3.63.9 \(\int \frac {1}{8} (5 x+15 x^2+10 x^3+e^{2 x} (30 x+45 x^2)+e^x (-25 x-55 x^2-15 x^3)+(e^{2 x} (-25 x-30 x^2)+e^x (10 x+20 x^2+5 x^3)) \log (2 x)+e^{2 x} (5 x+5 x^2) \log ^2(2 x)) \, dx\)

Optimal. Leaf size=25 \[ \frac {5}{16} x^2 \left (1+x-e^x (3-\log (2 x))\right )^2 \]

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Rubi [F]  time = 0.73, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x + 15*x^2 + 10*x^3 + E^(2*x)*(30*x + 45*x^2) + E^x*(-25*x - 55*x^2 - 15*x^3) + (E^(2*x)*(-25*x - 30*x^
2) + E^x*(10*x + 20*x^2 + 5*x^3))*Log[2*x] + E^(2*x)*(5*x + 5*x^2)*Log[2*x]^2)/8,x]

[Out]

(-5*E^(2*x))/32 + (5*x^2)/16 - (15*E^x*x^2)/8 + (45*E^(2*x)*x^2)/16 + (5*x^3)/8 - (15*E^x*x^3)/8 + (5*x^4)/16
+ (5*ExpIntegralEi[2*x])/32 - (5*E^(2*x)*Log[2*x])/32 + (5*E^(2*x)*x*Log[2*x])/16 + (5*E^x*x^2*Log[2*x])/8 - (
15*E^(2*x)*x^2*Log[2*x])/8 + (5*E^x*x^3*Log[2*x])/8 + (5*Defer[Int][E^(2*x)*x*Log[2*x]^2, x])/8 + (5*Defer[Int
][E^(2*x)*x^2*Log[2*x]^2, x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx\\ &=\frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}+\frac {1}{8} \int e^{2 x} \left (30 x+45 x^2\right ) \, dx+\frac {1}{8} \int e^x \left (-25 x-55 x^2-15 x^3\right ) \, dx+\frac {1}{8} \int \left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x) \, dx+\frac {1}{8} \int e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x) \, dx\\ &=\frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {1}{8} \int e^{2 x} x (30+45 x) \, dx+\frac {1}{8} \int e^x x \left (-25-55 x-15 x^2\right ) \, dx-\frac {1}{8} \int e^x \left (e^x \left (\frac {5}{2}-\frac {5}{4 x}-15 x\right )+5 x (1+x)\right ) \, dx+\frac {1}{8} \int e^{2 x} x (5+5 x) \log ^2(2 x) \, dx\\ &=\frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {1}{8} \int \left (30 e^{2 x} x+45 e^{2 x} x^2\right ) \, dx+\frac {1}{8} \int \left (-25 e^x x-55 e^x x^2-15 e^x x^3\right ) \, dx-\frac {1}{8} \int \left (5 e^x x (1+x)-\frac {5 e^{2 x} \left (1-2 x+12 x^2\right )}{4 x}\right ) \, dx+\frac {1}{8} \int \left (5 e^{2 x} x \log ^2(2 x)+5 e^{2 x} x^2 \log ^2(2 x)\right ) \, dx\\ &=\frac {5 x^2}{16}+\frac {5 x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \frac {e^{2 x} \left (1-2 x+12 x^2\right )}{x} \, dx-\frac {5}{8} \int e^x x (1+x) \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{8} \int e^x x^3 \, dx-\frac {25}{8} \int e^x x \, dx+\frac {15}{4} \int e^{2 x} x \, dx+\frac {45}{8} \int e^{2 x} x^2 \, dx-\frac {55}{8} \int e^x x^2 \, dx\\ &=-\frac {25 e^x x}{8}+\frac {15}{8} e^{2 x} x+\frac {5 x^2}{16}-\frac {55 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \left (-2 e^{2 x}+\frac {e^{2 x}}{x}+12 e^{2 x} x\right ) \, dx-\frac {5}{8} \int \left (e^x x+e^x x^2\right ) \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{8} \int e^{2 x} \, dx+\frac {25 \int e^x \, dx}{8}-\frac {45}{8} \int e^{2 x} x \, dx+\frac {45}{8} \int e^x x^2 \, dx+\frac {55}{4} \int e^x x \, dx\\ &=\frac {25 e^x}{8}-\frac {15 e^{2 x}}{16}+\frac {85 e^x x}{8}-\frac {15}{16} e^{2 x} x+\frac {5 x^2}{16}-\frac {5 e^x x^2}{4}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{32} \int \frac {e^{2 x}}{x} \, dx-\frac {5}{16} \int e^{2 x} \, dx-\frac {5}{8} \int e^x x \, dx-\frac {5}{8} \int e^x x^2 \, dx+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx+\frac {15}{8} \int e^{2 x} x \, dx+\frac {45}{16} \int e^{2 x} \, dx-\frac {45}{4} \int e^x x \, dx-\frac {55 \int e^x \, dx}{4}\\ &=-\frac {85 e^x}{8}+\frac {5 e^{2 x}}{16}-\frac {5 e^x x}{4}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5 \int e^x \, dx}{8}+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {15}{16} \int e^{2 x} \, dx+\frac {5}{4} \int e^x x \, dx+\frac {45 \int e^x \, dx}{4}\\ &=\frac {5 e^x}{4}-\frac {5 e^{2 x}}{32}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx-\frac {5 \int e^x \, dx}{4}\\ &=-\frac {5 e^{2 x}}{32}+\frac {5 x^2}{16}-\frac {15 e^x x^2}{8}+\frac {45}{16} e^{2 x} x^2+\frac {5 x^3}{8}-\frac {15 e^x x^3}{8}+\frac {5 x^4}{16}+\frac {5 \text {Ei}(2 x)}{32}-\frac {5}{32} e^{2 x} \log (2 x)+\frac {5}{16} e^{2 x} x \log (2 x)+\frac {5}{8} e^x x^2 \log (2 x)-\frac {15}{8} e^{2 x} x^2 \log (2 x)+\frac {5}{8} e^x x^3 \log (2 x)+\frac {5}{8} \int e^{2 x} x \log ^2(2 x) \, dx+\frac {5}{8} \int e^{2 x} x^2 \log ^2(2 x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.00 \begin {gather*} \frac {5}{16} x^2 \left (1-3 e^x+x+e^x \log (2 x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x + 15*x^2 + 10*x^3 + E^(2*x)*(30*x + 45*x^2) + E^x*(-25*x - 55*x^2 - 15*x^3) + (E^(2*x)*(-25*x -
30*x^2) + E^x*(10*x + 20*x^2 + 5*x^3))*Log[2*x] + E^(2*x)*(5*x + 5*x^2)*Log[2*x]^2)/8,x]

[Out]

(5*x^2*(1 - 3*E^x + x + E^x*Log[2*x])^2)/16

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fricas [B]  time = 0.73, size = 78, normalized size = 3.12 \begin {gather*} \frac {5}{16} \, x^{2} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {45}{16} \, x^{2} e^{\left (2 \, x\right )} + \frac {5}{16} \, x^{2} - \frac {15}{8} \, {\left (x^{3} + x^{2}\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{2} e^{\left (2 \, x\right )} - {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2+(5*x^3+20*x^2+10*x)*exp(x))*log(2*x
)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15*x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm="fricas")

[Out]

5/16*x^2*e^(2*x)*log(2*x)^2 + 5/16*x^4 + 5/8*x^3 + 45/16*x^2*e^(2*x) + 5/16*x^2 - 15/8*(x^3 + x^2)*e^x - 5/8*(
3*x^2*e^(2*x) - (x^3 + x^2)*e^x)*log(2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5}{8} \, {\left (x^{2} + x\right )} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{4} \, x^{3} + \frac {15}{8} \, x^{2} + \frac {15}{8} \, {\left (3 \, x^{2} + 2 \, x\right )} e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (3 \, x^{3} + 11 \, x^{2} + 5 \, x\right )} e^{x} - \frac {5}{8} \, {\left ({\left (6 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} - {\left (x^{3} + 4 \, x^{2} + 2 \, x\right )} e^{x}\right )} \log \left (2 \, x\right ) + \frac {5}{8} \, x\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2+(5*x^3+20*x^2+10*x)*exp(x))*log(2*x
)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15*x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm="giac")

[Out]

integrate(5/8*(x^2 + x)*e^(2*x)*log(2*x)^2 + 5/4*x^3 + 15/8*x^2 + 15/8*(3*x^2 + 2*x)*e^(2*x) - 5/8*(3*x^3 + 11
*x^2 + 5*x)*e^x - 5/8*((6*x^2 + 5*x)*e^(2*x) - (x^3 + 4*x^2 + 2*x)*e^x)*log(2*x) + 5/8*x, x)

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maple [B]  time = 0.30, size = 90, normalized size = 3.60

method	result	size
risch	\(\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{x} \ln \left (2 x \right ) x^{3}}{8}+\frac {5 \,{\mathrm e}^{x} \ln \left (2 x \right ) x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}+\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{4}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{2}}{16}\)	\(90\)
default	\(\left (\frac {5 \ln \relax (2)}{8}-\frac {15}{8}\right ) x^{2} {\mathrm e}^{x}+\left (\frac {5 \ln \relax (2)}{8}-\frac {15}{8}\right ) x^{3} {\mathrm e}^{x}+\frac {5 x^{2} {\mathrm e}^{x} \ln \relax (x )}{8}+\frac {5 x^{3} {\mathrm e}^{x} \ln \relax (x )}{8}+\left (\frac {5 \ln \relax (2)^{2}}{16}-\frac {15 \ln \relax (2)}{8}+\frac {45}{16}\right ) x^{2} {\mathrm e}^{2 x}+\left (\frac {5 \ln \relax (2)}{8}-\frac {15}{8}\right ) x^{2} {\mathrm e}^{2 x} \ln \relax (x )+\frac {5 \ln \relax (x )^{2} {\mathrm e}^{2 x} x^{2}}{16}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(5*x^2+5*x)*exp(x)^2*ln(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2+(5*x^3+20*x^2+10*x)*exp(x))*ln(2*x)+1/8*(4
5*x^2+30*x)*exp(x)^2+1/8*(-15*x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x,method=_RETURNVERBOSE)

[Out]

5/16*exp(2*x)*ln(2*x)^2*x^2-15/8*exp(2*x)*ln(2*x)*x^2+5/8*exp(x)*ln(2*x)*x^3+5/8*exp(x)*ln(2*x)*x^2-15/8*exp(x
)*x^2+45/16*exp(2*x)*x^2-15/8*exp(x)*x^3+5/16*x^4+5/8*x^3+5/16*x^2

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maxima [B]  time = 0.53, size = 125, normalized size = 5.00 \begin {gather*} \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {5}{16} \, x^{2} + \frac {5}{32} \, {\left (4 \, x^{2} {\left (\log \relax (2) - 3\right )} \log \relax (x) + 2 \, x^{2} \log \relax (x)^{2} + 2 \, {\left (\log \relax (2)^{2} - 6 \, \log \relax (2)\right )} x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {15}{32} \, {\left (6 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {5}{8} \, {\left (x^{3} \log \relax (2) + x^{2} {\left (\log \relax (2) - 1\right )} + {\left (x^{3} + x^{2}\right )} \log \relax (x) + x - 1\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{3} + 2 \, x^{2} + x - 1\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2+(5*x^3+20*x^2+10*x)*exp(x))*log(2*x
)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15*x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm="maxima")

[Out]

5/16*x^4 + 5/8*x^3 + 5/16*x^2 + 5/32*(4*x^2*(log(2) - 3)*log(x) + 2*x^2*log(x)^2 + 2*(log(2)^2 - 6*log(2))*x^2
+ 6*x - 3)*e^(2*x) + 15/32*(6*x^2 - 2*x + 1)*e^(2*x) + 5/8*(x^3*log(2) + x^2*(log(2) - 1) + (x^3 + x^2)*log(x
) + x - 1)*e^x - 5/8*(3*x^3 + 2*x^2 + x - 1)*e^x

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mupad [B]  time = 4.19, size = 21, normalized size = 0.84 \begin {gather*} \frac {5\,x^2\,{\left (x-3\,{\mathrm {e}}^x+\ln \left (2\,x\right )\,{\mathrm {e}}^x+1\right )}^2}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x)/8 + (exp(2*x)*(30*x + 45*x^2))/8 + (15*x^2)/8 + (5*x^3)/4 - (exp(x)*(25*x + 55*x^2 + 15*x^3))/8 - (l
og(2*x)*(exp(2*x)*(25*x + 30*x^2) - exp(x)*(10*x + 20*x^2 + 5*x^3)))/8 + (log(2*x)^2*exp(2*x)*(5*x + 5*x^2))/8
,x)

[Out]

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

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sympy [B]  time = 0.48, size = 88, normalized size = 3.52 \begin {gather*} \frac {5 x^{4}}{16} + \frac {5 x^{3}}{8} + \frac {5 x^{2}}{16} + \frac {\left (40 x^{2} \log {\left (2 x \right )}^{2} - 240 x^{2} \log {\left (2 x \right )} + 360 x^{2}\right ) e^{2 x}}{128} + \frac {\left (80 x^{3} \log {\left (2 x \right )} - 240 x^{3} + 80 x^{2} \log {\left (2 x \right )} - 240 x^{2}\right ) e^{x}}{128} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(5*x**2+5*x)*exp(x)**2*ln(2*x)**2+1/8*((-30*x**2-25*x)*exp(x)**2+(5*x**3+20*x**2+10*x)*exp(x))*l
n(2*x)+1/8*(45*x**2+30*x)*exp(x)**2+1/8*(-15*x**3-55*x**2-25*x)*exp(x)+5/4*x**3+15/8*x**2+5/8*x,x)

[Out]

5*x**4/16 + 5*x**3/8 + 5*x**2/16 + (40*x**2*log(2*x)**2 - 240*x**2*log(2*x) + 360*x**2)*exp(2*x)/128 + (80*x**
3*log(2*x) - 240*x**3 + 80*x**2*log(2*x) - 240*x**2)*exp(x)/128

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