3.404 \(\int e^{\frac{e}{c+d x}} (a+b x) \, dx\)

Optimal. Leaf size=125 \[ \frac{e (b c-a d) \text{Ei}\left (\frac{e}{c+d x}\right )}{d^2}-\frac{(c+d x) (b c-a d) e^{\frac{e}{c+d x}}}{d^2}-\frac{b e^2 \text{Ei}\left (\frac{e}{c+d x}\right )}{2 d^2}+\frac{b e (c+d x) e^{\frac{e}{c+d x}}}{2 d^2}+\frac{b (c+d x)^2 e^{\frac{e}{c+d x}}}{2 d^2} \]

[Out]

-(((b*c - a*d)*E^(e/(c + d*x))*(c + d*x))/d^2) + (b*e*E^(e/(c + d*x))*(c + d*x))/(2*d^2) + (b*E^(e/(c + d*x))*
(c + d*x)^2)/(2*d^2) + ((b*c - a*d)*e*ExpIntegralEi[e/(c + d*x)])/d^2 - (b*e^2*ExpIntegralEi[e/(c + d*x)])/(2*
d^2)

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Rubi [A]  time = 0.128417, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2226, 2206, 2210, 2214} \[ \frac{e (b c-a d) \text{Ei}\left (\frac{e}{c+d x}\right )}{d^2}-\frac{(c+d x) (b c-a d) e^{\frac{e}{c+d x}}}{d^2}-\frac{b e^2 \text{Ei}\left (\frac{e}{c+d x}\right )}{2 d^2}+\frac{b e (c+d x) e^{\frac{e}{c+d x}}}{2 d^2}+\frac{b (c+d x)^2 e^{\frac{e}{c+d x}}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(e/(c + d*x))*(a + b*x),x]

[Out]

-(((b*c - a*d)*E^(e/(c + d*x))*(c + d*x))/d^2) + (b*e*E^(e/(c + d*x))*(c + d*x))/(2*d^2) + (b*E^(e/(c + d*x))*
(c + d*x)^2)/(2*d^2) + ((b*c - a*d)*e*ExpIntegralEi[e/(c + d*x)])/d^2 - (b*e^2*ExpIntegralEi[e/(c + d*x)])/(2*
d^2)

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

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

Mathematica [A]  time = 0.0935075, size = 91, normalized size = 0.73 \[ \frac{d x e^{\frac{e}{c+d x}} (2 a d+b (d x+e))-e (2 a d+b (e-2 c)) \text{Ei}\left (\frac{e}{c+d x}\right )}{2 d^2}+\frac{c e^{\frac{e}{c+d x}} (2 a d+b (e-c))}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(e/(c + d*x))*(a + b*x),x]

[Out]

(c*(2*a*d + b*(-c + e))*E^(e/(c + d*x)))/(2*d^2) + (d*E^(e/(c + d*x))*x*(2*a*d + b*(e + d*x)) - e*(2*a*d + b*(
-2*c + e))*ExpIntegralEi[e/(c + d*x)])/(2*d^2)

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Maple [A]  time = 0.006, size = 150, normalized size = 1.2 \begin{align*} -{\frac{e}{d} \left ( a \left ( -{\frac{dx+c}{e}{{\rm e}^{{\frac{e}{dx+c}}}}}-{\it Ei} \left ( 1,-{\frac{e}{dx+c}} \right ) \right ) +{\frac{be}{d} \left ( -{\frac{ \left ( dx+c \right ) ^{2}}{2\,{e}^{2}}{{\rm e}^{{\frac{e}{dx+c}}}}}-{\frac{dx+c}{2\,e}{{\rm e}^{{\frac{e}{dx+c}}}}}-{\frac{1}{2}{\it Ei} \left ( 1,-{\frac{e}{dx+c}} \right ) } \right ) }-{\frac{bc}{d} \left ( -{\frac{dx+c}{e}{{\rm e}^{{\frac{e}{dx+c}}}}}-{\it Ei} \left ( 1,-{\frac{e}{dx+c}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e/(d*x+c))*(b*x+a),x)

[Out]

-1/d*e*(a*(-(d*x+c)/e*exp(e/(d*x+c))-Ei(1,-e/(d*x+c)))+b/d*e*(-1/2*exp(e/(d*x+c))*(d*x+c)^2/e^2-1/2*(d*x+c)/e*
exp(e/(d*x+c))-1/2*Ei(1,-e/(d*x+c)))-b/d*c*(-(d*x+c)/e*exp(e/(d*x+c))-Ei(1,-e/(d*x+c))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b d x^{2} +{\left (2 \, a d + b e\right )} x\right )} e^{\left (\frac{e}{d x + c}\right )}}{2 \, d} + \int -\frac{{\left (b c^{2} e -{\left (2 \, a d^{2} e -{\left (2 \, c d e - d e^{2}\right )} b\right )} x\right )} e^{\left (\frac{e}{d x + c}\right )}}{2 \,{\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e/(d*x+c))*(b*x+a),x, algorithm="maxima")

[Out]

1/2*(b*d*x^2 + (2*a*d + b*e)*x)*e^(e/(d*x + c))/d + integrate(-1/2*(b*c^2*e - (2*a*d^2*e - (2*c*d*e - d*e^2)*b
)*x)*e^(e/(d*x + c))/(d^3*x^2 + 2*c*d^2*x + c^2*d), x)

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Fricas [A]  time = 1.55061, size = 178, normalized size = 1.42 \begin{align*} -\frac{{\left (b e^{2} - 2 \,{\left (b c - a d\right )} e\right )}{\rm Ei}\left (\frac{e}{d x + c}\right ) -{\left (b d^{2} x^{2} - b c^{2} + 2 \, a c d + b c e +{\left (2 \, a d^{2} + b d e\right )} x\right )} e^{\left (\frac{e}{d x + c}\right )}}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e/(d*x+c))*(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x\right ) e^{\frac{e}{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e/(d*x+c))*(b*x+a),x)

[Out]

Integral((a + b*x)*exp(e/(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e/(d*x+c))*(b*x+a),x, algorithm="giac")

[Out]

integrate((b*x + a)*e^(e/(d*x + c)), x)