Integrand size = 18, antiderivative size = 82 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\frac {\left (c d^2-b d e+a e^2\right ) (d+e x)^{1+m}}{e^3 (1+m)}-\frac {(2 c d-b e) (d+e x)^{2+m}}{e^3 (2+m)}+\frac {c (d+e x)^{3+m}}{e^3 (3+m)} \]
(a*e^2-b*d*e+c*d^2)*(e*x+d)^(1+m)/e^3/(1+m)-(-b*e+2*c*d)*(e*x+d)^(2+m)/e^3 /(2+m)+c*(e*x+d)^(3+m)/e^3/(3+m)
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\frac {\left (c d^2-e (b d-a e)\right ) (d+e x)^{1+m}}{e^3 (1+m)}-\frac {(2 c d-b e) (d+e x)^{2+m}}{e^3 (2+m)}+\frac {c (d+e x)^{3+m}}{e^3 (3+m)} \]
((c*d^2 - e*(b*d - a*e))*(d + e*x)^(1 + m))/(e^3*(1 + m)) - ((2*c*d - b*e) *(d + e*x)^(2 + m))/(e^3*(2 + m)) + (c*(d + e*x)^(3 + m))/(e^3*(3 + m))
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int \left (a+b x+c x^2\right ) (d+e x)^m \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (\frac {(d+e x)^m \left (a e^2-b d e+c d^2\right )}{e^2}+\frac {(b e-2 c d) (d+e x)^{m+1}}{e^2}+\frac {c (d+e x)^{m+2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^{m+1} \left (a e^2-b d e+c d^2\right )}{e^3 (m+1)}-\frac {(2 c d-b e) (d+e x)^{m+2}}{e^3 (m+2)}+\frac {c (d+e x)^{m+3}}{e^3 (m+3)}\) |
((c*d^2 - b*d*e + a*e^2)*(d + e*x)^(1 + m))/(e^3*(1 + m)) - ((2*c*d - b*e) *(d + e*x)^(2 + m))/(e^3*(2 + m)) + (c*(d + e*x)^(3 + m))/(e^3*(3 + m))
3.26.52.3.1 Defintions of rubi rules used
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 = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.65
method | result | size |
gosper | \(\frac {\left (e x +d \right )^{1+m} \left (c \,e^{2} m^{2} x^{2}+b \,e^{2} m^{2} x +3 c \,e^{2} m \,x^{2}+a \,e^{2} m^{2}+4 b \,e^{2} m x -2 c d e m x +2 c \,e^{2} x^{2}+5 a \,e^{2} m -b d e m +3 b \,e^{2} x -2 c d e x +6 a \,e^{2}-3 b d e +2 c \,d^{2}\right )}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}\) | \(135\) |
norman | \(\frac {c \,x^{3} {\mathrm e}^{m \ln \left (e x +d \right )}}{3+m}+\frac {d \left (a \,e^{2} m^{2}+5 a \,e^{2} m -b d e m +6 a \,e^{2}-3 b d e +2 c \,d^{2}\right ) {\mathrm e}^{m \ln \left (e x +d \right )}}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (b e m +c d m +3 b e \right ) x^{2} {\mathrm e}^{m \ln \left (e x +d \right )}}{e \left (m^{2}+5 m +6\right )}+\frac {\left (a \,e^{2} m^{2}+b d e \,m^{2}+5 a \,e^{2} m +3 b d e m -2 c \,d^{2} m +6 a \,e^{2}\right ) x \,{\mathrm e}^{m \ln \left (e x +d \right )}}{e^{2} \left (m^{3}+6 m^{2}+11 m +6\right )}\) | \(199\) |
risch | \(\frac {\left (c \,e^{3} m^{2} x^{3}+b \,e^{3} m^{2} x^{2}+c d \,e^{2} m^{2} x^{2}+3 c \,e^{3} m \,x^{3}+a \,e^{3} m^{2} x +b d \,e^{2} m^{2} x +4 b \,e^{3} m \,x^{2}+c d \,e^{2} m \,x^{2}+2 c \,x^{3} e^{3}+a d \,e^{2} m^{2}+5 a \,e^{3} m x +3 b d \,e^{2} m x +3 b \,e^{3} x^{2}-2 c \,d^{2} e m x +5 a d \,e^{2} m +6 e^{3} x a -b \,d^{2} e m +6 a d \,e^{2}-3 b \,d^{2} e +2 c \,d^{3}\right ) \left (e x +d \right )^{m}}{\left (2+m \right ) \left (3+m \right ) \left (1+m \right ) e^{3}}\) | \(207\) |
parallelrisch | \(\frac {x^{3} \left (e x +d \right )^{m} c \,e^{3} m^{2}+3 x^{3} \left (e x +d \right )^{m} c \,e^{3} m +x^{2} \left (e x +d \right )^{m} b \,e^{3} m^{2}+x^{2} \left (e x +d \right )^{m} c d \,e^{2} m^{2}+2 x^{3} \left (e x +d \right )^{m} c \,e^{3}+4 x^{2} \left (e x +d \right )^{m} b \,e^{3} m +x^{2} \left (e x +d \right )^{m} c d \,e^{2} m +x \left (e x +d \right )^{m} a \,e^{3} m^{2}+x \left (e x +d \right )^{m} b d \,e^{2} m^{2}+3 x^{2} \left (e x +d \right )^{m} b \,e^{3}+5 x \left (e x +d \right )^{m} a \,e^{3} m +3 x \left (e x +d \right )^{m} b d \,e^{2} m -2 x \left (e x +d \right )^{m} c \,d^{2} e m +\left (e x +d \right )^{m} a d \,e^{2} m^{2}+6 x \left (e x +d \right )^{m} a \,e^{3}+5 \left (e x +d \right )^{m} a d \,e^{2} m -\left (e x +d \right )^{m} b \,d^{2} e m +6 \left (e x +d \right )^{m} a d \,e^{2}-3 \left (e x +d \right )^{m} b \,d^{2} e +2 \left (e x +d \right )^{m} c \,d^{3}}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}\) | \(340\) |
1/e^3*(e*x+d)^(1+m)/(m^3+6*m^2+11*m+6)*(c*e^2*m^2*x^2+b*e^2*m^2*x+3*c*e^2* m*x^2+a*e^2*m^2+4*b*e^2*m*x-2*c*d*e*m*x+2*c*e^2*x^2+5*a*e^2*m-b*d*e*m+3*b* e^2*x-2*c*d*e*x+6*a*e^2-3*b*d*e+2*c*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.45 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\frac {{\left (a d e^{2} m^{2} + 2 \, c d^{3} - 3 \, b d^{2} e + 6 \, a d e^{2} + {\left (c e^{3} m^{2} + 3 \, c e^{3} m + 2 \, c e^{3}\right )} x^{3} + {\left (3 \, b e^{3} + {\left (c d e^{2} + b e^{3}\right )} m^{2} + {\left (c d e^{2} + 4 \, b e^{3}\right )} m\right )} x^{2} - {\left (b d^{2} e - 5 \, a d e^{2}\right )} m + {\left (6 \, a e^{3} + {\left (b d e^{2} + a e^{3}\right )} m^{2} - {\left (2 \, c d^{2} e - 3 \, b d e^{2} - 5 \, a e^{3}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \]
(a*d*e^2*m^2 + 2*c*d^3 - 3*b*d^2*e + 6*a*d*e^2 + (c*e^3*m^2 + 3*c*e^3*m + 2*c*e^3)*x^3 + (3*b*e^3 + (c*d*e^2 + b*e^3)*m^2 + (c*d*e^2 + 4*b*e^3)*m)*x ^2 - (b*d^2*e - 5*a*d*e^2)*m + (6*a*e^3 + (b*d*e^2 + a*e^3)*m^2 - (2*c*d^2 *e - 3*b*d*e^2 - 5*a*e^3)*m)*x)*(e*x + d)^m/(e^3*m^3 + 6*e^3*m^2 + 11*e^3* m + 6*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 1416 vs. \(2 (70) = 140\).
Time = 0.61 (sec) , antiderivative size = 1416, normalized size of antiderivative = 17.27 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\text {Too large to display} \]
Piecewise((d**m*(a*x + b*x**2/2 + c*x**3/3), Eq(e, 0)), (-a*e**2/(2*d**2*e **3 + 4*d*e**4*x + 2*e**5*x**2) - b*d*e/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5 *x**2) - 2*b*e**2*x/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 2*c*d**2*lo g(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 3*c*d**2/(2*d**2*e** 3 + 4*d*e**4*x + 2*e**5*x**2) + 4*c*d*e*x*log(d/e + x)/(2*d**2*e**3 + 4*d* e**4*x + 2*e**5*x**2) + 4*c*d*e*x/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 2*c*e**2*x**2*log(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2), Eq (m, -3)), (-a*e**2/(d*e**3 + e**4*x) + b*d*e*log(d/e + x)/(d*e**3 + e**4*x ) + b*d*e/(d*e**3 + e**4*x) + b*e**2*x*log(d/e + x)/(d*e**3 + e**4*x) - 2* c*d**2*log(d/e + x)/(d*e**3 + e**4*x) - 2*c*d**2/(d*e**3 + e**4*x) - 2*c*d *e*x*log(d/e + x)/(d*e**3 + e**4*x) + c*e**2*x**2/(d*e**3 + e**4*x), Eq(m, -2)), (a*log(d/e + x)/e - b*d*log(d/e + x)/e**2 + b*x/e + c*d**2*log(d/e + x)/e**3 - c*d*x/e**2 + c*x**2/(2*e), Eq(m, -1)), (a*d*e**2*m**2*(d + e*x )**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 5*a*d*e**2*m*(d + e* x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 6*a*d*e**2*(d + e*x )**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + a*e**3*m**2*x*(d + e *x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 5*a*e**3*m*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 6*a*e**3*x*(d + e *x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) - b*d**2*e*m*(d + e* x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) - 3*b*d**2*e*(d + ...
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.61 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\frac {{\left (e^{2} {\left (m + 1\right )} x^{2} + d e m x - d^{2}\right )} {\left (e x + d\right )}^{m} b}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} + \frac {{\left (e x + d\right )}^{m + 1} a}{e {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} e^{3} x^{3} + {\left (m^{2} + m\right )} d e^{2} x^{2} - 2 \, d^{2} e m x + 2 \, d^{3}\right )} {\left (e x + d\right )}^{m} c}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} e^{3}} \]
(e^2*(m + 1)*x^2 + d*e*m*x - d^2)*(e*x + d)^m*b/((m^2 + 3*m + 2)*e^2) + (e *x + d)^(m + 1)*a/(e*(m + 1)) + ((m^2 + 3*m + 2)*e^3*x^3 + (m^2 + m)*d*e^2 *x^2 - 2*d^2*e*m*x + 2*d^3)*(e*x + d)^m*c/((m^3 + 6*m^2 + 11*m + 6)*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (82) = 164\).
Time = 0.26 (sec) , antiderivative size = 350, normalized size of antiderivative = 4.27 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx=\frac {{\left (e x + d\right )}^{m} c e^{3} m^{2} x^{3} + {\left (e x + d\right )}^{m} c d e^{2} m^{2} x^{2} + {\left (e x + d\right )}^{m} b e^{3} m^{2} x^{2} + 3 \, {\left (e x + d\right )}^{m} c e^{3} m x^{3} + {\left (e x + d\right )}^{m} b d e^{2} m^{2} x + {\left (e x + d\right )}^{m} a e^{3} m^{2} x + {\left (e x + d\right )}^{m} c d e^{2} m x^{2} + 4 \, {\left (e x + d\right )}^{m} b e^{3} m x^{2} + 2 \, {\left (e x + d\right )}^{m} c e^{3} x^{3} + {\left (e x + d\right )}^{m} a d e^{2} m^{2} - 2 \, {\left (e x + d\right )}^{m} c d^{2} e m x + 3 \, {\left (e x + d\right )}^{m} b d e^{2} m x + 5 \, {\left (e x + d\right )}^{m} a e^{3} m x + 3 \, {\left (e x + d\right )}^{m} b e^{3} x^{2} - {\left (e x + d\right )}^{m} b d^{2} e m + 5 \, {\left (e x + d\right )}^{m} a d e^{2} m + 6 \, {\left (e x + d\right )}^{m} a e^{3} x + 2 \, {\left (e x + d\right )}^{m} c d^{3} - 3 \, {\left (e x + d\right )}^{m} b d^{2} e + 6 \, {\left (e x + d\right )}^{m} a d e^{2}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \]
((e*x + d)^m*c*e^3*m^2*x^3 + (e*x + d)^m*c*d*e^2*m^2*x^2 + (e*x + d)^m*b*e ^3*m^2*x^2 + 3*(e*x + d)^m*c*e^3*m*x^3 + (e*x + d)^m*b*d*e^2*m^2*x + (e*x + d)^m*a*e^3*m^2*x + (e*x + d)^m*c*d*e^2*m*x^2 + 4*(e*x + d)^m*b*e^3*m*x^2 + 2*(e*x + d)^m*c*e^3*x^3 + (e*x + d)^m*a*d*e^2*m^2 - 2*(e*x + d)^m*c*d^2 *e*m*x + 3*(e*x + d)^m*b*d*e^2*m*x + 5*(e*x + d)^m*a*e^3*m*x + 3*(e*x + d) ^m*b*e^3*x^2 - (e*x + d)^m*b*d^2*e*m + 5*(e*x + d)^m*a*d*e^2*m + 6*(e*x + d)^m*a*e^3*x + 2*(e*x + d)^m*c*d^3 - 3*(e*x + d)^m*b*d^2*e + 6*(e*x + d)^m *a*d*e^2)/(e^3*m^3 + 6*e^3*m^2 + 11*e^3*m + 6*e^3)
Time = 10.07 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.45 \[ \int (d+e x)^m \left (a+b x+c x^2\right ) \, dx={\left (d+e\,x\right )}^m\,\left (\frac {c\,x^3\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}+\frac {d\,\left (2\,c\,d^2-b\,d\,e\,m-3\,b\,d\,e+a\,e^2\,m^2+5\,a\,e^2\,m+6\,a\,e^2\right )}{e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {x\,\left (-2\,c\,d^2\,e\,m+b\,d\,e^2\,m^2+3\,b\,d\,e^2\,m+a\,e^3\,m^2+5\,a\,e^3\,m+6\,a\,e^3\right )}{e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {x^2\,\left (m+1\right )\,\left (3\,b\,e+b\,e\,m+c\,d\,m\right )}{e\,\left (m^3+6\,m^2+11\,m+6\right )}\right ) \]
(d + e*x)^m*((c*x^3*(3*m + m^2 + 2))/(11*m + 6*m^2 + m^3 + 6) + (d*(6*a*e^ 2 + 2*c*d^2 + a*e^2*m^2 - 3*b*d*e + 5*a*e^2*m - b*d*e*m))/(e^3*(11*m + 6*m ^2 + m^3 + 6)) + (x*(6*a*e^3 + a*e^3*m^2 + 5*a*e^3*m + 3*b*d*e^2*m - 2*c*d ^2*e*m + b*d*e^2*m^2))/(e^3*(11*m + 6*m^2 + m^3 + 6)) + (x^2*(m + 1)*(3*b* e + b*e*m + c*d*m))/(e*(11*m + 6*m^2 + m^3 + 6)))