^2))/b - (a^2 *d^5)/b^4 + (10*c^2*d^3)/b^2))/b + (a^2*((2*a*d^5)/b^3 - (5*c*d^4)/b^2))/b ^2) - x^7*((2*a*d^5)/(7*b^3) - (5*c*d^4)/(7*b^2)) + x^4*((a*((2*a*d^5)/b^3 - (5*c*d^4)/b^2))/(2*b) - (a^2*d^5)/(4*b^4) + (5*c^2*d^3)/(2*b^2)) + (d^5 *x^10)/(10*b^2) - (x*(a^5*d^5 - b^5*c^5 - 10*a^2*b^3*c^3*d^2 + 10*a^3*b^2* c^2*d^3 + 5*a*b^4*c^4*d - 5*a^4*b*c*d^4))/(3*a*(a*b^5 + b^6*x^3)) + (log(b ^(1/3)*x + a^(1/3))*(a*d - b*c)^4*(13*a*d + 2*b*c))/(9*a^(5/3)*b^(16/3)) - (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*( a*d - b*c)^4*(13*a*d + 2*b*c))/(9*a^(5/3)*b^(16/3)) + (log(3^(1/2)*a^(1/3) *1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^4*(13*a*d + 2*b*c))/(9*a^(5/3)*b^(16/3))
Time = 0.20 (sec) , antiderivative size = 1349, normalized size of antiderivative = 4.22 \[ \int \frac {\left (c+d x^3\right )^5}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((d*x^3+c)^5/(b*x^3+a)^2,x)
Output:
( - 1820*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3) ))*a**6*d**5 + 7000*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1 /3)*sqrt(3)))*a**5*b*c*d**4 - 1820*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b** (1/3)*x)/(a**(1/3)*sqrt(3)))*a**5*b*d**5*x**3 - 9800*a**(1/3)*sqrt(3)*atan ((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**4*b**2*c**2*d**3 + 7000* a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**4*b **2*c*d**4*x**3 + 5600*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a* *(1/3)*sqrt(3)))*a**3*b**3*c**3*d**2 - 9800*a**(1/3)*sqrt(3)*atan((a**(1/3 ) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**3*b**3*c**2*d**3*x**3 - 700*a**(1 /3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*b**4*c **4*d + 5600*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqr t(3)))*a**2*b**4*c**3*d**2*x**3 - 280*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2* b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**5*c**5 - 700*a**(1/3)*sqrt(3)*atan((a **(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b**5*c**4*d*x**3 - 280*a**(1 /3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**6*c**5*x **3 - 910*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a** 6*d**5 + 3500*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2) *a**5*b*c*d**4 - 910*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3 )*x**2)*a**5*b*d**5*x**3 - 4900*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)* x + b**(2/3)*x**2)*a**4*b**2*c**2*d**3 + 3500*a**(1/3)*log(a**(2/3) - b...