3.42.44 \(\int \frac {25-260 x+130 x^2-15 x^3+e^4 (-5+40 x-10 x^2)+e^x (5+15 x-30 x^2+5 x^3)+(5 x-5 e^x x) \log (-\frac {4}{x})}{x} \, dx\)

Optimal. Leaf size=30 \[ 5 \left (-5+e^4-e^x+x\right ) \left (5-(4-x)^2+\log \left (-\frac {4}{x}\right )\right ) \]

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Rubi [B]  time = 0.32, antiderivative size = 81, normalized size of antiderivative = 2.70, number of steps used = 20, number of rules used = 8, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {14, 6742, 2199, 2194, 2178, 2176, 2554, 2295} \begin {gather*} -5 x^3+5 e^x x^2+5 \left (13-e^4\right ) x^2-40 e^x x-20 \left (13-2 e^4\right ) x+5 x+55 e^x+5 x \log \left (-\frac {4}{x}\right )-5 e^x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 - 260*x + 130*x^2 - 15*x^3 + E^4*(-5 + 40*x - 10*x^2) + E^x*(5 + 15*x - 30*x^2 + 5*x^3) + (5*x - 5*E^x
*x)*Log[-4/x])/x,x]

[Out]

55*E^x + 5*x - 40*E^x*x - 20*(13 - 2*E^4)*x + 5*E^x*x^2 + 5*(13 - E^4)*x^2 - 5*x^3 - 5*E^x*Log[-4/x] + 5*x*Log
[-4/x] + 5*(5 - E^4)*Log[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 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 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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {5 e^x \left (1+3 x-6 x^2+x^3-x \log \left (-\frac {4}{x}\right )\right )}{x}+\frac {5 \left (5 \left (1-\frac {e^4}{5}\right )-52 \left (1-\frac {2 e^4}{13}\right ) x+26 \left (1-\frac {e^4}{13}\right ) x^2-3 x^3+x \log \left (-\frac {4}{x}\right )\right )}{x}\right ) \, dx\\ &=5 \int \frac {e^x \left (1+3 x-6 x^2+x^3-x \log \left (-\frac {4}{x}\right )\right )}{x} \, dx+5 \int \frac {5 \left (1-\frac {e^4}{5}\right )-52 \left (1-\frac {2 e^4}{13}\right ) x+26 \left (1-\frac {e^4}{13}\right ) x^2-3 x^3+x \log \left (-\frac {4}{x}\right )}{x} \, dx\\ &=5 \int \left (\frac {5-e^4-4 \left (13-2 e^4\right ) x+2 \left (13-e^4\right ) x^2-3 x^3}{x}+\log \left (-\frac {4}{x}\right )\right ) \, dx+5 \int \left (\frac {e^x \left (1+3 x-6 x^2+x^3\right )}{x}-e^x \log \left (-\frac {4}{x}\right )\right ) \, dx\\ &=5 \int \frac {5-e^4-4 \left (13-2 e^4\right ) x+2 \left (13-e^4\right ) x^2-3 x^3}{x} \, dx+5 \int \frac {e^x \left (1+3 x-6 x^2+x^3\right )}{x} \, dx+5 \int \log \left (-\frac {4}{x}\right ) \, dx-5 \int e^x \log \left (-\frac {4}{x}\right ) \, dx\\ &=5 x-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )-5 \int \frac {e^x}{x} \, dx+5 \int \left (-4 \left (13-2 e^4\right )+\frac {5-e^4}{x}+2 \left (13-e^4\right ) x-3 x^2\right ) \, dx+5 \int \left (3 e^x+\frac {e^x}{x}-6 e^x x+e^x x^2\right ) \, dx\\ &=5 x-20 \left (13-2 e^4\right ) x+5 \left (13-e^4\right ) x^2-5 x^3-5 \text {Ei}(x)-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)+5 \int \frac {e^x}{x} \, dx+5 \int e^x x^2 \, dx+15 \int e^x \, dx-30 \int e^x x \, dx\\ &=15 e^x+5 x-30 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)-10 \int e^x x \, dx+30 \int e^x \, dx\\ &=45 e^x+5 x-40 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)+10 \int e^x \, dx\\ &=55 e^x+5 x-40 e^x x-20 \left (13-2 e^4\right ) x+5 e^x x^2+5 \left (13-e^4\right ) x^2-5 x^3-5 e^x \log \left (-\frac {4}{x}\right )+5 x \log \left (-\frac {4}{x}\right )+5 \left (5-e^4\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.18, size = 64, normalized size = 2.13 \begin {gather*} 5 \left (\left (-51+8 e^4\right ) x+\left (13-e^4\right ) x^2-x^3+e^x \left (11-8 x+x^2\right )+\left (-e^x+x\right ) \log \left (-\frac {4}{x}\right )+\left (5-e^4\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 260*x + 130*x^2 - 15*x^3 + E^4*(-5 + 40*x - 10*x^2) + E^x*(5 + 15*x - 30*x^2 + 5*x^3) + (5*x -
5*E^x*x)*Log[-4/x])/x,x]

[Out]

5*((-51 + 8*E^4)*x + (13 - E^4)*x^2 - x^3 + E^x*(11 - 8*x + x^2) + (-E^x + x)*Log[-4/x] + (5 - E^4)*Log[x])

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fricas [B]  time = 0.65, size = 54, normalized size = 1.80 \begin {gather*} -5 \, x^{3} + 65 \, x^{2} - 5 \, {\left (x^{2} - 8 \, x\right )} e^{4} + 5 \, {\left (x^{2} - 8 \, x + 11\right )} e^{x} + 5 \, {\left (x + e^{4} - e^{x} - 5\right )} \log \left (-\frac {4}{x}\right ) - 255 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(x)*x+5*x)*log(-4/x)+(5*x^3-30*x^2+15*x+5)*exp(x)+(-10*x^2+40*x-5)*exp(4)-15*x^3+130*x^2-260
*x+25)/x,x, algorithm="fricas")

[Out]

-5*x^3 + 65*x^2 - 5*(x^2 - 8*x)*e^4 + 5*(x^2 - 8*x + 11)*e^x + 5*(x + e^4 - e^x - 5)*log(-4/x) - 255*x

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giac [B]  time = 0.14, size = 71, normalized size = 2.37 \begin {gather*} -5 \, x^{3} - 5 \, x^{2} e^{4} + 5 \, x^{2} e^{x} + 65 \, x^{2} + 40 \, x e^{4} - 40 \, x e^{x} - 5 \, e^{4} \log \relax (x) + 5 \, x \log \left (-\frac {4}{x}\right ) - 5 \, e^{x} \log \left (-\frac {4}{x}\right ) - 255 \, x + 55 \, e^{x} + 25 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(x)*x+5*x)*log(-4/x)+(5*x^3-30*x^2+15*x+5)*exp(x)+(-10*x^2+40*x-5)*exp(4)-15*x^3+130*x^2-260
*x+25)/x,x, algorithm="giac")

[Out]

-5*x^3 - 5*x^2*e^4 + 5*x^2*e^x + 65*x^2 + 40*x*e^4 - 40*x*e^x - 5*e^4*log(x) + 5*x*log(-4/x) - 5*e^x*log(-4/x)
- 255*x + 55*e^x + 25*log(x)

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maple [B]  time = 0.10, size = 73, normalized size = 2.43

method	result	size
norman	\(\left (-255+40 \,{\mathrm e}^{4}\right ) x +\left (-5 \,{\mathrm e}^{4}+65\right ) x^{2}+\left (5 \,{\mathrm e}^{4}-25\right ) \ln \left (-\frac {4}{x}\right )-5 x^{3}-40 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} \ln \left (-\frac {4}{x}\right )+5 \ln \left (-\frac {4}{x}\right ) x +55 \,{\mathrm e}^{x}\)	\(73\)
default	\(-5 \,{\mathrm e}^{x} \left (\ln \left (-\frac {4}{x}\right )+\ln \relax (x )\right )+55 \,{\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} \ln \relax (x )-40 \,{\mathrm e}^{x} x -5 x^{2} {\mathrm e}^{4}-5 x^{3}+40 x \,{\mathrm e}^{4}+65 x^{2}-255 x -5 \,{\mathrm e}^{4} \ln \relax (x )+25 \ln \relax (x )+5 \ln \left (-\frac {4}{x}\right ) x\)	\(81\)
risch	\(\left (5 \,{\mathrm e}^{x}-5 x \right ) \ln \relax (x )-5 i {\mathrm e}^{x} \pi +5 i \pi x +5 i \pi x \mathrm {csgn}\left (\frac {i}{x}\right )^{3}-5 i \pi x \mathrm {csgn}\left (\frac {i}{x}\right )^{2}+5 i {\mathrm e}^{x} \pi \mathrm {csgn}\left (\frac {i}{x}\right )^{2}-5 x^{2} {\mathrm e}^{4}-5 x^{3}+40 x \,{\mathrm e}^{4}+10 x \ln \relax (2)+65 x^{2}-255 x -5 \,{\mathrm e}^{4} \ln \relax (x )+25 \ln \relax (x )-5 i {\mathrm e}^{x} \pi \mathrm {csgn}\left (\frac {i}{x}\right )^{3}-10 \,{\mathrm e}^{x} \ln \relax (2)+55 \,{\mathrm e}^{x}-40 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{2}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*exp(x)*x+5*x)*ln(-4/x)+(5*x^3-30*x^2+15*x+5)*exp(x)+(-10*x^2+40*x-5)*exp(4)-15*x^3+130*x^2-260*x+25)/
x,x,method=_RETURNVERBOSE)

[Out]

(-255+40*exp(4))*x+(-5*exp(4)+65)*x^2+(5*exp(4)-25)*ln(-4/x)-5*x^3-40*exp(x)*x+5*exp(x)*x^2-5*exp(x)*ln(-4/x)+
5*ln(-4/x)*x+55*exp(x)

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maxima [B]  time = 0.41, size = 78, normalized size = 2.60 \begin {gather*} -5 \, x^{3} - 5 \, x^{2} e^{4} + 65 \, x^{2} + 40 \, x e^{4} + 5 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 30 \, {\left (x - 1\right )} e^{x} - 5 \, e^{4} \log \relax (x) + 5 \, x \log \left (-\frac {4}{x}\right ) - 5 \, e^{x} \log \left (-\frac {4}{x}\right ) - 255 \, x + 15 \, e^{x} + 25 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(x)*x+5*x)*log(-4/x)+(5*x^3-30*x^2+15*x+5)*exp(x)+(-10*x^2+40*x-5)*exp(4)-15*x^3+130*x^2-260
*x+25)/x,x, algorithm="maxima")

[Out]

-5*x^3 - 5*x^2*e^4 + 65*x^2 + 40*x*e^4 + 5*(x^2 - 2*x + 2)*e^x - 30*(x - 1)*e^x - 5*e^4*log(x) + 5*x*log(-4/x)
- 5*e^x*log(-4/x) - 255*x + 15*e^x + 25*log(x)

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mupad [B]  time = 3.04, size = 65, normalized size = 2.17 \begin {gather*} 55\,{\mathrm {e}}^x+x\,\left (40\,{\mathrm {e}}^4+5\,\ln \left (-\frac {4}{x}\right )-40\,{\mathrm {e}}^x-255\right )+x^2\,\left (5\,{\mathrm {e}}^x-5\,{\mathrm {e}}^4+65\right )+\ln \left (\frac {1}{x}\right )\,\left (5\,{\mathrm {e}}^4-25\right )-5\,x^3-5\,{\mathrm {e}}^x\,\ln \left (-\frac {4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((130*x^2 - exp(4)*(10*x^2 - 40*x + 5) - 260*x - 15*x^3 + exp(x)*(15*x - 30*x^2 + 5*x^3 + 5) + log(-4/x)*(5
*x - 5*x*exp(x)) + 25)/x,x)

[Out]

55*exp(x) + x*(40*exp(4) + 5*log(-4/x) - 40*exp(x) - 255) + x^2*(5*exp(x) - 5*exp(4) + 65) + log(1/x)*(5*exp(4
) - 25) - 5*x^3 - 5*exp(x)*log(-4/x)

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sympy [B]  time = 0.46, size = 65, normalized size = 2.17 \begin {gather*} - 5 x^{3} - x^{2} \left (-65 + 5 e^{4}\right ) + 5 x \log {\left (- \frac {4}{x} \right )} - x \left (255 - 40 e^{4}\right ) + \left (5 x^{2} - 40 x - 5 \log {\left (- \frac {4}{x} \right )} + 55\right ) e^{x} - 5 \left (-5 + e^{4}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(x)*x+5*x)*ln(-4/x)+(5*x**3-30*x**2+15*x+5)*exp(x)+(-10*x**2+40*x-5)*exp(4)-15*x**3+130*x**2
-260*x+25)/x,x)

[Out]

-5*x**3 - x**2*(-65 + 5*exp(4)) + 5*x*log(-4/x) - x*(255 - 40*exp(4)) + (5*x**2 - 40*x - 5*log(-4/x) + 55)*exp
(x) - 5*(-5 + exp(4))*log(x)

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