3.626 \(\int \frac{e^{a+b x+c x^2} (b+2 c x)}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=72 \[ \frac{1}{2} \text{Ei}\left (c x^2+b x+a\right )-\frac{e^{a+b x+c x^2}}{2 \left (a+b x+c x^2\right )}-\frac{e^{a+b x+c x^2}}{2 \left (a+b x+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.241148, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {6707, 2177, 2178} \[ \frac{1}{2} \text{Ei}\left (c x^2+b x+a\right )-\frac{e^{a+b x+c x^2}}{2 \left (a+b x+c x^2\right )}-\frac{e^{a+b x+c x^2}}{2 \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^3,x]

[Out]

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

Rule 6707

Int[(F_)^(v_)*(u_)*(w_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Dist[q, Subst[Int[x^m*F^x,
x], x, v], x] /;  !FalseQ[q]] /; FreeQ[{F, m}, x] && EqQ[w, v]

Rule 2177

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] &&  !$UseGamma ===
True

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

Mathematica [A]  time = 0.0688787, size = 50, normalized size = 0.69 \[ \frac{1}{2} \left (\text{Ei}(a+x (b+c x))-\frac{e^{a+x (b+c x)} \left (a+b x+c x^2+1\right )}{(a+x (b+c x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.041, size = 70, normalized size = 1. \begin{align*} -{\frac{{{\rm e}^{c{x}^{2}+bx+a}}}{2\, \left ( c{x}^{2}+bx+a \right ) ^{2}}}-{\frac{{{\rm e}^{c{x}^{2}+bx+a}}}{2\,c{x}^{2}+2\,bx+2\,a}}-{\frac{{\it Ei} \left ( 1,-c{x}^{2}-bx-a \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.906188, size = 252, normalized size = 3.5 \begin{align*} \frac{{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}{\rm Ei}\left (c x^{2} + b x + a\right ) -{\left (c x^{2} + b x + a + 1\right )} e^{\left (c x^{2} + b x + a\right )}}{2 \,{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x**2+b*x+a)*(2*c*x+b)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

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