3.89.28 \(\int \frac {16+48 x+48 x^2+16 x^3+e^4 (-16+16 x+72 x^2+56 x^3+16 x^4)+e^8 (4-20 x+16 x^3+13 x^4+3 x^5)+(e^4 (8+24 x+24 x^2+8 x^3)+e^8 (-4+4 x+18 x^2+14 x^3+4 x^4)) \log (4)+e^8 (1+3 x+3 x^2+x^3) \log ^2(4)}{e^8 (1+3 x+3 x^2+x^3)} \, dx\)

Optimal. Leaf size=21 \[ x \left (1+\frac {4}{e^4}+x-\frac {3}{1+x}+\log (4)\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.24, antiderivative size = 82, normalized size of antiderivative = 3.90, number of steps used = 3, number of rules used = 2, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {12, 2074} \begin {gather*} x^3+\frac {2 x^2 \left (4+e^4 (1+\log (4))\right )}{e^4}-\frac {9}{(x+1)^2}+\frac {x \left (16-e^8 \left (5-\log ^2(4)-\log (16)\right )+8 e^4 (1+\log (4))\right )}{e^8}+\frac {3 \left (8+e^4 (3+\log (16))\right )}{e^4 (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + 48*x + 48*x^2 + 16*x^3 + E^4*(-16 + 16*x + 72*x^2 + 56*x^3 + 16*x^4) + E^8*(4 - 20*x + 16*x^3 + 13*x
^4 + 3*x^5) + (E^4*(8 + 24*x + 24*x^2 + 8*x^3) + E^8*(-4 + 4*x + 18*x^2 + 14*x^3 + 4*x^4))*Log[4] + E^8*(1 + 3
*x + 3*x^2 + x^3)*Log[4]^2)/(E^8*(1 + 3*x + 3*x^2 + x^3)),x]

[Out]

x^3 - 9/(1 + x)^2 + (2*x^2*(4 + E^4*(1 + Log[4])))/E^4 + (x*(16 + 8*E^4*(1 + Log[4]) - E^8*(5 - Log[4]^2 - Log
[16])))/E^8 + (3*(8 + E^4*(3 + Log[16])))/(E^4*(1 + 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 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {16+48 x+48 x^2+16 x^3+e^4 \left (-16+16 x+72 x^2+56 x^3+16 x^4\right )+e^8 \left (4-20 x+16 x^3+13 x^4+3 x^5\right )+\left (e^4 \left (8+24 x+24 x^2+8 x^3\right )+e^8 \left (-4+4 x+18 x^2+14 x^3+4 x^4\right )\right ) \log (4)+e^8 \left (1+3 x+3 x^2+x^3\right ) \log ^2(4)}{1+3 x+3 x^2+x^3} \, dx}{e^8}\\ &=\frac {\int \left (3 e^8 x^2+\frac {18 e^8}{(1+x)^3}+4 e^4 x \left (4+e^4 (1+\log (4))\right )+\frac {3 e^4 \left (-8-e^4 (3+\log (16))\right )}{(1+x)^2}+16 \left (1+\frac {1}{2} e^4 \left (1+\log (4)+\frac {1}{8} e^4 \left (-5+\log ^2(4)+\log (16)\right )\right )\right )\right ) \, dx}{e^8}\\ &=x^3-\frac {9}{(1+x)^2}+\frac {2 x^2 \left (4+e^4 (1+\log (4))\right )}{e^4}+\frac {x \left (16+8 e^4 (1+\log (4))-e^8 \left (5-\log ^2(4)-\log (16)\right )\right )}{e^8}+\frac {3 \left (8+e^4 (3+\log (16))\right )}{e^4 (1+x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.15, size = 99, normalized size = 4.71 \begin {gather*} \frac {16 x}{e^8}-\frac {9}{(1+x)^2}+\frac {24}{e^4 (1+x)}+\frac {8 (1+x)^2}{e^4}+(1+x)^3+\frac {8 (1+x) (-1+\log (4))}{e^4}+(1+x) \left (-6+14 \log (4)+\log ^2(4)-4 \log (256)\right )-\frac {3 (-3+2 \log (4)-\log (256))}{1+x}+\frac {1}{2} (1+x)^2 (-2+\log (256)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 48*x + 48*x^2 + 16*x^3 + E^4*(-16 + 16*x + 72*x^2 + 56*x^3 + 16*x^4) + E^8*(4 - 20*x + 16*x^3
+ 13*x^4 + 3*x^5) + (E^4*(8 + 24*x + 24*x^2 + 8*x^3) + E^8*(-4 + 4*x + 18*x^2 + 14*x^3 + 4*x^4))*Log[4] + E^8*
(1 + 3*x + 3*x^2 + x^3)*Log[4]^2)/(E^8*(1 + 3*x + 3*x^2 + x^3)),x]

[Out]

(16*x)/E^8 - 9/(1 + x)^2 + 24/(E^4*(1 + x)) + (8*(1 + x)^2)/E^4 + (1 + x)^3 + (8*(1 + x)*(-1 + Log[4]))/E^4 +
(1 + x)*(-6 + 14*Log[4] + Log[4]^2 - 4*Log[256]) - (3*(-3 + 2*Log[4] - Log[256]))/(1 + x) + ((1 + x)^2*(-2 + L
og[256]))/2

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 127, normalized size = 6.05 \begin {gather*} \frac {{\left (4 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{8} \log \relax (2)^{2} + 16 \, x^{3} + 32 \, x^{2} + {\left (x^{5} + 4 \, x^{4} - 8 \, x^{2} + 4 \, x\right )} e^{8} + 8 \, {\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 4 \, x + 3\right )} e^{4} + 4 \, {\left ({\left (x^{4} + 3 \, x^{3} + 3 \, x^{2} + 4 \, x + 3\right )} e^{8} + 4 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{4}\right )} \log \relax (2) + 16 \, x\right )} e^{\left (-8\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+3*x^2+3*x+1)*exp(4)^2*log(2)^2+2*((4*x^4+14*x^3+18*x^2+4*x-4)*exp(4)^2+(8*x^3+24*x^2+24*x+8)
*exp(4))*log(2)+(3*x^5+13*x^4+16*x^3-20*x+4)*exp(4)^2+(16*x^4+56*x^3+72*x^2+16*x-16)*exp(4)+16*x^3+48*x^2+48*x
+16)/(x^3+3*x^2+3*x+1)/exp(4)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.18, size = 104, normalized size = 4.95 \begin {gather*} {\left (x^{3} e^{8} + 4 \, x^{2} e^{8} \log \relax (2) + 4 \, x e^{8} \log \relax (2)^{2} + 2 \, x^{2} e^{8} + 8 \, x^{2} e^{4} + 4 \, x e^{8} \log \relax (2) + 16 \, x e^{4} \log \relax (2) - 5 \, x e^{8} + 8 \, x e^{4} + 16 \, x + \frac {3 \, {\left (4 \, x e^{8} \log \relax (2) + 3 \, x e^{8} + 8 \, x e^{4} + 4 \, e^{8} \log \relax (2) + 8 \, e^{4}\right )}}{{\left (x + 1\right )}^{2}}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+3*x^2+3*x+1)*exp(4)^2*log(2)^2+2*((4*x^4+14*x^3+18*x^2+4*x-4)*exp(4)^2+(8*x^3+24*x^2+24*x+8)
*exp(4))*log(2)+(3*x^5+13*x^4+16*x^3-20*x+4)*exp(4)^2+(16*x^4+56*x^3+72*x^2+16*x-16)*exp(4)+16*x^3+48*x^2+48*x
+16)/(x^3+3*x^2+3*x+1)/exp(4)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.14, size = 97, normalized size = 4.62

method	result	size
risch	\(4 x \ln \relax (2)^{2}+4 x^{2} \ln \relax (2)+x^{3}+4 x \ln \relax (2)+2 x^{2}+16 \,{\mathrm e}^{-4} x \ln \relax (2)-5 x +8 \,{\mathrm e}^{-4} x^{2}+8 x \,{\mathrm e}^{-4}+16 x \,{\mathrm e}^{-8}+\frac {{\mathrm e}^{-8} \left (\left (9 \,{\mathrm e}^{8}+12 \,{\mathrm e}^{8} \ln \relax (2)+24 \,{\mathrm e}^{4}\right ) x +12 \,{\mathrm e}^{8} \ln \relax (2)+24 \,{\mathrm e}^{4}\right )}{x^{2}+2 x +1}\)	\(97\)
default	\({\mathrm e}^{-8} \left (4 x \,{\mathrm e}^{8} \ln \relax (2)^{2}+4 x^{2} {\mathrm e}^{8} \ln \relax (2)+x^{3} {\mathrm e}^{8}+16 x \,{\mathrm e}^{4} \ln \relax (2)+4 \,{\mathrm e}^{8} \ln \relax (2) x +8 x^{2} {\mathrm e}^{4}+2 x^{2} {\mathrm e}^{8}+8 x \,{\mathrm e}^{4}-5 x \,{\mathrm e}^{8}+16 x -\frac {9 \,{\mathrm e}^{8}}{\left (x +1\right )^{2}}-\frac {-9 \,{\mathrm e}^{8}-12 \,{\mathrm e}^{8} \ln \relax (2)-24 \,{\mathrm e}^{4}}{x +1}\right )\)	\(103\)
norman	\(\frac {\left (x^{5} {\mathrm e}^{4}-4 \left (3 \,{\mathrm e}^{8} \ln \relax (2)^{2}+2 \,{\mathrm e}^{8} \ln \relax (2)+12 \,{\mathrm e}^{4} \ln \relax (2)-5 \,{\mathrm e}^{8}+4 \,{\mathrm e}^{4}+12\right ) {\mathrm e}^{-4} x +\left (4 \,{\mathrm e}^{4} \ln \relax (2)+4 \,{\mathrm e}^{4}+8\right ) x^{4}+4 \left ({\mathrm e}^{8} \ln \relax (2)^{2}+3 \,{\mathrm e}^{8} \ln \relax (2)+4 \,{\mathrm e}^{4} \ln \relax (2)+6 \,{\mathrm e}^{4}+4\right ) {\mathrm e}^{-4} x^{3}-8 \left ({\mathrm e}^{8} \ln \relax (2)^{2}+4 \,{\mathrm e}^{4} \ln \relax (2)-{\mathrm e}^{8}+4\right ) {\mathrm e}^{-4}\right ) {\mathrm e}^{-4}}{\left (x +1\right )^{2}}\)	\(144\)
gosper	\(\frac {\left (4 \ln \relax (2)^{2} {\mathrm e}^{8} x^{3}+4 \ln \relax (2) {\mathrm e}^{8} x^{4}+{\mathrm e}^{8} x^{5}+12 \ln \relax (2) {\mathrm e}^{8} x^{3}+4 x^{4} {\mathrm e}^{8}-12 x \,{\mathrm e}^{8} \ln \relax (2)^{2}+16 x^{3} {\mathrm e}^{4} \ln \relax (2)+8 x^{4} {\mathrm e}^{4}-8 \,{\mathrm e}^{8} \ln \relax (2)^{2}-8 \,{\mathrm e}^{8} \ln \relax (2) x +24 x^{3} {\mathrm e}^{4}-48 x \,{\mathrm e}^{4} \ln \relax (2)+20 x \,{\mathrm e}^{8}+16 x^{3}-32 \,{\mathrm e}^{4} \ln \relax (2)+8 \,{\mathrm e}^{8}-16 x \,{\mathrm e}^{4}-48 x -32\right ) {\mathrm e}^{-8}}{x^{2}+2 x +1}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(x^3+3*x^2+3*x+1)*exp(4)^2*ln(2)^2+2*((4*x^4+14*x^3+18*x^2+4*x-4)*exp(4)^2+(8*x^3+24*x^2+24*x+8)*exp(4)
)*ln(2)+(3*x^5+13*x^4+16*x^3-20*x+4)*exp(4)^2+(16*x^4+56*x^3+72*x^2+16*x-16)*exp(4)+16*x^3+48*x^2+48*x+16)/(x^
3+3*x^2+3*x+1)/exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

4*x*ln(2)^2+4*x^2*ln(2)+x^3+4*x*ln(2)+2*x^2+16*exp(-4)*x*ln(2)-5*x+8*exp(-4)*x^2+8*x*exp(-4)+16*x*exp(-8)+exp(
-8)*((9*exp(8)+12*exp(8)*ln(2)+24*exp(4))*x+12*exp(8)*ln(2)+24*exp(4))/(x^2+2*x+1)

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 99, normalized size = 4.71 \begin {gather*} {\left (x^{3} e^{8} + 2 \, {\left (2 \, e^{8} \log \relax (2) + e^{8} + 4 \, e^{4}\right )} x^{2} + {\left (4 \, e^{8} \log \relax (2)^{2} + 4 \, {\left (e^{8} + 4 \, e^{4}\right )} \log \relax (2) - 5 \, e^{8} + 8 \, e^{4} + 16\right )} x + \frac {3 \, {\left ({\left (4 \, e^{8} \log \relax (2) + 3 \, e^{8} + 8 \, e^{4}\right )} x + 4 \, e^{8} \log \relax (2) + 8 \, e^{4}\right )}}{x^{2} + 2 \, x + 1}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+3*x^2+3*x+1)*exp(4)^2*log(2)^2+2*((4*x^4+14*x^3+18*x^2+4*x-4)*exp(4)^2+(8*x^3+24*x^2+24*x+8)
*exp(4))*log(2)+(3*x^5+13*x^4+16*x^3-20*x+4)*exp(4)^2+(16*x^4+56*x^3+72*x^2+16*x-16)*exp(4)+16*x^3+48*x^2+48*x
+16)/(x^3+3*x^2+3*x+1)/exp(4)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.34, size = 124, normalized size = 5.90 \begin {gather*} \frac {x\,\left (9\,{\mathrm {e}}^4+12\,{\mathrm {e}}^4\,\ln \relax (2)+24\right )+12\,{\mathrm {e}}^4\,\ln \relax (2)+24}{{\mathrm {e}}^4\,x^2+2\,{\mathrm {e}}^4\,x+{\mathrm {e}}^4}+x^2\,\left (\frac {{\mathrm {e}}^{-8}\,\left (16\,{\mathrm {e}}^4+13\,{\mathrm {e}}^8+8\,{\mathrm {e}}^8\,\ln \relax (2)\right )}{2}-\frac {9}{2}\right )+x\,\left ({\mathrm {e}}^{-8}\,\left (56\,{\mathrm {e}}^4+16\,{\mathrm {e}}^8+4\,{\mathrm {e}}^8\,{\ln \relax (2)}^2+4\,{\mathrm {e}}^4\,\ln \relax (2)\,\left (7\,{\mathrm {e}}^4+4\right )+16\right )-3\,{\mathrm {e}}^{-8}\,\left (16\,{\mathrm {e}}^4+13\,{\mathrm {e}}^8+8\,{\mathrm {e}}^8\,\ln \relax (2)\right )+18\right )+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-8)*(48*x + 2*log(2)*(exp(4)*(24*x + 24*x^2 + 8*x^3 + 8) + exp(8)*(4*x + 18*x^2 + 14*x^3 + 4*x^4 - 4)
) + exp(8)*(16*x^3 - 20*x + 13*x^4 + 3*x^5 + 4) + exp(4)*(16*x + 72*x^2 + 56*x^3 + 16*x^4 - 16) + 48*x^2 + 16*
x^3 + 4*exp(8)*log(2)^2*(3*x + 3*x^2 + x^3 + 1) + 16))/(3*x + 3*x^2 + x^3 + 1),x)

[Out]

(x*(9*exp(4) + 12*exp(4)*log(2) + 24) + 12*exp(4)*log(2) + 24)/(exp(4) + 2*x*exp(4) + x^2*exp(4)) + x^2*((exp(
-8)*(16*exp(4) + 13*exp(8) + 8*exp(8)*log(2)))/2 - 9/2) + x*(exp(-8)*(56*exp(4) + 16*exp(8) + 4*exp(8)*log(2)^
2 + 4*exp(4)*log(2)*(7*exp(4) + 4) + 16) - 3*exp(-8)*(16*exp(4) + 13*exp(8) + 8*exp(8)*log(2)) + 18) + x^3

________________________________________________________________________________________

sympy [B]  time = 0.68, size = 95, normalized size = 4.52 \begin {gather*} x^{3} + x^{2} \left (\frac {8}{e^{4}} + 2 + 4 \log {\relax (2 )}\right ) + x \left (-5 + \frac {16}{e^{8}} + \frac {8}{e^{4}} + \frac {16 \log {\relax (2 )}}{e^{4}} + 4 \log {\relax (2 )}^{2} + 4 \log {\relax (2 )}\right ) + \frac {x \left (24 + 12 e^{4} \log {\relax (2 )} + 9 e^{4}\right ) + 24 + 12 e^{4} \log {\relax (2 )}}{x^{2} e^{4} + 2 x e^{4} + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x**3+3*x**2+3*x+1)*exp(4)**2*ln(2)**2+2*((4*x**4+14*x**3+18*x**2+4*x-4)*exp(4)**2+(8*x**3+24*x**
2+24*x+8)*exp(4))*ln(2)+(3*x**5+13*x**4+16*x**3-20*x+4)*exp(4)**2+(16*x**4+56*x**3+72*x**2+16*x-16)*exp(4)+16*
x**3+48*x**2+48*x+16)/(x**3+3*x**2+3*x+1)/exp(4)**2,x)

[Out]

x**3 + x**2*(8*exp(-4) + 2 + 4*log(2)) + x*(-5 + 16*exp(-8) + 8*exp(-4) + 16*exp(-4)*log(2) + 4*log(2)**2 + 4*
log(2)) + (x*(24 + 12*exp(4)*log(2) + 9*exp(4)) + 24 + 12*exp(4)*log(2))/(x**2*exp(4) + 2*x*exp(4) + exp(4))

________________________________________________________________________________________