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]

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

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Rubi [A]  time = 0.13, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.40, size = 83, normalized size = 0.66 \[ -\frac {{\left (b e^{2} - 2 \, {\left (b c - a d\right )} e\right )} {\rm Ei}\left (\frac {e}{d x + c}\right ) - {\left (b d^{2} x^{2} - b c^{2} + 2 \, a c d + b c e + {\left (2 \, a d^{2} + b d e\right )} x\right )} e^{\left (\frac {e}{d x + c}\right )}}{2 \, d^{2}} \]

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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giac [A]  time = 0.44, size = 171, normalized size = 1.37 \[ \frac {{\left (d x + c\right )}^{2} {\left (\frac {2 \, b c {\rm Ei}\left (\frac {e}{d x + c}\right ) e^{4}}{{\left (d x + c\right )}^{2}} - \frac {2 \, a d {\rm Ei}\left (\frac {e}{d x + c}\right ) e^{4}}{{\left (d x + c\right )}^{2}} + b e^{\left (\frac {e}{d x + c} + 3\right )} - \frac {2 \, b c e^{\left (\frac {e}{d x + c} + 3\right )}}{d x + c} + \frac {2 \, a d e^{\left (\frac {e}{d x + c} + 3\right )}}{d x + c} - \frac {b {\rm Ei}\left (\frac {e}{d x + c}\right ) e^{5}}{{\left (d x + c\right )}^{2}} + \frac {b e^{\left (\frac {e}{d x + c} + 4\right )}}{d x + c}\right )} e^{\left (-3\right )}}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d*x + c)^2*(2*b*c*Ei(e/(d*x + c))*e^4/(d*x + c)^2 - 2*a*d*Ei(e/(d*x + c))*e^4/(d*x + c)^2 + b*e^(e/(d*x +
c) + 3) - 2*b*c*e^(e/(d*x + c) + 3)/(d*x + c) + 2*a*d*e^(e/(d*x + c) + 3)/(d*x + c) - b*Ei(e/(d*x + c))*e^5/(
d*x + c)^2 + b*e^(e/(d*x + c) + 4)/(d*x + c))*e^(-3)/d^2

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maple [A]  time = 0.02, size = 150, normalized size = 1.20 \[ -\frac {\left (\left (-\Ei \left (1, -\frac {e}{d x +c}\right )-\frac {\left (d x +c \right ) {\mathrm e}^{\frac {e}{d x +c}}}{e}\right ) a -\frac {\left (-\Ei \left (1, -\frac {e}{d x +c}\right )-\frac {\left (d x +c \right ) {\mathrm e}^{\frac {e}{d x +c}}}{e}\right ) b c}{d}+\frac {\left (-\frac {\Ei \left (1, -\frac {e}{d x +c}\right )}{2}-\frac {\left (d x +c \right ) {\mathrm e}^{\frac {e}{d x +c}}}{2 e}-\frac {\left (d x +c \right )^{2} {\mathrm e}^{\frac {e}{d x +c}}}{2 e^{2}}\right ) b e}{d}\right ) e}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(1/(d*x+c)*e)*(b*x+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b d x^{2} + {\left (2 \, a d + b e\right )} x\right )} e^{\left (\frac {e}{d x + c}\right )}}{2 \, d} + \int -\frac {{\left (b c^{2} e - {\left (2 \, a d^{2} e - {\left (2 \, c d e - d e^{2}\right )} b\right )} x\right )} e^{\left (\frac {e}{d x + c}\right )}}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}}\,{d x} \]

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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mupad [B]  time = 3.67, size = 153, normalized size = 1.22 \[ \frac {\frac {{\mathrm {e}}^{\frac {e}{c+d\,x}}\,\left (2\,a\,c^2\,d-b\,c^3+b\,c^2\,e\right )}{2\,d^2}+x\,{\mathrm {e}}^{\frac {e}{c+d\,x}}\,\left (2\,a\,c-\frac {\frac {b\,c^2}{2}-b\,c\,e}{d}\right )+x^2\,{\mathrm {e}}^{\frac {e}{c+d\,x}}\,\left (a\,d+\frac {b\,c}{2}+\frac {b\,e}{2}\right )+\frac {b\,d\,x^3\,{\mathrm {e}}^{\frac {e}{c+d\,x}}}{2}}{c+d\,x}-\frac {\mathrm {ei}\left (\frac {e}{c+d\,x}\right )\,\left (b\,e^2+2\,a\,d\,e-2\,b\,c\,e\right )}{2\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e/(c + d*x))*(a + b*x),x)

[Out]

((exp(e/(c + d*x))*(2*a*c^2*d - b*c^3 + b*c^2*e))/(2*d^2) + x*exp(e/(c + d*x))*(2*a*c - ((b*c^2)/2 - b*c*e)/d)
+ x^2*exp(e/(c + d*x))*(a*d + (b*c)/2 + (b*e)/2) + (b*d*x^3*exp(e/(c + d*x)))/2)/(c + d*x) - (ei(e/(c + d*x))
*(b*e^2 + 2*a*d*e - 2*b*c*e))/(2*d^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x\right ) e^{\frac {e}{c + d x}}\, dx \]

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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