3.47 \(\int 5^x x \, dx\)

Optimal. Leaf size=19 \[ \frac {5^x x}{\log (5)}-\frac {5^x}{\log ^2(5)} \]

[Out]

-5^x/ln(5)^2+5^x*x/ln(5)

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2176, 2194} \[ \frac {5^x x}{\log (5)}-\frac {5^x}{\log ^2(5)} \]

Antiderivative was successfully verified.

[In]

Int[5^x*x,x]

[Out]

-(5^x/Log[5]^2) + (5^x*x)/Log[5]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {align*} \int 5^x x \, dx &=\frac {5^x x}{\log (5)}-\frac {\int 5^x \, dx}{\log (5)}\\ &=-\frac {5^x}{\log ^2(5)}+\frac {5^x x}{\log (5)}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.74 \[ \frac {5^x (x \log (5)-1)}{\log ^2(5)} \]

Antiderivative was successfully verified.

[In]

Integrate[5^x*x,x]

[Out]

(5^x*(-1 + x*Log[5]))/Log[5]^2

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fricas [A]  time = 0.42, size = 14, normalized size = 0.74 \[ \frac {{\left (x \log \relax (5) - 1\right )} 5^{x}}{\log \relax (5)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5^x*x,x, algorithm="fricas")

[Out]

(x*log(5) - 1)*5^x/log(5)^2

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giac [A]  time = 0.86, size = 14, normalized size = 0.74 \[ \frac {{\left (x \log \relax (5) - 1\right )} 5^{x}}{\log \relax (5)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5^x*x,x, algorithm="giac")

[Out]

(x*log(5) - 1)*5^x/log(5)^2

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maple [A]  time = 0.01, size = 15, normalized size = 0.79 \[ \frac {\left (\ln \relax (5) x -1\right ) 5^{x}}{\ln \relax (5)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5^x*x,x)

[Out]

(ln(5)*x-1)*5^x/ln(5)^2

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maxima [A]  time = 0.97, size = 14, normalized size = 0.74 \[ \frac {{\left (x \log \relax (5) - 1\right )} 5^{x}}{\log \relax (5)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5^x*x,x, algorithm="maxima")

[Out]

(x*log(5) - 1)*5^x/log(5)^2

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mupad [B]  time = 0.02, size = 14, normalized size = 0.74 \[ \frac {5^x\,\left (x\,\ln \relax (5)-1\right )}{{\ln \relax (5)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5^x*x,x)

[Out]

(5^x*(x*log(5) - 1))/log(5)^2

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sympy [A]  time = 0.10, size = 14, normalized size = 0.74 \[ \frac {5^{x} \left (x \log {\relax (5 )} - 1\right )}{\log {\relax (5 )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5**x*x,x)

[Out]

5**x*(x*log(5) - 1)/log(5)**2

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