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]

-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(c*x^2+b*x+a)

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Rubi [A]  time = 0.24, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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 )^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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.40, size = 111, normalized size = 1.54 \[ \frac {{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} {\rm Ei}\left (c x^{2} + b x + a\right ) - {\left (c x^{2} + b x + a + 1\right )} e^{\left (c x^{2} + b x + a\right )}}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{3}}\,{d x} \]

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)^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)

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maple [A]  time = 0.03, size = 70, normalized size = 0.97 \[ -\frac {\Ei \left (1, -c \,x^{2}-b x -a \right )}{2}-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{2 \left (c \,x^{2}+b x +a \right )^{2}}-\frac {{\mathrm e}^{c \,x^{2}+b x +a}}{2 \left (c \,x^{2}+b x +a \right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{3}}\,{d x} \]

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)^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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mupad [B]  time = 4.04, size = 62, normalized size = 0.86 \[ -\frac {\mathrm {expint}\left (-c\,x^2-b\,x-a\right )}{2}-{\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (\frac {1}{2\,\left (c\,x^2+b\,x+a\right )}+\frac {1}{2\,{\left (c\,x^2+b\,x+a\right )}^2}\right ) \]

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)^3,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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