3.28.28 \(\int \frac {4 x^2+6 x^3+e (2 x+3 x^2)+e^{2 x} (20 x^2+e (-10+20 x))+2 x^2 \log (x)}{2 x^2} \, dx\)

Optimal. Leaf size=25 \[ (e+x) \left (\frac {3 x}{2}+\frac {5 e^{2 x}+x}{x}+\log (x)\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 48, normalized size of antiderivative = 1.92, number of steps used = 14, number of rules used = 8, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {12, 14, 2199, 2194, 2177, 2178, 76, 2295} \begin {gather*} \frac {3 x^2}{2}+\frac {1}{2} (4+3 e) x-x+5 e^{2 x}+\frac {5 e^{2 x+1}}{x}+x \log (x)+e \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + 6*x^3 + E*(2*x + 3*x^2) + E^(2*x)*(20*x^2 + E*(-10 + 20*x)) + 2*x^2*Log[x])/(2*x^2),x]

[Out]

5*E^(2*x) + (5*E^(1 + 2*x))/x - x + ((4 + 3*E)*x)/2 + (3*x^2)/2 + E*Log[x] + x*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {4 x^2+6 x^3+e \left (2 x+3 x^2\right )+e^{2 x} \left (20 x^2+e (-10+20 x)\right )+2 x^2 \log (x)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {10 e^{2 x} \left (-e+2 e x+2 x^2\right )}{x^2}+\frac {2 e+4 \left (1+\frac {3 e}{4}\right ) x+6 x^2+2 x \log (x)}{x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {2 e+4 \left (1+\frac {3 e}{4}\right ) x+6 x^2+2 x \log (x)}{x} \, dx+5 \int \frac {e^{2 x} \left (-e+2 e x+2 x^2\right )}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {(e+2 x) (2+3 x)}{x}+2 \log (x)\right ) \, dx+5 \int \left (2 e^{2 x}-\frac {e^{1+2 x}}{x^2}+\frac {2 e^{1+2 x}}{x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {(e+2 x) (2+3 x)}{x} \, dx-5 \int \frac {e^{1+2 x}}{x^2} \, dx+10 \int e^{2 x} \, dx+10 \int \frac {e^{1+2 x}}{x} \, dx+\int \log (x) \, dx\\ &=5 e^{2 x}+\frac {5 e^{1+2 x}}{x}-x+10 e \text {Ei}(2 x)+x \log (x)+\frac {1}{2} \int \left (4+3 e+\frac {2 e}{x}+6 x\right ) \, dx-10 \int \frac {e^{1+2 x}}{x} \, dx\\ &=5 e^{2 x}+\frac {5 e^{1+2 x}}{x}-x+\frac {1}{2} (4+3 e) x+\frac {3 x^2}{2}+e \log (x)+x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 42, normalized size = 1.68 \begin {gather*} 5 e^{2 x}+\frac {5 e^{1+2 x}}{x}+x+\frac {3 e x}{2}+\frac {3 x^2}{2}+e \log (x)+x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + 6*x^3 + E*(2*x + 3*x^2) + E^(2*x)*(20*x^2 + E*(-10 + 20*x)) + 2*x^2*Log[x])/(2*x^2),x]

[Out]

5*E^(2*x) + (5*E^(1 + 2*x))/x + x + (3*E*x)/2 + (3*x^2)/2 + E*Log[x] + x*Log[x]

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fricas [A]  time = 0.54, size = 45, normalized size = 1.80 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} e + 2 \, x^{2} + 10 \, {\left (x + e\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x^{2} + x e\right )} \log \relax (x)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^2*log(x)+((20*x-10)*exp(1)+20*x^2)*exp(x)^2+(3*x^2+2*x)*exp(1)+6*x^3+4*x^2)/x^2,x, algorith
m="fricas")

[Out]

1/2*(3*x^3 + 3*x^2*e + 2*x^2 + 10*(x + e)*e^(2*x) + 2*(x^2 + x*e)*log(x))/x

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giac [A]  time = 0.21, size = 52, normalized size = 2.08 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} e + 2 \, x^{2} \log \relax (x) + 2 \, x e \log \relax (x) + 2 \, x^{2} + 10 \, x e^{\left (2 \, x\right )} + 10 \, e^{\left (2 \, x + 1\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^2*log(x)+((20*x-10)*exp(1)+20*x^2)*exp(x)^2+(3*x^2+2*x)*exp(1)+6*x^3+4*x^2)/x^2,x, algorith
m="giac")

[Out]

1/2*(3*x^3 + 3*x^2*e + 2*x^2*log(x) + 2*x*e*log(x) + 2*x^2 + 10*x*e^(2*x) + 10*e^(2*x + 1))/x

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maple [B]  time = 0.04, size = 51, normalized size = 2.04

method	result	size
risch	\(x \ln \relax (x )+\frac {2 x \,{\mathrm e} \ln \relax (x )+3 x^{2} {\mathrm e}+10 \,{\mathrm e}^{2 x +1}+3 x^{3}+10 x \,{\mathrm e}^{2 x}+2 x^{2}}{2 x}\)	\(51\)
default	\(\frac {3 x^{2}}{2}+\frac {3 x \,{\mathrm e}}{2}+x +{\mathrm e} \ln \relax (x )+5 \,{\mathrm e}^{2 x}-5 \,{\mathrm e} \left (-\frac {{\mathrm e}^{2 x}}{x}-2 \expIntegralEi \left (1, -2 x \right )\right )-10 \,{\mathrm e} \expIntegralEi \left (1, -2 x \right )+x \ln \relax (x )\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*x^2*ln(x)+((20*x-10)*exp(1)+20*x^2)*exp(x)^2+(3*x^2+2*x)*exp(1)+6*x^3+4*x^2)/x^2,x,method=_RETURNVE
RBOSE)

[Out]

x*ln(x)+1/2*(2*x*exp(1)*ln(x)+3*x^2*exp(1)+10*exp(2*x+1)+3*x^3+10*x*exp(2*x)+2*x^2)/x

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maxima [C]  time = 0.47, size = 44, normalized size = 1.76 \begin {gather*} \frac {3}{2} \, x^{2} + \frac {3}{2} \, x e + 10 \, {\rm Ei}\left (2 \, x\right ) e - 10 \, e \Gamma \left (-1, -2 \, x\right ) + x \log \relax (x) + e \log \relax (x) + x + 5 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^2*log(x)+((20*x-10)*exp(1)+20*x^2)*exp(x)^2+(3*x^2+2*x)*exp(1)+6*x^3+4*x^2)/x^2,x, algorith
m="maxima")

[Out]

3/2*x^2 + 3/2*x*e + 10*Ei(2*x)*e - 10*e*gamma(-1, -2*x) + x*log(x) + e*log(x) + x + 5*e^(2*x)

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mupad [B]  time = 2.09, size = 38, normalized size = 1.52 \begin {gather*} x+5\,{\mathrm {e}}^{2\,x}+\frac {3\,x\,\mathrm {e}}{2}+\mathrm {e}\,\ln \relax (x)+x\,\ln \relax (x)+\frac {5\,{\mathrm {e}}^{2\,x+1}}{x}+\frac {3\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*log(x) + (exp(2*x)*(20*x^2 + exp(1)*(20*x - 10)))/2 + (exp(1)*(2*x + 3*x^2))/2 + 2*x^2 + 3*x^3)/x^2,x
)

[Out]

x + 5*exp(2*x) + (3*x*exp(1))/2 + exp(1)*log(x) + x*log(x) + (5*exp(2*x + 1))/x + (3*x^2)/2

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sympy [A]  time = 0.33, size = 42, normalized size = 1.68 \begin {gather*} \frac {3 x^{2}}{2} + x \log {\relax (x )} + \frac {x \left (2 + 3 e\right )}{2} + e \log {\relax (x )} + \frac {\left (5 x + 5 e\right ) e^{2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x**2*ln(x)+((20*x-10)*exp(1)+20*x**2)*exp(x)**2+(3*x**2+2*x)*exp(1)+6*x**3+4*x**2)/x**2,x)

[Out]

3*x**2/2 + x*log(x) + x*(2 + 3*E)/2 + E*log(x) + (5*x + 5*E)*exp(2*x)/x

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