Integrand size = 17, antiderivative size = 165 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=\frac {c^2 \left (10 c d^2+3 a e^2\right ) x}{e^6}-\frac {2 c^3 d x^2}{e^5}+\frac {c^3 x^3}{3 e^4}-\frac {\left (c d^2+a e^2\right )^3}{3 e^7 (d+e x)^3}+\frac {3 c d \left (c d^2+a e^2\right )^2}{e^7 (d+e x)^2}-\frac {3 c \left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right )}{e^7 (d+e x)}-\frac {4 c^2 d \left (5 c d^2+3 a e^2\right ) \log (d+e x)}{e^7} \]
c^2*(3*a*e^2+10*c*d^2)*x/e^6-2*c^3*d*x^2/e^5+1/3*c^3*x^3/e^4-1/3*(a*e^2+c* d^2)^3/e^7/(e*x+d)^3+3*c*d*(a*e^2+c*d^2)^2/e^7/(e*x+d)^2-3*c*(a*e^2+c*d^2) *(a*e^2+5*c*d^2)/e^7/(e*x+d)-4*c^2*d*(3*a*e^2+5*c*d^2)*ln(e*x+d)/e^7
Time = 0.07 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=\frac {-a^3 e^6-3 a^2 c e^4 \left (d^2+3 d e x+3 e^2 x^2\right )+3 a c^2 e^2 \left (-13 d^4-27 d^3 e x-9 d^2 e^2 x^2+9 d e^3 x^3+3 e^4 x^4\right )+c^3 \left (-37 d^6-51 d^5 e x+39 d^4 e^2 x^2+73 d^3 e^3 x^3+15 d^2 e^4 x^4-3 d e^5 x^5+e^6 x^6\right )-12 c^2 d \left (5 c d^2+3 a e^2\right ) (d+e x)^3 \log (d+e x)}{3 e^7 (d+e x)^3} \]
(-(a^3*e^6) - 3*a^2*c*e^4*(d^2 + 3*d*e*x + 3*e^2*x^2) + 3*a*c^2*e^2*(-13*d ^4 - 27*d^3*e*x - 9*d^2*e^2*x^2 + 9*d*e^3*x^3 + 3*e^4*x^4) + c^3*(-37*d^6 - 51*d^5*e*x + 39*d^4*e^2*x^2 + 73*d^3*e^3*x^3 + 15*d^2*e^4*x^4 - 3*d*e^5* x^5 + e^6*x^6) - 12*c^2*d*(5*c*d^2 + 3*a*e^2)*(d + e*x)^3*Log[d + e*x])/(3 *e^7*(d + e*x)^3)
Time = 0.36 (sec) , antiderivative size = 165, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \int \left (-\frac {4 c^2 d \left (3 a e^2+5 c d^2\right )}{e^6 (d+e x)}+\frac {c^2 \left (3 a e^2+10 c d^2\right )}{e^6}+\frac {3 c \left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right )}{e^6 (d+e x)^2}-\frac {6 c d \left (a e^2+c d^2\right )^2}{e^6 (d+e x)^3}+\frac {\left (a e^2+c d^2\right )^3}{e^6 (d+e x)^4}-\frac {4 c^3 d x}{e^5}+\frac {c^3 x^2}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 c^2 d \left (3 a e^2+5 c d^2\right ) \log (d+e x)}{e^7}+\frac {c^2 x \left (3 a e^2+10 c d^2\right )}{e^6}-\frac {3 c \left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right )}{e^7 (d+e x)}+\frac {3 c d \left (a e^2+c d^2\right )^2}{e^7 (d+e x)^2}-\frac {\left (a e^2+c d^2\right )^3}{3 e^7 (d+e x)^3}-\frac {2 c^3 d x^2}{e^5}+\frac {c^3 x^3}{3 e^4}\) |
(c^2*(10*c*d^2 + 3*a*e^2)*x)/e^6 - (2*c^3*d*x^2)/e^5 + (c^3*x^3)/(3*e^4) - (c*d^2 + a*e^2)^3/(3*e^7*(d + e*x)^3) + (3*c*d*(c*d^2 + a*e^2)^2)/(e^7*(d + e*x)^2) - (3*c*(c*d^2 + a*e^2)*(5*c*d^2 + a*e^2))/(e^7*(d + e*x)) - (4* c^2*d*(5*c*d^2 + 3*a*e^2)*Log[d + e*x])/e^7
3.5.81.3.1 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 = 2.36 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.20
| method | result | size |
| default | \(\frac {c^{2} \left (\frac {1}{3} c \,e^{2} x^{3}-2 c d e \,x^{2}+3 a \,e^{2} x +10 c \,d^{2} x \right )}{e^{6}}-\frac {3 c \left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right )}{e^{7} \left (e x +d \right )}-\frac {e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a +c^{3} d^{6}}{3 e^{7} \left (e x +d \right )^{3}}+\frac {3 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{e^{7} \left (e x +d \right )^{2}}-\frac {4 c^{2} d \left (3 e^{2} a +5 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(198\) |
| norman | \(\frac {\frac {c^{2} \left (3 e^{2} a +5 c \,d^{2}\right ) x^{4}}{e^{3}}-\frac {e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +66 d^{4} e^{2} c^{2} a +110 c^{3} d^{6}}{3 e^{7}}+\frac {c^{3} x^{6}}{3 e}-\frac {3 \left (e^{4} a^{2} c +12 d^{2} e^{2} c^{2} a +20 d^{4} c^{3}\right ) x^{2}}{e^{5}}-\frac {c^{3} d \,x^{5}}{e^{2}}-\frac {3 d \left (e^{4} a^{2} c +18 d^{2} e^{2} c^{2} a +30 d^{4} c^{3}\right ) x}{e^{6}}}{\left (e x +d \right )^{3}}-\frac {4 c^{2} d \left (3 e^{2} a +5 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(203\) |
| risch | \(\frac {c^{3} x^{3}}{3 e^{4}}-\frac {2 c^{3} d \,x^{2}}{e^{5}}+\frac {3 c^{2} a x}{e^{4}}+\frac {10 c^{3} d^{2} x}{e^{6}}+\frac {\left (-3 e^{5} a^{2} c -18 d^{2} e^{3} c^{2} a -15 d^{4} e \,c^{3}\right ) x^{2}-3 c d \left (a^{2} e^{4}+10 a c \,d^{2} e^{2}+9 c^{2} d^{4}\right ) x -\frac {e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +39 d^{4} e^{2} c^{2} a +37 c^{3} d^{6}}{3 e}}{e^{6} \left (e x +d \right )^{3}}-\frac {12 c^{2} d \ln \left (e x +d \right ) a}{e^{5}}-\frac {20 \ln \left (e x +d \right ) c^{3} d^{3}}{e^{7}}\) | \(203\) |
| parallelrisch | \(-\frac {108 \ln \left (e x +d \right ) x a \,c^{2} d^{3} e^{3}+66 d^{4} e^{2} c^{2} a +3 d^{2} e^{4} a^{2} c -15 x^{4} c^{3} d^{2} e^{4}+180 x^{2} c^{3} d^{4} e^{2}+270 x \,c^{3} d^{5} e +110 c^{3} d^{6}+36 \ln \left (e x +d \right ) x^{3} a \,c^{2} d \,e^{5}+9 x \,a^{2} c d \,e^{5}+162 x a \,c^{2} d^{3} e^{3}+180 \ln \left (e x +d \right ) x \,c^{3} d^{5} e -x^{6} c^{3} e^{6}+108 \ln \left (e x +d \right ) x^{2} a \,c^{2} d^{2} e^{4}+e^{6} a^{3}+36 \ln \left (e x +d \right ) a \,c^{2} d^{4} e^{2}+180 \ln \left (e x +d \right ) x^{2} c^{3} d^{4} e^{2}+108 x^{2} a \,c^{2} d^{2} e^{4}+60 \ln \left (e x +d \right ) x^{3} c^{3} d^{3} e^{3}+3 x^{5} c^{3} d \,e^{5}-9 x^{4} a \,c^{2} e^{6}+9 x^{2} a^{2} c \,e^{6}+60 \ln \left (e x +d \right ) c^{3} d^{6}}{3 e^{7} \left (e x +d \right )^{3}}\) | \(324\) |
c^2/e^6*(1/3*c*e^2*x^3-2*c*d*e*x^2+3*a*e^2*x+10*c*d^2*x)-3/e^7*c*(a^2*e^4+ 6*a*c*d^2*e^2+5*c^2*d^4)/(e*x+d)-1/3*(a^3*e^6+3*a^2*c*d^2*e^4+3*a*c^2*d^4* e^2+c^3*d^6)/e^7/(e*x+d)^3+3*c*d/e^7*(a^2*e^4+2*a*c*d^2*e^2+c^2*d^4)/(e*x+ d)^2-4*c^2*d*(3*a*e^2+5*c*d^2)*ln(e*x+d)/e^7
Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (161) = 322\).
Time = 0.25 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=\frac {c^{3} e^{6} x^{6} - 3 \, c^{3} d e^{5} x^{5} - 37 \, c^{3} d^{6} - 39 \, a c^{2} d^{4} e^{2} - 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6} + 3 \, {\left (5 \, c^{3} d^{2} e^{4} + 3 \, a c^{2} e^{6}\right )} x^{4} + {\left (73 \, c^{3} d^{3} e^{3} + 27 \, a c^{2} d e^{5}\right )} x^{3} + 3 \, {\left (13 \, c^{3} d^{4} e^{2} - 9 \, a c^{2} d^{2} e^{4} - 3 \, a^{2} c e^{6}\right )} x^{2} - 3 \, {\left (17 \, c^{3} d^{5} e + 27 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5}\right )} x - 12 \, {\left (5 \, c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + {\left (5 \, c^{3} d^{3} e^{3} + 3 \, a c^{2} d e^{5}\right )} x^{3} + 3 \, {\left (5 \, c^{3} d^{4} e^{2} + 3 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 3 \, {\left (5 \, c^{3} d^{5} e + 3 \, a c^{2} d^{3} e^{3}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{10} x^{3} + 3 \, d e^{9} x^{2} + 3 \, d^{2} e^{8} x + d^{3} e^{7}\right )}} \]
1/3*(c^3*e^6*x^6 - 3*c^3*d*e^5*x^5 - 37*c^3*d^6 - 39*a*c^2*d^4*e^2 - 3*a^2 *c*d^2*e^4 - a^3*e^6 + 3*(5*c^3*d^2*e^4 + 3*a*c^2*e^6)*x^4 + (73*c^3*d^3*e ^3 + 27*a*c^2*d*e^5)*x^3 + 3*(13*c^3*d^4*e^2 - 9*a*c^2*d^2*e^4 - 3*a^2*c*e ^6)*x^2 - 3*(17*c^3*d^5*e + 27*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x - 12*(5*c^ 3*d^6 + 3*a*c^2*d^4*e^2 + (5*c^3*d^3*e^3 + 3*a*c^2*d*e^5)*x^3 + 3*(5*c^3*d ^4*e^2 + 3*a*c^2*d^2*e^4)*x^2 + 3*(5*c^3*d^5*e + 3*a*c^2*d^3*e^3)*x)*log(e *x + d))/(e^10*x^3 + 3*d*e^9*x^2 + 3*d^2*e^8*x + d^3*e^7)
Time = 1.06 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=- \frac {2 c^{3} d x^{2}}{e^{5}} + \frac {c^{3} x^{3}}{3 e^{4}} - \frac {4 c^{2} d \left (3 a e^{2} + 5 c d^{2}\right ) \log {\left (d + e x \right )}}{e^{7}} + x \left (\frac {3 a c^{2}}{e^{4}} + \frac {10 c^{3} d^{2}}{e^{6}}\right ) + \frac {- a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} - 39 a c^{2} d^{4} e^{2} - 37 c^{3} d^{6} + x^{2} \left (- 9 a^{2} c e^{6} - 54 a c^{2} d^{2} e^{4} - 45 c^{3} d^{4} e^{2}\right ) + x \left (- 9 a^{2} c d e^{5} - 90 a c^{2} d^{3} e^{3} - 81 c^{3} d^{5} e\right )}{3 d^{3} e^{7} + 9 d^{2} e^{8} x + 9 d e^{9} x^{2} + 3 e^{10} x^{3}} \]
-2*c**3*d*x**2/e**5 + c**3*x**3/(3*e**4) - 4*c**2*d*(3*a*e**2 + 5*c*d**2)* log(d + e*x)/e**7 + x*(3*a*c**2/e**4 + 10*c**3*d**2/e**6) + (-a**3*e**6 - 3*a**2*c*d**2*e**4 - 39*a*c**2*d**4*e**2 - 37*c**3*d**6 + x**2*(-9*a**2*c* e**6 - 54*a*c**2*d**2*e**4 - 45*c**3*d**4*e**2) + x*(-9*a**2*c*d*e**5 - 90 *a*c**2*d**3*e**3 - 81*c**3*d**5*e))/(3*d**3*e**7 + 9*d**2*e**8*x + 9*d*e* *9*x**2 + 3*e**10*x**3)
Time = 0.22 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=-\frac {37 \, c^{3} d^{6} + 39 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + 9 \, {\left (5 \, c^{3} d^{4} e^{2} + 6 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} x^{2} + 9 \, {\left (9 \, c^{3} d^{5} e + 10 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x}{3 \, {\left (e^{10} x^{3} + 3 \, d e^{9} x^{2} + 3 \, d^{2} e^{8} x + d^{3} e^{7}\right )}} + \frac {c^{3} e^{2} x^{3} - 6 \, c^{3} d e x^{2} + 3 \, {\left (10 \, c^{3} d^{2} + 3 \, a c^{2} e^{2}\right )} x}{3 \, e^{6}} - \frac {4 \, {\left (5 \, c^{3} d^{3} + 3 \, a c^{2} d e^{2}\right )} \log \left (e x + d\right )}{e^{7}} \]
-1/3*(37*c^3*d^6 + 39*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6 + 9*(5*c^3 *d^4*e^2 + 6*a*c^2*d^2*e^4 + a^2*c*e^6)*x^2 + 9*(9*c^3*d^5*e + 10*a*c^2*d^ 3*e^3 + a^2*c*d*e^5)*x)/(e^10*x^3 + 3*d*e^9*x^2 + 3*d^2*e^8*x + d^3*e^7) + 1/3*(c^3*e^2*x^3 - 6*c^3*d*e*x^2 + 3*(10*c^3*d^2 + 3*a*c^2*e^2)*x)/e^6 - 4*(5*c^3*d^3 + 3*a*c^2*d*e^2)*log(e*x + d)/e^7
Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=-\frac {4 \, {\left (5 \, c^{3} d^{3} + 3 \, a c^{2} d e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{7}} - \frac {37 \, c^{3} d^{6} + 39 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + 9 \, {\left (5 \, c^{3} d^{4} e^{2} + 6 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} x^{2} + 9 \, {\left (9 \, c^{3} d^{5} e + 10 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x}{3 \, {\left (e x + d\right )}^{3} e^{7}} + \frac {c^{3} e^{8} x^{3} - 6 \, c^{3} d e^{7} x^{2} + 30 \, c^{3} d^{2} e^{6} x + 9 \, a c^{2} e^{8} x}{3 \, e^{12}} \]
-4*(5*c^3*d^3 + 3*a*c^2*d*e^2)*log(abs(e*x + d))/e^7 - 1/3*(37*c^3*d^6 + 3 9*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6 + 9*(5*c^3*d^4*e^2 + 6*a*c^2*d ^2*e^4 + a^2*c*e^6)*x^2 + 9*(9*c^3*d^5*e + 10*a*c^2*d^3*e^3 + a^2*c*d*e^5) *x)/((e*x + d)^3*e^7) + 1/3*(c^3*e^8*x^3 - 6*c^3*d*e^7*x^2 + 30*c^3*d^2*e^ 6*x + 9*a*c^2*e^8*x)/e^12
Time = 9.35 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^4} \, dx=x\,\left (\frac {3\,a\,c^2}{e^4}+\frac {10\,c^3\,d^2}{e^6}\right )-\frac {x^2\,\left (3\,a^2\,c\,e^5+18\,a\,c^2\,d^2\,e^3+15\,c^3\,d^4\,e\right )+\frac {a^3\,e^6+3\,a^2\,c\,d^2\,e^4+39\,a\,c^2\,d^4\,e^2+37\,c^3\,d^6}{3\,e}+x\,\left (3\,a^2\,c\,d\,e^4+30\,a\,c^2\,d^3\,e^2+27\,c^3\,d^5\right )}{d^3\,e^6+3\,d^2\,e^7\,x+3\,d\,e^8\,x^2+e^9\,x^3}-\frac {\ln \left (d+e\,x\right )\,\left (20\,c^3\,d^3+12\,a\,c^2\,d\,e^2\right )}{e^7}+\frac {c^3\,x^3}{3\,e^4}-\frac {2\,c^3\,d\,x^2}{e^5} \]
x*((3*a*c^2)/e^4 + (10*c^3*d^2)/e^6) - (x^2*(3*a^2*c*e^5 + 15*c^3*d^4*e + 18*a*c^2*d^2*e^3) + (a^3*e^6 + 37*c^3*d^6 + 39*a*c^2*d^4*e^2 + 3*a^2*c*d^2 *e^4)/(3*e) + x*(27*c^3*d^5 + 30*a*c^2*d^3*e^2 + 3*a^2*c*d*e^4))/(d^3*e^6 + e^9*x^3 + 3*d^2*e^7*x + 3*d*e^8*x^2) - (log(d + e*x)*(20*c^3*d^3 + 12*a* c^2*d*e^2))/e^7 + (c^3*x^3)/(3*e^4) - (2*c^3*d*x^2)/e^5