Integrand size = 17, antiderivative size = 264 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=-\frac {8 c d \left (c d^2+a e^2\right )^3 x}{e^8}+\frac {2 c \left (c d^2+a e^2\right )^2 \left (7 c d^2+a e^2\right ) (d+e x)^2}{e^9}-\frac {8 c^2 d \left (c d^2+a e^2\right ) \left (7 c d^2+3 a e^2\right ) (d+e x)^3}{3 e^9}+\frac {c^2 \left (35 c^2 d^4+30 a c d^2 e^2+3 a^2 e^4\right ) (d+e x)^4}{2 e^9}-\frac {8 c^3 d \left (7 c d^2+3 a e^2\right ) (d+e x)^5}{5 e^9}+\frac {2 c^3 \left (7 c d^2+a e^2\right ) (d+e x)^6}{3 e^9}-\frac {8 c^4 d (d+e x)^7}{7 e^9}+\frac {c^4 (d+e x)^8}{8 e^9}+\frac {\left (c d^2+a e^2\right )^4 \log (d+e x)}{e^9} \]
-8*c*d*(a*e^2+c*d^2)^3*x/e^8+2*c*(a*e^2+c*d^2)^2*(a*e^2+7*c*d^2)*(e*x+d)^2 /e^9-8/3*c^2*d*(a*e^2+c*d^2)*(3*a*e^2+7*c*d^2)*(e*x+d)^3/e^9+1/2*c^2*(3*a^ 2*e^4+30*a*c*d^2*e^2+35*c^2*d^4)*(e*x+d)^4/e^9-8/5*c^3*d*(3*a*e^2+7*c*d^2) *(e*x+d)^5/e^9+2/3*c^3*(a*e^2+7*c*d^2)*(e*x+d)^6/e^9-8/7*c^4*d*(e*x+d)^7/e ^9+1/8*c^4*(e*x+d)^8/e^9+(a*e^2+c*d^2)^4*ln(e*x+d)/e^9
Time = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=\frac {c x \left (1680 a^3 e^6 (-2 d+e x)+420 a^2 c e^4 \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )+56 a c^2 e^2 \left (-60 d^5+30 d^4 e x-20 d^3 e^2 x^2+15 d^2 e^3 x^3-12 d e^4 x^4+10 e^5 x^5\right )+c^3 \left (-840 d^7+420 d^6 e x-280 d^5 e^2 x^2+210 d^4 e^3 x^3-168 d^3 e^4 x^4+140 d^2 e^5 x^5-120 d e^6 x^6+105 e^7 x^7\right )\right )}{840 e^8}+\frac {\left (c d^2+a e^2\right )^4 \log (d+e x)}{e^9} \]
(c*x*(1680*a^3*e^6*(-2*d + e*x) + 420*a^2*c*e^4*(-12*d^3 + 6*d^2*e*x - 4*d *e^2*x^2 + 3*e^3*x^3) + 56*a*c^2*e^2*(-60*d^5 + 30*d^4*e*x - 20*d^3*e^2*x^ 2 + 15*d^2*e^3*x^3 - 12*d*e^4*x^4 + 10*e^5*x^5) + c^3*(-840*d^7 + 420*d^6* e*x - 280*d^5*e^2*x^2 + 210*d^4*e^3*x^3 - 168*d^3*e^4*x^4 + 140*d^2*e^5*x^ 5 - 120*d*e^6*x^6 + 105*e^7*x^7)))/(840*e^8) + ((c*d^2 + a*e^2)^4*Log[d + e*x])/e^9
Time = 0.52 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {476, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \int \left (\frac {2 c^2 (d+e x)^3 \left (3 a^2 e^4+30 a c d^2 e^2+35 c^2 d^4\right )}{e^8}+\frac {4 c^3 (d+e x)^5 \left (a e^2+7 c d^2\right )}{e^8}-\frac {8 c^3 d (d+e x)^4 \left (3 a e^2+7 c d^2\right )}{e^8}+\frac {8 c^2 d (d+e x)^2 \left (-3 a e^2-7 c d^2\right ) \left (a e^2+c d^2\right )}{e^8}+\frac {4 c (d+e x) \left (a e^2+c d^2\right )^2 \left (a e^2+7 c d^2\right )}{e^8}+\frac {\left (a e^2+c d^2\right )^4}{e^8 (d+e x)}-\frac {8 c d \left (a e^2+c d^2\right )^3}{e^8}+\frac {c^4 (d+e x)^7}{e^8}-\frac {8 c^4 d (d+e x)^6}{e^8}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c^2 (d+e x)^4 \left (3 a^2 e^4+30 a c d^2 e^2+35 c^2 d^4\right )}{2 e^9}+\frac {2 c^3 (d+e x)^6 \left (a e^2+7 c d^2\right )}{3 e^9}-\frac {8 c^3 d (d+e x)^5 \left (3 a e^2+7 c d^2\right )}{5 e^9}-\frac {8 c^2 d (d+e x)^3 \left (a e^2+c d^2\right ) \left (3 a e^2+7 c d^2\right )}{3 e^9}+\frac {2 c (d+e x)^2 \left (a e^2+c d^2\right )^2 \left (a e^2+7 c d^2\right )}{e^9}+\frac {\left (a e^2+c d^2\right )^4 \log (d+e x)}{e^9}-\frac {8 c d x \left (a e^2+c d^2\right )^3}{e^8}+\frac {c^4 (d+e x)^8}{8 e^9}-\frac {8 c^4 d (d+e x)^7}{7 e^9}\) |
(-8*c*d*(c*d^2 + a*e^2)^3*x)/e^8 + (2*c*(c*d^2 + a*e^2)^2*(7*c*d^2 + a*e^2 )*(d + e*x)^2)/e^9 - (8*c^2*d*(c*d^2 + a*e^2)*(7*c*d^2 + 3*a*e^2)*(d + e*x )^3)/(3*e^9) + (c^2*(35*c^2*d^4 + 30*a*c*d^2*e^2 + 3*a^2*e^4)*(d + e*x)^4) /(2*e^9) - (8*c^3*d*(7*c*d^2 + 3*a*e^2)*(d + e*x)^5)/(5*e^9) + (2*c^3*(7*c *d^2 + a*e^2)*(d + e*x)^6)/(3*e^9) - (8*c^4*d*(d + e*x)^7)/(7*e^9) + (c^4* (d + e*x)^8)/(8*e^9) + ((c*d^2 + a*e^2)^4*Log[d + e*x])/e^9
3.5.95.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 2.34 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.16
| method | result | size |
| norman | \(\frac {c^{4} x^{8}}{8 e}+\frac {c \left (4 e^{6} a^{3}+6 d^{2} e^{4} a^{2} c +4 d^{4} e^{2} c^{2} a +c^{3} d^{6}\right ) x^{2}}{2 e^{7}}-\frac {d \,c^{4} x^{7}}{7 e^{2}}+\frac {c^{2} \left (6 a^{2} e^{4}+4 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x^{4}}{4 e^{5}}+\frac {c^{3} \left (4 e^{2} a +c \,d^{2}\right ) x^{6}}{6 e^{3}}-\frac {c d \left (4 e^{6} a^{3}+6 d^{2} e^{4} a^{2} c +4 d^{4} e^{2} c^{2} a +c^{3} d^{6}\right ) x}{e^{8}}-\frac {d \,c^{2} \left (6 a^{2} e^{4}+4 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x^{3}}{3 e^{6}}-\frac {d \,c^{3} \left (4 e^{2} a +c \,d^{2}\right ) x^{5}}{5 e^{4}}+\frac {\left (a^{4} e^{8}+4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}+4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \ln \left (e x +d \right )}{e^{9}}\) | \(307\) |
| risch | \(-\frac {d \,c^{4} x^{7}}{7 e^{2}}+\frac {2 c^{3} x^{6} a}{3 e}+\frac {c^{4} x^{6} d^{2}}{6 e^{3}}-\frac {c^{4} x^{5} d^{3}}{5 e^{4}}+\frac {3 c^{2} x^{4} a^{2}}{2 e}+\frac {c^{4} x^{4} d^{4}}{4 e^{5}}-\frac {c^{4} x^{3} d^{5}}{3 e^{6}}+\frac {2 c \,a^{3} x^{2}}{e}+\frac {c^{4} d^{6} x^{2}}{2 e^{7}}-\frac {c^{4} d^{7} x}{e^{8}}+\frac {\ln \left (e x +d \right ) c^{4} d^{8}}{e^{9}}+\frac {c^{4} x^{8}}{8 e}+\frac {4 \ln \left (e x +d \right ) a^{3} c \,d^{2}}{e^{3}}+\frac {6 \ln \left (e x +d \right ) a^{2} c^{2} d^{4}}{e^{5}}+\frac {4 \ln \left (e x +d \right ) a \,c^{3} d^{6}}{e^{7}}+\frac {\ln \left (e x +d \right ) a^{4}}{e}-\frac {4 c^{3} x^{5} d a}{5 e^{2}}+\frac {c^{3} x^{4} a \,d^{2}}{e^{3}}-\frac {2 c^{2} x^{3} a^{2} d}{e^{2}}-\frac {4 c^{3} x^{3} a \,d^{3}}{3 e^{4}}+\frac {3 c^{2} a^{2} d^{2} x^{2}}{e^{3}}+\frac {2 c^{3} a \,d^{4} x^{2}}{e^{5}}-\frac {4 c \,a^{3} d x}{e^{2}}-\frac {6 c^{2} a^{2} d^{3} x}{e^{4}}-\frac {4 c^{3} a \,d^{5} x}{e^{6}}\) | \(358\) |
| parallelrisch | \(\frac {840 \ln \left (e x +d \right ) a^{4} e^{8}-168 x^{5} c^{4} d^{3} e^{5}+1260 x^{4} a^{2} c^{2} e^{8}+210 x^{4} c^{4} d^{4} e^{4}-280 x^{3} c^{4} d^{5} e^{3}+1680 x^{2} a^{3} c \,e^{8}+420 x^{2} c^{4} d^{6} e^{2}-840 x \,c^{4} d^{7} e -120 d \,c^{4} x^{7} e^{7}+560 x^{6} a \,c^{3} e^{8}+140 x^{6} c^{4} d^{2} e^{6}-1120 x^{3} a \,c^{3} d^{3} e^{5}+2520 x^{2} a^{2} c^{2} d^{2} e^{6}+1680 x^{2} a \,c^{3} d^{4} e^{4}-3360 x \,a^{3} c d \,e^{7}-5040 x \,a^{2} c^{2} d^{3} e^{5}-3360 x a \,c^{3} d^{5} e^{3}+3360 \ln \left (e x +d \right ) a^{3} c \,d^{2} e^{6}+5040 \ln \left (e x +d \right ) a^{2} c^{2} d^{4} e^{4}+3360 \ln \left (e x +d \right ) a \,c^{3} d^{6} e^{2}+840 \ln \left (e x +d \right ) c^{4} d^{8}+105 x^{8} c^{4} e^{8}-1680 x^{3} a^{2} c^{2} d \,e^{7}-672 x^{5} a \,c^{3} d \,e^{7}+840 x^{4} a \,c^{3} d^{2} e^{6}}{840 e^{9}}\) | \(361\) |
| default | \(-\frac {c \left (-\frac {c^{3} x^{8} e^{7}}{8}+\frac {c^{3} d \,x^{7} e^{6}}{7}+\frac {\left (-e^{5} \left (2 e^{2} a +c \,d^{2}\right ) c^{2}-2 e^{7} c^{2} a \right ) x^{6}}{6}+\frac {\left (d \left (2 e^{2} a +c \,d^{2}\right ) e^{4} c^{2}+2 d \,e^{6} c^{2} a \right ) x^{5}}{5}+\frac {\left (-2 e^{5} \left (2 e^{2} a +c \,d^{2}\right ) a c -e^{3} c \left (2 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )\right ) x^{4}}{4}+\frac {\left (2 d \left (2 e^{2} a +c \,d^{2}\right ) a c \,e^{4}+d \,e^{2} c \left (2 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )\right ) x^{3}}{3}-\frac {e \left (2 e^{2} a +c \,d^{2}\right ) \left (2 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x^{2}}{2}+d \left (2 e^{2} a +c \,d^{2}\right ) \left (2 a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x \right )}{e^{8}}+\frac {\left (a^{4} e^{8}+4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}+4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \ln \left (e x +d \right )}{e^{9}}\) | \(366\) |
1/8/e*c^4*x^8+1/2*c/e^7*(4*a^3*e^6+6*a^2*c*d^2*e^4+4*a*c^2*d^4*e^2+c^3*d^6 )*x^2-1/7*d/e^2*c^4*x^7+1/4/e^5*c^2*(6*a^2*e^4+4*a*c*d^2*e^2+c^2*d^4)*x^4+ 1/6/e^3*c^3*(4*a*e^2+c*d^2)*x^6-c*d*(4*a^3*e^6+6*a^2*c*d^2*e^4+4*a*c^2*d^4 *e^2+c^3*d^6)/e^8*x-1/3*d/e^6*c^2*(6*a^2*e^4+4*a*c*d^2*e^2+c^2*d^4)*x^3-1/ 5*d/e^4*c^3*(4*a*e^2+c*d^2)*x^5+(a^4*e^8+4*a^3*c*d^2*e^6+6*a^2*c^2*d^4*e^4 +4*a*c^3*d^6*e^2+c^4*d^8)/e^9*ln(e*x+d)
Time = 0.25 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=\frac {105 \, c^{4} e^{8} x^{8} - 120 \, c^{4} d e^{7} x^{7} + 140 \, {\left (c^{4} d^{2} e^{6} + 4 \, a c^{3} e^{8}\right )} x^{6} - 168 \, {\left (c^{4} d^{3} e^{5} + 4 \, a c^{3} d e^{7}\right )} x^{5} + 210 \, {\left (c^{4} d^{4} e^{4} + 4 \, a c^{3} d^{2} e^{6} + 6 \, a^{2} c^{2} e^{8}\right )} x^{4} - 280 \, {\left (c^{4} d^{5} e^{3} + 4 \, a c^{3} d^{3} e^{5} + 6 \, a^{2} c^{2} d e^{7}\right )} x^{3} + 420 \, {\left (c^{4} d^{6} e^{2} + 4 \, a c^{3} d^{4} e^{4} + 6 \, a^{2} c^{2} d^{2} e^{6} + 4 \, a^{3} c e^{8}\right )} x^{2} - 840 \, {\left (c^{4} d^{7} e + 4 \, a c^{3} d^{5} e^{3} + 6 \, a^{2} c^{2} d^{3} e^{5} + 4 \, a^{3} c d e^{7}\right )} x + 840 \, {\left (c^{4} d^{8} + 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} + 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left (e x + d\right )}{840 \, e^{9}} \]
1/840*(105*c^4*e^8*x^8 - 120*c^4*d*e^7*x^7 + 140*(c^4*d^2*e^6 + 4*a*c^3*e^ 8)*x^6 - 168*(c^4*d^3*e^5 + 4*a*c^3*d*e^7)*x^5 + 210*(c^4*d^4*e^4 + 4*a*c^ 3*d^2*e^6 + 6*a^2*c^2*e^8)*x^4 - 280*(c^4*d^5*e^3 + 4*a*c^3*d^3*e^5 + 6*a^ 2*c^2*d*e^7)*x^3 + 420*(c^4*d^6*e^2 + 4*a*c^3*d^4*e^4 + 6*a^2*c^2*d^2*e^6 + 4*a^3*c*e^8)*x^2 - 840*(c^4*d^7*e + 4*a*c^3*d^5*e^3 + 6*a^2*c^2*d^3*e^5 + 4*a^3*c*d*e^7)*x + 840*(c^4*d^8 + 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^6 + a^4*e^8)*log(e*x + d))/e^9
Time = 0.29 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=- \frac {c^{4} d x^{7}}{7 e^{2}} + \frac {c^{4} x^{8}}{8 e} + x^{6} \cdot \left (\frac {2 a c^{3}}{3 e} + \frac {c^{4} d^{2}}{6 e^{3}}\right ) + x^{5} \left (- \frac {4 a c^{3} d}{5 e^{2}} - \frac {c^{4} d^{3}}{5 e^{4}}\right ) + x^{4} \cdot \left (\frac {3 a^{2} c^{2}}{2 e} + \frac {a c^{3} d^{2}}{e^{3}} + \frac {c^{4} d^{4}}{4 e^{5}}\right ) + x^{3} \left (- \frac {2 a^{2} c^{2} d}{e^{2}} - \frac {4 a c^{3} d^{3}}{3 e^{4}} - \frac {c^{4} d^{5}}{3 e^{6}}\right ) + x^{2} \cdot \left (\frac {2 a^{3} c}{e} + \frac {3 a^{2} c^{2} d^{2}}{e^{3}} + \frac {2 a c^{3} d^{4}}{e^{5}} + \frac {c^{4} d^{6}}{2 e^{7}}\right ) + x \left (- \frac {4 a^{3} c d}{e^{2}} - \frac {6 a^{2} c^{2} d^{3}}{e^{4}} - \frac {4 a c^{3} d^{5}}{e^{6}} - \frac {c^{4} d^{7}}{e^{8}}\right ) + \frac {\left (a e^{2} + c d^{2}\right )^{4} \log {\left (d + e x \right )}}{e^{9}} \]
-c**4*d*x**7/(7*e**2) + c**4*x**8/(8*e) + x**6*(2*a*c**3/(3*e) + c**4*d**2 /(6*e**3)) + x**5*(-4*a*c**3*d/(5*e**2) - c**4*d**3/(5*e**4)) + x**4*(3*a* *2*c**2/(2*e) + a*c**3*d**2/e**3 + c**4*d**4/(4*e**5)) + x**3*(-2*a**2*c** 2*d/e**2 - 4*a*c**3*d**3/(3*e**4) - c**4*d**5/(3*e**6)) + x**2*(2*a**3*c/e + 3*a**2*c**2*d**2/e**3 + 2*a*c**3*d**4/e**5 + c**4*d**6/(2*e**7)) + x*(- 4*a**3*c*d/e**2 - 6*a**2*c**2*d**3/e**4 - 4*a*c**3*d**5/e**6 - c**4*d**7/e **8) + (a*e**2 + c*d**2)**4*log(d + e*x)/e**9
Time = 0.20 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=\frac {105 \, c^{4} e^{7} x^{8} - 120 \, c^{4} d e^{6} x^{7} + 140 \, {\left (c^{4} d^{2} e^{5} + 4 \, a c^{3} e^{7}\right )} x^{6} - 168 \, {\left (c^{4} d^{3} e^{4} + 4 \, a c^{3} d e^{6}\right )} x^{5} + 210 \, {\left (c^{4} d^{4} e^{3} + 4 \, a c^{3} d^{2} e^{5} + 6 \, a^{2} c^{2} e^{7}\right )} x^{4} - 280 \, {\left (c^{4} d^{5} e^{2} + 4 \, a c^{3} d^{3} e^{4} + 6 \, a^{2} c^{2} d e^{6}\right )} x^{3} + 420 \, {\left (c^{4} d^{6} e + 4 \, a c^{3} d^{4} e^{3} + 6 \, a^{2} c^{2} d^{2} e^{5} + 4 \, a^{3} c e^{7}\right )} x^{2} - 840 \, {\left (c^{4} d^{7} + 4 \, a c^{3} d^{5} e^{2} + 6 \, a^{2} c^{2} d^{3} e^{4} + 4 \, a^{3} c d e^{6}\right )} x}{840 \, e^{8}} + \frac {{\left (c^{4} d^{8} + 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} + 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left (e x + d\right )}{e^{9}} \]
1/840*(105*c^4*e^7*x^8 - 120*c^4*d*e^6*x^7 + 140*(c^4*d^2*e^5 + 4*a*c^3*e^ 7)*x^6 - 168*(c^4*d^3*e^4 + 4*a*c^3*d*e^6)*x^5 + 210*(c^4*d^4*e^3 + 4*a*c^ 3*d^2*e^5 + 6*a^2*c^2*e^7)*x^4 - 280*(c^4*d^5*e^2 + 4*a*c^3*d^3*e^4 + 6*a^ 2*c^2*d*e^6)*x^3 + 420*(c^4*d^6*e + 4*a*c^3*d^4*e^3 + 6*a^2*c^2*d^2*e^5 + 4*a^3*c*e^7)*x^2 - 840*(c^4*d^7 + 4*a*c^3*d^5*e^2 + 6*a^2*c^2*d^3*e^4 + 4* a^3*c*d*e^6)*x)/e^8 + (c^4*d^8 + 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a ^3*c*d^2*e^6 + a^4*e^8)*log(e*x + d)/e^9
Time = 0.29 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=\frac {105 \, c^{4} e^{7} x^{8} - 120 \, c^{4} d e^{6} x^{7} + 140 \, c^{4} d^{2} e^{5} x^{6} + 560 \, a c^{3} e^{7} x^{6} - 168 \, c^{4} d^{3} e^{4} x^{5} - 672 \, a c^{3} d e^{6} x^{5} + 210 \, c^{4} d^{4} e^{3} x^{4} + 840 \, a c^{3} d^{2} e^{5} x^{4} + 1260 \, a^{2} c^{2} e^{7} x^{4} - 280 \, c^{4} d^{5} e^{2} x^{3} - 1120 \, a c^{3} d^{3} e^{4} x^{3} - 1680 \, a^{2} c^{2} d e^{6} x^{3} + 420 \, c^{4} d^{6} e x^{2} + 1680 \, a c^{3} d^{4} e^{3} x^{2} + 2520 \, a^{2} c^{2} d^{2} e^{5} x^{2} + 1680 \, a^{3} c e^{7} x^{2} - 840 \, c^{4} d^{7} x - 3360 \, a c^{3} d^{5} e^{2} x - 5040 \, a^{2} c^{2} d^{3} e^{4} x - 3360 \, a^{3} c d e^{6} x}{840 \, e^{8}} + \frac {{\left (c^{4} d^{8} + 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} + 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{9}} \]
1/840*(105*c^4*e^7*x^8 - 120*c^4*d*e^6*x^7 + 140*c^4*d^2*e^5*x^6 + 560*a*c ^3*e^7*x^6 - 168*c^4*d^3*e^4*x^5 - 672*a*c^3*d*e^6*x^5 + 210*c^4*d^4*e^3*x ^4 + 840*a*c^3*d^2*e^5*x^4 + 1260*a^2*c^2*e^7*x^4 - 280*c^4*d^5*e^2*x^3 - 1120*a*c^3*d^3*e^4*x^3 - 1680*a^2*c^2*d*e^6*x^3 + 420*c^4*d^6*e*x^2 + 1680 *a*c^3*d^4*e^3*x^2 + 2520*a^2*c^2*d^2*e^5*x^2 + 1680*a^3*c*e^7*x^2 - 840*c ^4*d^7*x - 3360*a*c^3*d^5*e^2*x - 5040*a^2*c^2*d^3*e^4*x - 3360*a^3*c*d*e^ 6*x)/e^8 + (c^4*d^8 + 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^ 6 + a^4*e^8)*log(abs(e*x + d))/e^9
Time = 0.07 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+c x^2\right )^4}{d+e x} \, dx=x^2\,\left (\frac {d^2\,\left (\frac {d^2\,\left (\frac {4\,a\,c^3}{e}+\frac {c^4\,d^2}{e^3}\right )}{e^2}+\frac {6\,a^2\,c^2}{e}\right )}{2\,e^2}+\frac {2\,a^3\,c}{e}\right )+x^4\,\left (\frac {d^2\,\left (\frac {4\,a\,c^3}{e}+\frac {c^4\,d^2}{e^3}\right )}{4\,e^2}+\frac {3\,a^2\,c^2}{2\,e}\right )+x^6\,\left (\frac {2\,a\,c^3}{3\,e}+\frac {c^4\,d^2}{6\,e^3}\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^4\,e^8+4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4+4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{e^9}+\frac {c^4\,x^8}{8\,e}-\frac {c^4\,d\,x^7}{7\,e^2}-\frac {d\,x^3\,\left (\frac {d^2\,\left (\frac {4\,a\,c^3}{e}+\frac {c^4\,d^2}{e^3}\right )}{e^2}+\frac {6\,a^2\,c^2}{e}\right )}{3\,e}-\frac {d\,x^5\,\left (\frac {4\,a\,c^3}{e}+\frac {c^4\,d^2}{e^3}\right )}{5\,e}-\frac {d\,x\,\left (\frac {d^2\,\left (\frac {d^2\,\left (\frac {4\,a\,c^3}{e}+\frac {c^4\,d^2}{e^3}\right )}{e^2}+\frac {6\,a^2\,c^2}{e}\right )}{e^2}+\frac {4\,a^3\,c}{e}\right )}{e} \]
x^2*((d^2*((d^2*((4*a*c^3)/e + (c^4*d^2)/e^3))/e^2 + (6*a^2*c^2)/e))/(2*e^ 2) + (2*a^3*c)/e) + x^4*((d^2*((4*a*c^3)/e + (c^4*d^2)/e^3))/(4*e^2) + (3* a^2*c^2)/(2*e)) + x^6*((2*a*c^3)/(3*e) + (c^4*d^2)/(6*e^3)) + (log(d + e*x )*(a^4*e^8 + c^4*d^8 + 4*a*c^3*d^6*e^2 + 4*a^3*c*d^2*e^6 + 6*a^2*c^2*d^4*e ^4))/e^9 + (c^4*x^8)/(8*e) - (c^4*d*x^7)/(7*e^2) - (d*x^3*((d^2*((4*a*c^3) /e + (c^4*d^2)/e^3))/e^2 + (6*a^2*c^2)/e))/(3*e) - (d*x^5*((4*a*c^3)/e + ( c^4*d^2)/e^3))/(5*e) - (d*x*((d^2*((d^2*((4*a*c^3)/e + (c^4*d^2)/e^3))/e^2 + (6*a^2*c^2)/e))/e^2 + (4*a^3*c)/e))/e