Integrand size = 20, antiderivative size = 91 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {(A b-a B) e (b d-a e) x}{b^3}+\frac {(A b-a B) (d+e x)^2}{2 b^2}+\frac {B (d+e x)^3}{3 b e}+\frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4} \] Output:
(A*b-B*a)*e*(-a*e+b*d)*x/b^3+1/2*(A*b-B*a)*(e*x+d)^2/b^2+1/3*B*(e*x+d)^3/b /e+(A*b-B*a)*(-a*e+b*d)^2*ln(b*x+a)/b^4
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {b x \left (6 a^2 B e^2-3 a b e (4 B d+2 A e+B e x)+b^2 \left (3 A e (4 d+e x)+2 B \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+6 (A b-a B) (b d-a e)^2 \log (a+b x)}{6 b^4} \] Input:
Integrate[((A + B*x)*(d + e*x)^2)/(a + b*x),x]
Output:
(b*x*(6*a^2*B*e^2 - 3*a*b*e*(4*B*d + 2*A*e + B*e*x) + b^2*(3*A*e*(4*d + e* x) + 2*B*(3*d^2 + 3*d*e*x + e^2*x^2))) + 6*(A*b - a*B)*(b*d - a*e)^2*Log[a + b*x])/(6*b^4)
Time = 0.23 (sec) , antiderivative size = 91, 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 = {86, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(A b-a B) (b d-a e)^2}{b^3 (a+b x)}+\frac {e (A b-a B) (b d-a e)}{b^3}+\frac {e (d+e x) (A b-a B)}{b^2}+\frac {B (d+e x)^2}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4}+\frac {e x (A b-a B) (b d-a e)}{b^3}+\frac {(d+e x)^2 (A b-a B)}{2 b^2}+\frac {B (d+e x)^3}{3 b e}\) |
Input:
Int[((A + B*x)*(d + e*x)^2)/(a + b*x),x]
Output:
((A*b - a*B)*e*(b*d - a*e)*x)/b^3 + ((A*b - a*B)*(d + e*x)^2)/(2*b^2) + (B *(d + e*x)^3)/(3*b*e) + ((A*b - a*B)*(b*d - a*e)^2*Log[a + b*x])/b^4
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.65
method | result | size |
norman | \(-\frac {\left (A a b \,e^{2}-2 A \,b^{2} d e -B \,a^{2} e^{2}+2 B a b d e -b^{2} B \,d^{2}\right ) x}{b^{3}}+\frac {B \,e^{2} x^{3}}{3 b}+\frac {e \left (A b e -B a e +2 B b d \right ) x^{2}}{2 b^{2}}+\frac {\left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(150\) |
default | \(-\frac {-\frac {1}{3} b^{2} B \,x^{3} e^{2}-\frac {1}{2} A \,b^{2} e^{2} x^{2}+\frac {1}{2} B a b \,e^{2} x^{2}-B \,b^{2} d e \,x^{2}+A a b \,e^{2} x -2 A \,b^{2} d e x -B \,a^{2} e^{2} x +2 B a b d e x -b^{2} B \,d^{2} x}{b^{3}}+\frac {\left (A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(164\) |
risch | \(\frac {B \,e^{2} x^{3}}{3 b}+\frac {A \,e^{2} x^{2}}{2 b}-\frac {B a \,e^{2} x^{2}}{2 b^{2}}+\frac {B d e \,x^{2}}{b}-\frac {A a \,e^{2} x}{b^{2}}+\frac {2 A d e x}{b}+\frac {B \,a^{2} e^{2} x}{b^{3}}-\frac {2 B a d e x}{b^{2}}+\frac {B \,d^{2} x}{b}+\frac {\ln \left (b x +a \right ) A \,a^{2} e^{2}}{b^{3}}-\frac {2 \ln \left (b x +a \right ) A a d e}{b^{2}}+\frac {\ln \left (b x +a \right ) A \,d^{2}}{b}-\frac {\ln \left (b x +a \right ) B \,a^{3} e^{2}}{b^{4}}+\frac {2 \ln \left (b x +a \right ) B \,a^{2} d e}{b^{3}}-\frac {\ln \left (b x +a \right ) B a \,d^{2}}{b^{2}}\) | \(197\) |
parallelrisch | \(\frac {2 B \,e^{2} x^{3} b^{3}+3 A \,x^{2} b^{3} e^{2}-3 B \,x^{2} a \,b^{2} e^{2}+6 B \,x^{2} b^{3} d e +6 A \ln \left (b x +a \right ) a^{2} b \,e^{2}-12 A \ln \left (b x +a \right ) a \,b^{2} d e +6 A \ln \left (b x +a \right ) b^{3} d^{2}-6 A x a \,b^{2} e^{2}+12 A x \,b^{3} d e -6 B \ln \left (b x +a \right ) a^{3} e^{2}+12 B \ln \left (b x +a \right ) a^{2} b d e -6 B \ln \left (b x +a \right ) a \,b^{2} d^{2}+6 B x \,a^{2} b \,e^{2}-12 B x a \,b^{2} d e +6 B x \,b^{3} d^{2}}{6 b^{4}}\) | \(198\) |
Input:
int((B*x+A)*(e*x+d)^2/(b*x+a),x,method=_RETURNVERBOSE)
Output:
-(A*a*b*e^2-2*A*b^2*d*e-B*a^2*e^2+2*B*a*b*d*e-B*b^2*d^2)/b^3*x+1/3*B/b*e^2 *x^3+1/2/b^2*e*(A*b*e-B*a*e+2*B*b*d)*x^2+(A*a^2*b*e^2-2*A*a*b^2*d*e+A*b^3* d^2-B*a^3*e^2+2*B*a^2*b*d*e-B*a*b^2*d^2)/b^4*ln(b*x+a)
Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{3} e^{2} x^{3} + 3 \, {\left (2 \, B b^{3} d e - {\left (B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{3} d^{2} - 2 \, {\left (B a b^{2} - A b^{3}\right )} d e + {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x - 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a),x, algorithm="fricas")
Output:
1/6*(2*B*b^3*e^2*x^3 + 3*(2*B*b^3*d*e - (B*a*b^2 - A*b^3)*e^2)*x^2 + 6*(B* b^3*d^2 - 2*(B*a*b^2 - A*b^3)*d*e + (B*a^2*b - A*a*b^2)*e^2)*x - 6*((B*a*b ^2 - A*b^3)*d^2 - 2*(B*a^2*b - A*a*b^2)*d*e + (B*a^3 - A*a^2*b)*e^2)*log(b *x + a))/b^4
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {B e^{2} x^{3}}{3 b} + x^{2} \left (\frac {A e^{2}}{2 b} - \frac {B a e^{2}}{2 b^{2}} + \frac {B d e}{b}\right ) + x \left (- \frac {A a e^{2}}{b^{2}} + \frac {2 A d e}{b} + \frac {B a^{2} e^{2}}{b^{3}} - \frac {2 B a d e}{b^{2}} + \frac {B d^{2}}{b}\right ) - \frac {\left (- A b + B a\right ) \left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)**2/(b*x+a),x)
Output:
B*e**2*x**3/(3*b) + x**2*(A*e**2/(2*b) - B*a*e**2/(2*b**2) + B*d*e/b) + x* (-A*a*e**2/b**2 + 2*A*d*e/b + B*a**2*e**2/b**3 - 2*B*a*d*e/b**2 + B*d**2/b ) - (-A*b + B*a)*(a*e - b*d)**2*log(a + b*x)/b**4
Time = 0.03 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} + 3 \, {\left (2 \, B b^{2} d e - {\left (B a b - A b^{2}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} - 2 \, {\left (B a b - A b^{2}\right )} d e + {\left (B a^{2} - A a b\right )} e^{2}\right )} x}{6 \, b^{3}} - \frac {{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a),x, algorithm="maxima")
Output:
1/6*(2*B*b^2*e^2*x^3 + 3*(2*B*b^2*d*e - (B*a*b - A*b^2)*e^2)*x^2 + 6*(B*b^ 2*d^2 - 2*(B*a*b - A*b^2)*d*e + (B*a^2 - A*a*b)*e^2)*x)/b^3 - ((B*a*b^2 - A*b^3)*d^2 - 2*(B*a^2*b - A*a*b^2)*d*e + (B*a^3 - A*a^2*b)*e^2)*log(b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {2 \, B b^{2} e^{2} x^{3} + 6 \, B b^{2} d e x^{2} - 3 \, B a b e^{2} x^{2} + 3 \, A b^{2} e^{2} x^{2} + 6 \, B b^{2} d^{2} x - 12 \, B a b d e x + 12 \, A b^{2} d e x + 6 \, B a^{2} e^{2} x - 6 \, A a b e^{2} x}{6 \, b^{3}} - \frac {{\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a),x, algorithm="giac")
Output:
1/6*(2*B*b^2*e^2*x^3 + 6*B*b^2*d*e*x^2 - 3*B*a*b*e^2*x^2 + 3*A*b^2*e^2*x^2 + 6*B*b^2*d^2*x - 12*B*a*b*d*e*x + 12*A*b^2*d*e*x + 6*B*a^2*e^2*x - 6*A*a *b*e^2*x)/b^3 - (B*a*b^2*d^2 - A*b^3*d^2 - 2*B*a^2*b*d*e + 2*A*a*b^2*d*e + B*a^3*e^2 - A*a^2*b*e^2)*log(abs(b*x + a))/b^4
Time = 0.97 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=x\,\left (\frac {B\,d^2+2\,A\,e\,d}{b}-\frac {a\,\left (\frac {A\,e^2+2\,B\,d\,e}{b}-\frac {B\,a\,e^2}{b^2}\right )}{b}\right )+x^2\,\left (\frac {A\,e^2+2\,B\,d\,e}{2\,b}-\frac {B\,a\,e^2}{2\,b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2\right )}{b^4}+\frac {B\,e^2\,x^3}{3\,b} \] Input:
int(((A + B*x)*(d + e*x)^2)/(a + b*x),x)
Output:
x*((B*d^2 + 2*A*d*e)/b - (a*((A*e^2 + 2*B*d*e)/b - (B*a*e^2)/b^2))/b) + x^ 2*((A*e^2 + 2*B*d*e)/(2*b) - (B*a*e^2)/(2*b^2)) + (log(a + b*x)*(A*b^3*d^2 - B*a^3*e^2 + A*a^2*b*e^2 - B*a*b^2*d^2 - 2*A*a*b^2*d*e + 2*B*a^2*b*d*e)) /b^4 + (B*e^2*x^3)/(3*b)
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.23 \[ \int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx=\frac {x \left (e^{2} x^{2}+3 d e x +3 d^{2}\right )}{3} \] Input:
int((B*x+A)*(e*x+d)^2/(b*x+a),x)
Output:
(x*(3*d**2 + 3*d*e*x + e**2*x**2))/3