3.7.82 \(\int (e^3 (5+e^2)+e^{32+2 x} (-5+e^2 (-1-2 x)-8 x)+9 x+2 e^2 x+(e^3+e^{32+2 x} (-1-2 x)+2 x) \log (x)) \, dx\)

Optimal. Leaf size=23 \[ x \left (e^3-e^{2 (16+x)}+x\right ) \left (4+e^2+\log (x)\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 133, normalized size of antiderivative = 5.78, number of steps used = 8, number of rules used = 5, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6, 2187, 2176, 2194, 2554} \begin {gather*} \frac {1}{2} \left (9+2 e^2\right ) x^2-\frac {x^2}{2}+x^2 \log (x)+e^3 \left (5+e^2\right ) x-e^3 x+\frac {1}{2} e^{2 x+32}-\frac {1}{2} e^{2 x+32} \left (2 \left (4+e^2\right ) x+e^2+5\right )+\frac {1}{2} \left (4+e^2\right ) e^{2 x+32}+e^3 x \log (x)+\frac {1}{2} e^{2 x+32} \log (x)-\frac {1}{2} e^{2 x+32} (2 x+1) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^3*(5 + E^2) + E^(32 + 2*x)*(-5 + E^2*(-1 - 2*x) - 8*x) + 9*x + 2*E^2*x + (E^3 + E^(32 + 2*x)*(-1 - 2*x)
+ 2*x)*Log[x],x]

[Out]

E^(32 + 2*x)/2 + (E^(32 + 2*x)*(4 + E^2))/2 - E^3*x + E^3*(5 + E^2)*x - x^2/2 + ((9 + 2*E^2)*x^2)/2 - (E^(32 +
2*x)*(5 + E^2 + 2*(4 + E^2)*x))/2 + (E^(32 + 2*x)*Log[x])/2 + E^3*x*Log[x] + x^2*Log[x] - (E^(32 + 2*x)*(1 +
2*x)*Log[x])/2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+\left (9+2 e^2\right ) x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx\\ &=e^3 \left (5+e^2\right ) x+\frac {1}{2} \left (9+2 e^2\right ) x^2+\int e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right ) \, dx+\int \left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x) \, dx\\ &=e^3 \left (5+e^2\right ) x+\frac {1}{2} \left (9+2 e^2\right ) x^2+\frac {1}{2} e^{32+2 x} \log (x)+e^3 x \log (x)+x^2 \log (x)-\frac {1}{2} e^{32+2 x} (1+2 x) \log (x)-\int \left (e^3-e^{32+2 x}+x\right ) \, dx+\int e^{32+2 x} \left (-5-e^2-2 \left (4+e^2\right ) x\right ) \, dx\\ &=-e^3 x+e^3 \left (5+e^2\right ) x-\frac {x^2}{2}+\frac {1}{2} \left (9+2 e^2\right ) x^2-\frac {1}{2} e^{32+2 x} \left (5+e^2+2 \left (4+e^2\right ) x\right )+\frac {1}{2} e^{32+2 x} \log (x)+e^3 x \log (x)+x^2 \log (x)-\frac {1}{2} e^{32+2 x} (1+2 x) \log (x)-\left (-4-e^2\right ) \int e^{32+2 x} \, dx+\int e^{32+2 x} \, dx\\ &=\frac {1}{2} e^{32+2 x}+\frac {1}{2} e^{32+2 x} \left (4+e^2\right )-e^3 x+e^3 \left (5+e^2\right ) x-\frac {x^2}{2}+\frac {1}{2} \left (9+2 e^2\right ) x^2-\frac {1}{2} e^{32+2 x} \left (5+e^2+2 \left (4+e^2\right ) x\right )+\frac {1}{2} e^{32+2 x} \log (x)+e^3 x \log (x)+x^2 \log (x)-\frac {1}{2} e^{32+2 x} (1+2 x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 26, normalized size = 1.13 \begin {gather*} -\left (\left (-e^3+e^{32+2 x}-x\right ) x \left (4+e^2+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^3*(5 + E^2) + E^(32 + 2*x)*(-5 + E^2*(-1 - 2*x) - 8*x) + 9*x + 2*E^2*x + (E^3 + E^(32 + 2*x)*(-1 -
2*x) + 2*x)*Log[x],x]

[Out]

-((-E^3 + E^(32 + 2*x) - x)*x*(4 + E^2 + Log[x]))

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fricas [B]  time = 0.76, size = 57, normalized size = 2.48 \begin {gather*} x^{2} e^{2} + 4 \, x^{2} + x e^{5} + 4 \, x e^{3} - {\left (x e^{2} + 4 \, x\right )} e^{\left (2 \, x + 32\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*exp(2*x+32)+2*x+exp(3))*log(x)+((-2*x-1)*exp(2)-8*x-5)*exp(2*x+32)+(exp(2)+5)*exp(3)+2*exp
(2)*x+9*x,x, algorithm="fricas")

[Out]

x^2*e^2 + 4*x^2 + x*e^5 + 4*x*e^3 - (x*e^2 + 4*x)*e^(2*x + 32) + (x^2 + x*e^3 - x*e^(2*x + 32))*log(x)

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giac [B]  time = 0.38, size = 75, normalized size = 3.26 \begin {gather*} x {\left (e^{2} + 5\right )} e^{3} + x^{2} e^{2} + 4 \, x^{2} - x e^{3} - x e^{\left (2 \, x + 34\right )} - \frac {1}{2} \, {\left (8 \, x + 1\right )} e^{\left (2 \, x + 32\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \relax (x) + \frac {1}{2} \, e^{\left (2 \, x + 32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*exp(2*x+32)+2*x+exp(3))*log(x)+((-2*x-1)*exp(2)-8*x-5)*exp(2*x+32)+(exp(2)+5)*exp(3)+2*exp
(2)*x+9*x,x, algorithm="giac")

[Out]

x*(e^2 + 5)*e^3 + x^2*e^2 + 4*x^2 - x*e^3 - x*e^(2*x + 34) - 1/2*(8*x + 1)*e^(2*x + 32) + (x^2 + x*e^3 - x*e^(
2*x + 32))*log(x) + 1/2*e^(2*x + 32)

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maple [B]  time = 0.06, size = 59, normalized size = 2.57

method	result	size
norman	\(x^{2} \ln \relax (x )+\left (4+{\mathrm e}^{2}\right ) x^{2}+\left (4 \,{\mathrm e}^{3}+{\mathrm e}^{2} {\mathrm e}^{3}\right ) x +x \,{\mathrm e}^{3} \ln \relax (x )+\left (-{\mathrm e}^{2}-4\right ) x \,{\mathrm e}^{2 x +32}-\ln \relax (x ) {\mathrm e}^{2 x +32} x\)	\(59\)
risch	\(x \,{\mathrm e}^{5}+4 x \,{\mathrm e}^{3}-x \,{\mathrm e}^{2 x +34}+x^{2} {\mathrm e}^{2}-\ln \relax (x ) {\mathrm e}^{2 x +32} x +x \,{\mathrm e}^{3} \ln \relax (x )+x^{2} \ln \relax (x )-4 \,{\mathrm e}^{2 x +32} x +4 x^{2}\)	\(63\)
default	\(\left ({\mathrm e}^{2}+5\right ) {\mathrm e}^{3} x -2 \,{\mathrm e}^{2 x +32} \left (2 x +32\right )+64 \,{\mathrm e}^{2 x +32}+\frac {31 \,{\mathrm e}^{2} {\mathrm e}^{2 x +32}}{2}-\frac {{\mathrm e}^{2} \left ({\mathrm e}^{2 x +32} \left (2 x +32\right )-{\mathrm e}^{2 x +32}\right )}{2}+x^{2} \ln \relax (x )+4 x^{2}+x \,{\mathrm e}^{3} \ln \relax (x )-x \,{\mathrm e}^{3}-\ln \relax (x ) {\mathrm e}^{2 x +32} x +x^{2} {\mathrm e}^{2}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-1)*exp(2*x+32)+2*x+exp(3))*ln(x)+((-2*x-1)*exp(2)-8*x-5)*exp(2*x+32)+(exp(2)+5)*exp(3)+2*exp(2)*x+9
*x,x,method=_RETURNVERBOSE)

[Out]

x^2*ln(x)+(4+exp(2))*x^2+(4*exp(3)+exp(2)*exp(3))*x+x*exp(3)*ln(x)+(-exp(2)-4)*x*exp(2*x+32)-ln(x)*exp(2*x+32)
*x

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maxima [B]  time = 0.60, size = 72, normalized size = 3.13 \begin {gather*} x {\left (e^{2} + 5\right )} e^{3} + x^{2} e^{2} + 4 \, x^{2} - x e^{3} - \frac {1}{2} \, {\left (2 \, x {\left (e^{34} + 4 \, e^{32}\right )} + e^{32}\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \relax (x) + \frac {1}{2} \, e^{\left (2 \, x + 32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*exp(2*x+32)+2*x+exp(3))*log(x)+((-2*x-1)*exp(2)-8*x-5)*exp(2*x+32)+(exp(2)+5)*exp(3)+2*exp
(2)*x+9*x,x, algorithm="maxima")

[Out]

x*(e^2 + 5)*e^3 + x^2*e^2 + 4*x^2 - x*e^3 - 1/2*(2*x*(e^34 + 4*e^32) + e^32)*e^(2*x) + (x^2 + x*e^3 - x*e^(2*x
+ 32))*log(x) + 1/2*e^(2*x + 32)

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mupad [B]  time = 0.88, size = 20, normalized size = 0.87 \begin {gather*} x\,\left (x+{\mathrm {e}}^3-{\mathrm {e}}^{2\,x+32}\right )\,\left ({\mathrm {e}}^2+\ln \relax (x)+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(9*x + log(x)*(2*x + exp(3) - exp(2*x + 32)*(2*x + 1)) - exp(2*x + 32)*(8*x + exp(2)*(2*x + 1) + 5) + 2*x*e
xp(2) + exp(3)*(exp(2) + 5),x)

[Out]

x*(x + exp(3) - exp(2*x + 32))*(exp(2) + log(x) + 4)

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sympy [B]  time = 0.37, size = 51, normalized size = 2.22 \begin {gather*} x^{2} \left (4 + e^{2}\right ) + x \left (4 e^{3} + e^{5}\right ) + \left (x^{2} + x e^{3}\right ) \log {\relax (x )} + \left (- x \log {\relax (x )} - x e^{2} - 4 x\right ) e^{2 x + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-1)*exp(2*x+32)+2*x+exp(3))*ln(x)+((-2*x-1)*exp(2)-8*x-5)*exp(2*x+32)+(exp(2)+5)*exp(3)+2*exp(
2)*x+9*x,x)

[Out]

x**2*(4 + exp(2)) + x*(4*exp(3) + exp(5)) + (x**2 + x*exp(3))*log(x) + (-x*log(x) - x*exp(2) - 4*x)*exp(2*x +
32)

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