3.58.91 \(\int \frac {1}{5} (5-10 x+30 x^2-25 x^4+(40 x-40 x^3) \log (4)+(10-15 x^2) \log ^2(4)+e^{x/5} (-5-x-20 x^3-x^4+(-30 x^2-2 x^3) \log (4)+(-10 x-x^2) \log ^2(4))) \, dx\)

Optimal. Leaf size=30 \[ 5+x-x \left (e^{x/5}-\frac {2}{x}+x\right ) \left (1+x (x+\log (4))^2\right ) \]

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Rubi [B]  time = 0.30, antiderivative size = 99, normalized size of antiderivative = 3.30, number of steps used = 34, number of rules used = 4, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -x^5-e^{x/5} x^4-2 x^4 \log (4)+2 x^3-x^3 \log ^2(4)-2 e^{x/5} x^3 \log (4)-x^2-e^{x/5} x^2 \log ^2(4)+4 x^2 \log (4)-e^{x/5} x+x+2 x \log ^2(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 10*x + 30*x^2 - 25*x^4 + (40*x - 40*x^3)*Log[4] + (10 - 15*x^2)*Log[4]^2 + E^(x/5)*(-5 - x - 20*x^3 -
x^4 + (-30*x^2 - 2*x^3)*Log[4] + (-10*x - x^2)*Log[4]^2))/5,x]

[Out]

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

Rule 12

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

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 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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (5-10 x+30 x^2-25 x^4+\left (40 x-40 x^3\right ) \log (4)+\left (10-15 x^2\right ) \log ^2(4)+e^{x/5} \left (-5-x-20 x^3-x^4+\left (-30 x^2-2 x^3\right ) \log (4)+\left (-10 x-x^2\right ) \log ^2(4)\right )\right ) \, dx\\ &=x-x^2+2 x^3-x^5+\frac {1}{5} \int e^{x/5} \left (-5-x-20 x^3-x^4+\left (-30 x^2-2 x^3\right ) \log (4)+\left (-10 x-x^2\right ) \log ^2(4)\right ) \, dx+\frac {1}{5} \log (4) \int \left (40 x-40 x^3\right ) \, dx+\frac {1}{5} \log ^2(4) \int \left (10-15 x^2\right ) \, dx\\ &=x-x^2+2 x^3-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)+\frac {1}{5} \int \left (-5 e^{x/5}-e^{x/5} x-20 e^{x/5} x^3-e^{x/5} x^4-2 e^{x/5} x^2 (15+x) \log (4)-e^{x/5} x (10+x) \log ^2(4)\right ) \, dx\\ &=x-x^2+2 x^3-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)-\frac {1}{5} \int e^{x/5} x \, dx-\frac {1}{5} \int e^{x/5} x^4 \, dx-4 \int e^{x/5} x^3 \, dx-\frac {1}{5} (2 \log (4)) \int e^{x/5} x^2 (15+x) \, dx-\frac {1}{5} \log ^2(4) \int e^{x/5} x (10+x) \, dx-\int e^{x/5} \, dx\\ &=-5 e^{x/5}+x-e^{x/5} x-x^2+2 x^3-20 e^{x/5} x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)+4 \int e^{x/5} x^3 \, dx+60 \int e^{x/5} x^2 \, dx-\frac {1}{5} (2 \log (4)) \int \left (15 e^{x/5} x^2+e^{x/5} x^3\right ) \, dx-\frac {1}{5} \log ^2(4) \int \left (10 e^{x/5} x+e^{x/5} x^2\right ) \, dx+\int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+300 e^{x/5} x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \log ^2(4)-60 \int e^{x/5} x^2 \, dx-600 \int e^{x/5} x \, dx-\frac {1}{5} (2 \log (4)) \int e^{x/5} x^3 \, dx-(6 \log (4)) \int e^{x/5} x^2 \, dx-\frac {1}{5} \log ^2(4) \int e^{x/5} x^2 \, dx-\left (2 \log ^2(4)\right ) \int e^{x/5} x \, dx\\ &=x-3001 e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-30 e^{x/5} x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-10 e^{x/5} x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)+600 \int e^{x/5} x \, dx+3000 \int e^{x/5} \, dx+(6 \log (4)) \int e^{x/5} x^2 \, dx+(60 \log (4)) \int e^{x/5} x \, dx+\left (2 \log ^2(4)\right ) \int e^{x/5} x \, dx+\left (10 \log ^2(4)\right ) \int e^{x/5} \, dx\\ &=15000 e^{x/5}+x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+300 e^{x/5} x \log (4)+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+50 e^{x/5} \log ^2(4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)-3000 \int e^{x/5} \, dx-(60 \log (4)) \int e^{x/5} x \, dx-(300 \log (4)) \int e^{x/5} \, dx-\left (10 \log ^2(4)\right ) \int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5-1500 e^{x/5} \log (4)+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)+(300 \log (4)) \int e^{x/5} \, dx\\ &=x-e^{x/5} x-x^2+2 x^3-e^{x/5} x^4-x^5+4 x^2 \log (4)-2 e^{x/5} x^3 \log (4)-2 x^4 \log (4)+2 x \log ^2(4)-e^{x/5} x^2 \log ^2(4)-x^3 \log ^2(4)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.17, size = 76, normalized size = 2.53 \begin {gather*} x-x^5-2 x^4 \log (4)+2 x \log ^2(4)-x^3 \left (-2+\log ^2(4)\right )-\frac {1}{5} e^{x/5} \left (5 x+5 x^4+10 x^3 \log (4)+5 x^2 \log ^2(4)\right )+x^2 (-1+\log (256)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 10*x + 30*x^2 - 25*x^4 + (40*x - 40*x^3)*Log[4] + (10 - 15*x^2)*Log[4]^2 + E^(x/5)*(-5 - x - 20
*x^3 - x^4 + (-30*x^2 - 2*x^3)*Log[4] + (-10*x - x^2)*Log[4]^2))/5,x]

[Out]

x - x^5 - 2*x^4*Log[4] + 2*x*Log[4]^2 - x^3*(-2 + Log[4]^2) - (E^(x/5)*(5*x + 5*x^4 + 10*x^3*Log[4] + 5*x^2*Lo
g[4]^2))/5 + x^2*(-1 + Log[256])

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fricas [B]  time = 0.85, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*(-x^2-10*x)*log(2)^2+2*(-2*x^3-30*x^2)*log(2)-x^4-20*x^3-x-5)*exp(1/5*x)+4/5*(-15*x^2+10)*log
(2)^2+2/5*(-40*x^3+40*x)*log(2)-5*x^4+6*x^2-2*x+1,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*(-x^2-10*x)*log(2)^2+2*(-2*x^3-30*x^2)*log(2)-x^4-20*x^3-x-5)*exp(1/5*x)+4/5*(-15*x^2+10)*log
(2)^2+2/5*(-40*x^3+40*x)*log(2)-5*x^4+6*x^2-2*x+1,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 79, normalized size = 2.63

method	result	size
risch	\(\frac {\left (-20 x^{2} \ln \relax (2)^{2}-20 x^{3} \ln \relax (2)-5 x^{4}-5 x \right ) {\mathrm e}^{\frac {x}{5}}}{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)-x^{5}+2 x^{3}-x^{2}+x\)	\(79\)
norman	\(\left (2-4 \ln \relax (2)^{2}\right ) x^{3}+\left (8 \ln \relax (2)^{2}+1\right ) x +\left (8 \ln \relax (2)-1\right ) x^{2}-x^{5}-x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 x^{4} \ln \relax (2)-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}\)	\(86\)
derivativedivides	\(x -x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}-x^{2}+2 x^{3}-x^{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)\)	\(88\)
default	\(x -x \,{\mathrm e}^{\frac {x}{5}}-x^{4} {\mathrm e}^{\frac {x}{5}}-4 \ln \relax (2) {\mathrm e}^{\frac {x}{5}} x^{3}-4 \ln \relax (2)^{2} {\mathrm e}^{\frac {x}{5}} x^{2}-x^{2}+2 x^{3}-x^{5}-4 x^{3} \ln \relax (2)^{2}+8 x \ln \relax (2)^{2}-4 x^{4} \ln \relax (2)+8 x^{2} \ln \relax (2)\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*(-x^2-10*x)*ln(2)^2+2*(-2*x^3-30*x^2)*ln(2)-x^4-20*x^3-x-5)*exp(1/5*x)+4/5*(-15*x^2+10)*ln(2)^2+2/5
*(-40*x^3+40*x)*ln(2)-5*x^4+6*x^2-2*x+1,x,method=_RETURNVERBOSE)

[Out]

1/5*(-20*x^2*ln(2)^2-20*x^3*ln(2)-5*x^4-5*x)*exp(1/5*x)-4*x^3*ln(2)^2+8*x*ln(2)^2-4*x^4*ln(2)+8*x^2*ln(2)-x^5+
2*x^3-x^2+x

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maxima [B]  time = 0.45, size = 70, normalized size = 2.33 \begin {gather*} -x^{5} + 2 \, x^{3} - 4 \, {\left (x^{3} - 2 \, x\right )} \log \relax (2)^{2} - x^{2} - {\left (x^{4} + 4 \, x^{3} \log \relax (2) + 4 \, x^{2} \log \relax (2)^{2} + x\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, {\left (x^{4} - 2 \, x^{2}\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*(-x^2-10*x)*log(2)^2+2*(-2*x^3-30*x^2)*log(2)-x^4-20*x^3-x-5)*exp(1/5*x)+4/5*(-15*x^2+10)*log
(2)^2+2/5*(-40*x^3+40*x)*log(2)-5*x^4+6*x^2-2*x+1,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.21, size = 86, normalized size = 2.87 \begin {gather*} x\,\left (8\,{\ln \relax (2)}^2+1\right )-x\,{\mathrm {e}}^{x/5}+x^2\,\left (8\,\ln \relax (2)-1\right )-x^4\,{\mathrm {e}}^{x/5}-4\,x^4\,\ln \relax (2)-x^3\,\left (4\,{\ln \relax (2)}^2-2\right )-x^5-4\,x^2\,{\mathrm {e}}^{x/5}\,{\ln \relax (2)}^2-4\,x^3\,{\mathrm {e}}^{x/5}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2)*(40*x - 40*x^3))/5 - 2*x - (exp(x/5)*(x + 4*log(2)^2*(10*x + x^2) + 2*log(2)*(30*x^2 + 2*x^3) +
20*x^3 + x^4 + 5))/5 - (4*log(2)^2*(15*x^2 - 10))/5 + 6*x^2 - 5*x^4 + 1,x)

[Out]

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

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sympy [B]  time = 0.18, size = 73, normalized size = 2.43 \begin {gather*} - x^{5} - 4 x^{4} \log {\relax (2 )} + x^{3} \left (2 - 4 \log {\relax (2 )}^{2}\right ) + x^{2} \left (-1 + 8 \log {\relax (2 )}\right ) + x \left (1 + 8 \log {\relax (2 )}^{2}\right ) + \left (- x^{4} - 4 x^{3} \log {\relax (2 )} - 4 x^{2} \log {\relax (2 )}^{2} - x\right ) e^{\frac {x}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*(-x**2-10*x)*ln(2)**2+2*(-2*x**3-30*x**2)*ln(2)-x**4-20*x**3-x-5)*exp(1/5*x)+4/5*(-15*x**2+10
)*ln(2)**2+2/5*(-40*x**3+40*x)*ln(2)-5*x**4+6*x**2-2*x+1,x)

[Out]

-x**5 - 4*x**4*log(2) + x**3*(2 - 4*log(2)**2) + x**2*(-1 + 8*log(2)) + x*(1 + 8*log(2)**2) + (-x**4 - 4*x**3*
log(2) - 4*x**2*log(2)**2 - x)*exp(x/5)

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