3.80.36 \(\int \frac {e^x (-1+x) \log (4)+(-10+10 e^8 x^2-40 e^4 x^3+30 x^4) \log (4)}{2 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (5+\frac {e^x}{2}+5 \left (e^4-x\right )^2 x^2\right ) \log (4)}{x} \]

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Rubi [A]  time = 0.07, antiderivative size = 45, normalized size of antiderivative = 1.55, number of steps used = 6, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2197} \begin {gather*} 5 x^3 \log (4)-10 e^4 x^2 \log (4)+5 e^8 x \log (4)+\frac {e^x \log (4)}{2 x}+\frac {5 \log (4)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-1 + x)*Log[4] + (-10 + 10*E^8*x^2 - 40*E^4*x^3 + 30*x^4)*Log[4])/(2*x^2),x]

[Out]

(5*Log[4])/x + (E^x*Log[4])/(2*x) + 5*E^8*x*Log[4] - 10*E^4*x^2*Log[4] + 5*x^3*Log[4]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^x (-1+x) \log (4)+\left (-10+10 e^8 x^2-40 e^4 x^3+30 x^4\right ) \log (4)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^x (-1+x) \log (4)}{x^2}+\frac {10 \left (-1+e^8 x^2-4 e^4 x^3+3 x^4\right ) \log (4)}{x^2}\right ) \, dx\\ &=\frac {1}{2} \log (4) \int \frac {e^x (-1+x)}{x^2} \, dx+(5 \log (4)) \int \frac {-1+e^8 x^2-4 e^4 x^3+3 x^4}{x^2} \, dx\\ &=\frac {e^x \log (4)}{2 x}+(5 \log (4)) \int \left (e^8-\frac {1}{x^2}-4 e^4 x+3 x^2\right ) \, dx\\ &=\frac {5 \log (4)}{x}+\frac {e^x \log (4)}{2 x}+5 e^8 x \log (4)-10 e^4 x^2 \log (4)+5 x^3 \log (4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 1.31 \begin {gather*} \frac {1}{2} \left (\frac {10}{x}+\frac {e^x}{x}+10 e^8 x-20 e^4 x^2+10 x^3\right ) \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 + x)*Log[4] + (-10 + 10*E^8*x^2 - 40*E^4*x^3 + 30*x^4)*Log[4])/(2*x^2),x]

[Out]

((10/x + E^x/x + 10*E^8*x - 20*E^4*x^2 + 10*x^3)*Log[4])/2

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fricas [A]  time = 0.55, size = 32, normalized size = 1.10 \begin {gather*} \frac {10 \, {\left (x^{4} - 2 \, x^{3} e^{4} + x^{2} e^{8} + 1\right )} \log \relax (2) + e^{x} \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(x-1)*log(2)*exp(x)+2*(10*x^2*exp(4)^2-40*x^3*exp(4)+30*x^4-10)*log(2))/x^2,x, algorithm="fri
cas")

[Out]

(10*(x^4 - 2*x^3*e^4 + x^2*e^8 + 1)*log(2) + e^x*log(2))/x

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giac [A]  time = 0.20, size = 39, normalized size = 1.34 \begin {gather*} \frac {10 \, x^{4} \log \relax (2) - 20 \, x^{3} e^{4} \log \relax (2) + 10 \, x^{2} e^{8} \log \relax (2) + e^{x} \log \relax (2) + 10 \, \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(x-1)*log(2)*exp(x)+2*(10*x^2*exp(4)^2-40*x^3*exp(4)+30*x^4-10)*log(2))/x^2,x, algorithm="gia
c")

[Out]

(10*x^4*log(2) - 20*x^3*e^4*log(2) + 10*x^2*e^8*log(2) + e^x*log(2) + 10*log(2))/x

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maple [A]  time = 0.06, size = 40, normalized size = 1.38

method	result	size
risch	\(10 x^{3} \ln \relax (2)+10 \,{\mathrm e}^{8} \ln \relax (2) x +\frac {10 \ln \relax (2)}{x}+\frac {\ln \relax (2) {\mathrm e}^{x}}{x}-20 x^{2} {\mathrm e}^{4} \ln \relax (2)\)	\(40\)
default	\(10 x^{3} \ln \relax (2)+10 \,{\mathrm e}^{8} \ln \relax (2) x +\frac {10 \ln \relax (2)}{x}+\frac {\ln \relax (2) {\mathrm e}^{x}}{x}-20 x^{2} {\mathrm e}^{4} \ln \relax (2)\)	\(42\)
norman	\(\frac {{\mathrm e}^{x} \ln \relax (2)+10 x^{4} \ln \relax (2)+10 x^{2} {\mathrm e}^{8} \ln \relax (2)-20 x^{3} {\mathrm e}^{4} \ln \relax (2)+10 \ln \relax (2)}{x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*(x-1)*ln(2)*exp(x)+2*(10*x^2*exp(4)^2-40*x^3*exp(4)+30*x^4-10)*ln(2))/x^2,x,method=_RETURNVERBOSE)

[Out]

10*x^3*ln(2)+10*exp(8)*ln(2)*x+10*ln(2)/x+ln(2)*exp(x)/x-20*x^2*exp(4)*ln(2)

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maxima [C]  time = 0.39, size = 45, normalized size = 1.55 \begin {gather*} 10 \, x^{3} \log \relax (2) - 20 \, x^{2} e^{4} \log \relax (2) + 10 \, x e^{8} \log \relax (2) + {\rm Ei}\relax (x) \log \relax (2) - \Gamma \left (-1, -x\right ) \log \relax (2) + \frac {10 \, \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(x-1)*log(2)*exp(x)+2*(10*x^2*exp(4)^2-40*x^3*exp(4)+30*x^4-10)*log(2))/x^2,x, algorithm="max
ima")

[Out]

10*x^3*log(2) - 20*x^2*e^4*log(2) + 10*x*e^8*log(2) + Ei(x)*log(2) - gamma(-1, -x)*log(2) + 10*log(2)/x

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mupad [B]  time = 4.69, size = 24, normalized size = 0.83 \begin {gather*} 10\,x\,\ln \relax (2)\,{\left (x-{\mathrm {e}}^4\right )}^2+\frac {\ln \relax (2)\,\left ({\mathrm {e}}^x+10\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(40*x^3*exp(4) - 10*x^2*exp(8) - 30*x^4 + 10) - exp(x)*log(2)*(x - 1))/x^2,x)

[Out]

10*x*log(2)*(x - exp(4))^2 + (log(2)*(exp(x) + 10))/x

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sympy [A]  time = 0.18, size = 44, normalized size = 1.52 \begin {gather*} 10 x^{3} \log {\relax (2 )} - 20 x^{2} e^{4} \log {\relax (2 )} + 10 x e^{8} \log {\relax (2 )} + \frac {e^{x} \log {\relax (2 )}}{x} + \frac {10 \log {\relax (2 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(x-1)*ln(2)*exp(x)+2*(10*x**2*exp(4)**2-40*x**3*exp(4)+30*x**4-10)*ln(2))/x**2,x)

[Out]

10*x**3*log(2) - 20*x**2*exp(4)*log(2) + 10*x*exp(8)*log(2) + exp(x)*log(2)/x + 10*log(2)/x

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