Integrand size = 18, antiderivative size = 223 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=\frac {4 a^2 d \left (3 b^4 c+7 a^4 d\right ) x}{b^9}-\frac {3 a d \left (2 b^4 c+7 a^4 d\right ) x^2}{2 b^8}+\frac {d \left (2 b^4 c+15 a^4 d\right ) x^3}{3 b^7}-\frac {5 a^3 d^2 x^4}{2 b^6}+\frac {6 a^2 d^2 x^5}{5 b^5}-\frac {a d^2 x^6}{2 b^4}+\frac {d^2 x^7}{7 b^3}+\frac {a \left (b^4 c+a^4 d\right )^2}{2 b^{10} (a+b x)^2}-\frac {\left (b^4 c+a^4 d\right ) \left (b^4 c+9 a^4 d\right )}{b^{10} (a+b x)}-\frac {4 a^3 d \left (5 b^4 c+9 a^4 d\right ) \log (a+b x)}{b^{10}} \] Output:
4*a^2*d*(7*a^4*d+3*b^4*c)*x/b^9-3/2*a*d*(7*a^4*d+2*b^4*c)*x^2/b^8+1/3*d*(1 5*a^4*d+2*b^4*c)*x^3/b^7-5/2*a^3*d^2*x^4/b^6+6/5*a^2*d^2*x^5/b^5-1/2*a*d^2 *x^6/b^4+1/7*d^2*x^7/b^3+1/2*a*(a^4*d+b^4*c)^2/b^10/(b*x+a)^2-(a^4*d+b^4*c )*(9*a^4*d+b^4*c)/b^10/(b*x+a)-4*a^3*d*(9*a^4*d+5*b^4*c)*ln(b*x+a)/b^10
Time = 0.06 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.93 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=\frac {840 a^2 b d \left (3 b^4 c+7 a^4 d\right ) x-315 a b^2 d \left (2 b^4 c+7 a^4 d\right ) x^2+70 b^3 d \left (2 b^4 c+15 a^4 d\right ) x^3-525 a^3 b^4 d^2 x^4+252 a^2 b^5 d^2 x^5-105 a b^6 d^2 x^6+30 b^7 d^2 x^7+\frac {105 a \left (b^4 c+a^4 d\right )^2}{(a+b x)^2}-\frac {210 \left (b^8 c^2+10 a^4 b^4 c d+9 a^8 d^2\right )}{a+b x}-840 a^3 d \left (5 b^4 c+9 a^4 d\right ) \log (a+b x)}{210 b^{10}} \] Input:
Integrate[(x*(c + d*x^4)^2)/(a + b*x)^3,x]
Output:
(840*a^2*b*d*(3*b^4*c + 7*a^4*d)*x - 315*a*b^2*d*(2*b^4*c + 7*a^4*d)*x^2 + 70*b^3*d*(2*b^4*c + 15*a^4*d)*x^3 - 525*a^3*b^4*d^2*x^4 + 252*a^2*b^5*d^2 *x^5 - 105*a*b^6*d^2*x^6 + 30*b^7*d^2*x^7 + (105*a*(b^4*c + a^4*d)^2)/(a + b*x)^2 - (210*(b^8*c^2 + 10*a^4*b^4*c*d + 9*a^8*d^2))/(a + b*x) - 840*a^3 *d*(5*b^4*c + 9*a^4*d)*Log[a + b*x])/(210*b^10)
Time = 0.52 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2123, 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.
| \(\displaystyle \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \int \left (\frac {\left (a^4 d+b^4 c\right ) \left (9 a^4 d+b^4 c\right )}{b^9 (a+b x)^2}-\frac {a \left (a^4 d+b^4 c\right )^2}{b^9 (a+b x)^3}-\frac {3 a d x \left (7 a^4 d+2 b^4 c\right )}{b^8}+\frac {d x^2 \left (15 a^4 d+2 b^4 c\right )}{b^7}-\frac {10 a^3 d^2 x^3}{b^6}+\frac {6 a^2 d^2 x^4}{b^5}-\frac {4 a^3 d \left (9 a^4 d+5 b^4 c\right )}{b^9 (a+b x)}+\frac {4 a^2 d \left (7 a^4 d+3 b^4 c\right )}{b^9}-\frac {3 a d^2 x^5}{b^4}+\frac {d^2 x^6}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (a^4 d+b^4 c\right ) \left (9 a^4 d+b^4 c\right )}{b^{10} (a+b x)}+\frac {a \left (a^4 d+b^4 c\right )^2}{2 b^{10} (a+b x)^2}-\frac {3 a d x^2 \left (7 a^4 d+2 b^4 c\right )}{2 b^8}+\frac {d x^3 \left (15 a^4 d+2 b^4 c\right )}{3 b^7}-\frac {5 a^3 d^2 x^4}{2 b^6}+\frac {6 a^2 d^2 x^5}{5 b^5}-\frac {4 a^3 d \left (9 a^4 d+5 b^4 c\right ) \log (a+b x)}{b^{10}}+\frac {4 a^2 d x \left (7 a^4 d+3 b^4 c\right )}{b^9}-\frac {a d^2 x^6}{2 b^4}+\frac {d^2 x^7}{7 b^3}\) |
Input:
Int[(x*(c + d*x^4)^2)/(a + b*x)^3,x]
Output:
(4*a^2*d*(3*b^4*c + 7*a^4*d)*x)/b^9 - (3*a*d*(2*b^4*c + 7*a^4*d)*x^2)/(2*b ^8) + (d*(2*b^4*c + 15*a^4*d)*x^3)/(3*b^7) - (5*a^3*d^2*x^4)/(2*b^6) + (6* a^2*d^2*x^5)/(5*b^5) - (a*d^2*x^6)/(2*b^4) + (d^2*x^7)/(7*b^3) + (a*(b^4*c + a^4*d)^2)/(2*b^10*(a + b*x)^2) - ((b^4*c + a^4*d)*(b^4*c + 9*a^4*d))/(b ^10*(a + b*x)) - (4*a^3*d*(5*b^4*c + 9*a^4*d)*Log[a + b*x])/b^10
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 0.41 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {d \left (\frac {1}{7} d \,x^{7} b^{6}-\frac {1}{2} a d \,x^{6} b^{5}+\frac {6}{5} a^{2} d \,x^{5} b^{4}-\frac {5}{2} a^{3} d \,x^{4} b^{3}+5 a^{4} b^{2} d \,x^{3}+\frac {2}{3} b^{6} c \,x^{3}-\frac {21}{2} a^{5} b d \,x^{2}-3 a \,b^{5} c \,x^{2}+28 a^{6} d x +12 a^{2} b^{4} c x \right )}{b^{9}}+\frac {a \left (a^{8} d^{2}+2 a^{4} b^{4} c d +b^{8} c^{2}\right )}{2 b^{10} \left (b x +a \right )^{2}}-\frac {9 a^{8} d^{2}+10 a^{4} b^{4} c d +b^{8} c^{2}}{b^{10} \left (b x +a \right )}-\frac {4 a^{3} d \left (9 a^{4} d +5 b^{4} c \right ) \ln \left (b x +a \right )}{b^{10}}\) | \(213\) |
| norman | \(\frac {-\frac {a \left (108 a^{8} d^{2}+60 a^{4} b^{4} c d +b^{8} c^{2}\right )}{2 b^{10}}+\frac {d^{2} x^{9}}{7 b}-\frac {\left (72 a^{8} d^{2}+40 a^{4} b^{4} c d +b^{8} c^{2}\right ) x}{b^{9}}-\frac {3 a \,d^{2} x^{8}}{14 b^{2}}+\frac {12 a^{2} d^{2} x^{7}}{35 b^{3}}-\frac {3 a^{3} d^{2} x^{6}}{5 b^{4}}+\frac {2 d \left (9 a^{4} d +5 b^{4} c \right ) x^{5}}{15 b^{5}}-\frac {a d \left (9 a^{4} d +5 b^{4} c \right ) x^{4}}{3 b^{6}}+\frac {4 d \,a^{2} \left (9 a^{4} d +5 b^{4} c \right ) x^{3}}{3 b^{7}}}{\left (b x +a \right )^{2}}-\frac {4 a^{3} d \left (9 a^{4} d +5 b^{4} c \right ) \ln \left (b x +a \right )}{b^{10}}\) | \(224\) |
| risch | \(\frac {d^{2} x^{7}}{7 b^{3}}-\frac {a \,d^{2} x^{6}}{2 b^{4}}+\frac {6 a^{2} d^{2} x^{5}}{5 b^{5}}-\frac {5 a^{3} d^{2} x^{4}}{2 b^{6}}+\frac {5 d^{2} a^{4} x^{3}}{b^{7}}+\frac {2 d c \,x^{3}}{3 b^{3}}-\frac {21 d^{2} a^{5} x^{2}}{2 b^{8}}-\frac {3 d a c \,x^{2}}{b^{4}}+\frac {28 d^{2} a^{6} x}{b^{9}}+\frac {12 d \,a^{2} c x}{b^{5}}+\frac {\left (-9 a^{8} d^{2}-10 a^{4} b^{4} c d -b^{8} c^{2}\right ) x -\frac {a \left (17 a^{8} d^{2}+18 a^{4} b^{4} c d +b^{8} c^{2}\right )}{2 b}}{b^{9} \left (b x +a \right )^{2}}-\frac {36 a^{7} d^{2} \ln \left (b x +a \right )}{b^{10}}-\frac {20 a^{3} d \ln \left (b x +a \right ) c}{b^{6}}\) | \(231\) |
| parallelrisch | \(-\frac {6300 a^{5} b^{4} c d +11340 a^{9} d^{2}+15120 \ln \left (b x +a \right ) x \,a^{8} b \,d^{2}+350 x^{4} a \,b^{8} c d -1400 x^{3} a^{2} b^{7} c d +4200 \ln \left (b x +a \right ) a^{5} b^{4} c d +8400 x \,a^{4} b^{5} c d +8400 \ln \left (b x +a \right ) x \,a^{4} b^{5} c d +4200 \ln \left (b x +a \right ) x^{2} a^{3} b^{6} c d +7560 \ln \left (b x +a \right ) x^{2} a^{7} b^{2} d^{2}+45 a \,d^{2} x^{8} b^{8}-72 a^{2} d^{2} x^{7} b^{7}+126 a^{3} d^{2} x^{6} b^{6}-252 x^{5} a^{4} b^{5} d^{2}-140 x^{5} b^{9} c d +630 x^{4} a^{5} b^{4} d^{2}-2520 x^{3} a^{6} b^{3} d^{2}+15120 x \,a^{8} b \,d^{2}+105 a \,b^{8} c^{2}+7560 \ln \left (b x +a \right ) a^{9} d^{2}-30 x^{9} d^{2} b^{9}+210 x \,b^{9} c^{2}}{210 b^{10} \left (b x +a \right )^{2}}\) | \(300\) |
Input:
int(x*(d*x^4+c)^2/(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
d/b^9*(1/7*d*x^7*b^6-1/2*a*d*x^6*b^5+6/5*a^2*d*x^5*b^4-5/2*a^3*d*x^4*b^3+5 *a^4*b^2*d*x^3+2/3*b^6*c*x^3-21/2*a^5*b*d*x^2-3*a*b^5*c*x^2+28*a^6*d*x+12* a^2*b^4*c*x)+1/2*a*(a^8*d^2+2*a^4*b^4*c*d+b^8*c^2)/b^10/(b*x+a)^2-1/b^10*( 9*a^8*d^2+10*a^4*b^4*c*d+b^8*c^2)/(b*x+a)-4*a^3*d*(9*a^4*d+5*b^4*c)*ln(b*x +a)/b^10
Time = 0.07 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.41 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=\frac {30 \, b^{9} d^{2} x^{9} - 45 \, a b^{8} d^{2} x^{8} + 72 \, a^{2} b^{7} d^{2} x^{7} - 126 \, a^{3} b^{6} d^{2} x^{6} - 105 \, a b^{8} c^{2} - 1890 \, a^{5} b^{4} c d - 1785 \, a^{9} d^{2} + 28 \, {\left (5 \, b^{9} c d + 9 \, a^{4} b^{5} d^{2}\right )} x^{5} - 70 \, {\left (5 \, a b^{8} c d + 9 \, a^{5} b^{4} d^{2}\right )} x^{4} + 280 \, {\left (5 \, a^{2} b^{7} c d + 9 \, a^{6} b^{3} d^{2}\right )} x^{3} + 735 \, {\left (6 \, a^{3} b^{6} c d + 13 \, a^{7} b^{2} d^{2}\right )} x^{2} - 210 \, {\left (b^{9} c^{2} - 2 \, a^{4} b^{5} c d - 19 \, a^{8} b d^{2}\right )} x - 840 \, {\left (5 \, a^{5} b^{4} c d + 9 \, a^{9} d^{2} + {\left (5 \, a^{3} b^{6} c d + 9 \, a^{7} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (5 \, a^{4} b^{5} c d + 9 \, a^{8} b d^{2}\right )} x\right )} \log \left (b x + a\right )}{210 \, {\left (b^{12} x^{2} + 2 \, a b^{11} x + a^{2} b^{10}\right )}} \] Input:
integrate(x*(d*x^4+c)^2/(b*x+a)^3,x, algorithm="fricas")
Output:
1/210*(30*b^9*d^2*x^9 - 45*a*b^8*d^2*x^8 + 72*a^2*b^7*d^2*x^7 - 126*a^3*b^ 6*d^2*x^6 - 105*a*b^8*c^2 - 1890*a^5*b^4*c*d - 1785*a^9*d^2 + 28*(5*b^9*c* d + 9*a^4*b^5*d^2)*x^5 - 70*(5*a*b^8*c*d + 9*a^5*b^4*d^2)*x^4 + 280*(5*a^2 *b^7*c*d + 9*a^6*b^3*d^2)*x^3 + 735*(6*a^3*b^6*c*d + 13*a^7*b^2*d^2)*x^2 - 210*(b^9*c^2 - 2*a^4*b^5*c*d - 19*a^8*b*d^2)*x - 840*(5*a^5*b^4*c*d + 9*a ^9*d^2 + (5*a^3*b^6*c*d + 9*a^7*b^2*d^2)*x^2 + 2*(5*a^4*b^5*c*d + 9*a^8*b* d^2)*x)*log(b*x + a))/(b^12*x^2 + 2*a*b^11*x + a^2*b^10)
Time = 0.55 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.15 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=- \frac {5 a^{3} d^{2} x^{4}}{2 b^{6}} - \frac {4 a^{3} d \left (9 a^{4} d + 5 b^{4} c\right ) \log {\left (a + b x \right )}}{b^{10}} + \frac {6 a^{2} d^{2} x^{5}}{5 b^{5}} - \frac {a d^{2} x^{6}}{2 b^{4}} + x^{3} \cdot \left (\frac {5 a^{4} d^{2}}{b^{7}} + \frac {2 c d}{3 b^{3}}\right ) + x^{2} \left (- \frac {21 a^{5} d^{2}}{2 b^{8}} - \frac {3 a c d}{b^{4}}\right ) + x \left (\frac {28 a^{6} d^{2}}{b^{9}} + \frac {12 a^{2} c d}{b^{5}}\right ) + \frac {- 17 a^{9} d^{2} - 18 a^{5} b^{4} c d - a b^{8} c^{2} + x \left (- 18 a^{8} b d^{2} - 20 a^{4} b^{5} c d - 2 b^{9} c^{2}\right )}{2 a^{2} b^{10} + 4 a b^{11} x + 2 b^{12} x^{2}} + \frac {d^{2} x^{7}}{7 b^{3}} \] Input:
integrate(x*(d*x**4+c)**2/(b*x+a)**3,x)
Output:
-5*a**3*d**2*x**4/(2*b**6) - 4*a**3*d*(9*a**4*d + 5*b**4*c)*log(a + b*x)/b **10 + 6*a**2*d**2*x**5/(5*b**5) - a*d**2*x**6/(2*b**4) + x**3*(5*a**4*d** 2/b**7 + 2*c*d/(3*b**3)) + x**2*(-21*a**5*d**2/(2*b**8) - 3*a*c*d/b**4) + x*(28*a**6*d**2/b**9 + 12*a**2*c*d/b**5) + (-17*a**9*d**2 - 18*a**5*b**4*c *d - a*b**8*c**2 + x*(-18*a**8*b*d**2 - 20*a**4*b**5*c*d - 2*b**9*c**2))/( 2*a**2*b**10 + 4*a*b**11*x + 2*b**12*x**2) + d**2*x**7/(7*b**3)
Time = 0.04 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.08 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=-\frac {a b^{8} c^{2} + 18 \, a^{5} b^{4} c d + 17 \, a^{9} d^{2} + 2 \, {\left (b^{9} c^{2} + 10 \, a^{4} b^{5} c d + 9 \, a^{8} b d^{2}\right )} x}{2 \, {\left (b^{12} x^{2} + 2 \, a b^{11} x + a^{2} b^{10}\right )}} + \frac {30 \, b^{6} d^{2} x^{7} - 105 \, a b^{5} d^{2} x^{6} + 252 \, a^{2} b^{4} d^{2} x^{5} - 525 \, a^{3} b^{3} d^{2} x^{4} + 70 \, {\left (2 \, b^{6} c d + 15 \, a^{4} b^{2} d^{2}\right )} x^{3} - 315 \, {\left (2 \, a b^{5} c d + 7 \, a^{5} b d^{2}\right )} x^{2} + 840 \, {\left (3 \, a^{2} b^{4} c d + 7 \, a^{6} d^{2}\right )} x}{210 \, b^{9}} - \frac {4 \, {\left (5 \, a^{3} b^{4} c d + 9 \, a^{7} d^{2}\right )} \log \left (b x + a\right )}{b^{10}} \] Input:
integrate(x*(d*x^4+c)^2/(b*x+a)^3,x, algorithm="maxima")
Output:
-1/2*(a*b^8*c^2 + 18*a^5*b^4*c*d + 17*a^9*d^2 + 2*(b^9*c^2 + 10*a^4*b^5*c* d + 9*a^8*b*d^2)*x)/(b^12*x^2 + 2*a*b^11*x + a^2*b^10) + 1/210*(30*b^6*d^2 *x^7 - 105*a*b^5*d^2*x^6 + 252*a^2*b^4*d^2*x^5 - 525*a^3*b^3*d^2*x^4 + 70* (2*b^6*c*d + 15*a^4*b^2*d^2)*x^3 - 315*(2*a*b^5*c*d + 7*a^5*b*d^2)*x^2 + 8 40*(3*a^2*b^4*c*d + 7*a^6*d^2)*x)/b^9 - 4*(5*a^3*b^4*c*d + 9*a^7*d^2)*log( b*x + a)/b^10
Time = 0.12 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.03 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=-\frac {4 \, {\left (5 \, a^{3} b^{4} c d + 9 \, a^{7} d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{10}} - \frac {a b^{8} c^{2} + 18 \, a^{5} b^{4} c d + 17 \, a^{9} d^{2} + 2 \, {\left (b^{9} c^{2} + 10 \, a^{4} b^{5} c d + 9 \, a^{8} b d^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{10}} + \frac {30 \, b^{18} d^{2} x^{7} - 105 \, a b^{17} d^{2} x^{6} + 252 \, a^{2} b^{16} d^{2} x^{5} - 525 \, a^{3} b^{15} d^{2} x^{4} + 140 \, b^{18} c d x^{3} + 1050 \, a^{4} b^{14} d^{2} x^{3} - 630 \, a b^{17} c d x^{2} - 2205 \, a^{5} b^{13} d^{2} x^{2} + 2520 \, a^{2} b^{16} c d x + 5880 \, a^{6} b^{12} d^{2} x}{210 \, b^{21}} \] Input:
integrate(x*(d*x^4+c)^2/(b*x+a)^3,x, algorithm="giac")
Output:
-4*(5*a^3*b^4*c*d + 9*a^7*d^2)*log(abs(b*x + a))/b^10 - 1/2*(a*b^8*c^2 + 1 8*a^5*b^4*c*d + 17*a^9*d^2 + 2*(b^9*c^2 + 10*a^4*b^5*c*d + 9*a^8*b*d^2)*x) /((b*x + a)^2*b^10) + 1/210*(30*b^18*d^2*x^7 - 105*a*b^17*d^2*x^6 + 252*a^ 2*b^16*d^2*x^5 - 525*a^3*b^15*d^2*x^4 + 140*b^18*c*d*x^3 + 1050*a^4*b^14*d ^2*x^3 - 630*a*b^17*c*d*x^2 - 2205*a^5*b^13*d^2*x^2 + 2520*a^2*b^16*c*d*x + 5880*a^6*b^12*d^2*x)/b^21
Time = 3.05 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.43 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=x^3\,\left (\frac {5\,a^4\,d^2}{b^7}+\frac {2\,c\,d}{3\,b^3}\right )-\frac {\frac {17\,a^9\,d^2+18\,a^5\,b^4\,c\,d+a\,b^8\,c^2}{2\,b}+x\,\left (9\,a^8\,d^2+10\,a^4\,b^4\,c\,d+b^8\,c^2\right )}{a^2\,b^9+2\,a\,b^{10}\,x+b^{11}\,x^2}+x\,\left (\frac {10\,a^6\,d^2}{b^9}-\frac {3\,a^2\,\left (\frac {15\,a^4\,d^2}{b^7}+\frac {2\,c\,d}{b^3}\right )}{b^2}+\frac {3\,a\,\left (\frac {3\,a\,\left (\frac {15\,a^4\,d^2}{b^7}+\frac {2\,c\,d}{b^3}\right )}{b}-\frac {24\,a^5\,d^2}{b^8}\right )}{b}\right )-x^2\,\left (\frac {3\,a\,\left (\frac {15\,a^4\,d^2}{b^7}+\frac {2\,c\,d}{b^3}\right )}{2\,b}-\frac {12\,a^5\,d^2}{b^8}\right )-\frac {\ln \left (a+b\,x\right )\,\left (36\,a^7\,d^2+20\,c\,a^3\,b^4\,d\right )}{b^{10}}+\frac {d^2\,x^7}{7\,b^3}-\frac {a\,d^2\,x^6}{2\,b^4}+\frac {6\,a^2\,d^2\,x^5}{5\,b^5}-\frac {5\,a^3\,d^2\,x^4}{2\,b^6} \] Input:
int((x*(c + d*x^4)^2)/(a + b*x)^3,x)
Output:
x^3*((5*a^4*d^2)/b^7 + (2*c*d)/(3*b^3)) - ((17*a^9*d^2 + a*b^8*c^2 + 18*a^ 5*b^4*c*d)/(2*b) + x*(9*a^8*d^2 + b^8*c^2 + 10*a^4*b^4*c*d))/(a^2*b^9 + b^ 11*x^2 + 2*a*b^10*x) + x*((10*a^6*d^2)/b^9 - (3*a^2*((15*a^4*d^2)/b^7 + (2 *c*d)/b^3))/b^2 + (3*a*((3*a*((15*a^4*d^2)/b^7 + (2*c*d)/b^3))/b - (24*a^5 *d^2)/b^8))/b) - x^2*((3*a*((15*a^4*d^2)/b^7 + (2*c*d)/b^3))/(2*b) - (12*a ^5*d^2)/b^8) - (log(a + b*x)*(36*a^7*d^2 + 20*a^3*b^4*c*d))/b^10 + (d^2*x^ 7)/(7*b^3) - (a*d^2*x^6)/(2*b^4) + (6*a^2*d^2*x^5)/(5*b^5) - (5*a^3*d^2*x^ 4)/(2*b^6)
Time = 0.14 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.43 \[ \int \frac {x \left (c+d x^4\right )^2}{(a+b x)^3} \, dx=\frac {-7560 \,\mathrm {log}\left (b x +a \right ) a^{10} d^{2}-15120 \,\mathrm {log}\left (b x +a \right ) a^{9} b \,d^{2} x -7560 \,\mathrm {log}\left (b x +a \right ) a^{8} b^{2} d^{2} x^{2}-4200 \,\mathrm {log}\left (b x +a \right ) a^{6} b^{4} c d -8400 \,\mathrm {log}\left (b x +a \right ) a^{5} b^{5} c d x -4200 \,\mathrm {log}\left (b x +a \right ) a^{4} b^{6} c d \,x^{2}-3780 a^{10} d^{2}+7560 a^{8} b^{2} d^{2} x^{2}+2520 a^{7} b^{3} d^{2} x^{3}-2100 a^{6} b^{4} c d -630 a^{6} b^{4} d^{2} x^{4}+252 a^{5} b^{5} d^{2} x^{5}+4200 a^{4} b^{6} c d \,x^{2}-126 a^{4} b^{6} d^{2} x^{6}+1400 a^{3} b^{7} c d \,x^{3}+72 a^{3} b^{7} d^{2} x^{7}-350 a^{2} b^{8} c d \,x^{4}-45 a^{2} b^{8} d^{2} x^{8}+140 a \,b^{9} c d \,x^{5}+30 a \,b^{9} d^{2} x^{9}+105 b^{10} c^{2} x^{2}}{210 a \,b^{10} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int(x*(d*x^4+c)^2/(b*x+a)^3,x)
Output:
( - 7560*log(a + b*x)*a**10*d**2 - 15120*log(a + b*x)*a**9*b*d**2*x - 7560 *log(a + b*x)*a**8*b**2*d**2*x**2 - 4200*log(a + b*x)*a**6*b**4*c*d - 8400 *log(a + b*x)*a**5*b**5*c*d*x - 4200*log(a + b*x)*a**4*b**6*c*d*x**2 - 378 0*a**10*d**2 + 7560*a**8*b**2*d**2*x**2 + 2520*a**7*b**3*d**2*x**3 - 2100* a**6*b**4*c*d - 630*a**6*b**4*d**2*x**4 + 252*a**5*b**5*d**2*x**5 + 4200*a **4*b**6*c*d*x**2 - 126*a**4*b**6*d**2*x**6 + 1400*a**3*b**7*c*d*x**3 + 72 *a**3*b**7*d**2*x**7 - 350*a**2*b**8*c*d*x**4 - 45*a**2*b**8*d**2*x**8 + 1 40*a*b**9*c*d*x**5 + 30*a*b**9*d**2*x**9 + 105*b**10*c**2*x**2)/(210*a*b** 10*(a**2 + 2*a*b*x + b**2*x**2))