\(\int \frac {e^{4 x} (50 x^2+100 x^3)+e^{-2 x+2 \log ^2(x)} (2 e^{4 x} x+4 e^{4 x} \log (x))+e^{-x+\log ^2(x)} (e^{4 x} (-10 x-30 x^2)-20 e^{4 x} x \log (x))}{x} \, dx\) [3977]

Optimal result

Integrand size = 90, antiderivative size = 24 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=e^{4 x} \left (-e^{-x+\log ^2(x)}+5 x\right )^2 \]

[Out]

exp(x)^4*(5*x-exp(ln(x)^2-x))^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2227, 2207, 2225, 6838, 2326} \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=25 e^{4 x} x^2+e^{2 \left (x+\log ^2(x)\right )}-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{\frac {2 \log (x)}{x}+3} \]

[In]

Int[(E^(4*x)*(50*x^2 + 100*x^3) + E^(-2*x + 2*Log[x]^2)*(2*E^(4*x)*x + 4*E^(4*x)*Log[x]) + E^(-x + Log[x]^2)*(
E^(4*x)*(-10*x - 30*x^2) - 20*E^(4*x)*x*Log[x]))/x,x]

[Out]

E^(2*(x + Log[x]^2)) + 25*E^(4*x)*x^2 - (10*E^(3*x + Log[x]^2)*(3*x + 2*Log[x]))/(3 + (2*Log[x])/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 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 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 2227

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] &&  !TrueQ[$UseGamma]

Rule 2326

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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (50 e^{4 x} x (1+2 x)+\frac {2 e^{2 \left (x+\log ^2(x)\right )} (x+2 \log (x))}{x}-10 e^{3 x+\log ^2(x)} (1+3 x+2 \log (x))\right ) \, dx \\ & = 2 \int \frac {e^{2 \left (x+\log ^2(x)\right )} (x+2 \log (x))}{x} \, dx-10 \int e^{3 x+\log ^2(x)} (1+3 x+2 \log (x)) \, dx+50 \int e^{4 x} x (1+2 x) \, dx \\ & = e^{2 \left (x+\log ^2(x)\right )}-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{3+\frac {2 \log (x)}{x}}+50 \int \left (e^{4 x} x+2 e^{4 x} x^2\right ) \, dx \\ & = e^{2 \left (x+\log ^2(x)\right )}-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{3+\frac {2 \log (x)}{x}}+50 \int e^{4 x} x \, dx+100 \int e^{4 x} x^2 \, dx \\ & = e^{2 \left (x+\log ^2(x)\right )}+\frac {25}{2} e^{4 x} x+25 e^{4 x} x^2-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{3+\frac {2 \log (x)}{x}}-\frac {25}{2} \int e^{4 x} \, dx-50 \int e^{4 x} x \, dx \\ & = -\frac {25 e^{4 x}}{8}+e^{2 \left (x+\log ^2(x)\right )}+25 e^{4 x} x^2-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{3+\frac {2 \log (x)}{x}}+\frac {25}{2} \int e^{4 x} \, dx \\ & = e^{2 \left (x+\log ^2(x)\right )}+25 e^{4 x} x^2-\frac {10 e^{3 x+\log ^2(x)} (3 x+2 \log (x))}{3+\frac {2 \log (x)}{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=e^{2 x} \left (e^{\log ^2(x)}-5 e^x x\right )^2 \]

[In]

Integrate[(E^(4*x)*(50*x^2 + 100*x^3) + E^(-2*x + 2*Log[x]^2)*(2*E^(4*x)*x + 4*E^(4*x)*Log[x]) + E^(-x + Log[x
]^2)*(E^(4*x)*(-10*x - 30*x^2) - 20*E^(4*x)*x*Log[x]))/x,x]

[Out]

E^(2*x)*(E^Log[x]^2 - 5*E^x*x)^2

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42

method	result	size
risch	\(25 x^{2} {\mathrm e}^{4 x}-10 x \,{\mathrm e}^{\ln \left (x \right )^{2}+3 x}+{\mathrm e}^{2 x +2 \ln \left (x \right )^{2}}\)	\(34\)
parallelrisch	\(25 x^{2} {\mathrm e}^{4 x}-10 \,{\mathrm e}^{\ln \left (x \right )^{2}-x} {\mathrm e}^{4 x} x +{\mathrm e}^{4 x} {\mathrm e}^{2 \ln \left (x \right )^{2}-2 x}\)	\(43\)

[In]

int(((4*exp(x)^4*ln(x)+2*x*exp(x)^4)*exp(ln(x)^2-x)^2+(-20*x*exp(x)^4*ln(x)+(-30*x^2-10*x)*exp(x)^4)*exp(ln(x)
^2-x)+(100*x^3+50*x^2)*exp(x)^4)/x,x,method=_RETURNVERBOSE)

[Out]

25*x^2*exp(4*x)-10*x*exp(ln(x)^2+3*x)+exp(2*x+2*ln(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=25 \, x^{2} e^{\left (4 \, x\right )} - 10 \, x e^{\left (\log \left (x\right )^{2} + 3 \, x\right )} + e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x\right )} \]

[In]

integrate(((4*exp(x)^4*log(x)+2*x*exp(x)^4)*exp(log(x)^2-x)^2+(-20*x*exp(x)^4*log(x)+(-30*x^2-10*x)*exp(x)^4)*
exp(log(x)^2-x)+(100*x^3+50*x^2)*exp(x)^4)/x,x, algorithm="fricas")

[Out]

25*x^2*e^(4*x) - 10*x*e^(log(x)^2 + 3*x) + e^(2*log(x)^2 + 2*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 8.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=25 x^{2} e^{4 x} - 10 x e^{4 x} e^{- x + \log {\left (x \right )}^{2}} + e^{4 x} e^{- 2 x + 2 \log {\left (x \right )}^{2}} \]

[In]

integrate(((4*exp(x)**4*ln(x)+2*x*exp(x)**4)*exp(ln(x)**2-x)**2+(-20*x*exp(x)**4*ln(x)+(-30*x**2-10*x)*exp(x)*
*4)*exp(ln(x)**2-x)+(100*x**3+50*x**2)*exp(x)**4)/x,x)

[Out]

25*x**2*exp(4*x) - 10*x*exp(4*x)*exp(-x + log(x)**2) + exp(4*x)*exp(-2*x + 2*log(x)**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=-10 \, x e^{\left (\log \left (x\right )^{2} + 3 \, x\right )} + \frac {25}{8} \, {\left (8 \, x^{2} - 4 \, x + 1\right )} e^{\left (4 \, x\right )} + \frac {25}{8} \, {\left (4 \, x - 1\right )} e^{\left (4 \, x\right )} + e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x\right )} \]

[In]

integrate(((4*exp(x)^4*log(x)+2*x*exp(x)^4)*exp(log(x)^2-x)^2+(-20*x*exp(x)^4*log(x)+(-30*x^2-10*x)*exp(x)^4)*
exp(log(x)^2-x)+(100*x^3+50*x^2)*exp(x)^4)/x,x, algorithm="maxima")

[Out]

-10*x*e^(log(x)^2 + 3*x) + 25/8*(8*x^2 - 4*x + 1)*e^(4*x) + 25/8*(4*x - 1)*e^(4*x) + e^(2*log(x)^2 + 2*x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx=25 \, x^{2} e^{\left (4 \, x\right )} - 10 \, x e^{\left (\log \left (x\right )^{2} + 3 \, x\right )} + e^{\left (2 \, \log \left (x\right )^{2} + 2 \, x\right )} \]

[In]

integrate(((4*exp(x)^4*log(x)+2*x*exp(x)^4)*exp(log(x)^2-x)^2+(-20*x*exp(x)^4*log(x)+(-30*x^2-10*x)*exp(x)^4)*
exp(log(x)^2-x)+(100*x^3+50*x^2)*exp(x)^4)/x,x, algorithm="giac")

[Out]

25*x^2*e^(4*x) - 10*x*e^(log(x)^2 + 3*x) + e^(2*log(x)^2 + 2*x)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{4 x} \left (50 x^2+100 x^3\right )+e^{-2 x+2 \log ^2(x)} \left (2 e^{4 x} x+4 e^{4 x} \log (x)\right )+e^{-x+\log ^2(x)} \left (e^{4 x} \left (-10 x-30 x^2\right )-20 e^{4 x} x \log (x)\right )}{x} \, dx={\mathrm {e}}^{2\,{\ln \left (x\right )}^2+2\,x}+25\,x^2\,{\mathrm {e}}^{4\,x}-10\,x\,{\mathrm {e}}^{{\ln \left (x\right )}^2+3\,x} \]

[In]

int((exp(4*x)*(50*x^2 + 100*x^3) + exp(2*log(x)^2 - 2*x)*(2*x*exp(4*x) + 4*exp(4*x)*log(x)) - exp(log(x)^2 - x
)*(exp(4*x)*(10*x + 30*x^2) + 20*x*exp(4*x)*log(x)))/x,x)

[Out]

exp(2*x + 2*log(x)^2) + 25*x^2*exp(4*x) - 10*x*exp(3*x + log(x)^2)