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

Optimal. Leaf size=186 \[ -\frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{e \left (b-\sqrt{b^2-4 a c}\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c^2}-\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c^2}+\frac{e^{d+e x}}{c e} \]

[Out]

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

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Rubi [A]  time = 0.407889, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2270, 2194, 2178} \[ -\frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) e^{d-\frac{e \left (b-\sqrt{b^2-4 a c}\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c^2}-\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )}{2 c^2}+\frac{e^{d+e x}}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.519179, size = 217, normalized size = 1.17 \[ -\frac{e^{d-\frac{e \left (\sqrt{b^2-4 a c}+b\right )}{2 c}} \left (e \left (b \sqrt{b^2-4 a c}+2 a c-b^2\right ) e^{\frac{e \sqrt{b^2-4 a c}}{c}} \text{Ei}\left (\frac{e \left (b+2 c x-\sqrt{b^2-4 a c}\right )}{2 c}\right )+e \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \text{Ei}\left (\frac{e \left (b+2 c x+\sqrt{b^2-4 a c}\right )}{2 c}\right )-2 c \sqrt{b^2-4 a c} e^{\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c}}\right )}{2 c^2 e \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.017, size = 1730, normalized size = 9.3 \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^2/(c*x^2+b*x+a),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.64275, size = 583, normalized size = 3.13 \begin{align*} -\frac{{\left ({\left (b^{3} - 4 \, a b c\right )} e -{\left (b^{2} c - 2 \, a c^{2}\right )} \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}\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 ({\left (b^{3} - 4 \, a b c\right )} e +{\left (b^{2} c - 2 \, a c^{2}\right )} \sqrt{\frac{{\left (b^{2} - 4 \, a c\right )} e^{2}}{c^{2}}}\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^{\left (e x + d\right )}}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} 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^2/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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