\(\int e^{-a-b x} x^m (a+b x)^3 \, dx\) [94]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=-\frac {a^3 e^{-a} x^m (b x)^{-m} \Gamma (1+m,b x)}{b}-\frac {3 a^2 e^{-a} x^m (b x)^{-m} \Gamma (2+m,b x)}{b}-\frac {3 a e^{-a} x^m (b x)^{-m} \Gamma (3+m,b x)}{b}-\frac {e^{-a} x^m (b x)^{-m} \Gamma (4+m,b x)}{b} \] Output:

-a^3*x^m*GAMMA(1+m,b*x)/b/exp(a)/((b*x)^m)-3*a^2*x^m*GAMMA(2+m,b*x)/b/exp( 
a)/((b*x)^m)-3*a*x^m*GAMMA(3+m,b*x)/b/exp(a)/((b*x)^m)-x^m*GAMMA(4+m,b*x)/ 
b/exp(a)/((b*x)^m)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.53 \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=-\frac {e^{-a} x^m (b x)^{-m} \left (a^3 \Gamma (1+m,b x)+3 a^2 \Gamma (2+m,b x)+3 a \Gamma (3+m,b x)+\Gamma (4+m,b x)\right )}{b} \] Input:

Integrate[E^(-a - b*x)*x^m*(a + b*x)^3,x]
 

Output:

-((x^m*(a^3*Gamma[1 + m, b*x] + 3*a^2*Gamma[2 + m, b*x] + 3*a*Gamma[3 + m, 
 b*x] + Gamma[4 + m, b*x]))/(b*E^a*(b*x)^m))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2629, 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.

\(\displaystyle \int x^m e^{-a-b x} (a+b x)^3 \, dx\)

\(\Big \downarrow \) 2629

\(\displaystyle \int \left (a^3 x^m e^{-a-b x}+3 a^2 b x^{m+1} e^{-a-b x}+b^3 x^{m+3} e^{-a-b x}+3 a b^2 x^{m+2} e^{-a-b x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3 e^{-a} x^m (b x)^{-m} \Gamma (m+1,b x)}{b}-\frac {3 a^2 e^{-a} x^m (b x)^{-m} \Gamma (m+2,b x)}{b}-\frac {3 a e^{-a} x^m (b x)^{-m} \Gamma (m+3,b x)}{b}-\frac {e^{-a} x^m (b x)^{-m} \Gamma (m+4,b x)}{b}\)

Input:

Int[E^(-a - b*x)*x^m*(a + b*x)^3,x]
 

Output:

-((a^3*x^m*Gamma[1 + m, b*x])/(b*E^a*(b*x)^m)) - (3*a^2*x^m*Gamma[2 + m, b 
*x])/(b*E^a*(b*x)^m) - (3*a*x^m*Gamma[3 + m, b*x])/(b*E^a*(b*x)^m) - (x^m* 
Gamma[4 + m, b*x])/(b*E^a*(b*x)^m)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2629

Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte 
grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ 
Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.59 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.88

method	result	size
meijerg	\(b^{-1-m} {\mathrm e}^{-a} \left (x^{m} b^{m} \left (m^{2}+5 m +6\right ) \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}, \frac {m}{2}+\frac {1}{2}, b x \right )-x^{m} b^{m} \left (b^{2} x^{2}+b m x +3 b x +m^{2}+5 m +6\right ) \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}+1, \frac {m}{2}+\frac {1}{2}, b x \right )\right )+3 b^{-1-m} {\mathrm e}^{-a} a \left (x^{m} b^{m} \left (2+m \right ) \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}, \frac {m}{2}+\frac {1}{2}, b x \right )-x^{m} b^{m} \left (b x +m +2\right ) \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}+1, \frac {m}{2}+\frac {1}{2}, b x \right )\right )+3 b^{-1-m} {\mathrm e}^{-a} a^{2} \left (x^{m} b^{m} \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}, \frac {m}{2}+\frac {1}{2}, b x \right )+\frac {x^{m} b^{m} \left (-2-m \right ) \left (b x \right )^{-\frac {m}{2}} {\mathrm e}^{-\frac {b x}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}+1, \frac {m}{2}+\frac {1}{2}, b x \right )}{2+m}\right )+\frac {{\mathrm e}^{-a -\frac {b x}{2}} a^{3} x^{m} \left (b x \right )^{-\frac {m}{2}} \operatorname {WhittakerM}\left (\frac {m}{2}, \frac {m}{2}+\frac {1}{2}, b x \right )}{b \left (1+m \right )}\)	\(334\)

Input:

int(exp(-b*x-a)*x^m*(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

b^(-1-m)*exp(-a)*(x^m*b^m*(m^2+5*m+6)*(b*x)^(-1/2*m)*exp(-1/2*b*x)*Whittak 
erM(1/2*m,1/2*m+1/2,b*x)-x^m*b^m*(b^2*x^2+b*m*x+3*b*x+m^2+5*m+6)*(b*x)^(-1 
/2*m)*exp(-1/2*b*x)*WhittakerM(1/2*m+1,1/2*m+1/2,b*x))+3*b^(-1-m)*exp(-a)* 
a*(x^m*b^m*(2+m)*(b*x)^(-1/2*m)*exp(-1/2*b*x)*WhittakerM(1/2*m,1/2*m+1/2,b 
*x)-x^m*b^m*(b*x+m+2)*(b*x)^(-1/2*m)*exp(-1/2*b*x)*WhittakerM(1/2*m+1,1/2* 
m+1/2,b*x))+3*b^(-1-m)*exp(-a)*a^2*(x^m*b^m*(b*x)^(-1/2*m)*exp(-1/2*b*x)*W 
hittakerM(1/2*m,1/2*m+1/2,b*x)+1/(2+m)*x^m*b^m*(-2-m)*(b*x)^(-1/2*m)*exp(- 
1/2*b*x)*WhittakerM(1/2*m+1,1/2*m+1/2,b*x))+exp(-a-1/2*b*x)/b*a^3/(1+m)*x^ 
m*(b*x)^(-1/2*m)*WhittakerM(1/2*m,1/2*m+1/2,b*x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09 \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=-\frac {{\left (b^{3} x^{3} + {\left (3 \, {\left (a + 1\right )} b^{2} + b^{2} m\right )} x^{2} + {\left ({\left (3 \, a + 5\right )} b m + b m^{2} + 3 \, {\left (a^{2} + 2 \, a + 2\right )} b\right )} x\right )} x^{m} e^{\left (-b x - a\right )} + {\left (a^{3} + 3 \, {\left (a + 2\right )} m^{2} + m^{3} + 3 \, a^{2} + {\left (3 \, a^{2} + 9 \, a + 11\right )} m + 6 \, a + 6\right )} e^{\left (-m \log \left (b\right ) - a\right )} \Gamma \left (m + 1, b x\right )}{b} \] Input:

integrate(exp(-b*x-a)*x^m*(b*x+a)^3,x, algorithm="fricas")
 

Output:

-((b^3*x^3 + (3*(a + 1)*b^2 + b^2*m)*x^2 + ((3*a + 5)*b*m + b*m^2 + 3*(a^2 
 + 2*a + 2)*b)*x)*x^m*e^(-b*x - a) + (a^3 + 3*(a + 2)*m^2 + m^3 + 3*a^2 + 
(3*a^2 + 9*a + 11)*m + 6*a + 6)*e^(-m*log(b) - a)*gamma(m + 1, b*x))/b
 

Sympy [A] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=\left (- \frac {a^{3} x^{m} \left (b x\right )^{- m} \Gamma \left (m + 1, b x\right )}{b} - 3 a^{2} x^{m + 1} \left (b x\right )^{- m - 1} \Gamma \left (m + 2, b x\right ) - 3 a b x^{m + 2} \left (b x\right )^{- m - 2} \Gamma \left (m + 3, b x\right ) - b^{2} x^{m + 3} \left (b x\right )^{- m - 3} \Gamma \left (m + 4, b x\right )\right ) e^{- a} \] Input:

integrate(exp(-b*x-a)*x**m*(b*x+a)**3,x)
                                                                                    
                                                                                    
 

Output:

(-a**3*x**m*uppergamma(m + 1, b*x)/(b*(b*x)**m) - 3*a**2*x**(m + 1)*(b*x)* 
*(-m - 1)*uppergamma(m + 2, b*x) - 3*a*b*x**(m + 2)*(b*x)**(-m - 2)*upperg 
amma(m + 3, b*x) - b**2*x**(m + 3)*(b*x)**(-m - 3)*uppergamma(m + 4, b*x)) 
*exp(-a)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06 \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=-\left (b x\right )^{-m - 4} b^{3} x^{m + 4} e^{\left (-a\right )} \Gamma \left (m + 4, b x\right ) - 3 \, \left (b x\right )^{-m - 3} a b^{2} x^{m + 3} e^{\left (-a\right )} \Gamma \left (m + 3, b x\right ) - 3 \, \left (b x\right )^{-m - 2} a^{2} b x^{m + 2} e^{\left (-a\right )} \Gamma \left (m + 2, b x\right ) - \left (b x\right )^{-m - 1} a^{3} x^{m + 1} e^{\left (-a\right )} \Gamma \left (m + 1, b x\right ) \] Input:

integrate(exp(-b*x-a)*x^m*(b*x+a)^3,x, algorithm="maxima")
 

Output:

-(b*x)^(-m - 4)*b^3*x^(m + 4)*e^(-a)*gamma(m + 4, b*x) - 3*(b*x)^(-m - 3)* 
a*b^2*x^(m + 3)*e^(-a)*gamma(m + 3, b*x) - 3*(b*x)^(-m - 2)*a^2*b*x^(m + 2 
)*e^(-a)*gamma(m + 2, b*x) - (b*x)^(-m - 1)*a^3*x^(m + 1)*e^(-a)*gamma(m + 
 1, b*x)
 

Giac [F]

\[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=\int { {\left (b x + a\right )}^{3} x^{m} e^{\left (-b x - a\right )} \,d x } \] Input:

integrate(exp(-b*x-a)*x^m*(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((b*x + a)^3*x^m*e^(-b*x - a), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=\int x^m\,{\mathrm {e}}^{-a-b\,x}\,{\left (a+b\,x\right )}^3 \,d x \] Input:

int(x^m*exp(- a - b*x)*(a + b*x)^3,x)
 

Output:

int(x^m*exp(- a - b*x)*(a + b*x)^3, x)
 

Reduce [F]

\[ \int e^{-a-b x} x^m (a+b x)^3 \, dx=\frac {-x^{m} b^{3} x^{3}-x^{m} b^{2} m \,x^{2}-x^{m} b \,m^{2} x -3 x^{m} a^{2} b x -3 x^{m} a \,b^{2} x^{2}-6 x^{m} a b x -5 x^{m} b m x +3 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a^{2} m^{2}+3 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a^{2} m +3 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a \,m^{3}+9 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a \,m^{2}+6 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a m +e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) a^{3} m -3 x^{m} a^{2} m -3 x^{m} a \,m^{2}-9 x^{m} a m -3 x^{m} b^{2} x^{2}-6 x^{m} b x -6 x^{m}+6 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) m^{3}+11 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) m^{2}+6 e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) m +e^{b x} \left (\int \frac {x^{m}}{e^{b x} x}d x \right ) m^{4}-3 x^{m} a b m x -x^{m} a^{3}-x^{m} m^{3}-3 x^{m} a^{2}-6 x^{m} a -6 x^{m} m^{2}-11 x^{m} m}{e^{b x +a} b} \] Input:

int(exp(-b*x-a)*x^m*(b*x+a)^3,x)
 

Output:

(e**(b*x)*int(x**m/(e**(b*x)*x),x)*a**3*m + 3*e**(b*x)*int(x**m/(e**(b*x)* 
x),x)*a**2*m**2 + 3*e**(b*x)*int(x**m/(e**(b*x)*x),x)*a**2*m + 3*e**(b*x)* 
int(x**m/(e**(b*x)*x),x)*a*m**3 + 9*e**(b*x)*int(x**m/(e**(b*x)*x),x)*a*m* 
*2 + 6*e**(b*x)*int(x**m/(e**(b*x)*x),x)*a*m + e**(b*x)*int(x**m/(e**(b*x) 
*x),x)*m**4 + 6*e**(b*x)*int(x**m/(e**(b*x)*x),x)*m**3 + 11*e**(b*x)*int(x 
**m/(e**(b*x)*x),x)*m**2 + 6*e**(b*x)*int(x**m/(e**(b*x)*x),x)*m - x**m*a* 
*3 - 3*x**m*a**2*b*x - 3*x**m*a**2*m - 3*x**m*a**2 - 3*x**m*a*b**2*x**2 - 
3*x**m*a*b*m*x - 6*x**m*a*b*x - 3*x**m*a*m**2 - 9*x**m*a*m - 6*x**m*a - x* 
*m*b**3*x**3 - x**m*b**2*m*x**2 - 3*x**m*b**2*x**2 - x**m*b*m**2*x - 5*x** 
m*b*m*x - 6*x**m*b*x - x**m*m**3 - 6*x**m*m**2 - 11*x**m*m - 6*x**m)/(e**( 
a + b*x)*b)