3.2.63 \(\int \frac {a^x x}{(1+b x)^2} \, dx\) [163]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2230, 2208, 2209} \begin {gather*} -\frac {a^{-1/b} \log (a) \text {Ei}\left (\frac {(b x+1) \log (a)}{b}\right )}{b^3}+\frac {a^{-1/b} \text {Ei}\left (\frac {(b x+1) \log (a)}{b}\right )}{b^2}+\frac {a^x}{b^2 (b x+1)} \end {gather*}

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2230

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] &&  !TrueQ[$UseGamma]

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*}

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Mathematica [A]
time = 0.13, size = 43, normalized size = 0.67 \begin {gather*} \frac {\frac {a^x b}{1+b x}+a^{-1/b} \text {Ei}\left (\frac {(1+b x) \log (a)}{b}\right ) (b-\log (a))}{b^3} \end {gather*}

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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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x*a^x/(1+b*x)^2,x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [A]
time = 0.04, size = 79, normalized size = 1.23

method	result	size
risch	\(-\frac {a^{-\frac {1}{b}} \expIntegral \left (1, -x \ln \left (a \right )-\frac {\ln \left (a \right )}{b}\right )}{b^{2}}+\frac {\ln \left (a \right ) a^{x}}{b^{3} \left (x \ln \left (a \right )+\frac {\ln \left (a \right )}{b}\right )}+\frac {\ln \left (a \right ) a^{-\frac {1}{b}} \expIntegral \left (1, -x \ln \left (a \right )-\frac {\ln \left (a \right )}{b}\right )}{b^{3}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.31, size = 54, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a^{x} x}{\left (b x + 1\right )^{2}}\, dx \end {gather*}

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] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a^x\,x}{{\left (b\,x+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^x*x)/(b*x + 1)^2,x)

[Out]

int((a^x*x)/(b*x + 1)^2, x)

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