3.99.67 \(\int \frac {2 x^2+e^{-4+5 x} (2-10 x+(-9+45 x+15 x^2) \log (2))}{x^2} \, dx\)

Optimal. Leaf size=25 \[ 3+2 x+\frac {e^{-4+5 x} (-2+3 (3+x) \log (2))}{x} \]

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Rubi [C]  time = 0.16, antiderivative size = 63, normalized size of antiderivative = 2.52, number of steps used = 8, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {14, 2199, 2177, 2178, 2194} \begin {gather*} -\frac {5 (2-\log (512)) \text {Ei}(5 x)}{e^4}+\frac {5 (2-9 \log (2)) \text {Ei}(5 x)}{e^4}+2 x+3 e^{5 x-4} \log (2)-\frac {e^{5 x-4} (2-\log (512))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + E^(-4 + 5*x)*(2 - 10*x + (-9 + 45*x + 15*x^2)*Log[2]))/x^2,x]

[Out]

2*x + (5*ExpIntegralEi[5*x]*(2 - 9*Log[2]))/E^4 + 3*E^(-4 + 5*x)*Log[2] - (E^(-4 + 5*x)*(2 - Log[512]))/x - (5
*ExpIntegralEi[5*x]*(2 - Log[512]))/E^4

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2+\frac {e^{-4+5 x} \left (2-9 \log (2)+15 x^2 \log (2)-5 x (2-\log (512))\right )}{x^2}\right ) \, dx\\ &=2 x+\int \frac {e^{-4+5 x} \left (2-9 \log (2)+15 x^2 \log (2)-5 x (2-\log (512))\right )}{x^2} \, dx\\ &=2 x+\int \left (\frac {e^{-4+5 x} (2-9 \log (2))}{x^2}+15 e^{-4+5 x} \log (2)+\frac {5 e^{-4+5 x} (-2+\log (512))}{x}\right ) \, dx\\ &=2 x+(2-9 \log (2)) \int \frac {e^{-4+5 x}}{x^2} \, dx+(15 \log (2)) \int e^{-4+5 x} \, dx-(5 (2-\log (512))) \int \frac {e^{-4+5 x}}{x} \, dx\\ &=2 x+3 e^{-4+5 x} \log (2)-\frac {e^{-4+5 x} (2-\log (512))}{x}-\frac {5 \text {Ei}(5 x) (2-\log (512))}{e^4}+(5 (2-9 \log (2))) \int \frac {e^{-4+5 x}}{x} \, dx\\ &=2 x+\frac {5 \text {Ei}(5 x) (2-9 \log (2))}{e^4}+3 e^{-4+5 x} \log (2)-\frac {e^{-4+5 x} (2-\log (512))}{x}-\frac {5 \text {Ei}(5 x) (2-\log (512))}{e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 34, normalized size = 1.36 \begin {gather*} 2 x+\frac {e^{-4+5 x} \left (15 x^2 \log (2)+x (-10+45 \log (2))\right )}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + E^(-4 + 5*x)*(2 - 10*x + (-9 + 45*x + 15*x^2)*Log[2]))/x^2,x]

[Out]

2*x + (E^(-4 + 5*x)*(15*x^2*Log[2] + x*(-10 + 45*Log[2])))/(5*x^2)

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fricas [A]  time = 0.67, size = 26, normalized size = 1.04 \begin {gather*} \frac {2 \, x^{2} + {\left (3 \, {\left (x + 3\right )} \log \relax (2) - 2\right )} e^{\left (5 \, x - 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2+45*x-9)*log(2)-10*x+2)*exp(5*x-4)+2*x^2)/x^2,x, algorithm="fricas")

[Out]

(2*x^2 + (3*(x + 3)*log(2) - 2)*e^(5*x - 4))/x

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giac [A]  time = 0.16, size = 37, normalized size = 1.48 \begin {gather*} \frac {{\left (2 \, x^{2} e^{4} + 3 \, x e^{\left (5 \, x\right )} \log \relax (2) + 9 \, e^{\left (5 \, x\right )} \log \relax (2) - 2 \, e^{\left (5 \, x\right )}\right )} e^{\left (-4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2+45*x-9)*log(2)-10*x+2)*exp(5*x-4)+2*x^2)/x^2,x, algorithm="giac")

[Out]

(2*x^2*e^4 + 3*x*e^(5*x)*log(2) + 9*e^(5*x)*log(2) - 2*e^(5*x))*e^(-4)/x

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maple [A]  time = 0.09, size = 26, normalized size = 1.04

method	result	size
risch	\(2 x +\frac {\left (3 x \ln \relax (2)+9 \ln \relax (2)-2\right ) {\mathrm e}^{5 x -4}}{x}\)	\(26\)
norman	\(\frac {\left (9 \ln \relax (2)-2\right ) {\mathrm e}^{5 x -4}+2 x^{2}+3 \ln \relax (2) {\mathrm e}^{5 x -4} x}{x}\)	\(35\)
derivativedivides	\(2 x -\frac {8}{5}-\frac {2 \,{\mathrm e}^{5 x -4}}{x}+\frac {9 \ln \relax (2) {\mathrm e}^{5 x -4}}{x}+3 \ln \relax (2) {\mathrm e}^{5 x -4}\)	\(40\)
default	\(2 x -\frac {8}{5}-\frac {2 \,{\mathrm e}^{5 x -4}}{x}+\frac {9 \ln \relax (2) {\mathrm e}^{5 x -4}}{x}+3 \ln \relax (2) {\mathrm e}^{5 x -4}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((15*x^2+45*x-9)*ln(2)-10*x+2)*exp(5*x-4)+2*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

2*x+(3*x*ln(2)+9*ln(2)-2)/x*exp(5*x-4)

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maxima [C]  time = 0.42, size = 52, normalized size = 2.08 \begin {gather*} 45 \, {\rm Ei}\left (5 \, x\right ) e^{\left (-4\right )} \log \relax (2) - 45 \, e^{\left (-4\right )} \Gamma \left (-1, -5 \, x\right ) \log \relax (2) - 10 \, {\rm Ei}\left (5 \, x\right ) e^{\left (-4\right )} + 10 \, e^{\left (-4\right )} \Gamma \left (-1, -5 \, x\right ) + 3 \, e^{\left (5 \, x - 4\right )} \log \relax (2) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x^2+45*x-9)*log(2)-10*x+2)*exp(5*x-4)+2*x^2)/x^2,x, algorithm="maxima")

[Out]

45*Ei(5*x)*e^(-4)*log(2) - 45*e^(-4)*gamma(-1, -5*x)*log(2) - 10*Ei(5*x)*e^(-4) + 10*e^(-4)*gamma(-1, -5*x) +
3*e^(5*x - 4)*log(2) + 2*x

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mupad [B]  time = 5.69, size = 27, normalized size = 1.08 \begin {gather*} 2\,x+{\mathrm {e}}^{5\,x-4}\,\ln \relax (8)+\frac {{\mathrm {e}}^{5\,x-4}\,\left (\ln \left (512\right )-2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5*x - 4)*(log(2)*(45*x + 15*x^2 - 9) - 10*x + 2) + 2*x^2)/x^2,x)

[Out]

2*x + exp(5*x - 4)*log(8) + (exp(5*x - 4)*(log(512) - 2))/x

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sympy [A]  time = 0.15, size = 24, normalized size = 0.96 \begin {gather*} 2 x + \frac {\left (3 x \log {\relax (2 )} - 2 + 9 \log {\relax (2 )}\right ) e^{5 x - 4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15*x**2+45*x-9)*ln(2)-10*x+2)*exp(5*x-4)+2*x**2)/x**2,x)

[Out]

2*x + (3*x*log(2) - 2 + 9*log(2))*exp(5*x - 4)/x

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