Integrand size = 29, antiderivative size = 74 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {b (b d-a e)^2 x}{e^3}-\frac {(b d-a e) (a+b x)^2}{2 e^2}+\frac {(a+b x)^3}{3 e}-\frac {(b d-a e)^3 \log (d+e x)}{e^4} \] Output:
b*(-a*e+b*d)^2*x/e^3-1/2*(-a*e+b*d)*(b*x+a)^2/e^2+1/3*(b*x+a)^3/e-(-a*e+b* d)^3*ln(e*x+d)/e^4
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {b e x \left (18 a^2 e^2+9 a b e (-2 d+e x)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (b d-a e)^3 \log (d+e x)}{6 e^4} \] Input:
Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x),x]
Output:
(b*e*x*(18*a^2*e^2 + 9*a*b*e*(-2*d + e*x) + b^2*(6*d^2 - 3*d*e*x + 2*e^2*x ^2)) - 6*(b*d - a*e)^3*Log[d + e*x])/(6*e^4)
Time = 0.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 49, 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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{d+e x}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{d+e x}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (\frac {(a e-b d)^3}{e^3 (d+e x)}+\frac {b (b d-a e)^2}{e^3}-\frac {b (a+b x) (b d-a e)}{e^2}+\frac {b (a+b x)^2}{e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(b d-a e)^3 \log (d+e x)}{e^4}+\frac {b x (b d-a e)^2}{e^3}-\frac {(a+b x)^2 (b d-a e)}{2 e^2}+\frac {(a+b x)^3}{3 e}\) |
Input:
Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x),x]
Output:
(b*(b*d - a*e)^2*x)/e^3 - ((b*d - a*e)*(a + b*x)^2)/(2*e^2) + (a + b*x)^3/ (3*e) - ((b*d - a*e)^3*Log[d + e*x])/e^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.77 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.45
method | result | size |
norman | \(\frac {b \left (3 e^{2} a^{2}-3 a b d e +b^{2} d^{2}\right ) x}{e^{3}}+\frac {b^{3} x^{3}}{3 e}+\frac {b^{2} \left (3 a e -b d \right ) x^{2}}{2 e^{2}}+\frac {\left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(107\) |
default | \(\frac {b \left (\frac {1}{3} b^{2} e^{2} x^{3}+\frac {3}{2} x^{2} a b \,e^{2}-\frac {1}{2} b^{2} d e \,x^{2}+3 a^{2} e^{2} x -3 a b d e x +b^{2} d^{2} x \right )}{e^{3}}+\frac {\left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(109\) |
risch | \(\frac {b^{3} x^{3}}{3 e}+\frac {3 b^{2} x^{2} a}{2 e}-\frac {b^{3} d \,x^{2}}{2 e^{2}}+\frac {3 b \,a^{2} x}{e}-\frac {3 b^{2} a d x}{e^{2}}+\frac {b^{3} d^{2} x}{e^{3}}+\frac {\ln \left (e x +d \right ) a^{3}}{e}-\frac {3 \ln \left (e x +d \right ) a^{2} b d}{e^{2}}+\frac {3 \ln \left (e x +d \right ) a \,b^{2} d^{2}}{e^{3}}-\frac {\ln \left (e x +d \right ) b^{3} d^{3}}{e^{4}}\) | \(133\) |
parallelrisch | \(\frac {2 e^{3} x^{3} b^{3}+9 x^{2} a \,b^{2} e^{3}-3 x^{2} b^{3} d \,e^{2}+6 \ln \left (e x +d \right ) a^{3} e^{3}-18 \ln \left (e x +d \right ) a^{2} b d \,e^{2}+18 \ln \left (e x +d \right ) a \,b^{2} d^{2} e -6 \ln \left (e x +d \right ) b^{3} d^{3}+18 a^{2} b \,e^{3} x -18 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x}{6 e^{4}}\) | \(133\) |
Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
b*(3*a^2*e^2-3*a*b*d*e+b^2*d^2)/e^3*x+1/3*b^3/e*x^3+1/2/e^2*b^2*(3*a*e-b*d )*x^2+(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/e^4*ln(e*x+d)
Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {2 \, b^{3} e^{3} x^{3} - 3 \, {\left (b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d),x, algorithm="fricas")
Output:
1/6*(2*b^3*e^3*x^3 - 3*(b^3*d*e^2 - 3*a*b^2*e^3)*x^2 + 6*(b^3*d^2*e - 3*a* b^2*d*e^2 + 3*a^2*b*e^3)*x - 6*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*log(e*x + d))/e^4
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {b^{3} x^{3}}{3 e} + x^{2} \cdot \left (\frac {3 a b^{2}}{2 e} - \frac {b^{3} d}{2 e^{2}}\right ) + x \left (\frac {3 a^{2} b}{e} - \frac {3 a b^{2} d}{e^{2}} + \frac {b^{3} d^{2}}{e^{3}}\right ) + \frac {\left (a e - b d\right )^{3} \log {\left (d + e x \right )}}{e^{4}} \] Input:
integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)/(e*x+d),x)
Output:
b**3*x**3/(3*e) + x**2*(3*a*b**2/(2*e) - b**3*d/(2*e**2)) + x*(3*a**2*b/e - 3*a*b**2*d/e**2 + b**3*d**2/e**3) + (a*e - b*d)**3*log(d + e*x)/e**4
Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {2 \, b^{3} e^{2} x^{3} - 3 \, {\left (b^{3} d e - 3 \, a b^{2} e^{2}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} - 3 \, a b^{2} d e + 3 \, a^{2} b e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d),x, algorithm="maxima")
Output:
1/6*(2*b^3*e^2*x^3 - 3*(b^3*d*e - 3*a*b^2*e^2)*x^2 + 6*(b^3*d^2 - 3*a*b^2* d*e + 3*a^2*b*e^2)*x)/e^3 - (b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3 *e^3)*log(e*x + d)/e^4
Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {2 \, b^{3} e^{2} x^{3} - 3 \, b^{3} d e x^{2} + 9 \, a b^{2} e^{2} x^{2} + 6 \, b^{3} d^{2} x - 18 \, a b^{2} d e x + 18 \, a^{2} b e^{2} x}{6 \, e^{3}} - \frac {{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d),x, algorithm="giac")
Output:
1/6*(2*b^3*e^2*x^3 - 3*b^3*d*e*x^2 + 9*a*b^2*e^2*x^2 + 6*b^3*d^2*x - 18*a* b^2*d*e*x + 18*a^2*b*e^2*x)/e^3 - (b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*log(abs(e*x + d))/e^4
Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=x^2\,\left (\frac {3\,a\,b^2}{2\,e}-\frac {b^3\,d}{2\,e^2}\right )+x\,\left (\frac {3\,a^2\,b}{e}-\frac {d\,\left (\frac {3\,a\,b^2}{e}-\frac {b^3\,d}{e^2}\right )}{e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{e^4}+\frac {b^3\,x^3}{3\,e} \] Input:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x))/(d + e*x),x)
Output:
x^2*((3*a*b^2)/(2*e) - (b^3*d)/(2*e^2)) + x*((3*a^2*b)/e - (d*((3*a*b^2)/e - (b^3*d)/e^2))/e) + (log(d + e*x)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3 *a^2*b*d*e^2))/e^4 + (b^3*x^3)/(3*e)
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{d+e x} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) a^{3} e^{3}-18 \,\mathrm {log}\left (e x +d \right ) a^{2} b d \,e^{2}+18 \,\mathrm {log}\left (e x +d \right ) a \,b^{2} d^{2} e -6 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{3}+18 a^{2} b \,e^{3} x -18 a \,b^{2} d \,e^{2} x +9 a \,b^{2} e^{3} x^{2}+6 b^{3} d^{2} e x -3 b^{3} d \,e^{2} x^{2}+2 b^{3} e^{3} x^{3}}{6 e^{4}} \] Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d),x)
Output:
(6*log(d + e*x)*a**3*e**3 - 18*log(d + e*x)*a**2*b*d*e**2 + 18*log(d + e*x )*a*b**2*d**2*e - 6*log(d + e*x)*b**3*d**3 + 18*a**2*b*e**3*x - 18*a*b**2* d*e**2*x + 9*a*b**2*e**3*x**2 + 6*b**3*d**2*e*x - 3*b**3*d*e**2*x**2 + 2*b **3*e**3*x**3)/(6*e**4)