3.103.87 \(\int (10 \log ^2(2)+(50 e^2+50 x) \log ^4(2)) \, dx\)

Optimal. Leaf size=15 \[ \left (1+5 \left (e^2+x\right ) \log ^2(2)\right )^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.40, number of steps used = 1, number of rules used = 0, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} 25 \left (x+e^2\right )^2 \log ^4(2)+10 x \log ^2(2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[10*Log[2]^2 + (50*E^2 + 50*x)*Log[2]^4,x]

[Out]

10*x*Log[2]^2 + 25*(E^2 + x)^2*Log[2]^4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=10 x \log ^2(2)+25 \left (e^2+x\right )^2 \log ^4(2)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 30, normalized size = 2.00 \begin {gather*} 10 \left (x \log ^2(2)+5 e^2 x \log ^4(2)+\frac {5}{2} x^2 \log ^4(2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[10*Log[2]^2 + (50*E^2 + 50*x)*Log[2]^4,x]

[Out]

10*(x*Log[2]^2 + 5*E^2*x*Log[2]^4 + (5*x^2*Log[2]^4)/2)

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fricas [A]  time = 1.00, size = 23, normalized size = 1.53 \begin {gather*} 25 \, {\left (x^{2} + 2 \, x e^{2}\right )} \log \relax (2)^{4} + 10 \, x \log \relax (2)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*exp(2)+50*x)*log(2)^4+10*log(2)^2,x, algorithm="fricas")

[Out]

25*(x^2 + 2*x*e^2)*log(2)^4 + 10*x*log(2)^2

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giac [A]  time = 0.15, size = 23, normalized size = 1.53 \begin {gather*} 25 \, {\left (x^{2} + 2 \, x e^{2}\right )} \log \relax (2)^{4} + 10 \, x \log \relax (2)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*exp(2)+50*x)*log(2)^4+10*log(2)^2,x, algorithm="giac")

[Out]

25*(x^2 + 2*x*e^2)*log(2)^4 + 10*x*log(2)^2

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maple [A]  time = 0.02, size = 25, normalized size = 1.67

method	result	size
gosper	\(5 \ln \relax (2)^{2} x \left (10 \,{\mathrm e}^{2} \ln \relax (2)^{2}+5 x \ln \relax (2)^{2}+2\right )\)	\(25\)
default	\(\ln \relax (2)^{4} \left (50 \,{\mathrm e}^{2} x +25 x^{2}\right )+10 x \ln \relax (2)^{2}\)	\(25\)
risch	\(50 \ln \relax (2)^{4} {\mathrm e}^{2} x +25 x^{2} \ln \relax (2)^{4}+10 x \ln \relax (2)^{2}\)	\(27\)
norman	\(\left (50 \ln \relax (2)^{4} {\mathrm e}^{2}+10 \ln \relax (2)^{2}\right ) x +25 x^{2} \ln \relax (2)^{4}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*exp(2)+50*x)*ln(2)^4+10*ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

5*ln(2)^2*x*(10*exp(2)*ln(2)^2+5*x*ln(2)^2+2)

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maxima [A]  time = 0.35, size = 23, normalized size = 1.53 \begin {gather*} 25 \, {\left (x^{2} + 2 \, x e^{2}\right )} \log \relax (2)^{4} + 10 \, x \log \relax (2)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*exp(2)+50*x)*log(2)^4+10*log(2)^2,x, algorithm="maxima")

[Out]

25*(x^2 + 2*x*e^2)*log(2)^4 + 10*x*log(2)^2

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mupad [B]  time = 6.77, size = 29, normalized size = 1.93 \begin {gather*} \frac {{\ln \relax (2)}^2\,\left ({\ln \relax (2)}^2\,\left (50\,x+50\,{\mathrm {e}}^2\right )+20\right )\,\left (50\,x+50\,{\mathrm {e}}^2\right )}{100} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(2)^4*(50*x + 50*exp(2)) + 10*log(2)^2,x)

[Out]

(log(2)^2*(log(2)^2*(50*x + 50*exp(2)) + 20)*(50*x + 50*exp(2)))/100

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sympy [A]  time = 0.06, size = 27, normalized size = 1.80 \begin {gather*} 25 x^{2} \log {\relax (2 )}^{4} + x \left (10 \log {\relax (2 )}^{2} + 50 e^{2} \log {\relax (2 )}^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*exp(2)+50*x)*ln(2)**4+10*ln(2)**2,x)

[Out]

25*x**2*log(2)**4 + x*(10*log(2)**2 + 50*exp(2)*log(2)**4)

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