3.50.91 \(\int \frac {e^4 x^2-e x^4+e^x (-48+16 x+(-16 x+8 x^2) \log (5)+(-x^2+x^3) \log ^2(5))}{e x^4} \, dx\)

Optimal. Leaf size=31 \[ -\frac {e^3}{x}-x+\frac {e^{-1+x} \left (\frac {4}{x}+\log (5)\right )^2}{x} \]

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Rubi [C]  time = 0.27, antiderivative size = 136, normalized size of antiderivative = 4.39, number of steps used = 17, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 14, 2199, 2177, 2178} \begin {gather*} -\frac {8 \text {Ei}(x)}{e}+\frac {\log ^2(5) \text {Ei}(x)}{e}+\frac {(8-\log (5)) \log (5) \text {Ei}(x)}{e}+\frac {8 (1-\log (5)) \text {Ei}(x)}{e}+\frac {16 e^{x-1}}{x^3}+\frac {8 e^{x-1}}{x^2}-\frac {8 e^{x-1} (1-\log (5))}{x^2}-x+\frac {8 e^{x-1}}{x}-\frac {e^3}{x}-\frac {e^{x-1} (8-\log (5)) \log (5)}{x}-\frac {8 e^{x-1} (1-\log (5))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*x^2 - E*x^4 + E^x*(-48 + 16*x + (-16*x + 8*x^2)*Log[5] + (-x^2 + x^3)*Log[5]^2))/(E*x^4),x]

[Out]

(16*E^(-1 + x))/x^3 + (8*E^(-1 + x))/x^2 - E^3/x + (8*E^(-1 + x))/x - x - (8*ExpIntegralEi[x])/E - (8*E^(-1 +
x)*(1 - Log[5]))/x^2 - (8*E^(-1 + x)*(1 - Log[5]))/x + (8*ExpIntegralEi[x]*(1 - Log[5]))/E - (E^(-1 + x)*(8 -
Log[5])*Log[5])/x + (ExpIntegralEi[x]*(8 - Log[5])*Log[5])/E + (ExpIntegralEi[x]*Log[5]^2)/E

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 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 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} &=\frac {\int \frac {e^4 x^2-e x^4+e^x \left (-48+16 x+\left (-16 x+8 x^2\right ) \log (5)+\left (-x^2+x^3\right ) \log ^2(5)\right )}{x^4} \, dx}{e}\\ &=\frac {\int \left (\frac {e^4-e x^2}{x^2}+\frac {e^x (4+x \log (5)) \left (-12+x (4-\log (5))+x^2 \log (5)\right )}{x^4}\right ) \, dx}{e}\\ &=\frac {\int \frac {e^4-e x^2}{x^2} \, dx}{e}+\frac {\int \frac {e^x (4+x \log (5)) \left (-12+x (4-\log (5))+x^2 \log (5)\right )}{x^4} \, dx}{e}\\ &=\frac {\int \left (-e+\frac {e^4}{x^2}\right ) \, dx}{e}+\frac {\int \left (-\frac {48 e^x}{x^4}-\frac {16 e^x (-1+\log (5))}{x^3}-\frac {e^x (-8+\log (5)) \log (5)}{x^2}+\frac {e^x \log ^2(5)}{x}\right ) \, dx}{e}\\ &=-\frac {e^3}{x}-x-\frac {48 \int \frac {e^x}{x^4} \, dx}{e}+\frac {(16 (1-\log (5))) \int \frac {e^x}{x^3} \, dx}{e}+\frac {((8-\log (5)) \log (5)) \int \frac {e^x}{x^2} \, dx}{e}+\frac {\log ^2(5) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}-\frac {e^3}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {16 \int \frac {e^x}{x^3} \, dx}{e}+\frac {(8 (1-\log (5))) \int \frac {e^x}{x^2} \, dx}{e}+\frac {((8-\log (5)) \log (5)) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {8 \int \frac {e^x}{x^2} \, dx}{e}+\frac {(8 (1-\log (5))) \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}+\frac {8 e^{-1+x}}{x}-x-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}+\frac {8 \text {Ei}(x) (1-\log (5))}{e}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}-\frac {8 \int \frac {e^x}{x} \, dx}{e}\\ &=\frac {16 e^{-1+x}}{x^3}+\frac {8 e^{-1+x}}{x^2}-\frac {e^3}{x}+\frac {8 e^{-1+x}}{x}-x-\frac {8 \text {Ei}(x)}{e}-\frac {8 e^{-1+x} (1-\log (5))}{x^2}-\frac {8 e^{-1+x} (1-\log (5))}{x}+\frac {8 \text {Ei}(x) (1-\log (5))}{e}-\frac {e^{-1+x} (8-\log (5)) \log (5)}{x}+\frac {\text {Ei}(x) (8-\log (5)) \log (5)}{e}+\frac {\text {Ei}(x) \log ^2(5)}{e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 47, normalized size = 1.52 \begin {gather*} \frac {16 e^{-1+x}}{x^3}-\frac {e^3}{x}-x+\frac {8 e^{-1+x} \log (5)}{x^2}+\frac {e^{-1+x} \log ^2(5)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*x^2 - E*x^4 + E^x*(-48 + 16*x + (-16*x + 8*x^2)*Log[5] + (-x^2 + x^3)*Log[5]^2))/(E*x^4),x]

[Out]

(16*E^(-1 + x))/x^3 - E^3/x - x + (8*E^(-1 + x)*Log[5])/x^2 + (E^(-1 + x)*Log[5]^2)/x

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fricas [A]  time = 0.90, size = 39, normalized size = 1.26 \begin {gather*} -\frac {{\left (x^{4} e + x^{2} e^{4} - {\left (x^{2} \log \relax (5)^{2} + 8 \, x \log \relax (5) + 16\right )} e^{x}\right )} e^{\left (-1\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-x^2)*log(5)^2+(8*x^2-16*x)*log(5)+16*x-48)*exp(x)+x^2*exp(1)*exp(3)-x^4*exp(1))/x^4/exp(1),x,
algorithm="fricas")

[Out]

-(x^4*e + x^2*e^4 - (x^2*log(5)^2 + 8*x*log(5) + 16)*e^x)*e^(-1)/x^3

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giac [A]  time = 0.15, size = 42, normalized size = 1.35 \begin {gather*} -\frac {{\left (x^{4} e - x^{2} e^{x} \log \relax (5)^{2} + x^{2} e^{4} - 8 \, x e^{x} \log \relax (5) - 16 \, e^{x}\right )} e^{\left (-1\right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-x^2)*log(5)^2+(8*x^2-16*x)*log(5)+16*x-48)*exp(x)+x^2*exp(1)*exp(3)-x^4*exp(1))/x^4/exp(1),x,
algorithm="giac")

[Out]

-(x^4*e - x^2*e^x*log(5)^2 + x^2*e^4 - 8*x*e^x*log(5) - 16*e^x)*e^(-1)/x^3

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maple [A]  time = 0.16, size = 35, normalized size = 1.13

method	result	size
risch	\(-x -\frac {{\mathrm e}^{3}}{x}+\frac {\left (x^{2} \ln \relax (5)^{2}+8 x \ln \relax (5)+16\right ) {\mathrm e}^{x -1}}{x^{3}}\)	\(35\)
default	\({\mathrm e}^{-1} \left (-\frac {{\mathrm e} \,{\mathrm e}^{3}}{x}+\frac {16 \,{\mathrm e}^{x}}{x^{3}}+\frac {8 \,{\mathrm e}^{x} \ln \relax (5)}{x^{2}}+\frac {\ln \relax (5)^{2} {\mathrm e}^{x}}{x}-x \,{\mathrm e}\right )\)	\(47\)
norman	\(\frac {{\mathrm e}^{-1} \ln \relax (5)^{2} x^{2} {\mathrm e}^{x}-x^{4}-x^{2} {\mathrm e}^{3}+16 \,{\mathrm e}^{x} {\mathrm e}^{-1}+8 \,{\mathrm e}^{-1} \ln \relax (5) x \,{\mathrm e}^{x}}{x^{3}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^3-x^2)*ln(5)^2+(8*x^2-16*x)*ln(5)+16*x-48)*exp(x)+x^2*exp(1)*exp(3)-x^4*exp(1))/x^4/exp(1),x,method=_
RETURNVERBOSE)

[Out]

-x-exp(3)/x+1/x^3*(x^2*ln(5)^2+8*x*ln(5)+16)*exp(x-1)

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maxima [C]  time = 0.39, size = 66, normalized size = 2.13 \begin {gather*} {\left ({\rm Ei}\relax (x) \log \relax (5)^{2} - \Gamma \left (-1, -x\right ) \log \relax (5)^{2} - x e + 8 \, \Gamma \left (-1, -x\right ) \log \relax (5) + 16 \, \Gamma \left (-2, -x\right ) \log \relax (5) - \frac {e^{4}}{x} - 16 \, \Gamma \left (-2, -x\right ) - 48 \, \Gamma \left (-3, -x\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-x^2)*log(5)^2+(8*x^2-16*x)*log(5)+16*x-48)*exp(x)+x^2*exp(1)*exp(3)-x^4*exp(1))/x^4/exp(1),x,
algorithm="maxima")

[Out]

(Ei(x)*log(5)^2 - gamma(-1, -x)*log(5)^2 - x*e + 8*gamma(-1, -x)*log(5) + 16*gamma(-2, -x)*log(5) - e^4/x - 16
*gamma(-2, -x) - 48*gamma(-3, -x))*e^(-1)

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mupad [B]  time = 3.25, size = 42, normalized size = 1.35 \begin {gather*} \frac {16\,{\mathrm {e}}^{x-1}-x^2\,\left ({\mathrm {e}}^3-{\mathrm {e}}^{x-1}\,{\ln \relax (5)}^2\right )+8\,x\,{\mathrm {e}}^{x-1}\,\ln \relax (5)}{x^3}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*(exp(x)*(log(5)*(16*x - 8*x^2) - 16*x + log(5)^2*(x^2 - x^3) + 48) + x^4*exp(1) - x^2*exp(4)))/x
^4,x)

[Out]

(16*exp(x - 1) - x^2*(exp(3) - exp(x - 1)*log(5)^2) + 8*x*exp(x - 1)*log(5))/x^3 - x

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sympy [A]  time = 0.17, size = 32, normalized size = 1.03 \begin {gather*} - x - \frac {e^{3}}{x} + \frac {\left (x^{2} \log {\relax (5 )}^{2} + 8 x \log {\relax (5 )} + 16\right ) e^{x}}{e x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**3-x**2)*ln(5)**2+(8*x**2-16*x)*ln(5)+16*x-48)*exp(x)+x**2*exp(1)*exp(3)-x**4*exp(1))/x**4/exp(
1),x)

[Out]

-x - exp(3)/x + (x**2*log(5)**2 + 8*x*log(5) + 16)*exp(-1)*exp(x)/x**3

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