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

Optimal. Leaf size=16 \[ 1-25 x \left (-e^{2 x}+x\right )+\log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.94, number of steps used = 6, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2176, 2194} \begin {gather*} -25 x^2-\frac {25 e^{2 x}}{2}+\frac {25}{2} e^{2 x} (2 x+1)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} 25 e^{2 x} x-25 x^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.48, size = 15, normalized size = 0.94 \begin {gather*} -25 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.15, size = 15, normalized size = 0.94 \begin {gather*} -25 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 16, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.35, size = 25, normalized size = 1.56 \begin {gather*} -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 \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.24, size = 15, normalized size = 0.94 \begin {gather*} \ln \relax (x)+25\,x\,{\mathrm {e}}^{2\,x}-25\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.10, size = 15, normalized size = 0.94 \begin {gather*} - 25 x^{2} + 25 x e^{2 x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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