3.61 \(\int \frac {e^{-a-b x} (a+b x)^3}{x^2} \, dx\)

Optimal. Leaf size=94 \[ e^{-a} a^3 (-b) \text {Ei}(-b x)-\frac {a^3 e^{-a-b x}}{x}+3 e^{-a} a^2 b \text {Ei}(-b x)-b^2 x e^{-a-b x}-3 a b e^{-a-b x}-b e^{-a-b x} \]

[Out]

-b*exp(-b*x-a)-3*a*b*exp(-b*x-a)-a^3*exp(-b*x-a)/x-b^2*exp(-b*x-a)*x+3*a^2*b*Ei(-b*x)/exp(a)-a^3*b*Ei(-b*x)/ex
p(a)

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Rubi [A]  time = 0.16, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2199, 2194, 2177, 2178, 2176} \[ e^{-a} a^3 (-b) \text {Ei}(-b x)+3 e^{-a} a^2 b \text {Ei}(-b x)-\frac {a^3 e^{-a-b x}}{x}-b^2 x e^{-a-b x}-3 a b e^{-a-b x}-b e^{-a-b x} \]

Antiderivative was successfully verified.

[In]

Int[(E^(-a - b*x)*(a + b*x)^3)/x^2,x]

[Out]

-(b*E^(-a - b*x)) - 3*a*b*E^(-a - b*x) - (a^3*E^(-a - b*x))/x - b^2*E^(-a - b*x)*x + (3*a^2*b*ExpIntegralEi[-(
b*x)])/E^a - (a^3*b*ExpIntegralEi[-(b*x)])/E^a

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 54, normalized size = 0.57 \[ \frac {e^{-a-b x} \left (-a^3-(a-3) a^2 b x e^{b x} \text {Ei}(-b x)-3 a b x-b x (b x+1)\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(-a - b*x)*(a + b*x)^3)/x^2,x]

[Out]

(E^(-a - b*x)*(-a^3 - 3*a*b*x - b*x*(1 + b*x) - (-3 + a)*a^2*b*E^(b*x)*x*ExpIntegralEi[-(b*x)]))/x

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fricas [A]  time = 0.40, size = 56, normalized size = 0.60 \[ -\frac {{\left (a^{3} - 3 \, a^{2}\right )} b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + {\left (b^{2} x^{2} + a^{3} + {\left (3 \, a + 1\right )} b x\right )} e^{\left (-b x - a\right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-b*x-a)*(b*x+a)^3/x^2,x, algorithm="fricas")

[Out]

-((a^3 - 3*a^2)*b*x*Ei(-b*x)*e^(-a) + (b^2*x^2 + a^3 + (3*a + 1)*b*x)*e^(-b*x - a))/x

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giac [A]  time = 0.45, size = 92, normalized size = 0.98 \[ -\frac {a^{3} b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 3 \, a^{2} b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + b^{2} x^{2} e^{\left (-b x - a\right )} + a^{3} e^{\left (-b x - a\right )} + 3 \, a b x e^{\left (-b x - a\right )} + b x e^{\left (-b x - a\right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-b*x-a)*(b*x+a)^3/x^2,x, algorithm="giac")

[Out]

-(a^3*b*x*Ei(-b*x)*e^(-a) - 3*a^2*b*x*Ei(-b*x)*e^(-a) + b^2*x^2*e^(-b*x - a) + a^3*e^(-b*x - a) + 3*a*b*x*e^(-
b*x - a) + b*x*e^(-b*x - a))/x

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maple [A]  time = 0.02, size = 92, normalized size = 0.98 \[ \left (-3 a^{2} \Ei \left (1, b x \right ) {\mathrm e}^{-a}-\left (-\Ei \left (1, b x \right ) {\mathrm e}^{-a}+\frac {{\mathrm e}^{-b x -a}}{b x}\right ) a^{3}-2 a \,{\mathrm e}^{-b x -a}+\left (-b x -a \right ) {\mathrm e}^{-b x -a}-{\mathrm e}^{-b x -a}\right ) b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-b*x-a)*(b*x+a)^3/x^2,x)

[Out]

b*((-b*x-a)*exp(-b*x-a)-exp(-b*x-a)-2*a*exp(-b*x-a)-a^3*(exp(-b*x-a)/b/x-exp(-a)*Ei(1,b*x))-3*a^2*exp(-a)*Ei(1
,b*x))

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maxima [A]  time = 0.97, size = 61, normalized size = 0.65 \[ -a^{3} b e^{\left (-a\right )} \Gamma \left (-1, b x\right ) + 3 \, a^{2} b {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - {\left (b x + 1\right )} b e^{\left (-b x - a\right )} - 3 \, a b e^{\left (-b x - a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-b*x-a)*(b*x+a)^3/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.64, size = 72, normalized size = 0.77 \[ a^3\,b\,{\mathrm {e}}^{-a}\,\left (\mathrm {expint}\left (b\,x\right )-\frac {{\mathrm {e}}^{-b\,x}}{b\,x}\right )-3\,a\,b\,{\mathrm {e}}^{-a-b\,x}-b\,{\mathrm {e}}^{-a-b\,x}\,\left (b\,x+1\right )-3\,a^2\,b\,{\mathrm {e}}^{-a}\,\mathrm {expint}\left (b\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- a - b*x)*(a + b*x)^3)/x^2,x)

[Out]

a^3*b*exp(-a)*(expint(b*x) - exp(-b*x)/(b*x)) - 3*a*b*exp(- a - b*x) - b*exp(- a - b*x)*(b*x + 1) - 3*a^2*b*ex
p(-a)*expint(b*x)

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sympy [A]  time = 5.87, size = 99, normalized size = 1.05 \[ - \frac {a^{3} e^{- a} \operatorname {E}_{2}\left (b x\right )}{x} + 3 a^{2} b e^{- a} \operatorname {Ei}{\left (- b x \right )} + 3 a b^{2} \left (\begin {cases} x & \text {for}\: b = 0 \\- \frac {e^{- b x}}{b} & \text {otherwise} \end {cases}\right ) e^{- a} + b^{3} x \left (\begin {cases} x & \text {for}\: b = 0 \\- \frac {e^{- b x}}{b} & \text {otherwise} \end {cases}\right ) e^{- a} - b^{3} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: b = 0 \\- \frac {\begin {cases} - \frac {e^{- b x}}{b} & \text {for}\: b \neq 0 \\x & \text {otherwise} \end {cases}}{b} & \text {otherwise} \end {cases}\right ) e^{- a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-b*x-a)*(b*x+a)**3/x**2,x)

[Out]

-a**3*exp(-a)*expint(2, b*x)/x + 3*a**2*b*exp(-a)*Ei(-b*x) + 3*a*b**2*Piecewise((x, Eq(b, 0)), (-exp(-b*x)/b,
True))*exp(-a) + b**3*x*Piecewise((x, Eq(b, 0)), (-exp(-b*x)/b, True))*exp(-a) - b**3*Piecewise((x**2/2, Eq(b,
0)), (-Piecewise((-exp(-b*x)/b, Ne(b, 0)), (x, True))/b, True))*exp(-a)

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