3.3.90 \(\int \frac {16 x+x^4+5 x^5+e^{2 x} (1+x+2 x^2)+e^x (-2 x^2-6 x^3-2 x^4)+(2 e^{2 x} x+4 x^4+e^x (-4 x^2-2 x^3)) \log (x)}{16 x} \, dx\)

Optimal. Leaf size=22 \[ -4+x+\frac {1}{16} \left (e^x-x^2\right )^2 (x+\log (x)) \]

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Rubi [B]  time = 0.23, antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 6, number of rules used = 4, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 14, 2304, 2288} \begin {gather*} \frac {x^5}{16}+\frac {1}{16} x^4 \log (x)-\frac {1}{8} e^x x \left (x^2+x \log (x)\right )+\frac {e^{2 x} \left (x^2+x \log (x)\right )}{16 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x + x^4 + 5*x^5 + E^(2*x)*(1 + x + 2*x^2) + E^x*(-2*x^2 - 6*x^3 - 2*x^4) + (2*E^(2*x)*x + 4*x^4 + E^x*
(-4*x^2 - 2*x^3))*Log[x])/(16*x),x]

[Out]

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

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2304

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {16 x+x^4+5 x^5+e^{2 x} \left (1+x+2 x^2\right )+e^x \left (-2 x^2-6 x^3-2 x^4\right )+\left (2 e^{2 x} x+4 x^4+e^x \left (-4 x^2-2 x^3\right )\right ) \log (x)}{x} \, dx\\ &=\frac {1}{16} \int \left (16+x^3+5 x^4+4 x^3 \log (x)-2 e^x x \left (1+3 x+x^2+2 \log (x)+x \log (x)\right )+\frac {e^{2 x} \left (1+x+2 x^2+2 x \log (x)\right )}{x}\right ) \, dx\\ &=x+\frac {x^4}{64}+\frac {x^5}{16}+\frac {1}{16} \int \frac {e^{2 x} \left (1+x+2 x^2+2 x \log (x)\right )}{x} \, dx-\frac {1}{8} \int e^x x \left (1+3 x+x^2+2 \log (x)+x \log (x)\right ) \, dx+\frac {1}{4} \int x^3 \log (x) \, dx\\ &=x+\frac {x^5}{16}+\frac {1}{16} x^4 \log (x)+\frac {e^{2 x} \left (x^2+x \log (x)\right )}{16 x}-\frac {1}{8} e^x x \left (x^2+x \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 39, normalized size = 1.77 \begin {gather*} \frac {1}{16} \left (x \left (16+e^{2 x}-2 e^x x^2+x^4\right )+\left (e^x-x^2\right )^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x + x^4 + 5*x^5 + E^(2*x)*(1 + x + 2*x^2) + E^x*(-2*x^2 - 6*x^3 - 2*x^4) + (2*E^(2*x)*x + 4*x^4
+ E^x*(-4*x^2 - 2*x^3))*Log[x])/(16*x),x]

[Out]

(x*(16 + E^(2*x) - 2*E^x*x^2 + x^4) + (E^x - x^2)^2*Log[x])/16

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fricas [B]  time = 0.66, size = 40, normalized size = 1.82 \begin {gather*} \frac {1}{16} \, x^{5} - \frac {1}{8} \, x^{3} e^{x} + \frac {1}{16} \, x e^{\left (2 \, x\right )} + \frac {1}{16} \, {\left (x^{4} - 2 \, x^{2} e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((2*x*exp(x)^2+(-2*x^3-4*x^2)*exp(x)+4*x^4)*log(x)+(2*x^2+x+1)*exp(x)^2+(-2*x^4-6*x^3-2*x^2)*ex
p(x)+5*x^5+x^4+16*x)/x,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.42, size = 45, normalized size = 2.05 \begin {gather*} \frac {1}{16} \, x^{5} + \frac {1}{16} \, x^{4} \log \relax (x) - \frac {1}{8} \, x^{3} e^{x} - \frac {1}{8} \, x^{2} e^{x} \log \relax (x) + \frac {1}{16} \, x e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (2 \, x\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((2*x*exp(x)^2+(-2*x^3-4*x^2)*exp(x)+4*x^4)*log(x)+(2*x^2+x+1)*exp(x)^2+(-2*x^4-6*x^3-2*x^2)*ex
p(x)+5*x^5+x^4+16*x)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 41, normalized size = 1.86

method	result	size
risch	\(\frac {\left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}+x^{4}\right ) \ln \relax (x )}{16}+\frac {x^{5}}{16}-\frac {{\mathrm e}^{x} x^{3}}{8}+\frac {x \,{\mathrm e}^{2 x}}{16}+x\)	\(41\)
default	\(x -\frac {{\mathrm e}^{x} x^{3}}{8}-\frac {x^{2} {\mathrm e}^{x} \ln \relax (x )}{8}+\frac {x \,{\mathrm e}^{2 x}}{16}+\frac {{\mathrm e}^{2 x} \ln \relax (x )}{16}+\frac {x^{5}}{16}+\frac {x^{4} \ln \relax (x )}{16}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*((2*x*exp(x)^2+(-2*x^3-4*x^2)*exp(x)+4*x^4)*ln(x)+(2*x^2+x+1)*exp(x)^2+(-2*x^4-6*x^3-2*x^2)*exp(x)+5*
x^5+x^4+16*x)/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{16} \, x^{5} + \frac {1}{16} \, x^{4} \log \relax (x) + \frac {1}{32} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {1}{8} \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - \frac {1}{8} \, {\left (x^{2} \log \relax (x) - x + 1\right )} e^{x} - \frac {3}{8} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - \frac {1}{8} \, {\left (x - 1\right )} e^{x} + \frac {1}{16} \, e^{\left (2 \, x\right )} \log \relax (x) + x + \frac {1}{16} \, {\rm Ei}\left (2 \, x\right ) + \frac {1}{32} \, e^{\left (2 \, x\right )} - \frac {1}{16} \, \int \frac {e^{\left (2 \, x\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((2*x*exp(x)^2+(-2*x^3-4*x^2)*exp(x)+4*x^4)*log(x)+(2*x^2+x+1)*exp(x)^2+(-2*x^4-6*x^3-2*x^2)*ex
p(x)+5*x^5+x^4+16*x)/x,x, algorithm="maxima")

[Out]

1/16*x^5 + 1/16*x^4*log(x) + 1/32*(2*x - 1)*e^(2*x) - 1/8*(x^3 - 3*x^2 + 6*x - 6)*e^x - 1/8*(x^2*log(x) - x +
1)*e^x - 3/8*(x^2 - 2*x + 2)*e^x - 1/8*(x - 1)*e^x + 1/16*e^(2*x)*log(x) + x + 1/16*Ei(2*x) + 1/32*e^(2*x) - 1
/16*integrate(e^(2*x)/x, x)

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mupad [B]  time = 0.46, size = 45, normalized size = 2.05 \begin {gather*} x+\frac {x\,{\mathrm {e}}^{2\,x}}{16}-\frac {x^3\,{\mathrm {e}}^x}{8}+\frac {x^4\,\ln \relax (x)}{16}+\frac {{\mathrm {e}}^{2\,x}\,\ln \relax (x)}{16}+\frac {x^5}{16}-\frac {x^2\,{\mathrm {e}}^x\,\ln \relax (x)}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - (exp(x)*(2*x^2 + 6*x^3 + 2*x^4))/16 + (exp(2*x)*(x + 2*x^2 + 1))/16 + x^4/16 + (5*x^5)/16 + (log(x)*(
2*x*exp(2*x) - exp(x)*(4*x^2 + 2*x^3) + 4*x^4))/16)/x,x)

[Out]

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

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sympy [B]  time = 0.47, size = 49, normalized size = 2.23 \begin {gather*} \frac {x^{5}}{16} + \frac {x^{4} \log {\relax (x )}}{16} + x + \frac {\left (8 x + 8 \log {\relax (x )}\right ) e^{2 x}}{128} + \frac {\left (- 16 x^{3} - 16 x^{2} \log {\relax (x )}\right ) e^{x}}{128} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((2*x*exp(x)**2+(-2*x**3-4*x**2)*exp(x)+4*x**4)*ln(x)+(2*x**2+x+1)*exp(x)**2+(-2*x**4-6*x**3-2*
x**2)*exp(x)+5*x**5+x**4+16*x)/x,x)

[Out]

x**5/16 + x**4*log(x)/16 + x + (8*x + 8*log(x))*exp(2*x)/128 + (-16*x**3 - 16*x**2*log(x))*exp(x)/128

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