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

Optimal. Leaf size=38 \[ \text {Ei}\left (c x^2+b x+a\right )-\frac {e^{a+b x+c x^2}}{a+b x+c x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {6707, 2177, 2178} \[ \text {Ei}\left (c x^2+b x+a\right )-\frac {e^{a+b x+c x^2}}{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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]

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 )^2} \, dx &=\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}}{a+b x+c x^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}}{a+b x+c x^2}+\text {Ei}\left (a+b x+c x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 35, normalized size = 0.92 \[ \text {Ei}(a+x (b+c x))-\frac {e^{a+x (b+c x)}}{a+x (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.39, size = 49, normalized size = 1.29 \[ \frac {{\left (c x^{2} + b x + a\right )} {\rm Ei}\left (c x^{2} + b x + a\right ) - e^{\left (c x^{2} + b x + a\right )}}{c x^{2} + b x + a} \]

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)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 45, normalized size = 1.18 \[ -\Ei \left (1, -c \,x^{2}-b x -a \right )-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{c \,x^{2}+b x +a} \]

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)^2,x)

[Out]

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

________________________________________________________________________________________

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

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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.99, size = 44, normalized size = 1.16 \[ -\mathrm {expint}\left (-c\,x^2-b\,x-a\right )-\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,{\mathrm {e}}^{c\,x^2}}{c\,x^2+b\,x+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- expint(- a - b*x - c*x^2) - (exp(b*x)*exp(a)*exp(c*x^2))/(a + b*x + c*x^2)

________________________________________________________________________________________

sympy [A]  time = 159.33, size = 24, normalized size = 0.63 \[ - \frac {\operatorname {E}_{2}\left (- a - b x - c x^{2}\right )}{a + b x + c x^{2}} \]

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)**2,x)

[Out]

-expint(2, -a - b*x - c*x**2)/(a + b*x + c*x**2)

________________________________________________________________________________________