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

Optimal. Leaf size=116 \[ -\frac{a^3 e^{-a} x^m (b x)^{-m} \text{Gamma}(m+1,b x)}{b}-\frac{3 a^2 e^{-a} x^m (b x)^{-m} \text{Gamma}(m+2,b x)}{b}-\frac{3 a e^{-a} x^m (b x)^{-m} \text{Gamma}(m+3,b x)}{b}-\frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+4,b x)}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.174801, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2199, 2181} \[ -\frac{a^3 e^{-a} x^m (b x)^{-m} \text{Gamma}(m+1,b x)}{b}-\frac{3 a^2 e^{-a} x^m (b x)^{-m} \text{Gamma}(m+2,b x)}{b}-\frac{3 a e^{-a} x^m (b x)^{-m} \text{Gamma}(m+3,b x)}{b}-\frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+4,b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(-a - b*x)*x^m*(a + b*x)^3,x]

[Out]

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

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int e^{-a-b x} x^m (a+b x)^3 \, dx &=\int \left (a^3 e^{-a-b x} x^m+3 a^2 b e^{-a-b x} x^{1+m}+3 a b^2 e^{-a-b x} x^{2+m}+b^3 e^{-a-b x} x^{3+m}\right ) \, dx\\ &=a^3 \int e^{-a-b x} x^m \, dx+\left (3 a^2 b\right ) \int e^{-a-b x} x^{1+m} \, dx+\left (3 a b^2\right ) \int e^{-a-b x} x^{2+m} \, dx+b^3 \int e^{-a-b x} x^{3+m} \, 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}\\ \end{align*}

Mathematica [A]  time = 0.0568417, size = 61, normalized size = 0.53 \[ -\frac{e^{-a} x^m (b x)^{-m} \left (a^3 \text{Gamma}(m+1,b x)+3 a^2 \text{Gamma}(m+2,b x)+3 a \text{Gamma}(m+3,b x)+\text{Gamma}(m+4,b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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))

________________________________________________________________________________________

Maple [C]  time = 0.08, size = 334, normalized size = 2.9 \begin{align*}{b}^{-m-1}{{\rm e}^{-a}} \left ({x}^{m}{b}^{m} \left ({m}^{2}+5\,m+6 \right ) \left ( bx \right ) ^{-{\frac{m}{2}}}{{\rm e}^{-{\frac{bx}{2}}}}{{\sl M}_{{\frac{m}{2}},\,{\frac{m}{2}}+{\frac{1}{2}}}\left (bx\right )}-{x}^{m}{b}^{m} \left ({b}^{2}{x}^{2}+bmx+3\,bx+{m}^{2}+5\,m+6 \right ) \left ( bx \right ) ^{-{\frac{m}{2}}}{{\rm e}^{-{\frac{bx}{2}}}}{{\sl M}_{{\frac{m}{2}}+1,\,{\frac{m}{2}}+{\frac{1}{2}}}\left (bx\right )} \right ) +3\,{b}^{-m-1}{{\rm e}^{-a}}a \left ({x}^{m}{b}^{m} \left ( 2+m \right ) \left ( bx \right ) ^{-m/2}{{\rm e}^{-1/2\,bx}}{{\sl M}_{m/2,\,m/2+1/2}\left (bx\right )}-{x}^{m}{b}^{m} \left ( bx+m+2 \right ) \left ( bx \right ) ^{-m/2}{{\rm e}^{-1/2\,bx}}{{\sl M}_{m/2+1,\,m/2+1/2}\left (bx\right )} \right ) +3\,{b}^{-m-1}{{\rm e}^{-a}}{a}^{2} \left ({x}^{m}{b}^{m} \left ( bx \right ) ^{-m/2}{{\rm e}^{-1/2\,bx}}{{\sl M}_{m/2,\,m/2+1/2}\left (bx\right )}+{\frac{{x}^{m}{b}^{m} \left ( -2-m \right ) \left ( bx \right ) ^{-m/2}{{\rm e}^{-1/2\,bx}}{{\sl M}_{m/2+1,\,m/2+1/2}\left (bx\right )}}{2+m}} \right ) +{\frac{{a}^{3}{x}^{m}}{b \left ( 1+m \right ) }{{\rm e}^{-a-{\frac{bx}{2}}}} \left ( bx \right ) ^{-{\frac{m}{2}}}{{\sl M}_{{\frac{m}{2}},\,{\frac{m}{2}}+{\frac{1}{2}}}\left (bx\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^(-m-1)*exp(-a)*(x^m*b^m*(m^2+5*m+6)*(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^
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^(-m-1)*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^(-m-1)*exp(-a)*a^2*(x^m*b^m*(b*x)^(-1/2*m)*exp(-1/2*b*
x)*WhittakerM(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)

________________________________________________________________________________________

Maxima [A]  time = 1.23557, size = 166, normalized size = 1.43 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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)

________________________________________________________________________________________

Fricas [A]  time = 1.56245, size = 297, normalized size = 2.56 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{3} x^{m} e^{\left (-b x - a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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