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

Optimal. Leaf size=111 \[ \frac {\sqrt {\pi } \sqrt {e} (b c-a d) \text {erfi}\left (\frac {\sqrt {e}}{c+d x}\right )}{d^2}-\frac {(c+d x) (b c-a d) e^{\frac {e}{(c+d x)^2}}}{d^2}-\frac {b e \text {Ei}\left (\frac {e}{(c+d x)^2}\right )}{2 d^2}+\frac {b (c+d x)^2 e^{\frac {e}{(c+d x)^2}}}{2 d^2} \]

[Out]

-(-a*d+b*c)*exp(e/(d*x+c)^2)*(d*x+c)/d^2+1/2*b*exp(e/(d*x+c)^2)*(d*x+c)^2/d^2-1/2*b*e*Ei(e/(d*x+c)^2)/d^2+(-a*
d+b*c)*erfi(e^(1/2)/(d*x+c))*e^(1/2)*Pi^(1/2)/d^2

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Rubi [A]  time = 0.13, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2226, 2206, 2211, 2204, 2214, 2210} \[ \frac {\sqrt {\pi } \sqrt {e} (b c-a d) \text {Erfi}\left (\frac {\sqrt {e}}{c+d x}\right )}{d^2}-\frac {(c+d x) (b c-a d) e^{\frac {e}{(c+d x)^2}}}{d^2}-\frac {b e \text {Ei}\left (\frac {e}{(c+d x)^2}\right )}{2 d^2}+\frac {b (c+d x)^2 e^{\frac {e}{(c+d x)^2}}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(e/(c + d*x)^2)*(a + b*x),x]

[Out]

-(((b*c - a*d)*E^(e/(c + d*x)^2)*(c + d*x))/d^2) + (b*E^(e/(c + d*x)^2)*(c + d*x)^2)/(2*d^2) + ((b*c - a*d)*Sq
rt[e]*Sqrt[Pi]*Erfi[Sqrt[e]/(c + d*x)])/d^2 - (b*e*ExpIntegralEi[e/(c + d*x)^2])/(2*d^2)

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2211

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 85, normalized size = 0.77 \[ -\frac {2 \sqrt {\pi } \sqrt {e} (a d-b c) \text {erfi}\left (\frac {\sqrt {e}}{c+d x}\right )+(c+d x) e^{\frac {e}{(c+d x)^2}} (-2 a d+b c-b d x)+b e \text {Ei}\left (\frac {e}{(c+d x)^2}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(E^(e/(c + d*x)^2)*(c + d*x)*(b*c - 2*a*d - b*d*x) + 2*(-(b*c) + a*d)*Sqrt[e]*Sqrt[Pi]*Erfi[Sqrt[e]/(c +
d*x)] + b*e*ExpIntegralEi[e/(c + d*x)^2])/d^2

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fricas [A]  time = 0.42, size = 122, normalized size = 1.10 \[ -\frac {b e {\rm Ei}\left (\frac {e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \, \sqrt {\pi } {\left (b c d - a d^{2}\right )} \sqrt {-\frac {e}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {e}{d^{2}}}}{d x + c}\right ) - {\left (b d^{2} x^{2} + 2 \, a d^{2} x - b c^{2} + 2 \, a c d\right )} e^{\left (\frac {e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*e*Ei(e/(d^2*x^2 + 2*c*d*x + c^2)) + 2*sqrt(pi)*(b*c*d - a*d^2)*sqrt(-e/d^2)*erf(d*sqrt(-e/d^2)/(d*x +
c)) - (b*d^2*x^2 + 2*a*d^2*x - b*c^2 + 2*a*c*d)*e^(e/(d^2*x^2 + 2*c*d*x + c^2)))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 140, normalized size = 1.26 \[ -\frac {\left (\frac {\sqrt {\pi }\, e \erf \left (\frac {\sqrt {-e}}{d x +c}\right )}{\sqrt {-e}}-\left (d x +c \right ) {\mathrm e}^{\frac {e}{\left (d x +c \right )^{2}}}\right ) a -\frac {\left (\frac {\sqrt {\pi }\, e \erf \left (\frac {\sqrt {-e}}{d x +c}\right )}{\sqrt {-e}}-\left (d x +c \right ) {\mathrm e}^{\frac {e}{\left (d x +c \right )^{2}}}\right ) b c}{d}+\frac {\left (-\frac {e \Ei \left (1, -\frac {e}{\left (d x +c \right )^{2}}\right )}{2}-\frac {\left (d x +c \right )^{2} {\mathrm e}^{\frac {e}{\left (d x +c \right )^{2}}}}{2}\right ) b}{d}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(1/(d*x+c)^2*e)*(b*x+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} e^{\left (\frac {e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )} + \int \frac {{\left (b d e x^{2} + 2 \, a d e x\right )} e^{\left (\frac {e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + 2*a*x)*e^(e/(d^2*x^2 + 2*c*d*x + c^2)) + integrate((b*d*e*x^2 + 2*a*d*e*x)*e^(e/(d^2*x^2 + 2*c*d*
x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{\frac {e}{{\left (c+d\,x\right )}^2}}\,\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e/(c + d*x)^2)*(a + b*x),x)

[Out]

int(exp(e/(c + d*x)^2)*(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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