3.100.68 \(\int \frac {1}{25} (10 x+16 x^3+e^{4 x} (6 x^5+4 x^6)+12 x^2 \log (2)+2 x \log ^2(2)+e^{2 x} (-20 x^4-8 x^5+(-8 x^3-4 x^4) \log (2))) \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{25} x^2 \left (5+\left (2 x-e^{2 x} x^2+\log (2)\right )^2\right ) \]

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Rubi [B]  time = 0.46, antiderivative size = 68, normalized size of antiderivative = 2.52, number of steps used = 43, number of rules used = 6, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6, 12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {1}{25} e^{4 x} x^6-\frac {4}{25} e^{2 x} x^5+\frac {4 x^4}{25}-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + 16*x^3 + E^(4*x)*(6*x^5 + 4*x^6) + 12*x^2*Log[2] + 2*x*Log[2]^2 + E^(2*x)*(-20*x^4 - 8*x^5 + (-8*x
^3 - 4*x^4)*Log[2]))/25,x]

[Out]

(4*x^4)/25 - (4*E^(2*x)*x^5)/25 + (E^(4*x)*x^6)/25 + (4*x^3*Log[2])/25 - (2*E^(2*x)*x^4*Log[2])/25 + (x^2*(5 +
Log[2]^2))/25

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2196

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{25} \left (16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )+x \left (10+2 \log ^2(2)\right )\right ) \, dx\\ &=\frac {1}{25} \int \left (16 x^3+e^{4 x} \left (6 x^5+4 x^6\right )+12 x^2 \log (2)+e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right )+x \left (10+2 \log ^2(2)\right )\right ) \, dx\\ &=\frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int e^{4 x} \left (6 x^5+4 x^6\right ) \, dx+\frac {1}{25} \int e^{2 x} \left (-20 x^4-8 x^5+\left (-8 x^3-4 x^4\right ) \log (2)\right ) \, dx\\ &=\frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int e^{4 x} x^5 (6+4 x) \, dx+\frac {1}{25} \int \left (-20 e^{2 x} x^4-8 e^{2 x} x^5-4 e^{2 x} x^3 (2+x) \log (2)\right ) \, dx\\ &=\frac {4 x^4}{25}+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {1}{25} \int \left (6 e^{4 x} x^5+4 e^{4 x} x^6\right ) \, dx-\frac {8}{25} \int e^{2 x} x^5 \, dx-\frac {4}{5} \int e^{2 x} x^4 \, dx-\frac {1}{25} (4 \log (2)) \int e^{2 x} x^3 (2+x) \, dx\\ &=\frac {4 x^4}{25}-\frac {2}{5} e^{2 x} x^4-\frac {4}{25} e^{2 x} x^5+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {4}{25} \int e^{4 x} x^6 \, dx+\frac {6}{25} \int e^{4 x} x^5 \, dx+\frac {4}{5} \int e^{2 x} x^4 \, dx+\frac {8}{5} \int e^{2 x} x^3 \, dx-\frac {1}{25} (4 \log (2)) \int \left (2 e^{2 x} x^3+e^{2 x} x^4\right ) \, dx\\ &=\frac {4}{5} e^{2 x} x^3+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {3}{50} e^{4 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {6}{25} \int e^{4 x} x^5 \, dx-\frac {3}{10} \int e^{4 x} x^4 \, dx-\frac {8}{5} \int e^{2 x} x^3 \, dx-\frac {12}{5} \int e^{2 x} x^2 \, dx-\frac {1}{25} (4 \log (2)) \int e^{2 x} x^4 \, dx-\frac {1}{25} (8 \log (2)) \int e^{2 x} x^3 \, dx\\ &=-\frac {6}{5} e^{2 x} x^2+\frac {4 x^4}{25}-\frac {3}{40} e^{4 x} x^4-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {4}{25} e^{2 x} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {3}{10} \int e^{4 x} x^3 \, dx+\frac {3}{10} \int e^{4 x} x^4 \, dx+\frac {12}{5} \int e^{2 x} x \, dx+\frac {12}{5} \int e^{2 x} x^2 \, dx+\frac {1}{25} (8 \log (2)) \int e^{2 x} x^3 \, dx+\frac {1}{25} (12 \log (2)) \int e^{2 x} x^2 \, dx\\ &=\frac {6}{5} e^{2 x} x+\frac {3}{40} e^{4 x} x^3+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {6}{25} e^{2 x} x^2 \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {9}{40} \int e^{4 x} x^2 \, dx-\frac {3}{10} \int e^{4 x} x^3 \, dx-\frac {6}{5} \int e^{2 x} \, dx-\frac {12}{5} \int e^{2 x} x \, dx-\frac {1}{25} (12 \log (2)) \int e^{2 x} x \, dx-\frac {1}{25} (12 \log (2)) \int e^{2 x} x^2 \, dx\\ &=-\frac {3 e^{2 x}}{5}-\frac {9}{160} e^{4 x} x^2+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6-\frac {6}{25} e^{2 x} x \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {9}{80} \int e^{4 x} x \, dx+\frac {9}{40} \int e^{4 x} x^2 \, dx+\frac {6}{5} \int e^{2 x} \, dx+\frac {1}{25} (6 \log (2)) \int e^{2 x} \, dx+\frac {1}{25} (12 \log (2)) \int e^{2 x} x \, dx\\ &=\frac {9}{320} e^{4 x} x+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {3}{25} e^{2 x} \log (2)+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )-\frac {9}{320} \int e^{4 x} \, dx-\frac {9}{80} \int e^{4 x} x \, dx-\frac {1}{25} (6 \log (2)) \int e^{2 x} \, dx\\ &=-\frac {9 e^{4 x}}{1280}+\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )+\frac {9}{320} \int e^{4 x} \, dx\\ &=\frac {4 x^4}{25}-\frac {4}{25} e^{2 x} x^5+\frac {1}{25} e^{4 x} x^6+\frac {4}{25} x^3 \log (2)-\frac {2}{25} e^{2 x} x^4 \log (2)+\frac {1}{25} x^2 \left (5+\log ^2(2)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 47, normalized size = 1.74 \begin {gather*} \frac {1}{25} x^2 \left (5+4 x^2+e^{4 x} x^4+\log ^2(2)-2 e^{2 x} x^2 (2 x+\log (2))+x \log (16)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 16*x^3 + E^(4*x)*(6*x^5 + 4*x^6) + 12*x^2*Log[2] + 2*x*Log[2]^2 + E^(2*x)*(-20*x^4 - 8*x^5 +
(-8*x^3 - 4*x^4)*Log[2]))/25,x]

[Out]

(x^2*(5 + 4*x^2 + E^(4*x)*x^4 + Log[2]^2 - 2*E^(2*x)*x^2*(2*x + Log[2]) + x*Log[16]))/25

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fricas [B]  time = 0.68, size = 54, normalized size = 2.00 \begin {gather*} \frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \relax (2) + \frac {1}{25} \, x^{2} \log \relax (2)^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \relax (2)\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^6+6*x^5)*exp(x)^4+1/25*((-4*x^4-8*x^3)*log(2)-8*x^5-20*x^4)*exp(x)^2+2/25*x*log(2)^2+12/25
*x^2*log(2)+16/25*x^3+2/5*x,x, algorithm="fricas")

[Out]

1/25*x^6*e^(4*x) + 4/25*x^4 + 4/25*x^3*log(2) + 1/25*x^2*log(2)^2 + 1/5*x^2 - 2/25*(2*x^5 + x^4*log(2))*e^(2*x
)

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giac [B]  time = 0.13, size = 54, normalized size = 2.00 \begin {gather*} \frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \relax (2) + \frac {1}{25} \, x^{2} \log \relax (2)^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \relax (2)\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^6+6*x^5)*exp(x)^4+1/25*((-4*x^4-8*x^3)*log(2)-8*x^5-20*x^4)*exp(x)^2+2/25*x*log(2)^2+12/25
*x^2*log(2)+16/25*x^3+2/5*x,x, algorithm="giac")

[Out]

1/25*x^6*e^(4*x) + 4/25*x^4 + 4/25*x^3*log(2) + 1/25*x^2*log(2)^2 + 1/5*x^2 - 2/25*(2*x^5 + x^4*log(2))*e^(2*x
)

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maple [B]  time = 0.05, size = 56, normalized size = 2.07

method	result	size
risch	\(\frac {x^{6} {\mathrm e}^{4 x}}{25}+\frac {\left (-2 x^{4} \ln \relax (2)-4 x^{5}\right ) {\mathrm e}^{2 x}}{25}+\frac {x^{2} \ln \relax (2)^{2}}{25}+\frac {4 x^{3} \ln \relax (2)}{25}+\frac {4 x^{4}}{25}+\frac {x^{2}}{5}\)	\(56\)
default	\(\frac {x^{2}}{5}+\frac {4 x^{4}}{25}+\frac {x^{2} \ln \relax (2)^{2}}{25}+\frac {4 x^{3} \ln \relax (2)}{25}-\frac {4 x^{5} {\mathrm e}^{2 x}}{25}-\frac {2 \,{\mathrm e}^{2 x} \ln \relax (2) x^{4}}{25}+\frac {x^{6} {\mathrm e}^{4 x}}{25}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(4*x^6+6*x^5)*exp(x)^4+1/25*((-4*x^4-8*x^3)*ln(2)-8*x^5-20*x^4)*exp(x)^2+2/25*x*ln(2)^2+12/25*x^2*ln(
2)+16/25*x^3+2/5*x,x,method=_RETURNVERBOSE)

[Out]

1/25*x^6*exp(4*x)+1/25*(-2*x^4*ln(2)-4*x^5)*exp(2*x)+1/25*x^2*ln(2)^2+4/25*x^3*ln(2)+4/25*x^4+1/5*x^2

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maxima [B]  time = 0.45, size = 54, normalized size = 2.00 \begin {gather*} \frac {1}{25} \, x^{6} e^{\left (4 \, x\right )} + \frac {4}{25} \, x^{4} + \frac {4}{25} \, x^{3} \log \relax (2) + \frac {1}{25} \, x^{2} \log \relax (2)^{2} + \frac {1}{5} \, x^{2} - \frac {2}{25} \, {\left (2 \, x^{5} + x^{4} \log \relax (2)\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x^6+6*x^5)*exp(x)^4+1/25*((-4*x^4-8*x^3)*log(2)-8*x^5-20*x^4)*exp(x)^2+2/25*x*log(2)^2+12/25
*x^2*log(2)+16/25*x^3+2/5*x,x, algorithm="maxima")

[Out]

1/25*x^6*e^(4*x) + 4/25*x^4 + 4/25*x^3*log(2) + 1/25*x^2*log(2)^2 + 1/5*x^2 - 2/25*(2*x^5 + x^4*log(2))*e^(2*x
)

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mupad [B]  time = 8.41, size = 54, normalized size = 2.00 \begin {gather*} \frac {x^6\,{\mathrm {e}}^{4\,x}}{25}-\frac {4\,x^5\,{\mathrm {e}}^{2\,x}}{25}+\frac {x^3\,\ln \left (16\right )}{25}+x^2\,\left (\frac {{\ln \relax (2)}^2}{25}+\frac {1}{5}\right )+\frac {4\,x^4}{25}-\frac {x^4\,{\mathrm {e}}^{2\,x}\,\ln \relax (4)}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x)/5 - (exp(2*x)*(log(2)*(8*x^3 + 4*x^4) + 20*x^4 + 8*x^5))/25 + (exp(4*x)*(6*x^5 + 4*x^6))/25 + (2*x*l
og(2)^2)/25 + (12*x^2*log(2))/25 + (16*x^3)/25,x)

[Out]

(x^6*exp(4*x))/25 - (4*x^5*exp(2*x))/25 + (x^3*log(16))/25 + x^2*(log(2)^2/25 + 1/5) + (4*x^4)/25 - (x^4*exp(2
*x)*log(4))/25

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sympy [B]  time = 0.18, size = 61, normalized size = 2.26 \begin {gather*} \frac {x^{6} e^{4 x}}{25} + \frac {4 x^{4}}{25} + \frac {4 x^{3} \log {\relax (2 )}}{25} + x^{2} \left (\frac {\log {\relax (2 )}^{2}}{25} + \frac {1}{5}\right ) + \frac {\left (- 100 x^{5} - 50 x^{4} \log {\relax (2 )}\right ) e^{2 x}}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(4*x**6+6*x**5)*exp(x)**4+1/25*((-4*x**4-8*x**3)*ln(2)-8*x**5-20*x**4)*exp(x)**2+2/25*x*ln(2)**
2+12/25*x**2*ln(2)+16/25*x**3+2/5*x,x)

[Out]

x**6*exp(4*x)/25 + 4*x**4/25 + 4*x**3*log(2)/25 + x**2*(log(2)**2/25 + 1/5) + (-100*x**5 - 50*x**4*log(2))*exp
(2*x)/625

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