Integrand size = 20, antiderivative size = 94 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=\frac {\left (3 c d^2+b e^2\right ) x}{e^4}-\frac {c d x^2}{e^3}+\frac {c x^3}{3 e^2}-\frac {c d^4+b d^2 e^2+a e^4}{e^5 (d+e x)}-\frac {2 d \left (2 c d^2+b e^2\right ) \log (d+e x)}{e^5} \] Output:
(b*e^2+3*c*d^2)*x/e^4-c*d*x^2/e^3+1/3*c*x^3/e^2-(a*e^4+b*d^2*e^2+c*d^4)/e^ 5/(e*x+d)-2*d*(b*e^2+2*c*d^2)*ln(e*x+d)/e^5
Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=\frac {3 e \left (3 c d^2+b e^2\right ) x-3 c d e^2 x^2+c e^3 x^3-\frac {3 \left (c d^4+b d^2 e^2+a e^4\right )}{d+e x}-6 \left (2 c d^3+b d e^2\right ) \log (d+e x)}{3 e^5} \] Input:
Integrate[(a + b*x^2 + c*x^4)/(d + e*x)^2,x]
Output:
(3*e*(3*c*d^2 + b*e^2)*x - 3*c*d*e^2*x^2 + c*e^3*x^3 - (3*(c*d^4 + b*d^2*e ^2 + a*e^4))/(d + e*x) - 6*(2*c*d^3 + b*d*e^2)*Log[d + e*x])/(3*e^5)
Time = 0.44 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2389, 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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {a e^4+b d^2 e^2+c d^4}{e^4 (d+e x)^2}-\frac {2 \left (b d e^2+2 c d^3\right )}{e^4 (d+e x)}+\frac {b e^2+3 c d^2}{e^4}-\frac {2 c d x}{e^3}+\frac {c x^2}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a e^4+b d^2 e^2+c d^4}{e^5 (d+e x)}-\frac {2 d \left (b e^2+2 c d^2\right ) \log (d+e x)}{e^5}+\frac {x \left (b e^2+3 c d^2\right )}{e^4}-\frac {c d x^2}{e^3}+\frac {c x^3}{3 e^2}\) |
Input:
Int[(a + b*x^2 + c*x^4)/(d + e*x)^2,x]
Output:
((3*c*d^2 + b*e^2)*x)/e^4 - (c*d*x^2)/e^3 + (c*x^3)/(3*e^2) - (c*d^4 + b*d ^2*e^2 + a*e^4)/(e^5*(d + e*x)) - (2*d*(2*c*d^2 + b*e^2)*Log[d + e*x])/e^5
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\frac {1}{3} c \,e^{2} x^{3}-c d e \,x^{2}+b \,e^{2} x +3 c \,d^{2} x}{e^{4}}-\frac {e^{4} a +b \,d^{2} e^{2}+c \,d^{4}}{e^{5} \left (e x +d \right )}-\frac {2 d \left (b \,e^{2}+2 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(92\) |
norman | \(\frac {\frac {\left (b \,e^{2}+2 c \,d^{2}\right ) x^{2}}{e^{3}}-\frac {e^{4} a +2 b \,d^{2} e^{2}+4 c \,d^{4}}{e^{5}}+\frac {c \,x^{4}}{3 e}-\frac {2 c d \,x^{3}}{3 e^{2}}}{e x +d}-\frac {2 d \left (b \,e^{2}+2 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(99\) |
risch | \(\frac {c \,x^{3}}{3 e^{2}}-\frac {c d \,x^{2}}{e^{3}}+\frac {b x}{e^{2}}+\frac {3 c \,d^{2} x}{e^{4}}-\frac {a}{e \left (e x +d \right )}-\frac {b \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {c \,d^{4}}{e^{5} \left (e x +d \right )}-\frac {2 d \ln \left (e x +d \right ) b}{e^{3}}-\frac {4 c \,d^{3} \ln \left (e x +d \right )}{e^{5}}\) | \(110\) |
parallelrisch | \(-\frac {-x^{4} c \,e^{4}+2 c d \,x^{3} e^{3}+6 \ln \left (e x +d \right ) x b d \,e^{3}+12 \ln \left (e x +d \right ) x c \,d^{3} e -3 x^{2} b \,e^{4}-6 x^{2} c \,d^{2} e^{2}+6 \ln \left (e x +d \right ) b \,d^{2} e^{2}+12 \ln \left (e x +d \right ) c \,d^{4}+3 e^{4} a +6 b \,d^{2} e^{2}+12 c \,d^{4}}{3 e^{5} \left (e x +d \right )}\) | \(130\) |
Input:
int((c*x^4+b*x^2+a)/(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/e^4*(1/3*c*e^2*x^3-c*d*e*x^2+b*e^2*x+3*c*d^2*x)-(a*e^4+b*d^2*e^2+c*d^4)/ e^5/(e*x+d)-2*d*(b*e^2+2*c*d^2)*ln(e*x+d)/e^5
Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=\frac {c e^{4} x^{4} - 2 \, c d e^{3} x^{3} - 3 \, c d^{4} - 3 \, b d^{2} e^{2} - 3 \, a e^{4} + 3 \, {\left (2 \, c d^{2} e^{2} + b e^{4}\right )} x^{2} + 3 \, {\left (3 \, c d^{3} e + b d e^{3}\right )} x - 6 \, {\left (2 \, c d^{4} + b d^{2} e^{2} + {\left (2 \, c d^{3} e + b d e^{3}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{6} x + d e^{5}\right )}} \] Input:
integrate((c*x^4+b*x^2+a)/(e*x+d)^2,x, algorithm="fricas")
Output:
1/3*(c*e^4*x^4 - 2*c*d*e^3*x^3 - 3*c*d^4 - 3*b*d^2*e^2 - 3*a*e^4 + 3*(2*c* d^2*e^2 + b*e^4)*x^2 + 3*(3*c*d^3*e + b*d*e^3)*x - 6*(2*c*d^4 + b*d^2*e^2 + (2*c*d^3*e + b*d*e^3)*x)*log(e*x + d))/(e^6*x + d*e^5)
Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=- \frac {c d x^{2}}{e^{3}} + \frac {c x^{3}}{3 e^{2}} - \frac {2 d \left (b e^{2} + 2 c d^{2}\right ) \log {\left (d + e x \right )}}{e^{5}} + x \left (\frac {b}{e^{2}} + \frac {3 c d^{2}}{e^{4}}\right ) + \frac {- a e^{4} - b d^{2} e^{2} - c d^{4}}{d e^{5} + e^{6} x} \] Input:
integrate((c*x**4+b*x**2+a)/(e*x+d)**2,x)
Output:
-c*d*x**2/e**3 + c*x**3/(3*e**2) - 2*d*(b*e**2 + 2*c*d**2)*log(d + e*x)/e* *5 + x*(b/e**2 + 3*c*d**2/e**4) + (-a*e**4 - b*d**2*e**2 - c*d**4)/(d*e**5 + e**6*x)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=-\frac {c d^{4} + b d^{2} e^{2} + a e^{4}}{e^{6} x + d e^{5}} + \frac {c e^{2} x^{3} - 3 \, c d e x^{2} + 3 \, {\left (3 \, c d^{2} + b e^{2}\right )} x}{3 \, e^{4}} - \frac {2 \, {\left (2 \, c d^{3} + b d e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \] Input:
integrate((c*x^4+b*x^2+a)/(e*x+d)^2,x, algorithm="maxima")
Output:
-(c*d^4 + b*d^2*e^2 + a*e^4)/(e^6*x + d*e^5) + 1/3*(c*e^2*x^3 - 3*c*d*e*x^ 2 + 3*(3*c*d^2 + b*e^2)*x)/e^4 - 2*(2*c*d^3 + b*d*e^2)*log(e*x + d)/e^5
Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.56 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=-\frac {1}{3} \, c {\left (\frac {{\left (e x + d\right )}^{3} {\left (\frac {6 \, d}{e x + d} - \frac {18 \, d^{2}}{{\left (e x + d\right )}^{2}} - 1\right )}}{e^{5}} - \frac {12 \, d^{3} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {3 \, d^{4}}{{\left (e x + d\right )} e^{5}}\right )} + b {\left (\frac {2 \, d \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {e x + d}{e^{3}} - \frac {d^{2}}{{\left (e x + d\right )} e^{3}}\right )} - \frac {a}{{\left (e x + d\right )} e} \] Input:
integrate((c*x^4+b*x^2+a)/(e*x+d)^2,x, algorithm="giac")
Output:
-1/3*c*((e*x + d)^3*(6*d/(e*x + d) - 18*d^2/(e*x + d)^2 - 1)/e^5 - 12*d^3* log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 + 3*d^4/((e*x + d)*e^5)) + b*(2 *d*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^3 + (e*x + d)/e^3 - d^2/((e*x + d)*e^3)) - a/((e*x + d)*e)
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=x\,\left (\frac {b}{e^2}+\frac {3\,c\,d^2}{e^4}\right )+\frac {c\,x^3}{3\,e^2}-\frac {\ln \left (d+e\,x\right )\,\left (4\,c\,d^3+2\,b\,d\,e^2\right )}{e^5}-\frac {c\,d^4+b\,d^2\,e^2+a\,e^4}{e\,\left (x\,e^5+d\,e^4\right )}-\frac {c\,d\,x^2}{e^3} \] Input:
int((a + b*x^2 + c*x^4)/(d + e*x)^2,x)
Output:
x*(b/e^2 + (3*c*d^2)/e^4) + (c*x^3)/(3*e^2) - (log(d + e*x)*(4*c*d^3 + 2*b *d*e^2))/e^5 - (a*e^4 + c*d^4 + b*d^2*e^2)/(e*(d*e^4 + e^5*x)) - (c*d*x^2) /e^3
Time = 0.15 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.50 \[ \int \frac {a+b x^2+c x^4}{(d+e x)^2} \, dx=\frac {-6 \,\mathrm {log}\left (e x +d \right ) b \,d^{3} e^{2}-6 \,\mathrm {log}\left (e x +d \right ) b \,d^{2} e^{3} x -12 \,\mathrm {log}\left (e x +d \right ) c \,d^{5}-12 \,\mathrm {log}\left (e x +d \right ) c \,d^{4} e x +3 a \,e^{5} x +6 b \,d^{2} e^{3} x +3 b d \,e^{4} x^{2}+12 c \,d^{4} e x +6 c \,d^{3} e^{2} x^{2}-2 c \,d^{2} e^{3} x^{3}+c d \,e^{4} x^{4}}{3 d \,e^{5} \left (e x +d \right )} \] Input:
int((c*x^4+b*x^2+a)/(e*x+d)^2,x)
Output:
( - 6*log(d + e*x)*b*d**3*e**2 - 6*log(d + e*x)*b*d**2*e**3*x - 12*log(d + e*x)*c*d**5 - 12*log(d + e*x)*c*d**4*e*x + 3*a*e**5*x + 6*b*d**2*e**3*x + 3*b*d*e**4*x**2 + 12*c*d**4*e*x + 6*c*d**3*e**2*x**2 - 2*c*d**2*e**3*x**3 + c*d*e**4*x**4)/(3*d*e**5*(d + e*x))