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

Optimal. Leaf size=24 \[ e^{4 x} \left (-e^{-x+\log ^2(x)}+5 x\right )^2 \]

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Rubi [B]  time = 0.26, antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 11, number of rules used = 6, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2196, 2176, 2194, 6706, 2288} \begin {gather*} 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} \end {gather*}

Antiderivative was successfully verified.

[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 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

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 6706

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 21, normalized size = 0.88 \begin {gather*} e^{2 x} \left (e^{\log ^2(x)}-5 e^x x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.72, size = 33, normalized size = 1.38 \begin {gather*} 25 \, x^{2} e^{\left (4 \, x\right )} - 10 \, x e^{\left (\log \relax (x)^{2} + 3 \, x\right )} + e^{\left (2 \, \log \relax (x)^{2} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.21, size = 33, normalized size = 1.38 \begin {gather*} 25 \, x^{2} e^{\left (4 \, x\right )} - 10 \, x e^{\left (\log \relax (x)^{2} + 3 \, x\right )} + e^{\left (2 \, \log \relax (x)^{2} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 34, normalized size = 1.42

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

Verification of antiderivative is not currently implemented for this CAS.

[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(3*x+ln(x)^2)+exp(2*x+2*ln(x)^2)

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maxima [B]  time = 0.44, size = 51, normalized size = 2.12 \begin {gather*} -10 \, x e^{\left (\log \relax (x)^{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 \relax (x)^{2} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.62, size = 33, normalized size = 1.38 \begin {gather*} {\mathrm {e}}^{2\,{\ln \relax (x)}^2+2\,x}+25\,x^2\,{\mathrm {e}}^{4\,x}-10\,x\,{\mathrm {e}}^{{\ln \relax (x)}^2+3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 18.51, size = 42, normalized size = 1.75 \begin {gather*} 25 x^{2} e^{4 x} - 10 x e^{4 x} e^{- x + \log {\relax (x )}^{2}} + e^{4 x} e^{- 2 x + 2 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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