3.36.63 \(\int \frac {2 e^{2+2 x} x^3+10 x \log (2)-200 \log ^2(2)+e^{1+x} (-x^3+(-20 x+20 x^2) \log (2))}{x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {\left (-\frac {x}{2}+x \left (e^{1+x}+\frac {10 \log (2)}{x}\right )\right )^2}{x^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 10, number of rules used = 6, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2194, 37, 2199, 2177, 2178} \begin {gather*} \frac {(x-20 \log (2))^2}{4 x^2}-e^{x+1}+e^{2 x+2}+\frac {20 e^{x+1} \log (2)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(2 + 2*x)*x^3 + 10*x*Log[2] - 200*Log[2]^2 + E^(1 + x)*(-x^3 + (-20*x + 20*x^2)*Log[2]))/x^3,x]

[Out]

-E^(1 + x) + E^(2 + 2*x) + (x - 20*Log[2])^2/(4*x^2) + (20*E^(1 + x)*Log[2])/x

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 e^{2+2 x}+\frac {10 (x-20 \log (2)) \log (2)}{x^3}-\frac {e^{1+x} \left (x^2+20 \log (2)-20 x \log (2)\right )}{x^2}\right ) \, dx\\ &=2 \int e^{2+2 x} \, dx+(10 \log (2)) \int \frac {x-20 \log (2)}{x^3} \, dx-\int \frac {e^{1+x} \left (x^2+20 \log (2)-20 x \log (2)\right )}{x^2} \, dx\\ &=e^{2+2 x}+\frac {(x-20 \log (2))^2}{4 x^2}-\int \left (e^{1+x}+\frac {20 e^{1+x} \log (2)}{x^2}-\frac {20 e^{1+x} \log (2)}{x}\right ) \, dx\\ &=e^{2+2 x}+\frac {(x-20 \log (2))^2}{4 x^2}-(20 \log (2)) \int \frac {e^{1+x}}{x^2} \, dx+(20 \log (2)) \int \frac {e^{1+x}}{x} \, dx-\int e^{1+x} \, dx\\ &=-e^{1+x}+e^{2+2 x}+\frac {(x-20 \log (2))^2}{4 x^2}+\frac {20 e^{1+x} \log (2)}{x}+20 e \text {Ei}(x) \log (2)-(20 \log (2)) \int \frac {e^{1+x}}{x} \, dx\\ &=-e^{1+x}+e^{2+2 x}+\frac {(x-20 \log (2))^2}{4 x^2}+\frac {20 e^{1+x} \log (2)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 43, normalized size = 1.59 \begin {gather*} -e^{1+x}+e^{2+2 x}-\frac {10 \log (2)}{x}+\frac {20 e^{1+x} \log (2)}{x}+\frac {100 \log ^2(2)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(2 + 2*x)*x^3 + 10*x*Log[2] - 200*Log[2]^2 + E^(1 + x)*(-x^3 + (-20*x + 20*x^2)*Log[2]))/x^3,x]

[Out]

-E^(1 + x) + E^(2 + 2*x) - (10*Log[2])/x + (20*E^(1 + x)*Log[2])/x + (100*Log[2]^2)/x^2

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fricas [A]  time = 0.72, size = 41, normalized size = 1.52 \begin {gather*} \frac {x^{2} e^{\left (2 \, x + 2\right )} - {\left (x^{2} - 20 \, x \log \relax (2)\right )} e^{\left (x + 1\right )} - 10 \, x \log \relax (2) + 100 \, \log \relax (2)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(x+1)^2+((20*x^2-20*x)*log(2)-x^3)*exp(x+1)-200*log(2)^2+10*x*log(2))/x^3,x, algorithm="fr
icas")

[Out]

(x^2*e^(2*x + 2) - (x^2 - 20*x*log(2))*e^(x + 1) - 10*x*log(2) + 100*log(2)^2)/x^2

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giac [A]  time = 0.13, size = 44, normalized size = 1.63 \begin {gather*} \frac {x^{2} e^{\left (2 \, x + 2\right )} - x^{2} e^{\left (x + 1\right )} + 20 \, x e^{\left (x + 1\right )} \log \relax (2) - 10 \, x \log \relax (2) + 100 \, \log \relax (2)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(x+1)^2+((20*x^2-20*x)*log(2)-x^3)*exp(x+1)-200*log(2)^2+10*x*log(2))/x^3,x, algorithm="gi
ac")

[Out]

(x^2*e^(2*x + 2) - x^2*e^(x + 1) + 20*x*e^(x + 1)*log(2) - 10*x*log(2) + 100*log(2)^2)/x^2

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maple [A]  time = 0.10, size = 40, normalized size = 1.48

method	result	size
risch	\(\frac {100 \ln \relax (2)^{2}-10 x \ln \relax (2)}{x^{2}}+{\mathrm e}^{2 x +2}+\frac {\left (20 \ln \relax (2)-x \right ) {\mathrm e}^{x +1}}{x}\)	\(40\)
derivativedivides	\(\frac {100 \ln \relax (2)^{2}}{x^{2}}-{\mathrm e}^{x +1}+\frac {20 \ln \relax (2) {\mathrm e}^{x +1}}{x}+{\mathrm e}^{2 x +2}-\frac {10 \ln \relax (2)}{x}\)	\(41\)
default	\(\frac {100 \ln \relax (2)^{2}}{x^{2}}-{\mathrm e}^{x +1}+\frac {20 \ln \relax (2) {\mathrm e}^{x +1}}{x}+{\mathrm e}^{2 x +2}-\frac {10 \ln \relax (2)}{x}\)	\(41\)
norman	\(\frac {x^{2} {\mathrm e}^{2 x +2}+100 \ln \relax (2)^{2}-10 x \ln \relax (2)-x^{2} {\mathrm e}^{x +1}+20 \,{\mathrm e}^{x +1} \ln \relax (2) x}{x^{2}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3*exp(x+1)^2+((20*x^2-20*x)*ln(2)-x^3)*exp(x+1)-200*ln(2)^2+10*x*ln(2))/x^3,x,method=_RETURNVERBOSE)

[Out]

(100*ln(2)^2-10*x*ln(2))/x^2+exp(2*x+2)+(20*ln(2)-x)/x*exp(x+1)

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maxima [C]  time = 0.50, size = 48, normalized size = 1.78 \begin {gather*} 20 \, {\rm Ei}\relax (x) e \log \relax (2) - 20 \, e \Gamma \left (-1, -x\right ) \log \relax (2) - \frac {10 \, \log \relax (2)}{x} + \frac {100 \, \log \relax (2)^{2}}{x^{2}} + e^{\left (2 \, x + 2\right )} - e^{\left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*exp(x+1)^2+((20*x^2-20*x)*log(2)-x^3)*exp(x+1)-200*log(2)^2+10*x*log(2))/x^3,x, algorithm="ma
xima")

[Out]

20*Ei(x)*e*log(2) - 20*e*gamma(-1, -x)*log(2) - 10*log(2)/x + 100*log(2)^2/x^2 + e^(2*x + 2) - e^(x + 1)

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mupad [B]  time = 0.12, size = 40, normalized size = 1.48 \begin {gather*} {\mathrm {e}}^{2\,x+2}-{\mathrm {e}}^{x+1}-\frac {x\,\left (10\,\ln \relax (2)-20\,{\mathrm {e}}^{x+1}\,\ln \relax (2)\right )-100\,{\ln \relax (2)}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + 1)*(log(2)*(20*x - 20*x^2) + x^3) - 10*x*log(2) + 200*log(2)^2 - 2*x^3*exp(2*x + 2))/x^3,x)

[Out]

exp(2*x + 2) - exp(x + 1) - (x*(10*log(2) - 20*exp(x + 1)*log(2)) - 100*log(2)^2)/x^2

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sympy [A]  time = 0.19, size = 37, normalized size = 1.37 \begin {gather*} \frac {x e^{2 x + 2} + \left (- x + 20 \log {\relax (2 )}\right ) e^{x + 1}}{x} + \frac {- 10 x \log {\relax (2 )} + 100 \log {\relax (2 )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3*exp(x+1)**2+((20*x**2-20*x)*ln(2)-x**3)*exp(x+1)-200*ln(2)**2+10*x*ln(2))/x**3,x)

[Out]

(x*exp(2*x + 2) + (-x + 20*log(2))*exp(x + 1))/x + (-10*x*log(2) + 100*log(2)**2)/x**2

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