Integrand size = 17, antiderivative size = 158 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=\frac {c \left (5 c^2 d^4+9 a c d^2 e^2+3 a^2 e^4\right ) x}{e^6}-\frac {c^2 d \left (2 c d^2+3 a e^2\right ) x^2}{e^5}+\frac {c^2 \left (c d^2+a e^2\right ) x^3}{e^4}-\frac {c^3 d x^4}{2 e^3}+\frac {c^3 x^5}{5 e^2}-\frac {\left (c d^2+a e^2\right )^3}{e^7 (d+e x)}-\frac {6 c d \left (c d^2+a e^2\right )^2 \log (d+e x)}{e^7} \]
c*(3*a^2*e^4+9*a*c*d^2*e^2+5*c^2*d^4)*x/e^6-c^2*d*(3*a*e^2+2*c*d^2)*x^2/e^ 5+c^2*(a*e^2+c*d^2)*x^3/e^4-1/2*c^3*d*x^4/e^3+1/5*c^3*x^5/e^2-(a*e^2+c*d^2 )^3/e^7/(e*x+d)-6*c*d*(a*e^2+c*d^2)^2*ln(e*x+d)/e^7
Time = 0.06 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=\frac {-10 a^3 e^6+30 a^2 c e^4 \left (-d^2+d e x+e^2 x^2\right )+10 a c^2 e^2 \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )+c^3 \left (-10 d^6+50 d^5 e x+30 d^4 e^2 x^2-10 d^3 e^3 x^3+5 d^2 e^4 x^4-3 d e^5 x^5+2 e^6 x^6\right )-60 c d \left (c d^2+a e^2\right )^2 (d+e x) \log (d+e x)}{10 e^7 (d+e x)} \]
(-10*a^3*e^6 + 30*a^2*c*e^4*(-d^2 + d*e*x + e^2*x^2) + 10*a*c^2*e^2*(-3*d^ 4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4) + c^3*(-10*d^6 + 50 *d^5*e*x + 30*d^4*e^2*x^2 - 10*d^3*e^3*x^3 + 5*d^2*e^4*x^4 - 3*d*e^5*x^5 + 2*e^6*x^6) - 60*c*d*(c*d^2 + a*e^2)^2*(d + e*x)*Log[d + e*x])/(10*e^7*(d + e*x))
Time = 0.35 (sec) , antiderivative size = 158, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \int \left (\frac {c \left (3 a^2 e^4+9 a c d^2 e^2+5 c^2 d^4\right )}{e^6}-\frac {2 c^2 d x \left (3 a e^2+2 c d^2\right )}{e^5}+\frac {3 c^2 x^2 \left (a e^2+c d^2\right )}{e^4}-\frac {6 c d \left (a e^2+c d^2\right )^2}{e^6 (d+e x)}+\frac {\left (a e^2+c d^2\right )^3}{e^6 (d+e x)^2}-\frac {2 c^3 d x^3}{e^3}+\frac {c^3 x^4}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c x \left (3 a^2 e^4+9 a c d^2 e^2+5 c^2 d^4\right )}{e^6}-\frac {c^2 d x^2 \left (3 a e^2+2 c d^2\right )}{e^5}+\frac {c^2 x^3 \left (a e^2+c d^2\right )}{e^4}-\frac {\left (a e^2+c d^2\right )^3}{e^7 (d+e x)}-\frac {6 c d \left (a e^2+c d^2\right )^2 \log (d+e x)}{e^7}-\frac {c^3 d x^4}{2 e^3}+\frac {c^3 x^5}{5 e^2}\) |
(c*(5*c^2*d^4 + 9*a*c*d^2*e^2 + 3*a^2*e^4)*x)/e^6 - (c^2*d*(2*c*d^2 + 3*a* e^2)*x^2)/e^5 + (c^2*(c*d^2 + a*e^2)*x^3)/e^4 - (c^3*d*x^4)/(2*e^3) + (c^3 *x^5)/(5*e^2) - (c*d^2 + a*e^2)^3/(e^7*(d + e*x)) - (6*c*d*(c*d^2 + a*e^2) ^2*Log[d + e*x])/e^7
3.5.79.3.1 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 = 2.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23
| method | result | size |
| default | \(\frac {c \left (\frac {1}{5} c^{2} x^{5} e^{4}-\frac {1}{2} x^{4} c^{2} d \,e^{3}+x^{3} a c \,e^{4}+x^{3} c^{2} d^{2} e^{2}-3 x^{2} a c d \,e^{3}-2 x^{2} c^{2} d^{3} e +3 a^{2} e^{4} x +9 a c \,d^{2} e^{2} x +5 c^{2} d^{4} x \right )}{e^{6}}-\frac {e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a +c^{3} d^{6}}{e^{7} \left (e x +d \right )}-\frac {6 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(194\) |
| norman | \(\frac {\frac {\left (e^{6} a^{3}+6 d^{2} e^{4} a^{2} c +12 d^{4} e^{2} c^{2} a +6 c^{3} d^{6}\right ) x}{d \,e^{6}}+\frac {c^{3} x^{6}}{5 e}+\frac {3 c \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x^{2}}{e^{5}}+\frac {c^{2} \left (2 e^{2} a +c \,d^{2}\right ) x^{4}}{2 e^{3}}-\frac {3 c^{3} d \,x^{5}}{10 e^{2}}-\frac {d \,c^{2} \left (2 e^{2} a +c \,d^{2}\right ) x^{3}}{e^{4}}}{e x +d}-\frac {6 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(201\) |
| risch | \(\frac {c^{3} x^{5}}{5 e^{2}}-\frac {c^{3} d \,x^{4}}{2 e^{3}}+\frac {c^{2} x^{3} a}{e^{2}}+\frac {c^{3} x^{3} d^{2}}{e^{4}}-\frac {3 c^{2} x^{2} a d}{e^{3}}-\frac {2 c^{3} x^{2} d^{3}}{e^{5}}+\frac {3 c \,a^{2} x}{e^{2}}+\frac {9 c^{2} a \,d^{2} x}{e^{4}}+\frac {5 c^{3} d^{4} x}{e^{6}}-\frac {a^{3}}{e \left (e x +d \right )}-\frac {3 d^{2} a^{2} c}{e^{3} \left (e x +d \right )}-\frac {3 d^{4} c^{2} a}{e^{5} \left (e x +d \right )}-\frac {c^{3} d^{6}}{e^{7} \left (e x +d \right )}-\frac {6 c d \ln \left (e x +d \right ) a^{2}}{e^{3}}-\frac {12 c^{2} d^{3} \ln \left (e x +d \right ) a}{e^{5}}-\frac {6 c^{3} d^{5} \ln \left (e x +d \right )}{e^{7}}\) | \(233\) |
| parallelrisch | \(-\frac {-2 x^{6} c^{3} e^{6}+3 x^{5} c^{3} d \,e^{5}-10 x^{4} a \,c^{2} e^{6}-5 x^{4} c^{3} d^{2} e^{4}+20 x^{3} a \,c^{2} d \,e^{5}+10 x^{3} c^{3} d^{3} e^{3}+60 \ln \left (e x +d \right ) x \,a^{2} c d \,e^{5}+120 \ln \left (e x +d \right ) x a \,c^{2} d^{3} e^{3}+60 \ln \left (e x +d \right ) x \,c^{3} d^{5} e -30 x^{2} a^{2} c \,e^{6}-60 x^{2} a \,c^{2} d^{2} e^{4}-30 x^{2} c^{3} d^{4} e^{2}+60 \ln \left (e x +d \right ) a^{2} c \,d^{2} e^{4}+120 \ln \left (e x +d \right ) a \,c^{2} d^{4} e^{2}+60 \ln \left (e x +d \right ) c^{3} d^{6}+10 e^{6} a^{3}+60 d^{2} e^{4} a^{2} c +120 d^{4} e^{2} c^{2} a +60 c^{3} d^{6}}{10 e^{7} \left (e x +d \right )}\) | \(273\) |
c/e^6*(1/5*c^2*x^5*e^4-1/2*x^4*c^2*d*e^3+x^3*a*c*e^4+x^3*c^2*d^2*e^2-3*x^2 *a*c*d*e^3-2*x^2*c^2*d^3*e+3*a^2*e^4*x+9*a*c*d^2*e^2*x+5*c^2*d^4*x)-(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)-6*c*d/e^7*(a^2*e^4 +2*a*c*d^2*e^2+c^2*d^4)*ln(e*x+d)
Time = 0.28 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=\frac {2 \, c^{3} e^{6} x^{6} - 3 \, c^{3} d e^{5} x^{5} - 10 \, c^{3} d^{6} - 30 \, a c^{2} d^{4} e^{2} - 30 \, a^{2} c d^{2} e^{4} - 10 \, a^{3} e^{6} + 5 \, {\left (c^{3} d^{2} e^{4} + 2 \, a c^{2} e^{6}\right )} x^{4} - 10 \, {\left (c^{3} d^{3} e^{3} + 2 \, a c^{2} d e^{5}\right )} x^{3} + 30 \, {\left (c^{3} d^{4} e^{2} + 2 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} x^{2} + 10 \, {\left (5 \, c^{3} d^{5} e + 9 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5}\right )} x - 60 \, {\left (c^{3} d^{6} + 2 \, a c^{2} d^{4} e^{2} + a^{2} c d^{2} e^{4} + {\left (c^{3} d^{5} e + 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \log \left (e x + d\right )}{10 \, {\left (e^{8} x + d e^{7}\right )}} \]
1/10*(2*c^3*e^6*x^6 - 3*c^3*d*e^5*x^5 - 10*c^3*d^6 - 30*a*c^2*d^4*e^2 - 30 *a^2*c*d^2*e^4 - 10*a^3*e^6 + 5*(c^3*d^2*e^4 + 2*a*c^2*e^6)*x^4 - 10*(c^3* d^3*e^3 + 2*a*c^2*d*e^5)*x^3 + 30*(c^3*d^4*e^2 + 2*a*c^2*d^2*e^4 + a^2*c*e ^6)*x^2 + 10*(5*c^3*d^5*e + 9*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x - 60*(c^3*d ^6 + 2*a*c^2*d^4*e^2 + a^2*c*d^2*e^4 + (c^3*d^5*e + 2*a*c^2*d^3*e^3 + a^2* c*d*e^5)*x)*log(e*x + d))/(e^8*x + d*e^7)
Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=- \frac {c^{3} d x^{4}}{2 e^{3}} + \frac {c^{3} x^{5}}{5 e^{2}} - \frac {6 c d \left (a e^{2} + c d^{2}\right )^{2} \log {\left (d + e x \right )}}{e^{7}} + x^{3} \left (\frac {a c^{2}}{e^{2}} + \frac {c^{3} d^{2}}{e^{4}}\right ) + x^{2} \left (- \frac {3 a c^{2} d}{e^{3}} - \frac {2 c^{3} d^{3}}{e^{5}}\right ) + x \left (\frac {3 a^{2} c}{e^{2}} + \frac {9 a c^{2} d^{2}}{e^{4}} + \frac {5 c^{3} d^{4}}{e^{6}}\right ) + \frac {- a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} - 3 a c^{2} d^{4} e^{2} - c^{3} d^{6}}{d e^{7} + e^{8} x} \]
-c**3*d*x**4/(2*e**3) + c**3*x**5/(5*e**2) - 6*c*d*(a*e**2 + c*d**2)**2*lo g(d + e*x)/e**7 + x**3*(a*c**2/e**2 + c**3*d**2/e**4) + x**2*(-3*a*c**2*d/ e**3 - 2*c**3*d**3/e**5) + x*(3*a**2*c/e**2 + 9*a*c**2*d**2/e**4 + 5*c**3* d**4/e**6) + (-a**3*e**6 - 3*a**2*c*d**2*e**4 - 3*a*c**2*d**4*e**2 - c**3* d**6)/(d*e**7 + e**8*x)
Time = 0.20 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=-\frac {c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}}{e^{8} x + d e^{7}} + \frac {2 \, c^{3} e^{4} x^{5} - 5 \, c^{3} d e^{3} x^{4} + 10 \, {\left (c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )} x^{3} - 10 \, {\left (2 \, c^{3} d^{3} e + 3 \, a c^{2} d e^{3}\right )} x^{2} + 10 \, {\left (5 \, c^{3} d^{4} + 9 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x}{10 \, e^{6}} - \frac {6 \, {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \log \left (e x + d\right )}{e^{7}} \]
-(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6)/(e^8*x + d*e^7) + 1/10*(2*c^3*e^4*x^5 - 5*c^3*d*e^3*x^4 + 10*(c^3*d^2*e^2 + a*c^2*e^4)*x^3 - 10*(2*c^3*d^3*e + 3*a*c^2*d*e^3)*x^2 + 10*(5*c^3*d^4 + 9*a*c^2*d^2*e^2 + 3*a^2*c*e^4)*x)/e^6 - 6*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*log(e*x + d)/e^7
Time = 0.28 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=\frac {{\left (2 \, c^{3} - \frac {15 \, c^{3} d}{e x + d} + \frac {10 \, {\left (5 \, c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )}}{{\left (e x + d\right )}^{2} e^{2}} - \frac {20 \, {\left (5 \, c^{3} d^{3} e^{3} + 3 \, a c^{2} d e^{5}\right )}}{{\left (e x + d\right )}^{3} e^{3}} + \frac {30 \, {\left (5 \, c^{3} d^{4} e^{4} + 6 \, a c^{2} d^{2} e^{6} + a^{2} c e^{8}\right )}}{{\left (e x + d\right )}^{4} e^{4}}\right )} {\left (e x + d\right )}^{5}}{10 \, e^{7}} + \frac {6 \, {\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{7}} - \frac {\frac {c^{3} d^{6} e^{5}}{e x + d} + \frac {3 \, a c^{2} d^{4} e^{7}}{e x + d} + \frac {3 \, a^{2} c d^{2} e^{9}}{e x + d} + \frac {a^{3} e^{11}}{e x + d}}{e^{12}} \]
1/10*(2*c^3 - 15*c^3*d/(e*x + d) + 10*(5*c^3*d^2*e^2 + a*c^2*e^4)/((e*x + d)^2*e^2) - 20*(5*c^3*d^3*e^3 + 3*a*c^2*d*e^5)/((e*x + d)^3*e^3) + 30*(5*c ^3*d^4*e^4 + 6*a*c^2*d^2*e^6 + a^2*c*e^8)/((e*x + d)^4*e^4))*(e*x + d)^5/e ^7 + 6*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^7 - (c^3*d^6*e^5/(e*x + d) + 3*a*c^2*d^4*e^7/(e*x + d) + 3 *a^2*c*d^2*e^9/(e*x + d) + a^3*e^11/(e*x + d))/e^12
Time = 9.40 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^2} \, dx=x^2\,\left (\frac {c^3\,d^3}{e^5}-\frac {d\,\left (\frac {3\,a\,c^2}{e^2}+\frac {3\,c^3\,d^2}{e^4}\right )}{e}\right )-x\,\left (\frac {d^2\,\left (\frac {3\,a\,c^2}{e^2}+\frac {3\,c^3\,d^2}{e^4}\right )}{e^2}-\frac {3\,a^2\,c}{e^2}+\frac {2\,d\,\left (\frac {2\,c^3\,d^3}{e^5}-\frac {2\,d\,\left (\frac {3\,a\,c^2}{e^2}+\frac {3\,c^3\,d^2}{e^4}\right )}{e}\right )}{e}\right )+x^3\,\left (\frac {a\,c^2}{e^2}+\frac {c^3\,d^2}{e^4}\right )-\frac {a^3\,e^6+3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2+c^3\,d^6}{e\,\left (x\,e^7+d\,e^6\right )}-\frac {\ln \left (d+e\,x\right )\,\left (6\,a^2\,c\,d\,e^4+12\,a\,c^2\,d^3\,e^2+6\,c^3\,d^5\right )}{e^7}+\frac {c^3\,x^5}{5\,e^2}-\frac {c^3\,d\,x^4}{2\,e^3} \]
x^2*((c^3*d^3)/e^5 - (d*((3*a*c^2)/e^2 + (3*c^3*d^2)/e^4))/e) - x*((d^2*(( 3*a*c^2)/e^2 + (3*c^3*d^2)/e^4))/e^2 - (3*a^2*c)/e^2 + (2*d*((2*c^3*d^3)/e ^5 - (2*d*((3*a*c^2)/e^2 + (3*c^3*d^2)/e^4))/e))/e) + x^3*((a*c^2)/e^2 + ( c^3*d^2)/e^4) - (a^3*e^6 + c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4)/(e *(d*e^6 + e^7*x)) - (log(d + e*x)*(6*c^3*d^5 + 12*a*c^2*d^3*e^2 + 6*a^2*c* d*e^4))/e^7 + (c^3*x^5)/(5*e^2) - (c^3*d*x^4)/(2*e^3)