3.61.26 \(\int \frac {4+e^{2 x} x+144 x^2+(-4+144 x^2+2 e^{2 x} x^2) \log (x)}{144 x^2} \, dx\)

Optimal. Leaf size=20 \[ \left (\frac {1}{144} \left (e^{2 x}+\frac {4}{x}\right )+x\right ) \log (x) \]

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Rubi [A]  time = 0.10, antiderivative size = 25, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 14, 2288, 2334} \begin {gather*} \frac {1}{144} e^{2 x} \log (x)+\frac {1}{36} \left (36 x+\frac {1}{x}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + E^(2*x)*x + 144*x^2 + (-4 + 144*x^2 + 2*E^(2*x)*x^2)*Log[x])/(144*x^2),x]

[Out]

(E^(2*x)*Log[x])/144 + ((x^(-1) + 36*x)*Log[x])/36

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 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{144} \int \frac {4+e^{2 x} x+144 x^2+\left (-4+144 x^2+2 e^{2 x} x^2\right ) \log (x)}{x^2} \, dx\\ &=\frac {1}{144} \int \left (\frac {e^{2 x} (1+2 x \log (x))}{x}+\frac {4 \left (1+36 x^2-\log (x)+36 x^2 \log (x)\right )}{x^2}\right ) \, dx\\ &=\frac {1}{144} \int \frac {e^{2 x} (1+2 x \log (x))}{x} \, dx+\frac {1}{36} \int \frac {1+36 x^2-\log (x)+36 x^2 \log (x)}{x^2} \, dx\\ &=\frac {1}{144} e^{2 x} \log (x)+\frac {1}{36} \int \left (\frac {1+36 x^2}{x^2}+\frac {\left (-1+36 x^2\right ) \log (x)}{x^2}\right ) \, dx\\ &=\frac {1}{144} e^{2 x} \log (x)+\frac {1}{36} \int \frac {1+36 x^2}{x^2} \, dx+\frac {1}{36} \int \frac {\left (-1+36 x^2\right ) \log (x)}{x^2} \, dx\\ &=\frac {1}{144} e^{2 x} \log (x)+\frac {1}{36} \left (\frac {1}{x}+36 x\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.25 \begin {gather*} \frac {1}{144} \left (e^{2 x} \log (x)+\frac {4 \log (x)}{x}+144 x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + E^(2*x)*x + 144*x^2 + (-4 + 144*x^2 + 2*E^(2*x)*x^2)*Log[x])/(144*x^2),x]

[Out]

(E^(2*x)*Log[x] + (4*Log[x])/x + 144*x*Log[x])/144

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fricas [A]  time = 0.61, size = 20, normalized size = 1.00 \begin {gather*} \frac {{\left (144 \, x^{2} + x e^{\left (2 \, x\right )} + 4\right )} \log \relax (x)}{144 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/144*((2*exp(x)^2*x^2+144*x^2-4)*log(x)+x*exp(x)^2+144*x^2+4)/x^2,x, algorithm="fricas")

[Out]

1/144*(144*x^2 + x*e^(2*x) + 4)*log(x)/x

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giac [A]  time = 0.14, size = 25, normalized size = 1.25 \begin {gather*} \frac {144 \, x^{2} \log \relax (x) + x e^{\left (2 \, x\right )} \log \relax (x) + 4 \, \log \relax (x)}{144 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/144*((2*exp(x)^2*x^2+144*x^2-4)*log(x)+x*exp(x)^2+144*x^2+4)/x^2,x, algorithm="giac")

[Out]

1/144*(144*x^2*log(x) + x*e^(2*x)*log(x) + 4*log(x))/x

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maple [A]  time = 0.05, size = 21, normalized size = 1.05

method	result	size
default	\(\frac {{\mathrm e}^{2 x} \ln \relax (x )}{144}+x \ln \relax (x )+\frac {\ln \relax (x )}{36 x}\)	\(21\)
risch	\(\frac {\left (x \,{\mathrm e}^{2 x}+144 x^{2}+4\right ) \ln \relax (x )}{144 x}\)	\(21\)
norman	\(\frac {x^{2} \ln \relax (x )+\frac {x \,{\mathrm e}^{2 x} \ln \relax (x )}{144}+\frac {\ln \relax (x )}{36}}{x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/144*((2*exp(x)^2*x^2+144*x^2-4)*ln(x)+x*exp(x)^2+144*x^2+4)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/144*exp(x)^2*ln(x)+x*ln(x)+1/36*ln(x)/x

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maxima [A]  time = 0.37, size = 20, normalized size = 1.00 \begin {gather*} x \log \relax (x) + \frac {1}{144} \, e^{\left (2 \, x\right )} \log \relax (x) + \frac {\log \relax (x)}{36 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/144*((2*exp(x)^2*x^2+144*x^2-4)*log(x)+x*exp(x)^2+144*x^2+4)/x^2,x, algorithm="maxima")

[Out]

x*log(x) + 1/144*e^(2*x)*log(x) + 1/36*log(x)/x

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mupad [B]  time = 4.37, size = 20, normalized size = 1.00 \begin {gather*} \frac {\ln \relax (x)\,\left (x\,{\mathrm {e}}^{2\,x}+144\,x^2+4\right )}{144\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(2*x))/144 + (log(x)*(2*x^2*exp(2*x) + 144*x^2 - 4))/144 + x^2 + 1/36)/x^2,x)

[Out]

(log(x)*(x*exp(2*x) + 144*x^2 + 4))/(144*x)

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sympy [A]  time = 0.34, size = 22, normalized size = 1.10 \begin {gather*} \frac {e^{2 x} \log {\relax (x )}}{144} + \frac {\left (36 x^{2} + 1\right ) \log {\relax (x )}}{36 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/144*((2*exp(x)**2*x**2+144*x**2-4)*ln(x)+x*exp(x)**2+144*x**2+4)/x**2,x)

[Out]

exp(2*x)*log(x)/144 + (36*x**2 + 1)*log(x)/(36*x)

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