\(\int \frac {e^{a+b x+c x^2} (b+2 c x)}{(a+b x+c x^2)^2} \, dx\) [625]

Optimal result

Integrand size = 31, antiderivative size = 38 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {e^{a+b x+c x^2}}{a+b x+c x^2}+\operatorname {ExpIntegralEi}\left (a+b x+c x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {6839, 2208, 2209} \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=\operatorname {ExpIntegralEi}\left (c x^2+b x+a\right )-\frac {e^{a+b x+c x^2}}{a+b x+c x^2} \]

[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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 6839

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*} \text {integral}& = \text {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}+\text {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] (verified)

Time = 0.38 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {e^{a+x (b+c x)}}{a+x (b+c x)}+\operatorname {ExpIntegralEi}(a+x (b+c x)) \]

[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)]

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{c \,x^{2}+b x +a}-\operatorname {Ei}_{1}\left (-c \,x^{2}-b x -a \right )\)	\(45\)
default	\(-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{c \,x^{2}+b x +a}-\operatorname {Ei}_{1}\left (-c \,x^{2}-b x -a \right )\)	\(45\)
risch	\(-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{c \,x^{2}+b x +a}-\operatorname {Ei}_{1}\left (-c \,x^{2}-b x -a \right )\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=\frac {{\left (c x^{2} + b x + a\right )} {\rm Ei}\left (c x^{2} + b x + a\right ) - e^{\left (c x^{2} + b x + a\right )}}{c x^{2} + b x + a} \]

[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)

Sympy [A] (verification not implemented)

Time = 77.95 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=- \frac {\operatorname {E}_{2}\left (- a - b x - c x^{2}\right )}{a + b x + c x^{2}} \]

[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)

Maxima [F]

\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=\int { \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{2}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^2} \, dx=-\mathrm {expint}\left (-c\,x^2-b\,x-a\right )-\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,{\mathrm {e}}^{c\,x^2}}{c\,x^2+b\,x+a} \]

[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)