\(\int \frac {3 x^3+4^{25 x} (-2 x^2-25 x^3 \log (4))+e^{2 \log ^2(x)} (x-25\ 4^{25 x} x \log (4)+(-4^{1+25 x}+4 x) \log (x))+e^{\log ^2(x)} (4 x^2+4^{25 x} (-2 x-50 x^2 \log (4))+(-4^{1+25 x} x+4 x^2) \log (x))}{x} \, dx\) [2856]

Optimal result

Integrand size = 115, antiderivative size = 31 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-\left (\left (-e^{\log ^2(x)}-x\right ) \left (-4^{25 x}+x\right ) \left (e^{\log ^2(x)}+x\right )\right ) \]

[Out]

-(x-exp(50*x*ln(2)))*(x+exp(ln(x)^2))*(-x-exp(ln(x)^2))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(31)=62\).

Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 4.74, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {14, 2227, 2207, 2225, 2326} \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=x^3-\frac {2^{50 x-1} x^2 \log (4)}{\log (2)}-\frac {2^{50 x-2} \log (4)}{625 \log ^3(2)}-\frac {x e^{\log ^2(x)} \left (2^{50 x+1} \log (x)-2 x \log (x)\right )}{\log (x)}+\frac {2^{50 x-1} x \log (4)}{25 \log ^2(2)}-\frac {e^{2 \log ^2(x)} \left (4^{25 x+1} \log (x)-4 x \log (x)\right )}{4 \log (x)}+\frac {2^{50 x-1}}{625 \log ^2(2)}-\frac {2^{50 x} x}{25 \log (2)} \]

[In]

Int[(3*x^3 + 4^(25*x)*(-2*x^2 - 25*x^3*Log[4]) + E^(2*Log[x]^2)*(x - 25*4^(25*x)*x*Log[4] + (-4^(1 + 25*x) + 4
*x)*Log[x]) + E^Log[x]^2*(4*x^2 + 4^(25*x)*(-2*x - 50*x^2*Log[4]) + (-(4^(1 + 25*x)*x) + 4*x^2)*Log[x]))/x,x]

[Out]

x^3 + 2^(-1 + 50*x)/(625*Log[2]^2) - (2^(50*x)*x)/(25*Log[2]) - (2^(-2 + 50*x)*Log[4])/(625*Log[2]^3) + (2^(-1
+ 50*x)*x*Log[4])/(25*Log[2]^2) - (2^(-1 + 50*x)*x^2*Log[4])/Log[2] - (E^(2*Log[x]^2)*(4^(1 + 25*x)*Log[x] -
4*x*Log[x]))/(4*Log[x]) - (E^Log[x]^2*x*(2^(1 + 50*x)*Log[x] - 2*x*Log[x]))/Log[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]

Rubi steps \begin{align*} \text {integral}& = \int \left (-x \left (2^{1+50 x}-3 x+25\ 2^{50 x} x \log (4)\right )-\frac {e^{2 \log ^2(x)} \left (-x+25\ 4^{25 x} x \log (4)+4^{1+25 x} \log (x)-4 x \log (x)\right )}{x}-2 e^{\log ^2(x)} \left (2^{50 x}-2 x+25\ 2^{50 x} x \log (4)+2^{1+50 x} \log (x)-2 x \log (x)\right )\right ) \, dx \\ & = -\left (2 \int e^{\log ^2(x)} \left (2^{50 x}-2 x+25\ 2^{50 x} x \log (4)+2^{1+50 x} \log (x)-2 x \log (x)\right ) \, dx\right )-\int x \left (2^{1+50 x}-3 x+25\ 2^{50 x} x \log (4)\right ) \, dx-\int \frac {e^{2 \log ^2(x)} \left (-x+25\ 4^{25 x} x \log (4)+4^{1+25 x} \log (x)-4 x \log (x)\right )}{x} \, dx \\ & = -\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}-\int \left (-3 x^2+2^{50 x} x (2+25 x \log (4))\right ) \, dx \\ & = x^3-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}-\int 2^{50 x} x (2+25 x \log (4)) \, dx \\ & = x^3-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}-\int \left (2^{1+50 x} x+25\ 2^{50 x} x^2 \log (4)\right ) \, dx \\ & = x^3-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}-(25 \log (4)) \int 2^{50 x} x^2 \, dx-\int 2^{1+50 x} x \, dx \\ & = x^3-\frac {2^{50 x} x}{25 \log (2)}-\frac {2^{-1+50 x} x^2 \log (4)}{\log (2)}-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}+\frac {\int 2^{1+50 x} \, dx}{50 \log (2)}+\frac {\log (4) \int 2^{50 x} x \, dx}{\log (2)} \\ & = x^3+\frac {2^{-1+50 x}}{625 \log ^2(2)}-\frac {2^{50 x} x}{25 \log (2)}+\frac {2^{-1+50 x} x \log (4)}{25 \log ^2(2)}-\frac {2^{-1+50 x} x^2 \log (4)}{\log (2)}-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)}-\frac {\log (4) \int 2^{50 x} \, dx}{50 \log ^2(2)} \\ & = x^3+\frac {2^{-1+50 x}}{625 \log ^2(2)}-\frac {2^{50 x} x}{25 \log (2)}-\frac {2^{-2+50 x} \log (4)}{625 \log ^3(2)}+\frac {2^{-1+50 x} x \log (4)}{25 \log ^2(2)}-\frac {2^{-1+50 x} x^2 \log (4)}{\log (2)}-\frac {e^{2 \log ^2(x)} \left (4^{1+25 x} \log (x)-4 x \log (x)\right )}{4 \log (x)}-\frac {e^{\log ^2(x)} x \left (2^{1+50 x} \log (x)-2 x \log (x)\right )}{\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-\left (\left (2^{50 x}-x\right ) \left (e^{\log ^2(x)}+x\right )^2\right ) \]

[In]

Integrate[(3*x^3 + 4^(25*x)*(-2*x^2 - 25*x^3*Log[4]) + E^(2*Log[x]^2)*(x - 25*4^(25*x)*x*Log[4] + (-4^(1 + 25*
x) + 4*x)*Log[x]) + E^Log[x]^2*(4*x^2 + 4^(25*x)*(-2*x - 50*x^2*Log[4]) + (-(4^(1 + 25*x)*x) + 4*x^2)*Log[x]))
/x,x]

[Out]

-((2^(50*x) - x)*(E^Log[x]^2 + x)^2)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68

method	result	size
risch	\(x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}+\left (x -{\mathrm e}^{50 x \ln \left (2\right )}\right ) {\mathrm e}^{2 \ln \left (x \right )^{2}}+2 \left (x -{\mathrm e}^{50 x \ln \left (2\right )}\right ) x \,{\mathrm e}^{\ln \left (x \right )^{2}}\)	\(52\)
default	\({\mathrm e}^{2 \ln \left (x \right )^{2}} x -{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{2 \ln \left (x \right )^{2}}+2 \,{\mathrm e}^{\ln \left (x \right )^{2}} x^{2}-2 \,{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{\ln \left (x \right )^{2}} x +x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}\)	\(64\)
parallelrisch	\({\mathrm e}^{2 \ln \left (x \right )^{2}} x -{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{2 \ln \left (x \right )^{2}}+2 \,{\mathrm e}^{\ln \left (x \right )^{2}} x^{2}-2 \,{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{\ln \left (x \right )^{2}} x +x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}\)	\(64\)

[In]

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

[Out]

x^3-x^2*exp(50*x*ln(2))+(x-exp(50*x*ln(2)))*exp(ln(x)^2)^2+2*(x-exp(50*x*ln(2)))*x*exp(ln(x)^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-2^{50 \, x} x^{2} + x^{3} - {\left (2^{50 \, x} - x\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} - 2 \, {\left (2^{50 \, x} x - x^{2}\right )} e^{\left (\log \left (x\right )^{2}\right )} \]

[In]

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

[Out]

-2^(50*x)*x^2 + x^3 - (2^(50*x) - x)*e^(2*log(x)^2) - 2*(2^(50*x)*x - x^2)*e^(log(x)^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).

Time = 93.40 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=x^{3} + 2 x^{2} e^{\log {\left (x \right )}^{2}} + x e^{2 \log {\left (x \right )}^{2}} + \left (- x^{2} - 2 x e^{\log {\left (x \right )}^{2}} - e^{2 \log {\left (x \right )}^{2}}\right ) e^{50 x \log {\left (2 \right )}} \]

[In]

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

[Out]

x**3 + 2*x**2*exp(log(x)**2) + x*exp(2*log(x)**2) + (-x**2 - 2*x*exp(log(x)**2) - exp(2*log(x)**2))*exp(50*x*l
og(2))

Maxima [F]

\[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=\int { \frac {3 \, x^{3} - 2 \, {\left (25 \, x^{3} \log \left (2\right ) + x^{2}\right )} 2^{50 \, x} - {\left (50 \cdot 2^{50 \, x} x \log \left (2\right ) + 4 \, {\left (2^{50 \, x} - x\right )} \log \left (x\right ) - x\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} - 2 \, {\left ({\left (50 \, x^{2} \log \left (2\right ) + x\right )} 2^{50 \, x} - 2 \, x^{2} + 2 \, {\left (2^{50 \, x} x - x^{2}\right )} \log \left (x\right )\right )} e^{\left (\log \left (x\right )^{2}\right )}}{x} \,d x } \]

[In]

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

[Out]

x^3 - 2*I*sqrt(pi)*erf(I*log(x) + I)*e^(-1) - 2*x*e^(50*x*log(2) + log(x)^2) - (2^(50*x) - x)*e^(2*log(x)^2) -
1/1250*(1250*x^2*log(2)^2 - 50*x*log(2) + 1)*2^(50*x)/log(2)^2 - 1/1250*(50*x*log(2) - 1)*2^(50*x)/log(2)^2 +
4*integrate(x*e^(log(x)^2)*log(x), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).

Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-2^{50 \, x} x^{2} + x^{3} + 2 \, x^{2} e^{\left (\log \left (x\right )^{2}\right )} - 2 \, x e^{\left (50 \, x \log \left (2\right ) + \log \left (x\right )^{2}\right )} + x e^{\left (2 \, \log \left (x\right )^{2}\right )} - e^{\left (50 \, x \log \left (2\right ) + 2 \, \log \left (x\right )^{2}\right )} \]

[In]

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

[Out]

-2^(50*x)*x^2 + x^3 + 2*x^2*e^(log(x)^2) - 2*x*e^(50*x*log(2) + log(x)^2) + x*e^(2*log(x)^2) - e^(50*x*log(2)
+ 2*log(x)^2)

Mupad [B] (verification not implemented)

Time = 10.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx={\mathrm {e}}^{2\,{\ln \left (x\right )}^2}\,\left (x-2^{50\,x}\right )-2^{50\,x}\,x^2+x^3+2\,x\,{\mathrm {e}}^{{\ln \left (x\right )}^2}\,\left (x-2^{50\,x}\right ) \]

[In]

int(-(exp(50*x*log(2))*(50*x^3*log(2) + 2*x^2) - exp(2*log(x)^2)*(x + log(x)*(4*x - 4*exp(50*x*log(2))) - 50*x
*exp(50*x*log(2))*log(2)) - 3*x^3 + exp(log(x)^2)*(log(x)*(4*x*exp(50*x*log(2)) - 4*x^2) - 4*x^2 + exp(50*x*lo
g(2))*(2*x + 100*x^2*log(2))))/x,x)

[Out]

exp(2*log(x)^2)*(x - 2^(50*x)) - 2^(50*x)*x^2 + x^3 + 2*x*exp(log(x)^2)*(x - 2^(50*x))