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

Optimal. Leaf size=107 \[ -\frac {d e e^{\frac {b e}{b c-a d}} \text {Ei}\left (-\frac {d e (a+b x)}{(b c-a d) (c+d x)}\right )}{(b c-a d)^2}-\frac {d e^{\frac {e}{c+d x}}}{b (b c-a d)}-\frac {e^{\frac {e}{c+d x}}}{b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.54, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2223, 6742, 2222, 2210, 2228, 2178, 2209} \[ -\frac {d e e^{\frac {b e}{b c-a d}} \text {Ei}\left (-\frac {d e (a+b x)}{(b c-a d) (c+d x)}\right )}{(b c-a d)^2}-\frac {d e^{\frac {e}{c+d x}}}{b (b c-a d)}-\frac {e^{\frac {e}{c+d x}}}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[E^(e/(c + d*x))/(a + b*x)^2,x]

[Out]

-((d*E^(e/(c + d*x)))/(b*(b*c - a*d))) - E^(e/(c + d*x))/(b*(a + b*x)) - (d*e*E^((b*e)/(b*c - a*d))*ExpIntegra
lEi[-((d*e*(a + b*x))/((b*c - a*d)*(c + d*x)))])/(b*c - a*d)^2

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 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 2222

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

Rule 2223

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m + 1)*
F^(a + b/(c + d*x)))/(f*(m + 1)), x] + Dist[(b*d*Log[F])/(f*(m + 1)), Int[((e + f*x)^(m + 1)*F^(a + b/(c + d*x
)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 105, normalized size = 0.98 \[ -\frac {d e e^{\frac {b e}{b c-a d}} \text {Ei}\left (\frac {e}{c+d x}-\frac {b e}{b c-a d}\right )}{(a d-b c)^2}-\frac {d e^{\frac {e}{c+d x}}}{b (b c-a d)}-\frac {e^{\frac {e}{c+d x}}}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*E^(e/(c + d*x)))/(b*(b*c - a*d))) - E^(e/(c + d*x))/(b*(a + b*x)) - (d*e*E^((b*e)/(b*c - a*d))*ExpIntegra
lEi[-((b*e)/(b*c - a*d)) + e/(c + d*x)])/(-(b*c) + a*d)^2

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fricas [A]  time = 0.42, size = 154, normalized size = 1.44 \[ -\frac {{\left (b d e x + a d e\right )} {\rm Ei}\left (-\frac {b d e x + a d e}{b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x}\right ) e^{\left (\frac {b e}{b c - a d}\right )} + {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} e^{\left (\frac {e}{d x + c}\right )}}{a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.34, size = 345, normalized size = 3.22 \[ -\frac {{\left (b {\rm Ei}\left (-\frac {b e - \frac {b c e}{d x + c} + \frac {a d e}{d x + c}}{b c - a d}\right ) e^{\left (\frac {b e}{b c - a d} + 3\right )} - \frac {b c {\rm Ei}\left (-\frac {b e - \frac {b c e}{d x + c} + \frac {a d e}{d x + c}}{b c - a d}\right ) e^{\left (\frac {b e}{b c - a d} + 3\right )}}{d x + c} + \frac {a d {\rm Ei}\left (-\frac {b e - \frac {b c e}{d x + c} + \frac {a d e}{d x + c}}{b c - a d}\right ) e^{\left (\frac {b e}{b c - a d} + 3\right )}}{d x + c} + b c e^{\left (\frac {e}{d x + c} + 2\right )} - a d e^{\left (\frac {e}{d x + c} + 2\right )}\right )} d e^{\left (-1\right )}}{b^{3} c^{2} e - \frac {b^{3} c^{3} e}{d x + c} - 2 \, a b^{2} c d e + \frac {3 \, a b^{2} c^{2} d e}{d x + c} + a^{2} b d^{2} e - \frac {3 \, a^{2} b c d^{2} e}{d x + c} + \frac {a^{3} d^{3} e}{d x + c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 97, normalized size = 0.91 \[ -\frac {\left (-\Ei \left (1, -\frac {b e}{a d -b c}-\frac {e}{d x +c}\right ) {\mathrm e}^{-\frac {b e}{a d -b c}}-\frac {{\mathrm e}^{\frac {e}{d x +c}}}{\frac {b e}{a d -b c}+\frac {e}{d x +c}}\right ) d e}{\left (a d -b c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e/(c + d*x))/(a + b*x)^2,x)

[Out]

int(exp(e/(c + d*x))/(a + b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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