3.11.69 \(\int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log (x^2)}{2 \log (4)} \, dx\)

Optimal. Leaf size=25 \[ \frac {3}{2} e^{\frac {5 (3+e) (2-x)}{\log (4)}} x \log \left (x^2\right ) \]

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Rubi [B]  time = 0.32, antiderivative size = 132, normalized size of antiderivative = 5.28, number of steps used = 8, number of rules used = 7, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2227, 2194, 6, 2187, 2176, 2554} \begin {gather*} \frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}+\frac {3 \log (4) e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log \left (x^2\right )}{10 (3+e)}+\frac {3 \log (4) e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}}}{5 (3+e)}-\frac {3 \log (4) e^{\frac {5 (3+e) (2-x)}{\log (4)}}}{5 (3+e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*Log[4] + E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*(-45*x - 15*E*x +
3*Log[4])*Log[x^2])/(2*Log[4]),x]

[Out]

(3*E^((10*(3 + E))/Log[4] - (5*(3 + E)*x)/Log[4])*Log[4])/(5*(3 + E)) - (3*E^((5*(3 + E)*(2 - x))/Log[4])*Log[
4])/(5*(3 + E)) + (3*E^((5*(3 + E)*(2 - x))/Log[4])*Log[4]*Log[x^2])/(10*(3 + E)) + (E^((5*(3 + E)*(2 - x))/Lo
g[4])*(15*(3 + E)*x - Log[64])*Log[x^2])/(10*(3 + E))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )\right ) \, dx}{2 \log (4)}\\ &=3 \int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \, dx+\frac {\int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right ) \, dx}{2 \log (4)}\\ &=3 \int e^{\frac {10 (3+e)-5 (3+e) x}{\log (4)}} \, dx+\frac {\int e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} ((-45-15 e) x+3 \log (4)) \log \left (x^2\right ) \, dx}{2 \log (4)}\\ &=-\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}-\frac {\int 6 e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \log (4) \, dx}{2 \log (4)}\\ &=-\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}-3 \int e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \, dx\\ &=\frac {3 e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}} \log (4)}{5 (3+e)}-\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4)}{5 (3+e)}+\frac {3 e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log (4) \log \left (x^2\right )}{10 (3+e)}+\frac {e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{10 (3+e)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 27, normalized size = 1.08 \begin {gather*} \frac {3 e^{-\frac {5 (3+e) (-2+x)}{\log (4)}} x \log (4) \log \left (x^2\right )}{\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*Log[4] + E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*(-45*x - 15*
E*x + 3*Log[4])*Log[x^2])/(2*Log[4]),x]

[Out]

(3*x*Log[4]*Log[x^2])/(E^((5*(3 + E)*(-2 + x))/Log[4])*Log[16])

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fricas [A]  time = 0.90, size = 25, normalized size = 1.00 \begin {gather*} \frac {3}{2} \, x e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \relax (2)}\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))*log(x^2)+12*log(2)*exp(1
/2*((-5*x+10)*exp(1)-15*x+30)/log(2)))/log(2),x, algorithm="fricas")

[Out]

3/2*x*e^(-5/2*((x - 2)*e + 3*x - 6)/log(2))*log(x^2)

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giac [B]  time = 0.60, size = 198, normalized size = 7.92 \begin {gather*} \frac {3 \, {\left (\frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \relax (2)} - \frac {15 \, x}{2 \, \log \relax (2)} + \frac {5 \, e}{\log \relax (2)} + \frac {15}{\log \relax (2)} + 1\right )} \log \relax (2)^{2}}{e^{2} + 3 \, e} - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \relax (2)}\right )} \log \relax (2)^{2}}{e + 3} + {\left (\frac {{\left (5 \, x e \log \relax (2) + 15 \, x \log \relax (2) + 2 \, \log \relax (2)^{2}\right )} e^{\left (-\frac {5 \, x e + 15 \, x - 10 \, e - 2 \, \log \relax (2) - 30}{2 \, \log \relax (2)}\right )}}{e^{2} + 6 \, e + 9} + \frac {{\left (15 \, x e \log \relax (2) - 2 \, e \log \relax (2)^{2} + 45 \, x \log \relax (2)\right )} e^{\left (-\frac {5 \, {\left (x e + 3 \, x - 2 \, e - 6\right )}}{2 \, \log \relax (2)}\right )}}{e^{2} + 6 \, e + 9}\right )} \log \left (x^{2}\right )\right )}}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))*log(x^2)+12*log(2)*exp(1
/2*((-5*x+10)*exp(1)-15*x+30)/log(2)))/log(2),x, algorithm="giac")

[Out]

3/10*(4*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2) + 1)*log(2)^2/(e^2 + 3*e) - 4*e^(-5/2*((x
- 2)*e + 3*x - 6)/log(2))*log(2)^2/(e + 3) + ((5*x*e*log(2) + 15*x*log(2) + 2*log(2)^2)*e^(-1/2*(5*x*e + 15*x
- 10*e - 2*log(2) - 30)/log(2))/(e^2 + 6*e + 9) + (15*x*e*log(2) - 2*e*log(2)^2 + 45*x*log(2))*e^(-5/2*(x*e +
3*x - 2*e - 6)/log(2))/(e^2 + 6*e + 9))*log(x^2))/log(2)

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maple [A]  time = 0.12, size = 28, normalized size = 1.12

method	result	size
norman	\(\frac {3 x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \relax (2)}} \ln \left (x^{2}\right )}{2}\)	\(28\)
risch	\(3 x \,{\mathrm e}^{-\frac {5 \left (x -2\right ) \left (3+{\mathrm e}\right )}{2 \ln \relax (2)}} \ln \relax (x )-\frac {3 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) x \,{\mathrm e}^{-\frac {5 \left (x -2\right ) \left (3+{\mathrm e}\right )}{2 \ln \relax (2)}}}{4}\)	\(78\)
default	\(\frac {\frac {24 \ln \relax (2)^{2} {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \relax (2)}}}{5 \left (3+{\mathrm e}\right )}+12 \ln \relax (2) x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \relax (2)}} \ln \relax (x )+6 \ln \relax (2) \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \relax (2)}}+\frac {24 \ln \relax (2)^{2} {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \relax (2)}}}{-5 \,{\mathrm e}-15}}{4 \ln \relax (2)}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((6*ln(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln(2))*ln(x^2)+12*ln(2)*exp(1/2*((-5*x+
10)*exp(1)-15*x+30)/ln(2)))/ln(2),x,method=_RETURNVERBOSE)

[Out]

3/2*x*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln(2))*ln(x^2)

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maxima [B]  time = 0.69, size = 130, normalized size = 5.20 \begin {gather*} \frac {3 \, {\left (5 \, x e^{\left (-\frac {5 \, x e}{2 \, \log \relax (2)} - \frac {15 \, x}{2 \, \log \relax (2)} + \frac {5 \, e}{\log \relax (2)} + \frac {15}{\log \relax (2)}\right )} \log \relax (2) \log \left (x^{2}\right ) - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \relax (2)}\right )} \log \relax (2)^{2}}{e + 3} + \frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \relax (2)} - \frac {15 \, x}{2 \, \log \relax (2)} + \frac {5 \, e}{\log \relax (2)} + \frac {15}{\log \relax (2)}\right )} \log \relax (2)}{\frac {e}{\log \relax (2)} + \frac {3}{\log \relax (2)}}\right )}}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))*log(x^2)+12*log(2)*exp(1
/2*((-5*x+10)*exp(1)-15*x+30)/log(2)))/log(2),x, algorithm="maxima")

[Out]

3/10*(5*x*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2))*log(2)*log(x^2) - 4*e^(-5/2*((x - 2)*e
+ 3*x - 6)/log(2))*log(2)^2/(e + 3) + 4*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2))*log(2)/(e
/log(2) + 3/log(2)))/log(2)

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mupad [B]  time = 0.84, size = 41, normalized size = 1.64 \begin {gather*} \frac {3\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{-\frac {15\,x}{2\,\ln \relax (2)}}\,{\mathrm {e}}^{-\frac {5\,x\,\mathrm {e}}{2\,\ln \relax (2)}}\,{\mathrm {e}}^{\frac {15}{\ln \relax (2)}}\,{\mathrm {e}}^{\frac {5\,\mathrm {e}}{\ln \relax (2)}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(-((15*x)/2 + (exp(1)*(5*x - 10))/2 - 15)/log(2))*log(2) - (log(x^2)*exp(-((15*x)/2 + (exp(1)*(5*x -
10))/2 - 15)/log(2))*(45*x - 6*log(2) + 15*x*exp(1)))/4)/log(2),x)

[Out]

(3*x*log(x^2)*exp(-(15*x)/(2*log(2)))*exp(-(5*x*exp(1))/(2*log(2)))*exp(15/log(2))*exp((5*exp(1))/log(2)))/2

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sympy [A]  time = 0.40, size = 31, normalized size = 1.24 \begin {gather*} \frac {3 x e^{\frac {- \frac {15 x}{2} + \frac {e \left (10 - 5 x\right )}{2} + 15}{\log {\relax (2 )}}} \log {\left (x^{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((6*ln(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln(2))*ln(x**2)+12*ln(2)*exp(1/2*
((-5*x+10)*exp(1)-15*x+30)/ln(2)))/ln(2),x)

[Out]

3*x*exp((-15*x/2 + E*(10 - 5*x)/2 + 15)/log(2))*log(x**2)/2

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