3.470 \(\int \frac{e^{d+e x} x}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=158 \[ \frac{\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{e \left (b-\sqrt{b^2-4 a c}\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c}+\frac{\left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c} \]

[Out]

((1 - b/Sqrt[b^2 - 4*a*c])*E^(d - ((b - Sqrt[b^2 - 4*a*c])*e)/(2*c))*ExpIntegralEi[(e*(b - Sqrt[b^2 - 4*a*c] +
 2*c*x))/(2*c)])/(2*c) + ((1 + b/Sqrt[b^2 - 4*a*c])*E^(d - ((b + Sqrt[b^2 - 4*a*c])*e)/(2*c))*ExpIntegralEi[(e
*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c)])/(2*c)

________________________________________________________________________________________

Rubi [A]  time = 0.214408, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2270, 2178} \[ \frac{\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{e \left (b-\sqrt{b^2-4 a c}\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c}+\frac{\left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - b/Sqrt[b^2 - 4*a*c])*E^(d - ((b - Sqrt[b^2 - 4*a*c])*e)/(2*c))*ExpIntegralEi[(e*(b - Sqrt[b^2 - 4*a*c] +
 2*c*x))/(2*c)])/(2*c) + ((1 + b/Sqrt[b^2 - 4*a*c])*E^(d - ((b + Sqrt[b^2 - 4*a*c])*e)/(2*c))*ExpIntegralEi[(e
*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c)])/(2*c)

Rule 2270

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_.) + (b_.)*(x_) + (c_)*(x_)^2), x_Symbol] :> Int[
ExpandIntegrand[F^(g*(d + e*x)^n), u^m/(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e, g, n}, x] && Poly
nomialQ[u, x] && IntegerQ[m]

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

Rubi steps

\begin{align*} \int \frac{e^{d+e x} x}{a+b x+c x^2} \, dx &=\int \left (\frac{\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d+e x}}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (1+\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d+e x}}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx\\ &=\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \int \frac{e^{d+e x}}{b-\sqrt{b^2-4 a c}+2 c x} \, dx+\left (1+\frac{b}{\sqrt{b^2-4 a c}}\right ) \int \frac{e^{d+e x}}{b+\sqrt{b^2-4 a c}+2 c x} \, dx\\ &=\frac{\left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}} \text{Ei}\left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{2 c}\right )}{2 c}+\frac{\left (1+\frac{b}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}} \text{Ei}\left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{2 c}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.178503, size = 153, normalized size = 0.97 \[ \frac{e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \left (\left (\sqrt{b^2-4 a c}-b\right ) e^{\frac{e \sqrt{b^2-4 a c}}{c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )+\left (\sqrt{b^2-4 a c}+b\right ) \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )\right )}{2 c \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(d - ((b + Sqrt[b^2 - 4*a*c])*e)/(2*c))*((-b + Sqrt[b^2 - 4*a*c])*E^((Sqrt[b^2 - 4*a*c]*e)/c)*ExpIntegralEi
[(e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c)] + (b + Sqrt[b^2 - 4*a*c])*ExpIntegralEi[(e*(b + Sqrt[b^2 - 4*a*c]
+ 2*c*x))/(2*c)]))/(2*c*Sqrt[b^2 - 4*a*c])

________________________________________________________________________________________

Maple [B]  time = 0.011, size = 685, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e*x+d)*x/(c*x^2+b*x+a),x)

[Out]

1/e^2*(-1/2*e^2*(-exp(1/2/c*(-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))*Ei(1,1/2*(-2*c*(e*x+d)-b*e+2*c*d+(-4*a*c*
e^2+b^2*e^2)^(1/2))/c)*b*e+2*exp(1/2/c*(-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))*Ei(1,1/2*(-2*c*(e*x+d)-b*e+2*c
*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*c*d+exp(-1/2*(b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*Ei(1,-1/2*(2*c*(e*x+d
)+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*b*e-2*exp(-1/2*(b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*Ei(1,-1/2*
(2*c*(e*x+d)+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*c*d+exp(1/2/c*(-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))*E
i(1,1/2*(-2*c*(e*x+d)-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*(-4*a*c*e^2+b^2*e^2)^(1/2)+exp(-1/2*(b*e-2*c*d+
(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*Ei(1,-1/2*(2*c*(e*x+d)+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*(-4*a*c*e^2+b^2
*e^2)^(1/2))/c/(-4*a*c*e^2+b^2*e^2)^(1/2)+d*e^2*(exp(1/2/c*(-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))*Ei(1,1/2*(
-2*c*(e*x+d)-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)-exp(-1/2*(b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c)*Ei(1,
-1/2*(2*c*(e*x+d)+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/c))/(-4*a*c*e^2+b^2*e^2)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x e^{\left (e x + d\right )}}{c e x^{2} + b e x + a e} + \int \frac{{\left (c x^{2} e^{d} - a e^{d}\right )} e^{\left (e x\right )}}{c^{2} e x^{4} + 2 \, b c e x^{3} + 2 \, a b e x + a^{2} e +{\left (b^{2} e + 2 \, a c e\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e*x+d)*x/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

x*e^(e*x + d)/(c*e*x^2 + b*e*x + a*e) + integrate((c*x^2*e^d - a*e^d)*e^(e*x)/(c^2*e*x^4 + 2*b*c*e*x^3 + 2*a*b
*e*x + a^2*e + (b^2*e + 2*a*c*e)*x^2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.53259, size = 491, normalized size = 3.11 \begin{align*} -\frac{{\left (b c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}} -{\left (b^{2} - 4 \, a c\right )} e\right )}{\rm Ei}\left (\frac{2 \, c e x + b e - c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}}{2 \, c}\right ) e^{\left (\frac{2 \, c d - b e + c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}}{2 \, c}\right )} -{\left (b c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}} +{\left (b^{2} - 4 \, a c\right )} e\right )}{\rm Ei}\left (\frac{2 \, c e x + b e + c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}}{2 \, c}\right ) e^{\left (\frac{2 \, c d - b e - c \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}}{2 \, c}\right )}}{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e*x+d)*x/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{d} \int \frac{x e^{e x}}{a + b x + c x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e*x+d)*x/(c*x**2+b*x+a),x)

[Out]

exp(d)*Integral(x*exp(e*x)/(a + b*x + c*x**2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(e*x+d)*x/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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