3.30.30 \(\int \frac {85+50 e^4-90 x+150 x^2+25 \log (3)+e^x (158-84 x+230 x^2+64 x^3+e^4 (100+24 x)+(50+12 x) \log (3))+e^{2 x} (48-8 x+83 x^2+46 x^3+6 x^4+e^4 (32+16 x+2 x^2)+(16+8 x+x^2) \log (3))}{25+e^x (40+10 x)+e^{2 x} (16+8 x+x^2)} \, dx\)

Optimal. Leaf size=34 \[ \left (-x+\frac {2}{5+e^x (4+x)}\right ) \left (-3+x-2 \left (e^4+x^2\right )-\log (3)\right ) \]

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Rubi [F]  time = 2.32, 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 {85+50 e^4-90 x+150 x^2+25 \log (3)+e^x \left (158-84 x+230 x^2+64 x^3+e^4 (100+24 x)+(50+12 x) \log (3)\right )+e^{2 x} \left (48-8 x+83 x^2+46 x^3+6 x^4+e^4 \left (32+16 x+2 x^2\right )+\left (16+8 x+x^2\right ) \log (3)\right )}{25+e^x (40+10 x)+e^{2 x} \left (16+8 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(85 + 50*E^4 - 90*x + 150*x^2 + 25*Log[3] + E^x*(158 - 84*x + 230*x^2 + 64*x^3 + E^4*(100 + 24*x) + (50 +
12*x)*Log[3]) + E^(2*x)*(48 - 8*x + 83*x^2 + 46*x^3 + 6*x^4 + E^4*(32 + 16*x + 2*x^2) + (16 + 8*x + x^2)*Log[3
]))/(25 + E^x*(40 + 10*x) + E^(2*x)*(16 + 8*x + x^2)),x]

[Out]

-x^2 + 2*x^3 + x*(3 + 2*E^4 + Log[3]) + 10*(6 - 2*E^4 - Log[3])*Defer[Int][(5 + 4*E^x + E^x*x)^(-2), x] - 10*D
efer[Int][x/(5 + 4*E^x + E^x*x)^2, x] - 20*Defer[Int][x^2/(5 + 4*E^x + E^x*x)^2, x] - 10*(39 + 2*E^4 + Log[3])
*Defer[Int][1/((4 + x)*(5 + 4*E^x + E^x*x)^2), x] - 2*(5 - 2*E^4 - Log[3])*Defer[Int][(5 + 4*E^x + E^x*x)^(-1)
, x] - 6*Defer[Int][x/(5 + 4*E^x + E^x*x), x] + 4*Defer[Int][x^2/(5 + 4*E^x + E^x*x), x] + 2*(39 + 2*E^4 + Log
[3])*Defer[Int][1/((4 + x)*(5 + 4*E^x + E^x*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-90 x+150 x^2+e^x \left (158-84 x+230 x^2+64 x^3+e^4 (100+24 x)+(50+12 x) \log (3)\right )+e^{2 x} \left (48-8 x+83 x^2+46 x^3+6 x^4+e^4 \left (32+16 x+2 x^2\right )+\left (16+8 x+x^2\right ) \log (3)\right )+85 \left (1+\frac {5}{17} \left (2 e^4+\log (3)\right )\right )}{\left (5+4 e^x+e^x x\right )^2} \, dx\\ &=\int \left (-2 x+6 x^2-\frac {10 (5+x) \left (3+2 e^4-x+2 x^2+\log (3)\right )}{(4+x) \left (5+4 e^x+e^x x\right )^2}+3 \left (1+\frac {1}{3} \left (2 e^4+\log (3)\right )\right )+\frac {2 \left (19+10 e^4+5 x^2+2 x^3-x \left (17-2 e^4-\log (3)\right )+\log (243)\right )}{(4+x) \left (5+4 e^x+e^x x\right )}\right ) \, dx\\ &=-x^2+2 x^3+x \left (3+2 e^4+\log (3)\right )+2 \int \frac {19+10 e^4+5 x^2+2 x^3-x \left (17-2 e^4-\log (3)\right )+\log (243)}{(4+x) \left (5+4 e^x+e^x x\right )} \, dx-10 \int \frac {(5+x) \left (3+2 e^4-x+2 x^2+\log (3)\right )}{(4+x) \left (5+4 e^x+e^x x\right )^2} \, dx\\ &=-x^2+2 x^3+x \left (3+2 e^4+\log (3)\right )+2 \int \left (-\frac {3 x}{5+4 e^x+e^x x}+\frac {2 x^2}{5+4 e^x+e^x x}-\frac {5 \left (1+\frac {1}{5} \left (-2 e^4-\log (3)\right )\right )}{5+4 e^x+e^x x}+\frac {39+2 e^4+\log (3)}{(4+x) \left (5+4 e^x+e^x x\right )}\right ) \, dx-10 \int \left (\frac {x}{\left (5+4 e^x+e^x x\right )^2}+\frac {2 x^2}{\left (5+4 e^x+e^x x\right )^2}-\frac {6 \left (1+\frac {1}{6} \left (-2 e^4-\log (3)\right )\right )}{\left (5+4 e^x+e^x x\right )^2}+\frac {39+2 e^4+\log (3)}{(4+x) \left (5+4 e^x+e^x x\right )^2}\right ) \, dx\\ &=-x^2+2 x^3+x \left (3+2 e^4+\log (3)\right )+4 \int \frac {x^2}{5+4 e^x+e^x x} \, dx-6 \int \frac {x}{5+4 e^x+e^x x} \, dx-10 \int \frac {x}{\left (5+4 e^x+e^x x\right )^2} \, dx-20 \int \frac {x^2}{\left (5+4 e^x+e^x x\right )^2} \, dx-\left (2 \left (5-2 e^4-\log (3)\right )\right ) \int \frac {1}{5+4 e^x+e^x x} \, dx+\left (10 \left (6-2 e^4-\log (3)\right )\right ) \int \frac {1}{\left (5+4 e^x+e^x x\right )^2} \, dx+\left (2 \left (39+2 e^4+\log (3)\right )\right ) \int \frac {1}{(4+x) \left (5+4 e^x+e^x x\right )} \, dx-\left (10 \left (39+2 e^4+\log (3)\right )\right ) \int \frac {1}{(4+x) \left (5+4 e^x+e^x x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 68, normalized size = 2.00 \begin {gather*} -x^2+2 x^3+x \left (3+2 e^4+\log (3)\right )-\frac {30+18 x^2+4 x^3+4 e^4 (5+x)+x (-4+\log (9))+\log (59049)}{(5+x) \left (5+e^x (4+x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(85 + 50*E^4 - 90*x + 150*x^2 + 25*Log[3] + E^x*(158 - 84*x + 230*x^2 + 64*x^3 + E^4*(100 + 24*x) +
(50 + 12*x)*Log[3]) + E^(2*x)*(48 - 8*x + 83*x^2 + 46*x^3 + 6*x^4 + E^4*(32 + 16*x + 2*x^2) + (16 + 8*x + x^2)
*Log[3]))/(25 + E^x*(40 + 10*x) + E^(2*x)*(16 + 8*x + x^2)),x]

[Out]

-x^2 + 2*x^3 + x*(3 + 2*E^4 + Log[3]) - (30 + 18*x^2 + 4*x^3 + 4*E^4*(5 + x) + x*(-4 + Log[9]) + Log[59049])/(
(5 + x)*(5 + E^x*(4 + x)))

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fricas [B]  time = 0.74, size = 86, normalized size = 2.53 \begin {gather*} \frac {10 \, x^{3} - 9 \, x^{2} + 2 \, {\left (5 \, x - 2\right )} e^{4} + {\left (2 \, x^{4} + 7 \, x^{3} - x^{2} + 2 \, {\left (x^{2} + 4 \, x\right )} e^{4} + {\left (x^{2} + 4 \, x\right )} \log \relax (3) + 12 \, x\right )} e^{x} + {\left (5 \, x - 2\right )} \log \relax (3) + 17 \, x - 6}{{\left (x + 4\right )} e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+8*x+16)*log(3)+(2*x^2+16*x+32)*exp(4)+6*x^4+46*x^3+83*x^2-8*x+48)*exp(x)^2+((12*x+50)*log(3)+
(24*x+100)*exp(4)+64*x^3+230*x^2-84*x+158)*exp(x)+25*log(3)+50*exp(4)+150*x^2-90*x+85)/((x^2+8*x+16)*exp(x)^2+
(10*x+40)*exp(x)+25),x, algorithm="fricas")

[Out]

(10*x^3 - 9*x^2 + 2*(5*x - 2)*e^4 + (2*x^4 + 7*x^3 - x^2 + 2*(x^2 + 4*x)*e^4 + (x^2 + 4*x)*log(3) + 12*x)*e^x
+ (5*x - 2)*log(3) + 17*x - 6)/((x + 4)*e^x + 5)

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giac [B]  time = 0.36, size = 103, normalized size = 3.03 \begin {gather*} \frac {2 \, x^{4} e^{x} + 7 \, x^{3} e^{x} + x^{2} e^{x} \log \relax (3) + 10 \, x^{3} + 2 \, x^{2} e^{\left (x + 4\right )} - x^{2} e^{x} + 4 \, x e^{x} \log \relax (3) - 9 \, x^{2} + 10 \, x e^{4} + 8 \, x e^{\left (x + 4\right )} + 12 \, x e^{x} + 5 \, x \log \relax (3) + 17 \, x - 4 \, e^{4} - 2 \, \log \relax (3) - 6}{x e^{x} + 4 \, e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+8*x+16)*log(3)+(2*x^2+16*x+32)*exp(4)+6*x^4+46*x^3+83*x^2-8*x+48)*exp(x)^2+((12*x+50)*log(3)+
(24*x+100)*exp(4)+64*x^3+230*x^2-84*x+158)*exp(x)+25*log(3)+50*exp(4)+150*x^2-90*x+85)/((x^2+8*x+16)*exp(x)^2+
(10*x+40)*exp(x)+25),x, algorithm="giac")

[Out]

(2*x^4*e^x + 7*x^3*e^x + x^2*e^x*log(3) + 10*x^3 + 2*x^2*e^(x + 4) - x^2*e^x + 4*x*e^x*log(3) - 9*x^2 + 10*x*e
^4 + 8*x*e^(x + 4) + 12*x*e^x + 5*x*log(3) + 17*x - 4*e^4 - 2*log(3) - 6)/(x*e^x + 4*e^x + 5)

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maple [A]  time = 0.11, size = 54, normalized size = 1.59

method	result	size
risch	\(2 x^{3}+x \ln \relax (3)+2 x \,{\mathrm e}^{4}-x^{2}+3 x -\frac {2 \left (2 x^{2}+\ln \relax (3)+2 \,{\mathrm e}^{4}-x +3\right )}{{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+5}\)	\(54\)
norman	\(\frac {\left (17+10 \,{\mathrm e}^{4}+5 \ln \relax (3)\right ) x +\left (-48-32 \,{\mathrm e}^{4}-16 \ln \relax (3)\right ) {\mathrm e}^{x}+\left (-1+2 \,{\mathrm e}^{4}+\ln \relax (3)\right ) x^{2} {\mathrm e}^{x}-9 x^{2}+10 x^{3}+7 \,{\mathrm e}^{x} x^{3}+2 \,{\mathrm e}^{x} x^{4}-66-22 \ln \relax (3)-44 \,{\mathrm e}^{4}}{{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+5}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+8*x+16)*ln(3)+(2*x^2+16*x+32)*exp(4)+6*x^4+46*x^3+83*x^2-8*x+48)*exp(x)^2+((12*x+50)*ln(3)+(24*x+10
0)*exp(4)+64*x^3+230*x^2-84*x+158)*exp(x)+25*ln(3)+50*exp(4)+150*x^2-90*x+85)/((x^2+8*x+16)*exp(x)^2+(10*x+40)
*exp(x)+25),x,method=_RETURNVERBOSE)

[Out]

2*x^3+x*ln(3)+2*x*exp(4)-x^2+3*x-2*(2*x^2+ln(3)+2*exp(4)-x+3)/(exp(x)*x+4*exp(x)+5)

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maxima [B]  time = 0.60, size = 80, normalized size = 2.35 \begin {gather*} \frac {10 \, x^{3} - 9 \, x^{2} + x {\left (10 \, e^{4} + 5 \, \log \relax (3) + 17\right )} + {\left (2 \, x^{4} + 7 \, x^{3} + x^{2} {\left (2 \, e^{4} + \log \relax (3) - 1\right )} + 4 \, x {\left (2 \, e^{4} + \log \relax (3) + 3\right )}\right )} e^{x} - 4 \, e^{4} - 2 \, \log \relax (3) - 6}{{\left (x + 4\right )} e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+8*x+16)*log(3)+(2*x^2+16*x+32)*exp(4)+6*x^4+46*x^3+83*x^2-8*x+48)*exp(x)^2+((12*x+50)*log(3)+
(24*x+100)*exp(4)+64*x^3+230*x^2-84*x+158)*exp(x)+25*log(3)+50*exp(4)+150*x^2-90*x+85)/((x^2+8*x+16)*exp(x)^2+
(10*x+40)*exp(x)+25),x, algorithm="maxima")

[Out]

(10*x^3 - 9*x^2 + x*(10*e^4 + 5*log(3) + 17) + (2*x^4 + 7*x^3 + x^2*(2*e^4 + log(3) - 1) + 4*x*(2*e^4 + log(3)
+ 3))*e^x - 4*e^4 - 2*log(3) - 6)/((x + 4)*e^x + 5)

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mupad [B]  time = 1.84, size = 89, normalized size = 2.62 \begin {gather*} \frac {7\,x^3\,{\mathrm {e}}^x+2\,x^4\,{\mathrm {e}}^x+{\mathrm {e}}^x\,\left (\frac {16\,{\mathrm {e}}^4}{5}+\frac {8\,\ln \relax (3)}{5}+\frac {24}{5}\right )+x\,\left (10\,{\mathrm {e}}^4+\ln \left (243\right )+17\right )-9\,x^2+10\,x^3+x\,{\mathrm {e}}^x\,\left (\frac {44\,{\mathrm {e}}^4}{5}+\frac {22\,\ln \relax (3)}{5}+\frac {66}{5}\right )+x^2\,{\mathrm {e}}^x\,\left (2\,{\mathrm {e}}^4+\ln \relax (3)-1\right )}{4\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*exp(4) - 90*x + 25*log(3) + exp(2*x)*(exp(4)*(16*x + 2*x^2 + 32) - 8*x + 83*x^2 + 46*x^3 + 6*x^4 + log
(3)*(8*x + x^2 + 16) + 48) + 150*x^2 + exp(x)*(log(3)*(12*x + 50) - 84*x + 230*x^2 + 64*x^3 + exp(4)*(24*x + 1
00) + 158) + 85)/(exp(x)*(10*x + 40) + exp(2*x)*(8*x + x^2 + 16) + 25),x)

[Out]

(7*x^3*exp(x) + 2*x^4*exp(x) + exp(x)*((16*exp(4))/5 + (8*log(3))/5 + 24/5) + x*(10*exp(4) + log(243) + 17) -
9*x^2 + 10*x^3 + x*exp(x)*((44*exp(4))/5 + (22*log(3))/5 + 66/5) + x^2*exp(x)*(2*exp(4) + log(3) - 1))/(4*exp(
x) + x*exp(x) + 5)

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sympy [A]  time = 0.27, size = 48, normalized size = 1.41 \begin {gather*} 2 x^{3} - x^{2} + x \left (\log {\relax (3 )} + 3 + 2 e^{4}\right ) + \frac {- 4 x^{2} + 2 x - 4 e^{4} - 6 - 2 \log {\relax (3 )}}{\left (x + 4\right ) e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+8*x+16)*ln(3)+(2*x**2+16*x+32)*exp(4)+6*x**4+46*x**3+83*x**2-8*x+48)*exp(x)**2+((12*x+50)*ln
(3)+(24*x+100)*exp(4)+64*x**3+230*x**2-84*x+158)*exp(x)+25*ln(3)+50*exp(4)+150*x**2-90*x+85)/((x**2+8*x+16)*ex
p(x)**2+(10*x+40)*exp(x)+25),x)

[Out]

2*x**3 - x**2 + x*(log(3) + 3 + 2*exp(4)) + (-4*x**2 + 2*x - 4*exp(4) - 6 - 2*log(3))/((x + 4)*exp(x) + 5)

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