3.85.53 \(\int \frac {1}{2} (e^x (10+2 \log ^2(2))-5 x^7 \log ^7(x)-5 x^7 \log ^8(x)) \, dx\)

Optimal. Leaf size=22 \[ e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x) \]

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Rubi [A]  time = 0.23, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 2194, 2305, 2304} \begin {gather*} e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(10 + 2*Log[2]^2) - 5*x^7*Log[x]^7 - 5*x^7*Log[x]^8)/2,x]

[Out]

E^x*(5 + Log[2]^2) - (5*x^8*Log[x]^8)/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (e^x \left (10+2 \log ^2(2)\right )-5 x^7 \log ^7(x)-5 x^7 \log ^8(x)\right ) \, dx\\ &=-\left (\frac {5}{2} \int x^7 \log ^7(x) \, dx\right )-\frac {5}{2} \int x^7 \log ^8(x) \, dx+\left (5+\log ^2(2)\right ) \int e^x \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^7(x)-\frac {5}{16} x^8 \log ^8(x)+\frac {35}{16} \int x^7 \log ^6(x) \, dx+\frac {5}{2} \int x^7 \log ^7(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )+\frac {35}{128} x^8 \log ^6(x)-\frac {5}{16} x^8 \log ^8(x)-\frac {105}{64} \int x^7 \log ^5(x) \, dx-\frac {35}{16} \int x^7 \log ^6(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {105}{512} x^8 \log ^5(x)-\frac {5}{16} x^8 \log ^8(x)+\frac {525}{512} \int x^7 \log ^4(x) \, dx+\frac {105}{64} \int x^7 \log ^5(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )+\frac {525 x^8 \log ^4(x)}{4096}-\frac {5}{16} x^8 \log ^8(x)-\frac {525 \int x^7 \log ^3(x) \, dx}{1024}-\frac {525}{512} \int x^7 \log ^4(x) \, dx\\ &=e^x \left (5+\log ^2(2)\right )-\frac {525 x^8 \log ^3(x)}{8192}-\frac {5}{16} x^8 \log ^8(x)+\frac {1575 \int x^7 \log ^2(x) \, dx}{8192}+\frac {525 \int x^7 \log ^3(x) \, dx}{1024}\\ &=e^x \left (5+\log ^2(2)\right )+\frac {1575 x^8 \log ^2(x)}{65536}-\frac {5}{16} x^8 \log ^8(x)-\frac {1575 \int x^7 \log (x) \, dx}{32768}-\frac {1575 \int x^7 \log ^2(x) \, dx}{8192}\\ &=\frac {1575 x^8}{2097152}+e^x \left (5+\log ^2(2)\right )-\frac {1575 x^8 \log (x)}{262144}-\frac {5}{16} x^8 \log ^8(x)+\frac {1575 \int x^7 \log (x) \, dx}{32768}\\ &=e^x \left (5+\log ^2(2)\right )-\frac {5}{16} x^8 \log ^8(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.14 \begin {gather*} 5 e^x+e^x \log ^2(2)-\frac {5}{16} x^8 \log ^8(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(10 + 2*Log[2]^2) - 5*x^7*Log[x]^7 - 5*x^7*Log[x]^8)/2,x]

[Out]

5*E^x + E^x*Log[2]^2 - (5*x^8*Log[x]^8)/16

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fricas [A]  time = 0.85, size = 19, normalized size = 0.86 \begin {gather*} -\frac {5}{16} \, x^{8} \log \relax (x)^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/2*x^7*log(x)^8-5/2*x^7*log(x)^7+1/2*(2*log(2)^2+10)*exp(x),x, algorithm="fricas")

[Out]

-5/16*x^8*log(x)^8 + (log(2)^2 + 5)*e^x

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giac [A]  time = 0.21, size = 19, normalized size = 0.86 \begin {gather*} -\frac {5}{16} \, x^{8} \log \relax (x)^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/2*x^7*log(x)^8-5/2*x^7*log(x)^7+1/2*(2*log(2)^2+10)*exp(x),x, algorithm="giac")

[Out]

-5/16*x^8*log(x)^8 + (log(2)^2 + 5)*e^x

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maple [A]  time = 0.03, size = 22, normalized size = 1.00

method	result	size
default	\(-\frac {5 x^{8} \ln \relax (x )^{8}}{16}+\ln \relax (2)^{2} {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\)	\(22\)
risch	\(-\frac {5 x^{8} \ln \relax (x )^{8}}{16}+\ln \relax (2)^{2} {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/2*x^7*ln(x)^8-5/2*x^7*ln(x)^7+1/2*(2*ln(2)^2+10)*exp(x),x,method=_RETURNVERBOSE)

[Out]

-5/16*x^8*ln(x)^8+ln(2)^2*exp(x)+5*exp(x)

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maxima [B]  time = 0.44, size = 110, normalized size = 5.00 \begin {gather*} -\frac {5}{2097152} \, {\left (131072 \, \log \relax (x)^{8} - 131072 \, \log \relax (x)^{7} + 114688 \, \log \relax (x)^{6} - 86016 \, \log \relax (x)^{5} + 53760 \, \log \relax (x)^{4} - 26880 \, \log \relax (x)^{3} + 10080 \, \log \relax (x)^{2} - 2520 \, \log \relax (x) + 315\right )} x^{8} - \frac {5}{2097152} \, {\left (131072 \, \log \relax (x)^{7} - 114688 \, \log \relax (x)^{6} + 86016 \, \log \relax (x)^{5} - 53760 \, \log \relax (x)^{4} + 26880 \, \log \relax (x)^{3} - 10080 \, \log \relax (x)^{2} + 2520 \, \log \relax (x) - 315\right )} x^{8} + {\left (\log \relax (2)^{2} + 5\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/2*x^7*log(x)^8-5/2*x^7*log(x)^7+1/2*(2*log(2)^2+10)*exp(x),x, algorithm="maxima")

[Out]

-5/2097152*(131072*log(x)^8 - 131072*log(x)^7 + 114688*log(x)^6 - 86016*log(x)^5 + 53760*log(x)^4 - 26880*log(
x)^3 + 10080*log(x)^2 - 2520*log(x) + 315)*x^8 - 5/2097152*(131072*log(x)^7 - 114688*log(x)^6 + 86016*log(x)^5
- 53760*log(x)^4 + 26880*log(x)^3 - 10080*log(x)^2 + 2520*log(x) - 315)*x^8 + (log(2)^2 + 5)*e^x

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mupad [B]  time = 5.30, size = 19, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^x\,\left ({\ln \relax (2)}^2+5\right )-\frac {5\,x^8\,{\ln \relax (x)}^8}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(2*log(2)^2 + 10))/2 - (5*x^7*log(x)^8)/2 - (5*x^7*log(x)^7)/2,x)

[Out]

exp(x)*(log(2)^2 + 5) - (5*x^8*log(x)^8)/16

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sympy [A]  time = 0.27, size = 20, normalized size = 0.91 \begin {gather*} - \frac {5 x^{8} \log {\relax (x )}^{8}}{16} + \left (\log {\relax (2 )}^{2} + 5\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/2*x**7*ln(x)**8-5/2*x**7*ln(x)**7+1/2*(2*ln(2)**2+10)*exp(x),x)

[Out]

-5*x**8*log(x)**8/16 + (log(2)**2 + 5)*exp(x)

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