3.163 \(\int \frac{a^x x}{(1+b x)^2} \, dx\)

Optimal. Leaf size=64 \[ \frac{a^{-1/b} \text{ExpIntegralEi}\left (\frac{\log (a) (b x+1)}{b}\right )}{b^2}-\frac{a^{-1/b} \log (a) \text{ExpIntegralEi}\left (\frac{\log (a) (b x+1)}{b}\right )}{b^3}+\frac{a^x}{b^2 (b x+1)} \]

[Out]

a^x/(b^2*(1 + b*x)) + ExpIntegralEi[((1 + b*x)*Log[a])/b]/(a^b^(-1)*b^2) - (ExpIntegralEi[((1 + b*x)*Log[a])/b
]*Log[a])/(a^b^(-1)*b^3)

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Rubi [A]  time = 0.0974545, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2199, 2177, 2178} \[ \frac{a^{-1/b} \text{ExpIntegralEi}\left (\frac{\log (a) (b x+1)}{b}\right )}{b^2}-\frac{a^{-1/b} \log (a) \text{ExpIntegralEi}\left (\frac{\log (a) (b x+1)}{b}\right )}{b^3}+\frac{a^x}{b^2 (b x+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a^x*x)/(1 + b*x)^2,x]

[Out]

a^x/(b^2*(1 + b*x)) + ExpIntegralEi[((1 + b*x)*Log[a])/b]/(a^b^(-1)*b^2) - (ExpIntegralEi[((1 + b*x)*Log[a])/b
]*Log[a])/(a^b^(-1)*b^3)

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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

Rubi steps

\begin{align*} \int \frac{a^x x}{(1+b x)^2} \, dx &=\int \left (-\frac{a^x}{b (1+b x)^2}+\frac{a^x}{b (1+b x)}\right ) \, dx\\ &=-\frac{\int \frac{a^x}{(1+b x)^2} \, dx}{b}+\frac{\int \frac{a^x}{1+b x} \, dx}{b}\\ &=\frac{a^x}{b^2 (1+b x)}+\frac{a^{-1/b} \text{Ei}\left (\frac{(1+b x) \log (a)}{b}\right )}{b^2}-\frac{\log (a) \int \frac{a^x}{1+b x} \, dx}{b^2}\\ &=\frac{a^x}{b^2 (1+b x)}+\frac{a^{-1/b} \text{Ei}\left (\frac{(1+b x) \log (a)}{b}\right )}{b^2}-\frac{a^{-1/b} \text{Ei}\left (\frac{(1+b x) \log (a)}{b}\right ) \log (a)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.126643, size = 43, normalized size = 0.67 \[ \frac{a^{-1/b} (b-\log (a)) \text{ExpIntegralEi}\left (\frac{\log (a) (b x+1)}{b}\right )+\frac{b a^x}{b x+1}}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^x*x)/(1 + b*x)^2,x]

[Out]

((a^x*b)/(1 + b*x) + (ExpIntegralEi[((1 + b*x)*Log[a])/b]*(b - Log[a]))/a^b^(-1))/b^3

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Maple [A]  time = 0.032, size = 79, normalized size = 1.2 \begin{align*} -{\frac{1}{{b}^{2}}{a}^{-{b}^{-1}}{\it Ei} \left ( 1,-x\ln \left ( a \right ) -{\frac{\ln \left ( a \right ) }{b}} \right ) }+{\frac{{a}^{x}\ln \left ( a \right ) }{{b}^{3}} \left ( x\ln \left ( a \right ) +{\frac{\ln \left ( a \right ) }{b}} \right ) ^{-1}}+{\frac{\ln \left ( a \right ) }{{b}^{3}}{a}^{-{b}^{-1}}{\it Ei} \left ( 1,-x\ln \left ( a \right ) -{\frac{\ln \left ( a \right ) }{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x*x/(b*x+1)^2,x)

[Out]

-1/b^2*a^(-1/b)*Ei(1,-x*ln(a)-ln(a)/b)+ln(a)/b^3*a^x/(x*ln(a)+ln(a)/b)+ln(a)/b^3*a^(-1/b)*Ei(1,-x*ln(a)-ln(a)/
b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{x} x}{b^{2} x^{2} \log \left (a\right ) + 2 \, b x \log \left (a\right ) + \log \left (a\right )} + \int \frac{{\left (b x - 1\right )} a^{x}}{b^{3} x^{3} \log \left (a\right ) + 3 \, b^{2} x^{2} \log \left (a\right ) + 3 \, b x \log \left (a\right ) + \log \left (a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*x/(b*x+1)^2,x, algorithm="maxima")

[Out]

a^x*x/(b^2*x^2*log(a) + 2*b*x*log(a) + log(a)) + integrate((b*x - 1)*a^x/(b^3*x^3*log(a) + 3*b^2*x^2*log(a) +
3*b*x*log(a) + log(a)), x)

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Fricas [A]  time = 1.76654, size = 117, normalized size = 1.83 \begin{align*} \frac{a^{x} b + \frac{{\left (b^{2} x -{\left (b x + 1\right )} \log \left (a\right ) + b\right )}{\rm Ei}\left (\frac{{\left (b x + 1\right )} \log \left (a\right )}{b}\right )}{a^{\left (\frac{1}{b}\right )}}}{b^{4} x + b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*x/(b*x+1)^2,x, algorithm="fricas")

[Out]

(a^x*b + (b^2*x - (b*x + 1)*log(a) + b)*Ei((b*x + 1)*log(a)/b)/a^(1/b))/(b^4*x + b^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a^{x} x}{\left (b x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a**x*x/(b*x+1)**2,x)

[Out]

Integral(a**x*x/(b*x + 1)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a^{x} x}{{\left (b x + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*x/(b*x+1)^2,x, algorithm="giac")

[Out]

integrate(a^x*x/(b*x + 1)^2, x)