\(\int \frac {1-50 x^2+e^{2 x} (25 x+50 x^2)}{x} \, dx\) [5180]

Optimal result

Integrand size = 26, antiderivative size = 16 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=1-25 x \left (-e^{2 x}+x\right )+\log (x) \]

[Out]

ln(x)-25*x*(x-exp(2*x))+1

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.94, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2207, 2225} \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=-25 x^2-\frac {25 e^{2 x}}{2}+\frac {25}{2} e^{2 x} (2 x+1)+\log (x) \]

[In]

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

[Out]

(-25*E^(2*x))/2 - 25*x^2 + (25*E^(2*x)*(1 + 2*x))/2 + 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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=25 e^{2 x} x-25 x^2+\log (x) \]

[In]

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

[Out]

25*E^(2*x)*x - 25*x^2 + Log[x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
norman	\(-25 x^{2}+25 x \,{\mathrm e}^{2 x}+\ln \left (x \right )\)	\(16\)
risch	\(-25 x^{2}+25 x \,{\mathrm e}^{2 x}+\ln \left (x \right )\)	\(16\)
parallelrisch	\(-25 x^{2}+25 x \,{\mathrm e}^{2 x}+\ln \left (x \right )\)	\(16\)
parts	\(-25 x^{2}+25 x \,{\mathrm e}^{2 x}+\ln \left (x \right )\)	\(16\)
derivativedivides	\(-25 x^{2}+\ln \left (2 x \right )+25 x \,{\mathrm e}^{2 x}\)	\(18\)
default	\(-25 x^{2}+\ln \left (2 x \right )+25 x \,{\mathrm e}^{2 x}\)	\(18\)

[In]

int(((50*x^2+25*x)*exp(2*x)-50*x^2+1)/x,x,method=_RETURNVERBOSE)

[Out]

-25*x^2+25*x*exp(2*x)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=-25 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x^2+25*x)*exp(2*x)-50*x^2+1)/x,x, algorithm="fricas")

[Out]

-25*x^2 + 25*x*e^(2*x) + log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=- 25 x^{2} + 25 x e^{2 x} + \log {\left (x \right )} \]

[In]

integrate(((50*x**2+25*x)*exp(2*x)-50*x**2+1)/x,x)

[Out]

-25*x**2 + 25*x*exp(2*x) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=-25 \, x^{2} + \frac {25}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {25}{2} \, e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x^2+25*x)*exp(2*x)-50*x^2+1)/x,x, algorithm="maxima")

[Out]

-25*x^2 + 25/2*(2*x - 1)*e^(2*x) + 25/2*e^(2*x) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=-25 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + \log \left (x\right ) \]

[In]

integrate(((50*x^2+25*x)*exp(2*x)-50*x^2+1)/x,x, algorithm="giac")

[Out]

-25*x^2 + 25*x*e^(2*x) + log(x)

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1-50 x^2+e^{2 x} \left (25 x+50 x^2\right )}{x} \, dx=\ln \left (x\right )+25\,x\,{\mathrm {e}}^{2\,x}-25\,x^2 \]

[In]

int((exp(2*x)*(25*x + 50*x^2) - 50*x^2 + 1)/x,x)

[Out]

log(x) + 25*x*exp(2*x) - 25*x^2