3.1.62 \(\int \frac {e^{-a-b x} (a+b x)^3}{x^3} \, dx\) [62]

Optimal. Leaf size=130 \[ -b^2 e^{-a-b x}-\frac {a^3 e^{-a-b x}}{2 x^2}-\frac {3 a^2 b e^{-a-b x}}{x}+\frac {a^3 b e^{-a-b x}}{2 x}+3 a b^2 e^{-a} \text {Ei}(-b x)-3 a^2 b^2 e^{-a} \text {Ei}(-b x)+\frac {1}{2} a^3 b^2 e^{-a} \text {Ei}(-b x) \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2230, 2225, 2208, 2209} \begin {gather*} \frac {1}{2} e^{-a} a^3 b^2 \text {Ei}(-b x)-\frac {a^3 e^{-a-b x}}{2 x^2}+\frac {a^3 b e^{-a-b x}}{2 x}-3 e^{-a} a^2 b^2 \text {Ei}(-b x)-\frac {3 a^2 b e^{-a-b x}}{x}+3 e^{-a} a b^2 \text {Ei}(-b x)-b^2 e^{-a-b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-a - b*x)*(a + b*x)^3)/x^3,x]

[Out]

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

Rule 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2225

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 2230

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] &&  !TrueQ[$UseGamma]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 68, normalized size = 0.52 \begin {gather*} \frac {e^{-a-b x} \left (-6 a^2 b x-2 b^2 x^2+a^3 (-1+b x)+a \left (6-6 a+a^2\right ) b^2 e^{b x} x^2 \text {Ei}(-b x)\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-a - b*x)*(-6*a^2*b*x - 2*b^2*x^2 + a^3*(-1 + b*x) + a*(6 - 6*a + a^2)*b^2*E^(b*x)*x^2*ExpIntegralEi[-(b*x
)]))/(2*x^2)

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Maple [A]
time = 0.07, size = 112, normalized size = 0.86

method	result	size
derivativedivides	\(-b^{2} \left ({\mathrm e}^{-b x -a}-a^{3} \left (-\frac {{\mathrm e}^{-b x -a}}{2 b^{2} x^{2}}+\frac {{\mathrm e}^{-b x -a}}{2 b x}-\frac {{\mathrm e}^{-a} \expIntegral \left (1, b x \right )}{2}\right )+3 a^{2} \left (\frac {{\mathrm e}^{-b x -a}}{b x}-{\mathrm e}^{-a} \expIntegral \left (1, b x \right )\right )+3 a \,{\mathrm e}^{-a} \expIntegral \left (1, b x \right )\right )\)	\(112\)
default	\(-b^{2} \left ({\mathrm e}^{-b x -a}-a^{3} \left (-\frac {{\mathrm e}^{-b x -a}}{2 b^{2} x^{2}}+\frac {{\mathrm e}^{-b x -a}}{2 b x}-\frac {{\mathrm e}^{-a} \expIntegral \left (1, b x \right )}{2}\right )+3 a^{2} \left (\frac {{\mathrm e}^{-b x -a}}{b x}-{\mathrm e}^{-a} \expIntegral \left (1, b x \right )\right )+3 a \,{\mathrm e}^{-a} \expIntegral \left (1, b x \right )\right )\)	\(112\)
risch	\(-b^{2} {\mathrm e}^{-b x -a}-\frac {a^{3} {\mathrm e}^{-b x -a}}{2 x^{2}}+\frac {a^{3} b \,{\mathrm e}^{-b x -a}}{2 x}-\frac {b^{2} a^{3} {\mathrm e}^{-a} \expIntegral \left (1, b x \right )}{2}-\frac {3 a^{2} b \,{\mathrm e}^{-b x -a}}{x}+3 b^{2} a^{2} {\mathrm e}^{-a} \expIntegral \left (1, b x \right )-3 b^{2} a \,{\mathrm e}^{-a} \expIntegral \left (1, b x \right )\)	\(118\)
meijerg	\({\mathrm e}^{-a} b^{2} \left (1-{\mathrm e}^{-b x}\right )+3 b^{2} {\mathrm e}^{-a} a \left (-\ln \left (b x \right )-\expIntegral \left (1, b x \right )+\ln \left (x \right )+\ln \left (b \right )\right )+3 b^{2} {\mathrm e}^{-a} a^{2} \left (\frac {-2 b x +2}{2 b x}-\frac {{\mathrm e}^{-b x}}{b x}+\ln \left (b x \right )+\expIntegral \left (1, b x \right )+1-\ln \left (x \right )-\ln \left (b \right )-\frac {1}{b x}\right )+{\mathrm e}^{-a} a^{3} b^{2} \left (\frac {9 b^{2} x^{2}-12 b x +6}{12 b^{2} x^{2}}-\frac {\left (-3 b x +3\right ) {\mathrm e}^{-b x}}{6 b^{2} x^{2}}-\frac {\ln \left (b x \right )}{2}-\frac {\expIntegral \left (1, b x \right )}{2}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (b \right )}{2}-\frac {1}{2 b^{2} x^{2}}+\frac {1}{b x}\right )\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 64, normalized size = 0.49 \begin {gather*} -a^{3} b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) - 3 \, a^{2} b^{2} e^{\left (-a\right )} \Gamma \left (-1, b x\right ) + 3 \, a b^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - b^{2} e^{\left (-b x - a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.45, size = 70, normalized size = 0.54 \begin {gather*} \frac {{\left (a^{3} - 6 \, a^{2} + 6 \, a\right )} b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - {\left (2 \, b^{2} x^{2} + a^{3} - {\left (a^{3} - 6 \, a^{2}\right )} b x\right )} e^{\left (-b x - a\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^3 - 6*a^2 + 6*a)*b^2*x^2*Ei(-b*x)*e^(-a) - (2*b^2*x^2 + a^3 - (a^3 - 6*a^2)*b*x)*e^(-b*x - a))/x^2

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Sympy [A]
time = 1.75, size = 56, normalized size = 0.43 \begin {gather*} \left (- \frac {a^{3} \operatorname {E}_{3}\left (b x\right )}{x^{2}} - \frac {3 a^{2} b \operatorname {E}_{2}\left (b x\right )}{x} + 3 a b^{2} \operatorname {Ei}{\left (- b x \right )} + b^{3} \left (\begin {cases} x & \text {for}\: b = 0 \\- \frac {e^{- b x}}{b} & \text {otherwise} \end {cases}\right )\right ) e^{- a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.79, size = 125, normalized size = 0.96 \begin {gather*} \frac {a^{3} b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 6 \, a^{2} b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 6 \, a b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + a^{3} b x e^{\left (-b x - a\right )} - 6 \, a^{2} b x e^{\left (-b x - a\right )} - 2 \, b^{2} x^{2} e^{\left (-b x - a\right )} - a^{3} e^{\left (-b x - a\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3*b^2*x^2*Ei(-b*x)*e^(-a) - 6*a^2*b^2*x^2*Ei(-b*x)*e^(-a) + 6*a*b^2*x^2*Ei(-b*x)*e^(-a) + a^3*b*x*e^(-b
*x - a) - 6*a^2*b*x*e^(-b*x - a) - 2*b^2*x^2*e^(-b*x - a) - a^3*e^(-b*x - a))/x^2

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Mupad [B]
time = 3.64, size = 100, normalized size = 0.77 \begin {gather*} 3\,a^2\,b^2\,{\mathrm {e}}^{-a}\,\left (\mathrm {expint}\left (b\,x\right )-\frac {{\mathrm {e}}^{-b\,x}}{b\,x}\right )-3\,a\,b^2\,{\mathrm {e}}^{-a}\,\mathrm {expint}\left (b\,x\right )-b^2\,{\mathrm {e}}^{-a-b\,x}+a^3\,b^2\,{\mathrm {e}}^{-a}\,\left ({\mathrm {e}}^{-b\,x}\,\left (\frac {1}{2\,b\,x}-\frac {1}{2\,b^2\,x^2}\right )-\frac {\mathrm {expint}\left (b\,x\right )}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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