3.20.56 \(\int \frac {4-144 x^3-576 x^3 \log (4)+e^{2 x} (-72 x-72 x^2+(-288 x-288 x^2) \log (4))+e^x (216 x^2+72 x^3+(864 x^2+288 x^3) \log (4))}{1+4 \log (4)} \, dx\)

Optimal. Leaf size=25 \[ -36 x^2 \left (-e^x+x\right )^2+\frac {x}{\frac {1}{4}+\log (4)} \]

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Rubi [B]  time = 0.40, antiderivative size = 79, normalized size of antiderivative = 3.16, number of steps used = 35, number of rules used = 5, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 12, 2196, 2176, 2194} \begin {gather*} -36 x^4+\frac {72 e^x x^3}{1+\log (256)}+\frac {288 e^x x^3 \log (4)}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}-\frac {144 e^{2 x} x^2 \log (4)}{1+\log (256)}+\frac {4 x}{1+\log (256)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 144*x^3 - 576*x^3*Log[4] + E^(2*x)*(-72*x - 72*x^2 + (-288*x - 288*x^2)*Log[4]) + E^x*(216*x^2 + 72*x
^3 + (864*x^2 + 288*x^3)*Log[4]))/(1 + 4*Log[4]),x]

[Out]

-36*x^4 + (4*x)/(1 + Log[256]) - (36*E^(2*x)*x^2)/(1 + Log[256]) + (72*E^x*x^3)/(1 + Log[256]) - (144*E^(2*x)*
x^2*Log[4])/(1 + Log[256]) + (288*E^x*x^3*Log[4])/(1 + Log[256])

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 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 {4+x^3 (-144-576 \log (4))+e^{2 x} \left (-72 x-72 x^2+\left (-288 x-288 x^2\right ) \log (4)\right )+e^x \left (216 x^2+72 x^3+\left (864 x^2+288 x^3\right ) \log (4)\right )}{1+4 \log (4)} \, dx\\ &=\frac {\int \left (4+x^3 (-144-576 \log (4))+e^{2 x} \left (-72 x-72 x^2+\left (-288 x-288 x^2\right ) \log (4)\right )+e^x \left (216 x^2+72 x^3+\left (864 x^2+288 x^3\right ) \log (4)\right )\right ) \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}+\frac {\int e^{2 x} \left (-72 x-72 x^2+\left (-288 x-288 x^2\right ) \log (4)\right ) \, dx}{1+\log (256)}+\frac {\int e^x \left (216 x^2+72 x^3+\left (864 x^2+288 x^3\right ) \log (4)\right ) \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}+\frac {\int \left (-72 e^{2 x} x-72 e^{2 x} x^2-288 e^{2 x} x (1+x) \log (4)\right ) \, dx}{1+\log (256)}+\frac {\int \left (216 e^x x^2+72 e^x x^3+288 e^x x^2 (3+x) \log (4)\right ) \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}-\frac {72 \int e^{2 x} x \, dx}{1+\log (256)}-\frac {72 \int e^{2 x} x^2 \, dx}{1+\log (256)}+\frac {72 \int e^x x^3 \, dx}{1+\log (256)}+\frac {216 \int e^x x^2 \, dx}{1+\log (256)}-\frac {(288 \log (4)) \int e^{2 x} x (1+x) \, dx}{1+\log (256)}+\frac {(288 \log (4)) \int e^x x^2 (3+x) \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}-\frac {36 e^{2 x} x}{1+\log (256)}+\frac {216 e^x x^2}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}+\frac {36 \int e^{2 x} \, dx}{1+\log (256)}+\frac {72 \int e^{2 x} x \, dx}{1+\log (256)}-\frac {216 \int e^x x^2 \, dx}{1+\log (256)}-\frac {432 \int e^x x \, dx}{1+\log (256)}-\frac {(288 \log (4)) \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx}{1+\log (256)}+\frac {(288 \log (4)) \int \left (3 e^x x^2+e^x x^3\right ) \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {18 e^{2 x}}{1+\log (256)}+\frac {4 x}{1+\log (256)}-\frac {432 e^x x}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}-\frac {36 \int e^{2 x} \, dx}{1+\log (256)}+\frac {432 \int e^x \, dx}{1+\log (256)}+\frac {432 \int e^x x \, dx}{1+\log (256)}-\frac {(288 \log (4)) \int e^{2 x} x \, dx}{1+\log (256)}-\frac {(288 \log (4)) \int e^{2 x} x^2 \, dx}{1+\log (256)}+\frac {(288 \log (4)) \int e^x x^3 \, dx}{1+\log (256)}+\frac {(864 \log (4)) \int e^x x^2 \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {432 e^x}{1+\log (256)}+\frac {4 x}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}-\frac {144 e^{2 x} x \log (4)}{1+\log (256)}+\frac {864 e^x x^2 \log (4)}{1+\log (256)}-\frac {144 e^{2 x} x^2 \log (4)}{1+\log (256)}+\frac {288 e^x x^3 \log (4)}{1+\log (256)}-\frac {432 \int e^x \, dx}{1+\log (256)}+\frac {(144 \log (4)) \int e^{2 x} \, dx}{1+\log (256)}+\frac {(288 \log (4)) \int e^{2 x} x \, dx}{1+\log (256)}-\frac {(864 \log (4)) \int e^x x^2 \, dx}{1+\log (256)}-\frac {(1728 \log (4)) \int e^x x \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}+\frac {72 e^{2 x} \log (4)}{1+\log (256)}-\frac {1728 e^x x \log (4)}{1+\log (256)}-\frac {144 e^{2 x} x^2 \log (4)}{1+\log (256)}+\frac {288 e^x x^3 \log (4)}{1+\log (256)}-\frac {(144 \log (4)) \int e^{2 x} \, dx}{1+\log (256)}+\frac {(1728 \log (4)) \int e^x \, dx}{1+\log (256)}+\frac {(1728 \log (4)) \int e^x x \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}+\frac {1728 e^x \log (4)}{1+\log (256)}-\frac {144 e^{2 x} x^2 \log (4)}{1+\log (256)}+\frac {288 e^x x^3 \log (4)}{1+\log (256)}-\frac {(1728 \log (4)) \int e^x \, dx}{1+\log (256)}\\ &=-36 x^4+\frac {4 x}{1+\log (256)}-\frac {36 e^{2 x} x^2}{1+\log (256)}+\frac {72 e^x x^3}{1+\log (256)}-\frac {144 e^{2 x} x^2 \log (4)}{1+\log (256)}+\frac {288 e^x x^3 \log (4)}{1+\log (256)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.36 \begin {gather*} 4 \left (-9 e^{2 x} x^2+18 e^x x^3-9 x^4+\frac {x}{1+\log (256)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 144*x^3 - 576*x^3*Log[4] + E^(2*x)*(-72*x - 72*x^2 + (-288*x - 288*x^2)*Log[4]) + E^x*(216*x^2
+ 72*x^3 + (864*x^2 + 288*x^3)*Log[4]))/(1 + 4*Log[4]),x]

[Out]

4*(-9*E^(2*x)*x^2 + 18*E^x*x^3 - 9*x^4 + x/(1 + Log[256]))

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fricas [B]  time = 1.56, size = 58, normalized size = 2.32 \begin {gather*} -\frac {4 \, {\left (72 \, x^{4} \log \relax (2) + 9 \, x^{4} + 9 \, {\left (8 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} - 18 \, {\left (8 \, x^{3} \log \relax (2) + x^{3}\right )} e^{x} - x\right )}}{8 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-288*x^2-288*x)*log(2)-72*x^2-72*x)*exp(x)^2+(2*(288*x^3+864*x^2)*log(2)+72*x^3+216*x^2)*exp(x)
-1152*x^3*log(2)-144*x^3+4)/(8*log(2)+1),x, algorithm="fricas")

[Out]

-4*(72*x^4*log(2) + 9*x^4 + 9*(8*x^2*log(2) + x^2)*e^(2*x) - 18*(8*x^3*log(2) + x^3)*e^x - x)/(8*log(2) + 1)

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giac [B]  time = 0.23, size = 58, normalized size = 2.32 \begin {gather*} -\frac {4 \, {\left (72 \, x^{4} \log \relax (2) + 9 \, x^{4} + 9 \, {\left (8 \, x^{2} \log \relax (2) + x^{2}\right )} e^{\left (2 \, x\right )} - 18 \, {\left (8 \, x^{3} \log \relax (2) + x^{3}\right )} e^{x} - x\right )}}{8 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-288*x^2-288*x)*log(2)-72*x^2-72*x)*exp(x)^2+(2*(288*x^3+864*x^2)*log(2)+72*x^3+216*x^2)*exp(x)
-1152*x^3*log(2)-144*x^3+4)/(8*log(2)+1),x, algorithm="giac")

[Out]

-4*(72*x^4*log(2) + 9*x^4 + 9*(8*x^2*log(2) + x^2)*e^(2*x) - 18*(8*x^3*log(2) + x^3)*e^x - x)/(8*log(2) + 1)

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maple [A]  time = 0.05, size = 34, normalized size = 1.36

method	result	size
norman	\(-36 x^{4}+\frac {4 x}{8 \ln \relax (2)+1}+72 \,{\mathrm e}^{x} x^{3}-36 \,{\mathrm e}^{2 x} x^{2}\)	\(34\)
risch	\(-\frac {288 x^{4} \ln \relax (2)}{8 \ln \relax (2)+1}-\frac {36 x^{4}}{8 \ln \relax (2)+1}+\frac {4 x}{8 \ln \relax (2)+1}-36 \,{\mathrm e}^{2 x} x^{2}+72 \,{\mathrm e}^{x} x^{3}\)	\(57\)
default	\(\frac {4 x -36 \,{\mathrm e}^{2 x} x^{2}-288 x^{2} \ln \relax (2) {\mathrm e}^{2 x}+72 \,{\mathrm e}^{x} x^{3}+576 x^{3} \ln \relax (2) {\mathrm e}^{x}-36 x^{4}-288 x^{4} \ln \relax (2)}{8 \ln \relax (2)+1}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(-288*x^2-288*x)*ln(2)-72*x^2-72*x)*exp(x)^2+(2*(288*x^3+864*x^2)*ln(2)+72*x^3+216*x^2)*exp(x)-1152*x^
3*ln(2)-144*x^3+4)/(8*ln(2)+1),x,method=_RETURNVERBOSE)

[Out]

-36*x^4+4/(8*ln(2)+1)*x+72*exp(x)*x^3-36*exp(x)^2*x^2

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maxima [B]  time = 0.56, size = 52, normalized size = 2.08 \begin {gather*} \frac {4 \, {\left (18 \, x^{3} {\left (8 \, \log \relax (2) + 1\right )} e^{x} - 72 \, x^{4} \log \relax (2) - 9 \, x^{4} - 9 \, x^{2} {\left (8 \, \log \relax (2) + 1\right )} e^{\left (2 \, x\right )} + x\right )}}{8 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-288*x^2-288*x)*log(2)-72*x^2-72*x)*exp(x)^2+(2*(288*x^3+864*x^2)*log(2)+72*x^3+216*x^2)*exp(x)
-1152*x^3*log(2)-144*x^3+4)/(8*log(2)+1),x, algorithm="maxima")

[Out]

4*(18*x^3*(8*log(2) + 1)*e^x - 72*x^4*log(2) - 9*x^4 - 9*x^2*(8*log(2) + 1)*e^(2*x) + x)/(8*log(2) + 1)

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mupad [B]  time = 1.17, size = 51, normalized size = 2.04 \begin {gather*} \frac {4\,x-x^4\,\left (288\,\ln \relax (2)+36\right )+x^3\,{\mathrm {e}}^x\,\left (576\,\ln \relax (2)+72\right )-x^2\,{\mathrm {e}}^{2\,x}\,\left (288\,\ln \relax (2)+36\right )}{8\,\ln \relax (2)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x)*(72*x + 2*log(2)*(288*x + 288*x^2) + 72*x^2) - exp(x)*(2*log(2)*(864*x^2 + 288*x^3) + 216*x^2 +
72*x^3) + 1152*x^3*log(2) + 144*x^3 - 4)/(8*log(2) + 1),x)

[Out]

(4*x - x^4*(288*log(2) + 36) + x^3*exp(x)*(576*log(2) + 72) - x^2*exp(2*x)*(288*log(2) + 36))/(8*log(2) + 1)

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sympy [A]  time = 0.16, size = 32, normalized size = 1.28 \begin {gather*} - 36 x^{4} + 72 x^{3} e^{x} - 36 x^{2} e^{2 x} + \frac {4 x}{1 + 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-288*x**2-288*x)*ln(2)-72*x**2-72*x)*exp(x)**2+(2*(288*x**3+864*x**2)*ln(2)+72*x**3+216*x**2)*e
xp(x)-1152*x**3*ln(2)-144*x**3+4)/(8*ln(2)+1),x)

[Out]

-36*x**4 + 72*x**3*exp(x) - 36*x**2*exp(2*x) + 4*x/(1 + 8*log(2))

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