Integrand size = 20, antiderivative size = 60 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {a c (e x)^{2+m}}{e^2 (2+m)}+\frac {(b c+a d) (e x)^{3+m}}{e^3 (3+m)}+\frac {b d (e x)^{4+m}}{e^4 (4+m)} \] Output:
a*c*(e*x)^(2+m)/e^2/(2+m)+(a*d+b*c)*(e*x)^(3+m)/e^3/(3+m)+b*d*(e*x)^(4+m)/ e^4/(4+m)
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {x^2 (e x)^m (a (4+m) (c (3+m)+d (2+m) x)+b (2+m) x (c (4+m)+d (3+m) x))}{(2+m) (3+m) (4+m)} \] Input:
Integrate[(e*x)^m*(c + d*x)*(a*x + b*x^2),x]
Output:
(x^2*(e*x)^m*(a*(4 + m)*(c*(3 + m) + d*(2 + m)*x) + b*(2 + m)*x*(c*(4 + m) + d*(3 + m)*x)))/((2 + m)*(3 + m)*(4 + m))
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {9, 85, 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.
\(\displaystyle \int \left (a x+b x^2\right ) (c+d x) (e x)^m \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {\int (e x)^{m+1} (a+b x) (c+d x)dx}{e}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\int \left (a c (e x)^{m+1}+\frac {(b c+a d) (e x)^{m+2}}{e}+\frac {b d (e x)^{m+3}}{e^2}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(e x)^{m+3} (a d+b c)}{e^2 (m+3)}+\frac {a c (e x)^{m+2}}{e (m+2)}+\frac {b d (e x)^{m+4}}{e^3 (m+4)}}{e}\) |
Input:
Int[(e*x)^m*(c + d*x)*(a*x + b*x^2),x]
Output:
((a*c*(e*x)^(2 + m))/(e*(2 + m)) + ((b*c + a*d)*(e*x)^(3 + m))/(e^2*(3 + m )) + (b*d*(e*x)^(4 + m))/(e^3*(4 + m)))/e
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02
method | result | size |
norman | \(\frac {\left (a d +b c \right ) x^{3} {\mathrm e}^{m \ln \left (e x \right )}}{3+m}+\frac {a c \,x^{2} {\mathrm e}^{m \ln \left (e x \right )}}{2+m}+\frac {b d \,x^{4} {\mathrm e}^{m \ln \left (e x \right )}}{4+m}\) | \(61\) |
gosper | \(\frac {\left (e x \right )^{m} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +5 b d m \,x^{2}+a c \,m^{2}+6 a d m x +6 b c m x +6 b d \,x^{2}+7 a c m +8 a d x +8 c b x +12 a c \right ) x^{2}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right )}\) | \(101\) |
risch | \(\frac {\left (e x \right )^{m} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +5 b d m \,x^{2}+a c \,m^{2}+6 a d m x +6 b c m x +6 b d \,x^{2}+7 a c m +8 a d x +8 c b x +12 a c \right ) x^{2}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right )}\) | \(101\) |
orering | \(\frac {\left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +5 b d m \,x^{2}+a c \,m^{2}+6 a d m x +6 b c m x +6 b d \,x^{2}+7 a c m +8 a d x +8 c b x +12 a c \right ) x \left (e x \right )^{m} \left (b \,x^{2}+a x \right )}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (b x +a \right )}\) | \(115\) |
parallelrisch | \(\frac {x^{4} \left (e x \right )^{m} b d \,m^{2}+5 x^{4} \left (e x \right )^{m} b d m +x^{3} \left (e x \right )^{m} a d \,m^{2}+x^{3} \left (e x \right )^{m} b c \,m^{2}+6 x^{4} \left (e x \right )^{m} b d +6 x^{3} \left (e x \right )^{m} a d m +6 x^{3} \left (e x \right )^{m} b c m +x^{2} \left (e x \right )^{m} a c \,m^{2}+8 x^{3} \left (e x \right )^{m} a d +8 x^{3} \left (e x \right )^{m} b c +7 x^{2} \left (e x \right )^{m} a c m +12 x^{2} \left (e x \right )^{m} a c}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right )}\) | \(174\) |
Input:
int((e*x)^m*(d*x+c)*(b*x^2+a*x),x,method=_RETURNVERBOSE)
Output:
(a*d+b*c)/(3+m)*x^3*exp(m*ln(e*x))+a*c/(2+m)*x^2*exp(m*ln(e*x))+b*d/(4+m)* x^4*exp(m*ln(e*x))
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {{\left ({\left (b d m^{2} + 5 \, b d m + 6 \, b d\right )} x^{4} + {\left ({\left (b c + a d\right )} m^{2} + 8 \, b c + 8 \, a d + 6 \, {\left (b c + a d\right )} m\right )} x^{3} + {\left (a c m^{2} + 7 \, a c m + 12 \, a c\right )} x^{2}\right )} \left (e x\right )^{m}}{m^{3} + 9 \, m^{2} + 26 \, m + 24} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a*x),x, algorithm="fricas")
Output:
((b*d*m^2 + 5*b*d*m + 6*b*d)*x^4 + ((b*c + a*d)*m^2 + 8*b*c + 8*a*d + 6*(b *c + a*d)*m)*x^3 + (a*c*m^2 + 7*a*c*m + 12*a*c)*x^2)*(e*x)^m/(m^3 + 9*m^2 + 26*m + 24)
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (53) = 106\).
Time = 0.29 (sec) , antiderivative size = 425, normalized size of antiderivative = 7.08 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\begin {cases} \frac {- \frac {a c}{2 x^{2}} - \frac {a d}{x} - \frac {b c}{x} + b d \log {\left (x \right )}}{e^{4}} & \text {for}\: m = -4 \\\frac {- \frac {a c}{x} + a d \log {\left (x \right )} + b c \log {\left (x \right )} + b d x}{e^{3}} & \text {for}\: m = -3 \\\frac {a c \log {\left (x \right )} + a d x + b c x + \frac {b d x^{2}}{2}}{e^{2}} & \text {for}\: m = -2 \\\frac {a c m^{2} x^{2} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {7 a c m x^{2} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {12 a c x^{2} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {a d m^{2} x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 a d m x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {8 a d x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {b c m^{2} x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 b c m x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {8 b c x^{3} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {b d m^{2} x^{4} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {5 b d m x^{4} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 b d x^{4} \left (e x\right )^{m}}{m^{3} + 9 m^{2} + 26 m + 24} & \text {otherwise} \end {cases} \] Input:
integrate((e*x)**m*(d*x+c)*(b*x**2+a*x),x)
Output:
Piecewise(((-a*c/(2*x**2) - a*d/x - b*c/x + b*d*log(x))/e**4, Eq(m, -4)), ((-a*c/x + a*d*log(x) + b*c*log(x) + b*d*x)/e**3, Eq(m, -3)), ((a*c*log(x) + a*d*x + b*c*x + b*d*x**2/2)/e**2, Eq(m, -2)), (a*c*m**2*x**2*(e*x)**m/( m**3 + 9*m**2 + 26*m + 24) + 7*a*c*m*x**2*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + 12*a*c*x**2*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + a*d*m**2*x**3*(e *x)**m/(m**3 + 9*m**2 + 26*m + 24) + 6*a*d*m*x**3*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + 8*a*d*x**3*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + b*c*m**2* x**3*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + 6*b*c*m*x**3*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + 8*b*c*x**3*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + b* d*m**2*x**4*(e*x)**m/(m**3 + 9*m**2 + 26*m + 24) + 5*b*d*m*x**4*(e*x)**m/( m**3 + 9*m**2 + 26*m + 24) + 6*b*d*x**4*(e*x)**m/(m**3 + 9*m**2 + 26*m + 2 4), True))
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {b d e^{m} x^{4} x^{m}}{m + 4} + \frac {b c e^{m} x^{3} x^{m}}{m + 3} + \frac {a d e^{m} x^{3} x^{m}}{m + 3} + \frac {a c e^{m} x^{2} x^{m}}{m + 2} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a*x),x, algorithm="maxima")
Output:
b*d*e^m*x^4*x^m/(m + 4) + b*c*e^m*x^3*x^m/(m + 3) + a*d*e^m*x^3*x^m/(m + 3 ) + a*c*e^m*x^2*x^m/(m + 2)
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (60) = 120\).
Time = 0.13 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.88 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {\left (e x\right )^{m} b d m^{2} x^{4} + \left (e x\right )^{m} b c m^{2} x^{3} + \left (e x\right )^{m} a d m^{2} x^{3} + 5 \, \left (e x\right )^{m} b d m x^{4} + \left (e x\right )^{m} a c m^{2} x^{2} + 6 \, \left (e x\right )^{m} b c m x^{3} + 6 \, \left (e x\right )^{m} a d m x^{3} + 6 \, \left (e x\right )^{m} b d x^{4} + 7 \, \left (e x\right )^{m} a c m x^{2} + 8 \, \left (e x\right )^{m} b c x^{3} + 8 \, \left (e x\right )^{m} a d x^{3} + 12 \, \left (e x\right )^{m} a c x^{2}}{m^{3} + 9 \, m^{2} + 26 \, m + 24} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a*x),x, algorithm="giac")
Output:
((e*x)^m*b*d*m^2*x^4 + (e*x)^m*b*c*m^2*x^3 + (e*x)^m*a*d*m^2*x^3 + 5*(e*x) ^m*b*d*m*x^4 + (e*x)^m*a*c*m^2*x^2 + 6*(e*x)^m*b*c*m*x^3 + 6*(e*x)^m*a*d*m *x^3 + 6*(e*x)^m*b*d*x^4 + 7*(e*x)^m*a*c*m*x^2 + 8*(e*x)^m*b*c*x^3 + 8*(e* x)^m*a*d*x^3 + 12*(e*x)^m*a*c*x^2)/(m^3 + 9*m^2 + 26*m + 24)
Time = 8.62 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx={\left (e\,x\right )}^m\,\left (\frac {x^3\,\left (a\,d+b\,c\right )\,\left (m^2+6\,m+8\right )}{m^3+9\,m^2+26\,m+24}+\frac {a\,c\,x^2\,\left (m^2+7\,m+12\right )}{m^3+9\,m^2+26\,m+24}+\frac {b\,d\,x^4\,\left (m^2+5\,m+6\right )}{m^3+9\,m^2+26\,m+24}\right ) \] Input:
int((a*x + b*x^2)*(e*x)^m*(c + d*x),x)
Output:
(e*x)^m*((x^3*(a*d + b*c)*(6*m + m^2 + 8))/(26*m + 9*m^2 + m^3 + 24) + (a* c*x^2*(7*m + m^2 + 12))/(26*m + 9*m^2 + m^3 + 24) + (b*d*x^4*(5*m + m^2 + 6))/(26*m + 9*m^2 + m^3 + 24))
Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.68 \[ \int (e x)^m (c+d x) \left (a x+b x^2\right ) \, dx=\frac {x^{m} e^{m} x^{2} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +5 b d m \,x^{2}+a c \,m^{2}+6 a d m x +6 b c m x +6 b d \,x^{2}+7 a c m +8 a d x +8 b c x +12 a c \right )}{m^{3}+9 m^{2}+26 m +24} \] Input:
int((e*x)^m*(d*x+c)*(b*x^2+a*x),x)
Output:
(x**m*e**m*x**2*(a*c*m**2 + 7*a*c*m + 12*a*c + a*d*m**2*x + 6*a*d*m*x + 8* a*d*x + b*c*m**2*x + 6*b*c*m*x + 8*b*c*x + b*d*m**2*x**2 + 5*b*d*m*x**2 + 6*b*d*x**2))/(m**3 + 9*m**2 + 26*m + 24)