3.12.27 \(\int \frac {5160 x-7380 x^2+2200 x^3+100 x^4+e^{2 x} (20000+800 x)+e^x (10320+400 x-5000 x^2-200 x^3)+(2064-1920 x-80 x^2+e^x (4000+160 x)) \log (625+50 x+x^2)}{625+25 x} \, dx\)

Optimal. Leaf size=28 \[ 16 \left (-e^x+\frac {1}{4} (-2+x) x-\frac {1}{5} \log \left ((25+x)^2\right )\right )^2 \]

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Rubi [A]  time = 0.30, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{25} \left (5 (2-x) x+20 e^x+4 \log \left ((x+25)^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5160*x - 7380*x^2 + 2200*x^3 + 100*x^4 + E^(2*x)*(20000 + 800*x) + E^x*(10320 + 400*x - 5000*x^2 - 200*x^
3) + (2064 - 1920*x - 80*x^2 + E^x*(4000 + 160*x))*Log[625 + 50*x + x^2])/(625 + 25*x),x]

[Out]

(20*E^x + 5*(2 - x)*x + 4*Log[(25 + x)^2])^2/25

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (129-120 x-5 x^2+10 e^x (25+x)\right ) \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )}{25 (25+x)} \, dx\\ &=\frac {4}{25} \int \frac {\left (129-120 x-5 x^2+10 e^x (25+x)\right ) \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )}{25+x} \, dx\\ &=\frac {1}{25} \left (20 e^x+5 (2-x) x+4 \log \left ((25+x)^2\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{25} \left (20 e^x-5 (-2+x) x+4 \log \left ((25+x)^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5160*x - 7380*x^2 + 2200*x^3 + 100*x^4 + E^(2*x)*(20000 + 800*x) + E^x*(10320 + 400*x - 5000*x^2 -
200*x^3) + (2064 - 1920*x - 80*x^2 + E^x*(4000 + 160*x))*Log[625 + 50*x + x^2])/(625 + 25*x),x]

[Out]

(20*E^x - 5*(-2 + x)*x + 4*Log[(25 + x)^2])^2/25

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fricas [B]  time = 0.71, size = 66, normalized size = 2.36 \begin {gather*} x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, {\left (x^{2} - 2 \, x\right )} e^{x} - \frac {8}{5} \, {\left (x^{2} - 2 \, x - 4 \, e^{x}\right )} \log \left (x^{2} + 50 \, x + 625\right ) + \frac {16}{25} \, \log \left (x^{2} + 50 \, x + 625\right )^{2} + 16 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((160*x+4000)*exp(x)-80*x^2-1920*x+2064)*log(x^2+50*x+625)+(800*x+20000)*exp(x)^2+(-200*x^3-5000*x^
2+400*x+10320)*exp(x)+100*x^4+2200*x^3-7380*x^2+5160*x)/(25*x+625),x, algorithm="fricas")

[Out]

x^4 - 4*x^3 + 4*x^2 - 8*(x^2 - 2*x)*e^x - 8/5*(x^2 - 2*x - 4*e^x)*log(x^2 + 50*x + 625) + 16/25*log(x^2 + 50*x
+ 625)^2 + 16*e^(2*x)

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giac [B]  time = 0.36, size = 84, normalized size = 3.00 \begin {gather*} x^{4} - 4 \, x^{3} - 8 \, x^{2} e^{x} - \frac {8}{5} \, x^{2} \log \left (x^{2} + 50 \, x + 625\right ) + 4 \, x^{2} + 16 \, x e^{x} + \frac {16}{5} \, x \log \left (x^{2} + 50 \, x + 625\right ) + \frac {32}{5} \, e^{x} \log \left (x^{2} + 50 \, x + 625\right ) + \frac {16}{25} \, \log \left (x^{2} + 50 \, x + 625\right )^{2} + 16 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((160*x+4000)*exp(x)-80*x^2-1920*x+2064)*log(x^2+50*x+625)+(800*x+20000)*exp(x)^2+(-200*x^3-5000*x^
2+400*x+10320)*exp(x)+100*x^4+2200*x^3-7380*x^2+5160*x)/(25*x+625),x, algorithm="giac")

[Out]

x^4 - 4*x^3 - 8*x^2*e^x - 8/5*x^2*log(x^2 + 50*x + 625) + 4*x^2 + 16*x*e^x + 16/5*x*log(x^2 + 50*x + 625) + 32
/5*e^x*log(x^2 + 50*x + 625) + 16/25*log(x^2 + 50*x + 625)^2 + 16*e^(2*x)

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maple [B]  time = 0.25, size = 107, normalized size = 3.82

method	result	size
default	\(\frac {32 \left (\ln \left (\left (x +25\right )^{2}\right )-2 \ln \left (x +25\right )\right ) {\mathrm e}^{x}}{5}+16 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x} x^{2}+\frac {64 \ln \left (x +25\right ) {\mathrm e}^{x}}{5}+16 \,{\mathrm e}^{2 x}+x^{4}-4 x^{3}+4 x^{2}-\frac {8 \ln \left (x^{2}+50 x +625\right ) x^{2}}{5}+\frac {16 \ln \left (x^{2}+50 x +625\right ) x}{5}+\frac {64 \ln \left (x +25\right ) \ln \left (x^{2}+50 x +625\right )}{25}-\frac {64 \ln \left (x +25\right )^{2}}{25}\)	\(107\)
risch	\(\frac {64 \ln \left (x +25\right )^{2}}{25}+\left (-\frac {16 x^{2}}{5}+\frac {32 x}{5}+\frac {64 \,{\mathrm e}^{x}}{5}\right ) \ln \left (x +25\right )+\frac {64 i \ln \left (x +25\right ) \pi \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{25}-\frac {8 i \pi x \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{5}-\frac {32 i \ln \left (x +25\right ) \pi \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{25}-\frac {16 i \pi \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3} {\mathrm e}^{x}}{5}-\frac {16 i \pi \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right ) {\mathrm e}^{x}}{5}+\frac {4 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{5}+x^{4}-4 x^{3}+4 x^{2}-\frac {8 i \pi x \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{5}-\frac {32 i \ln \left (x +25\right ) \pi \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{3}}{25}-\frac {8 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{5}+16 \,{\mathrm e}^{2 x}+\frac {32 i \pi \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2} {\mathrm e}^{x}}{5}+\frac {4 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x +25\right )\right )^{2} \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )}{5}+\frac {16 i \pi x \,\mathrm {csgn}\left (i \left (x +25\right )\right ) \mathrm {csgn}\left (i \left (x +25\right )^{2}\right )^{2}}{5}+16 \,{\mathrm e}^{x} x -8 \,{\mathrm e}^{x} x^{2}\)	\(325\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((160*x+4000)*exp(x)-80*x^2-1920*x+2064)*ln(x^2+50*x+625)+(800*x+20000)*exp(x)^2+(-200*x^3-5000*x^2+400*x
+10320)*exp(x)+100*x^4+2200*x^3-7380*x^2+5160*x)/(25*x+625),x,method=_RETURNVERBOSE)

[Out]

32/5*(ln((x+25)^2)-2*ln(x+25))*exp(x)+16*exp(x)*x-8*exp(x)*x^2+64/5*ln(x+25)*exp(x)+16*exp(x)^2+x^4-4*x^3+4*x^
2-8/5*ln(x^2+50*x+625)*x^2+16/5*ln(x^2+50*x+625)*x+64/25*ln(x+25)*ln(x^2+50*x+625)-64/25*ln(x+25)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, {\left (x^{2} - 2 \, x\right )} e^{x} - \frac {2064}{5} \, e^{\left (-25\right )} E_{1}\left (-x - 25\right ) - \frac {16}{5} \, {\left (x^{2} - 2 \, x - 4 \, e^{x} - 675\right )} \log \left (x + 25\right ) + \frac {64}{25} \, \log \left (x + 25\right )^{2} + 16 \, e^{\left (2 \, x\right )} - \frac {2064}{5} \, \int \frac {e^{x}}{x + 25}\,{d x} - 2160 \, \log \left (x + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((160*x+4000)*exp(x)-80*x^2-1920*x+2064)*log(x^2+50*x+625)+(800*x+20000)*exp(x)^2+(-200*x^3-5000*x^
2+400*x+10320)*exp(x)+100*x^4+2200*x^3-7380*x^2+5160*x)/(25*x+625),x, algorithm="maxima")

[Out]

x^4 - 4*x^3 + 4*x^2 - 8*(x^2 - 2*x)*e^x - 2064/5*e^(-25)*exp_integral_e(1, -x - 25) - 16/5*(x^2 - 2*x - 4*e^x
- 675)*log(x + 25) + 64/25*log(x + 25)^2 + 16*e^(2*x) - 2064/5*integrate(e^x/(x + 25), x) - 2160*log(x + 25)

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mupad [B]  time = 1.08, size = 68, normalized size = 2.43 \begin {gather*} 16\,{\mathrm {e}}^{2\,x}+\ln \left (x^2+50\,x+625\right )\,\left (\frac {16\,x}{5}+\frac {32\,{\mathrm {e}}^x}{5}-\frac {8\,x^2}{5}\right )+{\mathrm {e}}^x\,\left (16\,x-8\,x^2\right )+4\,x^2-4\,x^3+x^4+\frac {16\,{\ln \left (x^2+50\,x+625\right )}^2}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5160*x - log(50*x + x^2 + 625)*(1920*x - exp(x)*(160*x + 4000) + 80*x^2 - 2064) + exp(2*x)*(800*x + 20000
) - 7380*x^2 + 2200*x^3 + 100*x^4 + exp(x)*(400*x - 5000*x^2 - 200*x^3 + 10320))/(25*x + 625),x)

[Out]

16*exp(2*x) + log(50*x + x^2 + 625)*((16*x)/5 + (32*exp(x))/5 - (8*x^2)/5) + exp(x)*(16*x - 8*x^2) + 4*x^2 - 4
*x^3 + x^4 + (16*log(50*x + x^2 + 625)^2)/25

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sympy [A]  time = 0.48, size = 82, normalized size = 2.93 \begin {gather*} x^{4} - 4 x^{3} + 4 x^{2} + \left (- \frac {8 x^{2}}{5} + \frac {16 x}{5}\right ) \log {\left (x^{2} + 50 x + 625 \right )} + \frac {\left (- 40 x^{2} + 80 x + 32 \log {\left (x^{2} + 50 x + 625 \right )}\right ) e^{x}}{5} + 16 e^{2 x} + \frac {16 \log {\left (x^{2} + 50 x + 625 \right )}^{2}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((160*x+4000)*exp(x)-80*x**2-1920*x+2064)*ln(x**2+50*x+625)+(800*x+20000)*exp(x)**2+(-200*x**3-5000
*x**2+400*x+10320)*exp(x)+100*x**4+2200*x**3-7380*x**2+5160*x)/(25*x+625),x)

[Out]

x**4 - 4*x**3 + 4*x**2 + (-8*x**2/5 + 16*x/5)*log(x**2 + 50*x + 625) + (-40*x**2 + 80*x + 32*log(x**2 + 50*x +
625))*exp(x)/5 + 16*exp(2*x) + 16*log(x**2 + 50*x + 625)**2/25

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