\(\int (e^3 (5+e^2)+e^{32+2 x} (-5+e^2 (-1-2 x)-8 x)+9 x+2 e^2 x+(e^3+e^{32+2 x} (-1-2 x)+2 x) \log (x)) \, dx\) [682]

Optimal result

Integrand size = 64, antiderivative size = 23 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x \left (e^3-e^{2 (16+x)}+x\right ) \left (4+e^2+\log (x)\right ) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(133\) vs. \(2(23)=46\).

Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 5.78, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6, 2218, 2207, 2225, 2634} \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=\frac {1}{2} \left (9+2 e^2\right ) x^2-\frac {x^2}{2}+x^2 \log (x)+e^3 \left (5+e^2\right ) x-e^3 x+\frac {1}{2} e^{2 x+32}-\frac {1}{2} e^{2 x+32} \left (2 \left (4+e^2\right ) x+e^2+5\right )+\frac {1}{2} \left (4+e^2\right ) e^{2 x+32}+e^3 x \log (x)+\frac {1}{2} e^{2 x+32} \log (x)-\frac {1}{2} e^{2 x+32} (2 x+1) \log (x) \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2218

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 2225

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 2634

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=-\left (\left (-e^3+e^{32+2 x}-x\right ) x \left (4+e^2+\log (x)\right )\right ) \]

[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]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(58\) vs. \(2(20)=40\).

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.57

method	result	size
norman	\(x^{2} \ln \left (x \right )+\left (4+{\mathrm e}^{2}\right ) x^{2}+\left (4 \,{\mathrm e}^{3}+{\mathrm e}^{2} {\mathrm e}^{3}\right ) x +x \,{\mathrm e}^{3} \ln \left (x \right )+\left (-{\mathrm e}^{2}-4\right ) x \,{\mathrm e}^{2 x +32}-\ln \left (x \right ) {\mathrm e}^{2 x +32} x\)	\(59\)
risch	\(x \,{\mathrm e}^{5}+4 x \,{\mathrm e}^{3}-x \,{\mathrm e}^{2 x +34}+x^{2} {\mathrm e}^{2}-\ln \left (x \right ) {\mathrm e}^{2 x +32} x +x \,{\mathrm e}^{3} \ln \left (x \right )+x^{2} \ln \left (x \right )-4 \,{\mathrm e}^{2 x +32} x +4 x^{2}\)	\(63\)
parts	\(x^{2} \ln \left (x \right )+4 x^{2}+{\mathrm e}^{3} \left (x \ln \left (x \right )-x \right )+\left (-{\mathrm e}^{2}-4\right ) x \,{\mathrm e}^{2 x +32}-\ln \left (x \right ) {\mathrm e}^{2 x +32} x +x \,{\mathrm e}^{2} {\mathrm e}^{3}+x^{2} {\mathrm e}^{2}+5 x \,{\mathrm e}^{3}\)	\(66\)
parallelrisch	\(x^{2} {\mathrm e}^{2}-{\mathrm e}^{2 x +32} x \,{\mathrm e}^{2}+x \,{\mathrm e}^{3} \ln \left (x \right )+x^{2} \ln \left (x \right )-\ln \left (x \right ) {\mathrm e}^{2 x +32} x -x \,{\mathrm e}^{3}+4 x^{2}-4 \,{\mathrm e}^{2 x +32} x +\left ({\mathrm e}^{2}+5\right ) {\mathrm e}^{3} x\)	\(69\)
default	\(\left ({\mathrm e}^{2}+5\right ) {\mathrm e}^{3} x -2 \left (2 x +32\right ) {\mathrm e}^{2 x +32}+64 \,{\mathrm e}^{2 x +32}+\frac {31 \,{\mathrm e}^{2} {\mathrm e}^{2 x +32}}{2}-\frac {{\mathrm e}^{2} \left (\left (2 x +32\right ) {\mathrm e}^{2 x +32}-{\mathrm e}^{2 x +32}\right )}{2}+x^{2} \ln \left (x \right )+4 x^{2}+{\mathrm e}^{3} \left (x \ln \left (x \right )-x \right )-\ln \left (x \right ) {\mathrm e}^{2 x +32} x +x^{2} {\mathrm e}^{2}\)	\(105\)

[In]

int(((-1-2*x)*exp(2*x+32)+2*x+exp(3))*ln(x)+((-1-2*x)*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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.48 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x^{2} e^{2} + 4 \, x^{2} + x e^{5} + 4 \, x e^{3} - {\left (x e^{2} + 4 \, x\right )} e^{\left (2 \, x + 32\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \left (x\right ) \]

[In]

integrate(((-1-2*x)*exp(2*x+32)+2*x+exp(3))*log(x)+((-1-2*x)*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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).

Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x^{2} \cdot \left (4 + e^{2}\right ) + x \left (4 e^{3} + e^{5}\right ) + \left (x^{2} + x e^{3}\right ) \log {\left (x \right )} + \left (- x \log {\left (x \right )} - x e^{2} - 4 x\right ) e^{2 x + 32} \]

[In]

integrate(((-1-2*x)*exp(2*x+32)+2*x+exp(3))*ln(x)+((-1-2*x)*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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (20) = 40\).

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x {\left (e^{2} + 5\right )} e^{3} + x^{2} e^{2} + 4 \, x^{2} - x e^{3} - \frac {1}{2} \, {\left (2 \, x {\left (e^{34} + 4 \, e^{32}\right )} + e^{32}\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \left (x\right ) + \frac {1}{2} \, e^{\left (2 \, x + 32\right )} \]

[In]

integrate(((-1-2*x)*exp(2*x+32)+2*x+exp(3))*log(x)+((-1-2*x)*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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.26 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x {\left (e^{2} + 5\right )} e^{3} + x^{2} e^{2} + 4 \, x^{2} - x e^{3} - x e^{\left (2 \, x + 34\right )} - \frac {1}{2} \, {\left (8 \, x + 1\right )} e^{\left (2 \, x + 32\right )} + {\left (x^{2} + x e^{3} - x e^{\left (2 \, x + 32\right )}\right )} \log \left (x\right ) + \frac {1}{2} \, e^{\left (2 \, x + 32\right )} \]

[In]

integrate(((-1-2*x)*exp(2*x+32)+2*x+exp(3))*log(x)+((-1-2*x)*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)

Mupad [B] (verification not implemented)

Time = 7.78 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \left (e^3 \left (5+e^2\right )+e^{32+2 x} \left (-5+e^2 (-1-2 x)-8 x\right )+9 x+2 e^2 x+\left (e^3+e^{32+2 x} (-1-2 x)+2 x\right ) \log (x)\right ) \, dx=x\,\left (x+{\mathrm {e}}^3-{\mathrm {e}}^{2\,x+32}\right )\,\left ({\mathrm {e}}^2+\ln \left (x\right )+4\right ) \]

[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)