Integrand size = 19, antiderivative size = 93 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=-\frac {d (c d-b e)^2 x}{e^4}+\frac {(c d-b e)^2 x^2}{2 e^3}-\frac {c (c d-2 b e) x^3}{3 e^2}+\frac {c^2 x^4}{4 e}+\frac {d^2 (c d-b e)^2 \log (d+e x)}{e^5} \] Output:
-d*(-b*e+c*d)^2*x/e^4+1/2*(-b*e+c*d)^2*x^2/e^3-1/3*c*(-2*b*e+c*d)*x^3/e^2+ 1/4*c^2*x^4/e+d^2*(-b*e+c*d)^2*ln(e*x+d)/e^5
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=-\frac {d (c d-b e)^2 x}{e^4}+\frac {(-c d+b e)^2 x^2}{2 e^3}-\frac {c (c d-2 b e) x^3}{3 e^2}+\frac {c^2 x^4}{4 e}+\frac {\left (c^2 d^4-2 b c d^3 e+b^2 d^2 e^2\right ) \log (d+e x)}{e^5} \] Input:
Integrate[(b*x + c*x^2)^2/(d + e*x),x]
Output:
-((d*(c*d - b*e)^2*x)/e^4) + ((-(c*d) + b*e)^2*x^2)/(2*e^3) - (c*(c*d - 2* b*e)*x^3)/(3*e^2) + (c^2*x^4)/(4*e) + ((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^ 2)*Log[d + e*x])/e^5
Time = 0.45 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1140, 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.
\(\displaystyle \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (\frac {d^2 (c d-b e)^2}{e^4 (d+e x)}-\frac {d (c d-b e)^2}{e^4}+\frac {x (b e-c d)^2}{e^3}-\frac {c x^2 (c d-2 b e)}{e^2}+\frac {c^2 x^3}{e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 (c d-b e)^2 \log (d+e x)}{e^5}-\frac {d x (c d-b e)^2}{e^4}+\frac {x^2 (c d-b e)^2}{2 e^3}-\frac {c x^3 (c d-2 b e)}{3 e^2}+\frac {c^2 x^4}{4 e}\) |
Input:
Int[(b*x + c*x^2)^2/(d + e*x),x]
Output:
-((d*(c*d - b*e)^2*x)/e^4) + ((c*d - b*e)^2*x^2)/(2*e^3) - (c*(c*d - 2*b*e )*x^3)/(3*e^2) + (c^2*x^4)/(4*e) + (d^2*(c*d - b*e)^2*Log[d + e*x])/e^5
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 0.45 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.31
method | result | size |
norman | \(\frac {c^{2} x^{4}}{4 e}+\frac {\left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) x^{2}}{2 e^{3}}+\frac {c \left (2 b e -c d \right ) x^{3}}{3 e^{2}}-\frac {d \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) x}{e^{4}}+\frac {d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(122\) |
default | \(-\frac {-\frac {c^{2} x^{4} e^{3}}{4}+\frac {\left (-\left (b e -c d \right ) c \,e^{2}-e^{3} b c \right ) x^{3}}{3}+\frac {\left (-\left (b e -c d \right ) b \,e^{2}+c e \left (b d e -c \,d^{2}\right )\right ) x^{2}}{2}+\left (b e -c d \right ) \left (b d e -c \,d^{2}\right ) x}{e^{4}}+\frac {d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(135\) |
risch | \(\frac {c^{2} x^{4}}{4 e}+\frac {2 x^{3} b c}{3 e}-\frac {c^{2} d \,x^{3}}{3 e^{2}}+\frac {x^{2} b^{2}}{2 e}-\frac {x^{2} d b c}{e^{2}}+\frac {x^{2} d^{2} c^{2}}{2 e^{3}}-\frac {b^{2} d x}{e^{2}}+\frac {2 b c \,d^{2} x}{e^{3}}-\frac {c^{2} d^{3} x}{e^{4}}+\frac {d^{2} \ln \left (e x +d \right ) b^{2}}{e^{3}}-\frac {2 d^{3} \ln \left (e x +d \right ) b c}{e^{4}}+\frac {d^{4} \ln \left (e x +d \right ) c^{2}}{e^{5}}\) | \(152\) |
parallelrisch | \(\frac {3 c^{2} x^{4} e^{4}+8 x^{3} b c \,e^{4}-4 d \,c^{2} x^{3} e^{3}+6 x^{2} b^{2} e^{4}-12 x^{2} b c d \,e^{3}+6 x^{2} c^{2} d^{2} e^{2}+12 \ln \left (e x +d \right ) b^{2} d^{2} e^{2}-24 \ln \left (e x +d \right ) b c \,d^{3} e +12 \ln \left (e x +d \right ) c^{2} d^{4}-12 x \,b^{2} d \,e^{3}+24 x b c \,d^{2} e^{2}-12 x \,c^{2} d^{3} e}{12 e^{5}}\) | \(152\) |
Input:
int((c*x^2+b*x)^2/(e*x+d),x,method=_RETURNVERBOSE)
Output:
1/4*c^2*x^4/e+1/2/e^3*(b^2*e^2-2*b*c*d*e+c^2*d^2)*x^2+1/3*c/e^2*(2*b*e-c*d )*x^3-d*(b^2*e^2-2*b*c*d*e+c^2*d^2)/e^4*x+d^2*(b^2*e^2-2*b*c*d*e+c^2*d^2)/ e^5*ln(e*x+d)
Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.43 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {3 \, c^{2} e^{4} x^{4} - 4 \, {\left (c^{2} d e^{3} - 2 \, b c e^{4}\right )} x^{3} + 6 \, {\left (c^{2} d^{2} e^{2} - 2 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} - 12 \, {\left (c^{2} d^{3} e - 2 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x + 12 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \log \left (e x + d\right )}{12 \, e^{5}} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d),x, algorithm="fricas")
Output:
1/12*(3*c^2*e^4*x^4 - 4*(c^2*d*e^3 - 2*b*c*e^4)*x^3 + 6*(c^2*d^2*e^2 - 2*b *c*d*e^3 + b^2*e^4)*x^2 - 12*(c^2*d^3*e - 2*b*c*d^2*e^2 + b^2*d*e^3)*x + 1 2*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*log(e*x + d))/e^5
Time = 0.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.25 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {c^{2} x^{4}}{4 e} + \frac {d^{2} \left (b e - c d\right )^{2} \log {\left (d + e x \right )}}{e^{5}} + x^{3} \cdot \left (\frac {2 b c}{3 e} - \frac {c^{2} d}{3 e^{2}}\right ) + x^{2} \left (\frac {b^{2}}{2 e} - \frac {b c d}{e^{2}} + \frac {c^{2} d^{2}}{2 e^{3}}\right ) + x \left (- \frac {b^{2} d}{e^{2}} + \frac {2 b c d^{2}}{e^{3}} - \frac {c^{2} d^{3}}{e^{4}}\right ) \] Input:
integrate((c*x**2+b*x)**2/(e*x+d),x)
Output:
c**2*x**4/(4*e) + d**2*(b*e - c*d)**2*log(d + e*x)/e**5 + x**3*(2*b*c/(3*e ) - c**2*d/(3*e**2)) + x**2*(b**2/(2*e) - b*c*d/e**2 + c**2*d**2/(2*e**3)) + x*(-b**2*d/e**2 + 2*b*c*d**2/e**3 - c**2*d**3/e**4)
Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.41 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {3 \, c^{2} e^{3} x^{4} - 4 \, {\left (c^{2} d e^{2} - 2 \, b c e^{3}\right )} x^{3} + 6 \, {\left (c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3}\right )} x^{2} - 12 \, {\left (c^{2} d^{3} - 2 \, b c d^{2} e + b^{2} d e^{2}\right )} x}{12 \, e^{4}} + \frac {{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d),x, algorithm="maxima")
Output:
1/12*(3*c^2*e^3*x^4 - 4*(c^2*d*e^2 - 2*b*c*e^3)*x^3 + 6*(c^2*d^2*e - 2*b*c *d*e^2 + b^2*e^3)*x^2 - 12*(c^2*d^3 - 2*b*c*d^2*e + b^2*d*e^2)*x)/e^4 + (c ^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*log(e*x + d)/e^5
Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.49 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {3 \, c^{2} e^{3} x^{4} - 4 \, c^{2} d e^{2} x^{3} + 8 \, b c e^{3} x^{3} + 6 \, c^{2} d^{2} e x^{2} - 12 \, b c d e^{2} x^{2} + 6 \, b^{2} e^{3} x^{2} - 12 \, c^{2} d^{3} x + 24 \, b c d^{2} e x - 12 \, b^{2} d e^{2} x}{12 \, e^{4}} + \frac {{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d),x, algorithm="giac")
Output:
1/12*(3*c^2*e^3*x^4 - 4*c^2*d*e^2*x^3 + 8*b*c*e^3*x^3 + 6*c^2*d^2*e*x^2 - 12*b*c*d*e^2*x^2 + 6*b^2*e^3*x^2 - 12*c^2*d^3*x + 24*b*c*d^2*e*x - 12*b^2* d*e^2*x)/e^4 + (c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*log(abs(e*x + d))/e^5
Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.52 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=x^2\,\left (\frac {b^2}{2\,e}+\frac {d\,\left (\frac {c^2\,d}{e^2}-\frac {2\,b\,c}{e}\right )}{2\,e}\right )-x^3\,\left (\frac {c^2\,d}{3\,e^2}-\frac {2\,b\,c}{3\,e}\right )+\frac {c^2\,x^4}{4\,e}+\frac {\ln \left (d+e\,x\right )\,\left (b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4\right )}{e^5}-\frac {d\,x\,\left (\frac {b^2}{e}+\frac {d\,\left (\frac {c^2\,d}{e^2}-\frac {2\,b\,c}{e}\right )}{e}\right )}{e} \] Input:
int((b*x + c*x^2)^2/(d + e*x),x)
Output:
x^2*(b^2/(2*e) + (d*((c^2*d)/e^2 - (2*b*c)/e))/(2*e)) - x^3*((c^2*d)/(3*e^ 2) - (2*b*c)/(3*e)) + (c^2*x^4)/(4*e) + (log(d + e*x)*(c^2*d^4 + b^2*d^2*e ^2 - 2*b*c*d^3*e))/e^5 - (d*x*(b^2/e + (d*((c^2*d)/e^2 - (2*b*c)/e))/e))/e
Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.62 \[ \int \frac {\left (b x+c x^2\right )^2}{d+e x} \, dx=\frac {12 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{2} e^{2}-24 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e +12 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{4}-12 b^{2} d \,e^{3} x +6 b^{2} e^{4} x^{2}+24 b c \,d^{2} e^{2} x -12 b c d \,e^{3} x^{2}+8 b c \,e^{4} x^{3}-12 c^{2} d^{3} e x +6 c^{2} d^{2} e^{2} x^{2}-4 c^{2} d \,e^{3} x^{3}+3 c^{2} e^{4} x^{4}}{12 e^{5}} \] Input:
int((c*x^2+b*x)^2/(e*x+d),x)
Output:
(12*log(d + e*x)*b**2*d**2*e**2 - 24*log(d + e*x)*b*c*d**3*e + 12*log(d + e*x)*c**2*d**4 - 12*b**2*d*e**3*x + 6*b**2*e**4*x**2 + 24*b*c*d**2*e**2*x - 12*b*c*d*e**3*x**2 + 8*b*c*e**4*x**3 - 12*c**2*d**3*e*x + 6*c**2*d**2*e* *2*x**2 - 4*c**2*d*e**3*x**3 + 3*c**2*e**4*x**4)/(12*e**5)