3.14.38 \(\int \frac {1}{72} (-288 x+432 x^2+e^{4 x} (1+4 x)+(288-576 x) \log (x)+144 \log ^2(x)) \, dx\)

Optimal. Leaf size=21 \[ 2 x \left (\frac {e^{4 x}}{144}+(-x+\log (x))^2\right ) \]

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Rubi [B]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 3.29, number of steps used = 8, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {12, 2176, 2194, 2313, 9, 2296, 2295} \begin {gather*} 2 x^3-2 x^2+4 \left (x-x^2\right ) \log (x)+4 x-\frac {e^{4 x}}{288}+2 (1-x)^2+\frac {1}{288} e^{4 x} (4 x+1)+2 x \log ^2(x)-4 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-288*x + 432*x^2 + E^(4*x)*(1 + 4*x) + (288 - 576*x)*Log[x] + 144*Log[x]^2)/72,x]

[Out]

-1/288*E^(4*x) + 2*(1 - x)^2 + 4*x - 2*x^2 + 2*x^3 + (E^(4*x)*(1 + 4*x))/288 - 4*x*Log[x] + 4*(x - x^2)*Log[x]
+ 2*x*Log[x]^2

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, 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 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]

Rule 2295

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

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{72} \int \left (-288 x+432 x^2+e^{4 x} (1+4 x)+(288-576 x) \log (x)+144 \log ^2(x)\right ) \, dx\\ &=-2 x^2+2 x^3+\frac {1}{72} \int e^{4 x} (1+4 x) \, dx+\frac {1}{72} \int (288-576 x) \log (x) \, dx+2 \int \log ^2(x) \, dx\\ &=-2 x^2+2 x^3+\frac {1}{288} e^{4 x} (1+4 x)+4 \left (x-x^2\right ) \log (x)+2 x \log ^2(x)-\frac {1}{72} \int e^{4 x} \, dx-\frac {1}{72} \int 288 (1-x) \, dx-4 \int \log (x) \, dx\\ &=-\frac {e^{4 x}}{288}+2 (1-x)^2+4 x-2 x^2+2 x^3+\frac {1}{288} e^{4 x} (1+4 x)-4 x \log (x)+4 \left (x-x^2\right ) \log (x)+2 x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.43 \begin {gather*} \frac {1}{72} e^{4 x} x+2 x^3-4 x^2 \log (x)+2 x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-288*x + 432*x^2 + E^(4*x)*(1 + 4*x) + (288 - 576*x)*Log[x] + 144*Log[x]^2)/72,x]

[Out]

(E^(4*x)*x)/72 + 2*x^3 - 4*x^2*Log[x] + 2*x*Log[x]^2

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fricas [A]  time = 1.39, size = 27, normalized size = 1.29 \begin {gather*} 2 \, x^{3} - 4 \, x^{2} \log \relax (x) + 2 \, x \log \relax (x)^{2} + \frac {1}{72} \, x e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(x)^2+1/72*(-576*x+288)*log(x)+1/72*(4*x+1)*exp(2*x)^2+6*x^2-4*x,x, algorithm="fricas")

[Out]

2*x^3 - 4*x^2*log(x) + 2*x*log(x)^2 + 1/72*x*e^(4*x)

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giac [A]  time = 0.29, size = 27, normalized size = 1.29 \begin {gather*} 2 \, x^{3} - 4 \, x^{2} \log \relax (x) + 2 \, x \log \relax (x)^{2} + \frac {1}{72} \, x e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(x)^2+1/72*(-576*x+288)*log(x)+1/72*(4*x+1)*exp(2*x)^2+6*x^2-4*x,x, algorithm="giac")

[Out]

2*x^3 - 4*x^2*log(x) + 2*x*log(x)^2 + 1/72*x*e^(4*x)

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maple [A]  time = 0.03, size = 28, normalized size = 1.33

method	result	size
risch	\(2 x^{3}+2 x \ln \relax (x )^{2}-4 x^{2} \ln \relax (x )+\frac {x \,{\mathrm e}^{4 x}}{72}\)	\(28\)
default	\(2 x^{3}+2 x \ln \relax (x )^{2}-4 x^{2} \ln \relax (x )+\frac {x \,{\mathrm e}^{4 x}}{72}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*ln(x)^2+1/72*(-576*x+288)*ln(x)+1/72*(4*x+1)*exp(2*x)^2+6*x^2-4*x,x,method=_RETURNVERBOSE)

[Out]

2*x^3+2*x*ln(x)^2-4*x^2*ln(x)+1/72*x*exp(4*x)

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maxima [B]  time = 0.64, size = 40, normalized size = 1.90 \begin {gather*} 2 \, x^{3} + 2 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + \frac {1}{72} \, x e^{\left (4 \, x\right )} - 4 \, {\left (x^{2} - x\right )} \log \relax (x) - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*log(x)^2+1/72*(-576*x+288)*log(x)+1/72*(4*x+1)*exp(2*x)^2+6*x^2-4*x,x, algorithm="maxima")

[Out]

2*x^3 + 2*(log(x)^2 - 2*log(x) + 2)*x + 1/72*x*e^(4*x) - 4*(x^2 - x)*log(x) - 4*x

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mupad [B]  time = 0.98, size = 24, normalized size = 1.14 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^{4\,x}+144\,{\ln \relax (x)}^2-288\,x\,\ln \relax (x)+144\,x^2\right )}{72} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*log(x)^2 - 4*x - (log(x)*(576*x - 288))/72 + (exp(4*x)*(4*x + 1))/72 + 6*x^2,x)

[Out]

(x*(exp(4*x) + 144*log(x)^2 - 288*x*log(x) + 144*x^2))/72

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sympy [A]  time = 0.31, size = 29, normalized size = 1.38 \begin {gather*} 2 x^{3} - 4 x^{2} \log {\relax (x )} + \frac {x e^{4 x}}{72} + 2 x \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*ln(x)**2+1/72*(-576*x+288)*ln(x)+1/72*(4*x+1)*exp(2*x)**2+6*x**2-4*x,x)

[Out]

2*x**3 - 4*x**2*log(x) + x*exp(4*x)/72 + 2*x*log(x)**2

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