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

Optimal. Leaf size=145 \[ \frac{\sqrt{d} e^{a+\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (-\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}-\frac{\sqrt{d} e^{a-\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (\frac{b \left (\sqrt{d} x+\sqrt{-c}\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}+\frac{e^a b \text{Ei}(b x)}{c}-\frac{e^{a+b x}}{c x} \]

[Out]

-(E^(a + b*x)/(c*x)) + (b*E^a*ExpIntegralEi[b*x])/c + (Sqrt[d]*E^(a + (b*Sqrt[-c])/Sqrt[d])*ExpIntegralEi[-((b
*(Sqrt[-c] - Sqrt[d]*x))/Sqrt[d])])/(2*(-c)^(3/2)) - (Sqrt[d]*E^(a - (b*Sqrt[-c])/Sqrt[d])*ExpIntegralEi[(b*(S
qrt[-c] + Sqrt[d]*x))/Sqrt[d]])/(2*(-c)^(3/2))

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Rubi [A]  time = 0.35187, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2271, 2177, 2178, 2269} \[ \frac{\sqrt{d} e^{a+\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (-\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}-\frac{\sqrt{d} e^{a-\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (\frac{b \left (\sqrt{d} x+\sqrt{-c}\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}+\frac{e^a b \text{Ei}(b x)}{c}-\frac{e^{a+b x}}{c x} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)/(x^2*(c + d*x^2)),x]

[Out]

-(E^(a + b*x)/(c*x)) + (b*E^a*ExpIntegralEi[b*x])/c + (Sqrt[d]*E^(a + (b*Sqrt[-c])/Sqrt[d])*ExpIntegralEi[-((b
*(Sqrt[-c] - Sqrt[d]*x))/Sqrt[d])])/(2*(-c)^(3/2)) - (Sqrt[d]*E^(a - (b*Sqrt[-c])/Sqrt[d])*ExpIntegralEi[(b*(S
qrt[-c] + Sqrt[d]*x))/Sqrt[d]])/(2*(-c)^(3/2))

Rule 2271

Int[((F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))*(u_)^(m_.))/((a_) + (c_)*(x_)^2), x_Symbol] :> Int[ExpandIntegran
d[F^(g*(d + e*x)^n), u^m/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x] && PolynomialQ[u, x] && Intege
rQ[m]

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 2269

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d +
e*x)^n), 1/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x]

Rubi steps

\begin{align*} \int \frac{e^{a+b x}}{x^2 \left (c+d x^2\right )} \, dx &=\int \left (\frac{e^{a+b x}}{c x^2}-\frac{d e^{a+b x}}{c \left (c+d x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{e^{a+b x}}{x^2} \, dx}{c}-\frac{d \int \frac{e^{a+b x}}{c+d x^2} \, dx}{c}\\ &=-\frac{e^{a+b x}}{c x}+\frac{b \int \frac{e^{a+b x}}{x} \, dx}{c}-\frac{d \int \left (\frac{\sqrt{-c} e^{a+b x}}{2 c \left (\sqrt{-c}-\sqrt{d} x\right )}+\frac{\sqrt{-c} e^{a+b x}}{2 c \left (\sqrt{-c}+\sqrt{d} x\right )}\right ) \, dx}{c}\\ &=-\frac{e^{a+b x}}{c x}+\frac{b e^a \text{Ei}(b x)}{c}-\frac{d \int \frac{e^{a+b x}}{\sqrt{-c}-\sqrt{d} x} \, dx}{2 (-c)^{3/2}}-\frac{d \int \frac{e^{a+b x}}{\sqrt{-c}+\sqrt{d} x} \, dx}{2 (-c)^{3/2}}\\ &=-\frac{e^{a+b x}}{c x}+\frac{b e^a \text{Ei}(b x)}{c}+\frac{\sqrt{d} e^{a+\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (-\frac{b \left (\sqrt{-c}-\sqrt{d} x\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}-\frac{\sqrt{d} e^{a-\frac{b \sqrt{-c}}{\sqrt{d}}} \text{Ei}\left (\frac{b \left (\sqrt{-c}+\sqrt{d} x\right )}{\sqrt{d}}\right )}{2 (-c)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.243512, size = 133, normalized size = 0.92 \[ \frac{e^a \left (i \sqrt{d} x e^{\frac{i b \sqrt{c}}{\sqrt{d}}} \text{Ei}\left (b \left (x-\frac{i \sqrt{c}}{\sqrt{d}}\right )\right )-i \sqrt{d} x e^{-\frac{i b \sqrt{c}}{\sqrt{d}}} \text{Ei}\left (b \left (x+\frac{i \sqrt{c}}{\sqrt{d}}\right )\right )+2 b \sqrt{c} x \text{Ei}(b x)-2 \sqrt{c} e^{b x}\right )}{2 c^{3/2} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^a*(-2*Sqrt[c]*E^(b*x) + 2*b*Sqrt[c]*x*ExpIntegralEi[b*x] + I*Sqrt[d]*E^((I*b*Sqrt[c])/Sqrt[d])*x*ExpIntegra
lEi[b*(((-I)*Sqrt[c])/Sqrt[d] + x)] - (I*Sqrt[d]*x*ExpIntegralEi[b*((I*Sqrt[c])/Sqrt[d] + x)])/E^((I*b*Sqrt[c]
)/Sqrt[d])))/(2*c^(3/2)*x)

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Maple [A]  time = 0.023, size = 142, normalized size = 1. \begin{align*} b \left ( -{\frac{{{\rm e}^{bx+a}}}{bcx}}-{\frac{{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{c}}+{\frac{d}{2\,bc} \left ({{\rm e}^{{\frac{1}{d} \left ( b\sqrt{-cd}+ad \right ) }}}{\it Ei} \left ( 1,{\frac{1}{d} \left ( b\sqrt{-cd}-d \left ( bx+a \right ) +ad \right ) } \right ) -{{\rm e}^{-{\frac{1}{d} \left ( b\sqrt{-cd}-ad \right ) }}}{\it Ei} \left ( 1,-{\frac{1}{d} \left ( b\sqrt{-cd}+d \left ( bx+a \right ) -ad \right ) } \right ) \right ){\frac{1}{\sqrt{-cd}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (b x + a\right )}}{{\left (d x^{2} + c\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52924, size = 266, normalized size = 1.83 \begin{align*} \frac{2 \, b^{2} c x{\rm Ei}\left (b x\right ) e^{a} + \sqrt{-\frac{b^{2} c}{d}} d x{\rm Ei}\left (b x - \sqrt{-\frac{b^{2} c}{d}}\right ) e^{\left (a + \sqrt{-\frac{b^{2} c}{d}}\right )} - \sqrt{-\frac{b^{2} c}{d}} d x{\rm Ei}\left (b x + \sqrt{-\frac{b^{2} c}{d}}\right ) e^{\left (a - \sqrt{-\frac{b^{2} c}{d}}\right )} - 2 \, b c e^{\left (b x + a\right )}}{2 \, b c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int \frac{e^{b x}}{c x^{2} + d x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(a)*Integral(exp(b*x)/(c*x**2 + d*x**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (b x + a\right )}}{{\left (d x^{2} + c\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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