Integrand size = 22, antiderivative size = 60 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {a c (e x)^{3+m}}{e^3 (3+m)}+\frac {(b c+a d) (e x)^{4+m}}{e^4 (4+m)}+\frac {b d (e x)^{5+m}}{e^5 (5+m)} \] Output:
a*c*(e*x)^(3+m)/e^3/(3+m)+(a*d+b*c)*(e*x)^(4+m)/e^4/(4+m)+b*d*(e*x)^(5+m)/ e^5/(5+m)
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {x^3 (e x)^m (a (5+m) (c (4+m)+d (3+m) x)+b (3+m) x (c (5+m)+d (4+m) x))}{(3+m) (4+m) (5+m)} \] Input:
Integrate[(e*x)^m*(c + d*x)*(a*x^2 + b*x^3),x]
Output:
(x^3*(e*x)^m*(a*(5 + m)*(c*(4 + m) + d*(3 + m)*x) + b*(3 + m)*x*(c*(5 + m) + d*(4 + m)*x)))/((3 + m)*(4 + m)*(5 + m))
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.
\(\displaystyle \int \left (a x^2+b x^3\right ) (c+d x) (e x)^m \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {\int (e x)^{m+2} (a+b x) (c+d x)dx}{e^2}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\int \left (a c (e x)^{m+2}+\frac {(b c+a d) (e x)^{m+3}}{e}+\frac {b d (e x)^{m+4}}{e^2}\right )dx}{e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(e x)^{m+4} (a d+b c)}{e^2 (m+4)}+\frac {a c (e x)^{m+3}}{e (m+3)}+\frac {b d (e x)^{m+5}}{e^3 (m+5)}}{e^2}\) |
Input:
Int[(e*x)^m*(c + d*x)*(a*x^2 + b*x^3),x]
Output:
((a*c*(e*x)^(3 + m))/(e*(3 + m)) + ((b*c + a*d)*(e*x)^(4 + m))/(e^2*(4 + m )) + (b*d*(e*x)^(5 + m))/(e^3*(5 + m)))/e^2
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 | \(\frac {\left (a d +b c \right ) x^{4} {\mathrm e}^{m \ln \left (e x \right )}}{4+m}+\frac {a c \,x^{3} {\mathrm e}^{m \ln \left (e x \right )}}{3+m}+\frac {b d \,x^{5} {\mathrm e}^{m \ln \left (e x \right )}}{5+m}\) | \(61\) |
gosper | \(\frac {\left (e x \right )^{m} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +7 b d m \,x^{2}+a c \,m^{2}+8 a d m x +8 b c m x +12 b d \,x^{2}+9 a c m +15 a d x +15 c b x +20 a c \right ) x^{3}}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\) | \(101\) |
risch | \(\frac {\left (e x \right )^{m} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +7 b d m \,x^{2}+a c \,m^{2}+8 a d m x +8 b c m x +12 b d \,x^{2}+9 a c m +15 a d x +15 c b x +20 a c \right ) x^{3}}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\) | \(101\) |
orering | \(\frac {\left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +7 b d m \,x^{2}+a c \,m^{2}+8 a d m x +8 b c m x +12 b d \,x^{2}+9 a c m +15 a d x +15 c b x +20 a c \right ) x \left (e x \right )^{m} \left (b \,x^{3}+a \,x^{2}\right )}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right ) \left (b x +a \right )}\) | \(117\) |
parallelrisch | \(\frac {x^{5} \left (e x \right )^{m} b d \,m^{2}+7 x^{5} \left (e x \right )^{m} b d m +x^{4} \left (e x \right )^{m} a d \,m^{2}+x^{4} \left (e x \right )^{m} b c \,m^{2}+12 x^{5} \left (e x \right )^{m} b d +8 x^{4} \left (e x \right )^{m} a d m +8 x^{4} \left (e x \right )^{m} b c m +x^{3} \left (e x \right )^{m} a c \,m^{2}+15 x^{4} \left (e x \right )^{m} a d +15 x^{4} \left (e x \right )^{m} b c +9 x^{3} \left (e x \right )^{m} a c m +20 x^{3} \left (e x \right )^{m} a c}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\) | \(174\) |
Input:
int((e*x)^m*(d*x+c)*(b*x^3+a*x^2),x,method=_RETURNVERBOSE)
Output:
(a*d+b*c)/(4+m)*x^4*exp(m*ln(e*x))+a*c/(3+m)*x^3*exp(m*ln(e*x))+b*d/(5+m)* x^5*exp(m*ln(e*x))
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {{\left ({\left (b d m^{2} + 7 \, b d m + 12 \, b d\right )} x^{5} + {\left ({\left (b c + a d\right )} m^{2} + 15 \, b c + 15 \, a d + 8 \, {\left (b c + a d\right )} m\right )} x^{4} + {\left (a c m^{2} + 9 \, a c m + 20 \, a c\right )} x^{3}\right )} \left (e x\right )^{m}}{m^{3} + 12 \, m^{2} + 47 \, m + 60} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^3+a*x^2),x, algorithm="fricas")
Output:
((b*d*m^2 + 7*b*d*m + 12*b*d)*x^5 + ((b*c + a*d)*m^2 + 15*b*c + 15*a*d + 8 *(b*c + a*d)*m)*x^4 + (a*c*m^2 + 9*a*c*m + 20*a*c)*x^3)*(e*x)^m/(m^3 + 12* m^2 + 47*m + 60)
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (53) = 106\).
Time = 0.33 (sec) , antiderivative size = 425, normalized size of antiderivative = 7.08 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\begin {cases} \frac {- \frac {a c}{2 x^{2}} - \frac {a d}{x} - \frac {b c}{x} + b d \log {\left (x \right )}}{e^{5}} & \text {for}\: m = -5 \\\frac {- \frac {a c}{x} + a d \log {\left (x \right )} + b c \log {\left (x \right )} + b d x}{e^{4}} & \text {for}\: m = -4 \\\frac {a c \log {\left (x \right )} + a d x + b c x + \frac {b d x^{2}}{2}}{e^{3}} & \text {for}\: m = -3 \\\frac {a c m^{2} x^{3} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {9 a c m x^{3} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {20 a c x^{3} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {a d m^{2} x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {8 a d m x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {15 a d x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {b c m^{2} x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {8 b c m x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {15 b c x^{4} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {b d m^{2} x^{5} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {7 b d m x^{5} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {12 b d x^{5} \left (e x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} & \text {otherwise} \end {cases} \] Input:
integrate((e*x)**m*(d*x+c)*(b*x**3+a*x**2),x)
Output:
Piecewise(((-a*c/(2*x**2) - a*d/x - b*c/x + b*d*log(x))/e**5, Eq(m, -5)), ((-a*c/x + a*d*log(x) + b*c*log(x) + b*d*x)/e**4, Eq(m, -4)), ((a*c*log(x) + a*d*x + b*c*x + b*d*x**2/2)/e**3, Eq(m, -3)), (a*c*m**2*x**3*(e*x)**m/( m**3 + 12*m**2 + 47*m + 60) + 9*a*c*m*x**3*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 20*a*c*x**3*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + a*d*m**2*x**4 *(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 8*a*d*m*x**4*(e*x)**m/(m**3 + 12* m**2 + 47*m + 60) + 15*a*d*x**4*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + b* c*m**2*x**4*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 8*b*c*m*x**4*(e*x)**m/ (m**3 + 12*m**2 + 47*m + 60) + 15*b*c*x**4*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + b*d*m**2*x**5*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 7*b*d*m*x** 5*(e*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 12*b*d*x**5*(e*x)**m/(m**3 + 12* m**2 + 47*m + 60), True))
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {b d e^{m} x^{5} x^{m}}{m + 5} + \frac {b c e^{m} x^{4} x^{m}}{m + 4} + \frac {a d e^{m} x^{4} x^{m}}{m + 4} + \frac {a c e^{m} x^{3} x^{m}}{m + 3} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^3+a*x^2),x, algorithm="maxima")
Output:
b*d*e^m*x^5*x^m/(m + 5) + b*c*e^m*x^4*x^m/(m + 4) + a*d*e^m*x^4*x^m/(m + 4 ) + a*c*e^m*x^3*x^m/(m + 3)
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (60) = 120\).
Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.88 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {\left (e x\right )^{m} b d m^{2} x^{5} + \left (e x\right )^{m} b c m^{2} x^{4} + \left (e x\right )^{m} a d m^{2} x^{4} + 7 \, \left (e x\right )^{m} b d m x^{5} + \left (e x\right )^{m} a c m^{2} x^{3} + 8 \, \left (e x\right )^{m} b c m x^{4} + 8 \, \left (e x\right )^{m} a d m x^{4} + 12 \, \left (e x\right )^{m} b d x^{5} + 9 \, \left (e x\right )^{m} a c m x^{3} + 15 \, \left (e x\right )^{m} b c x^{4} + 15 \, \left (e x\right )^{m} a d x^{4} + 20 \, \left (e x\right )^{m} a c x^{3}}{m^{3} + 12 \, m^{2} + 47 \, m + 60} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^3+a*x^2),x, algorithm="giac")
Output:
((e*x)^m*b*d*m^2*x^5 + (e*x)^m*b*c*m^2*x^4 + (e*x)^m*a*d*m^2*x^4 + 7*(e*x) ^m*b*d*m*x^5 + (e*x)^m*a*c*m^2*x^3 + 8*(e*x)^m*b*c*m*x^4 + 8*(e*x)^m*a*d*m *x^4 + 12*(e*x)^m*b*d*x^5 + 9*(e*x)^m*a*c*m*x^3 + 15*(e*x)^m*b*c*x^4 + 15* (e*x)^m*a*d*x^4 + 20*(e*x)^m*a*c*x^3)/(m^3 + 12*m^2 + 47*m + 60)
Time = 8.84 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx={\left (e\,x\right )}^m\,\left (\frac {x^4\,\left (a\,d+b\,c\right )\,\left (m^2+8\,m+15\right )}{m^3+12\,m^2+47\,m+60}+\frac {a\,c\,x^3\,\left (m^2+9\,m+20\right )}{m^3+12\,m^2+47\,m+60}+\frac {b\,d\,x^5\,\left (m^2+7\,m+12\right )}{m^3+12\,m^2+47\,m+60}\right ) \] Input:
int((e*x)^m*(a*x^2 + b*x^3)*(c + d*x),x)
Output:
(e*x)^m*((x^4*(a*d + b*c)*(8*m + m^2 + 15))/(47*m + 12*m^2 + m^3 + 60) + ( a*c*x^3*(9*m + m^2 + 20))/(47*m + 12*m^2 + m^3 + 60) + (b*d*x^5*(7*m + m^2 + 12))/(47*m + 12*m^2 + m^3 + 60))
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.68 \[ \int (e x)^m (c+d x) \left (a x^2+b x^3\right ) \, dx=\frac {x^{m} e^{m} x^{3} \left (b d \,m^{2} x^{2}+a d \,m^{2} x +b c \,m^{2} x +7 b d m \,x^{2}+a c \,m^{2}+8 a d m x +8 b c m x +12 b d \,x^{2}+9 a c m +15 a d x +15 b c x +20 a c \right )}{m^{3}+12 m^{2}+47 m +60} \] Input:
int((e*x)^m*(d*x+c)*(b*x^3+a*x^2),x)
Output:
(x**m*e**m*x**3*(a*c*m**2 + 9*a*c*m + 20*a*c + a*d*m**2*x + 8*a*d*m*x + 15 *a*d*x + b*c*m**2*x + 8*b*c*m*x + 15*b*c*x + b*d*m**2*x**2 + 7*b*d*m*x**2 + 12*b*d*x**2))/(m**3 + 12*m**2 + 47*m + 60)