3.100.81 \(\int \frac {-400+24920 x+4996 x^2+250 x^3+(80-4992 x-500 x^2) \log (4)+(-4+250 x) \log ^2(4)+e^x (-10+490 x+274 x^2+25 x^3+(1-49 x-25 x^2) \log (4))}{100+20 x+x^2+(-20-2 x) \log (4)+\log ^2(4)} \, dx\)

Optimal. Leaf size=32 \[ x \left (1+\frac {\left (-x+25 x^2\right ) \left (5+\frac {e^x}{10+x-\log (4)}\right )}{x}\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 56, normalized size of antiderivative = 1.75, number of steps used = 10, number of rules used = 6, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6688, 2199, 2176, 2194, 2178, 2177} \begin {gather*} 125 x^2+25 e^x x-4 x-25 e^x+\frac {e^x (251-25 \log (4)) (10-\log (4))}{x+10-\log (4)}-2 e^x (113-25 \log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-400 + 24920*x + 4996*x^2 + 250*x^3 + (80 - 4992*x - 500*x^2)*Log[4] + (-4 + 250*x)*Log[4]^2 + E^x*(-10 +
490*x + 274*x^2 + 25*x^3 + (1 - 49*x - 25*x^2)*Log[4]))/(100 + 20*x + x^2 + (-20 - 2*x)*Log[4] + Log[4]^2),x]

[Out]

-25*E^x - 4*x + 25*E^x*x + 125*x^2 - 2*E^x*(113 - 25*Log[2]) + (E^x*(251 - 25*Log[4])*(10 - Log[4]))/(10 + x -
Log[4])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 \left (-4+250 x+\frac {e^x \left (-10+25 x^3+x^2 (274-25 \log (4))+49 x (10-\log (4))+\log (4)\right )}{(10+x-\log (4))^2}\right ) \, dx\\ &=-4 x+125 x^2+\int \frac {e^x \left (-10+25 x^3+x^2 (274-25 \log (4))+49 x (10-\log (4))+\log (4)\right )}{(10+x-\log (4))^2} \, dx\\ &=-4 x+125 x^2+\int \left (25 e^x x-226 e^x \left (1-\frac {25 \log (2)}{113}\right )+\frac {e^x (251-25 \log (4)) (10-\log (4))}{10+x-\log (4)}+\frac {e^x \left (-2510+501 \log (4)-25 \log ^2(4)\right )}{(10+x-\log (4))^2}\right ) \, dx\\ &=-4 x+125 x^2+25 \int e^x x \, dx-(2 (113-25 \log (2))) \int e^x \, dx-((251-25 \log (4)) (10-\log (4))) \int \frac {e^x}{(10+x-\log (4))^2} \, dx+((251-25 \log (4)) (10-\log (4))) \int \frac {e^x}{10+x-\log (4)} \, dx\\ &=-4 x+25 e^x x+125 x^2-2 e^x (113-25 \log (2))+\frac {4 \text {Ei}(10+x-\log (4)) (251-25 \log (4)) (10-\log (4))}{e^{10}}+\frac {e^x (251-25 \log (4)) (10-\log (4))}{10+x-\log (4)}-25 \int e^x \, dx-((251-25 \log (4)) (10-\log (4))) \int \frac {e^x}{10+x-\log (4)} \, dx\\ &=-25 e^x-4 x+25 e^x x+125 x^2-2 e^x (113-25 \log (2))+\frac {e^x (251-25 \log (4)) (10-\log (4))}{10+x-\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 25, normalized size = 0.78 \begin {gather*} x \left (-4+125 x+\frac {e^x (-1+25 x)}{10+x-\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-400 + 24920*x + 4996*x^2 + 250*x^3 + (80 - 4992*x - 500*x^2)*Log[4] + (-4 + 250*x)*Log[4]^2 + E^x*
(-10 + 490*x + 274*x^2 + 25*x^3 + (1 - 49*x - 25*x^2)*Log[4]))/(100 + 20*x + x^2 + (-20 - 2*x)*Log[4] + Log[4]
^2),x]

[Out]

x*(-4 + 125*x + (E^x*(-1 + 25*x))/(10 + x - Log[4]))

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fricas [A]  time = 0.60, size = 49, normalized size = 1.53 \begin {gather*} \frac {125 \, x^{3} + 1246 \, x^{2} + {\left (25 \, x^{2} - x\right )} e^{x} - 2 \, {\left (125 \, x^{2} - 4 \, x\right )} \log \relax (2) - 40 \, x}{x - 2 \, \log \relax (2) + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4)*log(2)^2+2*(-500*x^2-4992*x+
80)*log(2)+250*x^3+4996*x^2+24920*x-400)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="fricas")

[Out]

(125*x^3 + 1246*x^2 + (25*x^2 - x)*e^x - 2*(125*x^2 - 4*x)*log(2) - 40*x)/(x - 2*log(2) + 10)

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giac [A]  time = 0.21, size = 48, normalized size = 1.50 \begin {gather*} \frac {125 \, x^{3} + 25 \, x^{2} e^{x} - 250 \, x^{2} \log \relax (2) + 1246 \, x^{2} - x e^{x} + 8 \, x \log \relax (2) - 40 \, x}{x - 2 \, \log \relax (2) + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4)*log(2)^2+2*(-500*x^2-4992*x+
80)*log(2)+250*x^3+4996*x^2+24920*x-400)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="giac")

[Out]

(125*x^3 + 25*x^2*e^x - 250*x^2*log(2) + 1246*x^2 - x*e^x + 8*x*log(2) - 40*x)/(x - 2*log(2) + 10)

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maple [A]  time = 0.18, size = 51, normalized size = 1.59

method	result	size
norman	\(\frac {\left (250 \ln \relax (2)-1246\right ) x^{2}+{\mathrm e}^{x} x -125 x^{3}-25 \,{\mathrm e}^{x} x^{2}-16 \ln \relax (2)^{2}+160 \ln \relax (2)-400}{2 \ln \relax (2)-x -10}\)	\(51\)
default	\(-4 x +125 x^{2}+\frac {2510 \,{\mathrm e}^{x}}{10+x -2 \ln \relax (2)}-\frac {1002 \ln \relax (2) {\mathrm e}^{x}}{10+x -2 \ln \relax (2)}-251 \,{\mathrm e}^{x}+\frac {100 \,{\mathrm e}^{x} \ln \relax (2)^{2}}{10+x -2 \ln \relax (2)}+25 \,{\mathrm e}^{x} x +50 \,{\mathrm e}^{x} \ln \relax (2)\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(-25*x^2-49*x+1)*ln(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4)*ln(2)^2+2*(-500*x^2-4992*x+80)*ln(2
)+250*x^3+4996*x^2+24920*x-400)/(4*ln(2)^2+2*(-2*x-20)*ln(2)+x^2+20*x+100),x,method=_RETURNVERBOSE)

[Out]

((250*ln(2)-1246)*x^2+exp(x)*x-125*x^3-25*exp(x)*x^2-16*ln(2)^2+160*ln(2)-400)/(2*ln(2)-x-10)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -1000 \, {\left (\frac {2 \, {\left (\log \relax (2) - 5\right )}}{x - 2 \, \log \relax (2) + 10} - \log \left (x - 2 \, \log \relax (2) + 10\right )\right )} \log \relax (2)^{2} + 125 \, x^{2} + 1000 \, x {\left (\log \relax (2) - 5\right )} - 2 \, {\left (\log \relax (2) - 5\right )} \int \frac {e^{x}}{x^{2} - 4 \, x {\left (\log \relax (2) - 5\right )} + 4 \, \log \relax (2)^{2} - 40 \, \log \relax (2) + 100}\,{d x} - 1000 \, {\left (4 \, {\left (\log \relax (2) - 5\right )} \log \left (x - 2 \, \log \relax (2) + 10\right ) + x - \frac {4 \, {\left (\log \relax (2)^{2} - 10 \, \log \relax (2) + 25\right )}}{x - 2 \, \log \relax (2) + 10}\right )} \log \relax (2) + 9984 \, {\left (\frac {2 \, {\left (\log \relax (2) - 5\right )}}{x - 2 \, \log \relax (2) + 10} - \log \left (x - 2 \, \log \relax (2) + 10\right )\right )} \log \relax (2) - \frac {8 \, e^{\left (-10\right )} E_{2}\left (-x + 2 \, \log \relax (2) - 10\right ) \log \relax (2)}{x - 2 \, \log \relax (2) + 10} + 3000 \, {\left (\log \relax (2)^{2} - 10 \, \log \relax (2) + 25\right )} \log \left (x - 2 \, \log \relax (2) + 10\right ) + 19984 \, {\left (\log \relax (2) - 5\right )} \log \left (x - 2 \, \log \relax (2) + 10\right ) + 4996 \, x + \frac {{\left (25 \, x^{2} - x\right )} e^{x}}{x - 2 \, \log \relax (2) + 10} + \frac {40 \, e^{\left (-10\right )} E_{2}\left (-x + 2 \, \log \relax (2) - 10\right )}{x - 2 \, \log \relax (2) + 10} + \frac {16 \, \log \relax (2)^{2}}{x - 2 \, \log \relax (2) + 10} - \frac {2000 \, {\left (\log \relax (2)^{3} - 15 \, \log \relax (2)^{2} + 75 \, \log \relax (2) - 125\right )}}{x - 2 \, \log \relax (2) + 10} - \frac {19984 \, {\left (\log \relax (2)^{2} - 10 \, \log \relax (2) + 25\right )}}{x - 2 \, \log \relax (2) + 10} - \frac {49840 \, {\left (\log \relax (2) - 5\right )}}{x - 2 \, \log \relax (2) + 10} - \frac {160 \, \log \relax (2)}{x - 2 \, \log \relax (2) + 10} + \frac {400}{x - 2 \, \log \relax (2) + 10} + 24920 \, \log \left (x - 2 \, \log \relax (2) + 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-25*x^2-49*x+1)*log(2)+25*x^3+274*x^2+490*x-10)*exp(x)+4*(250*x-4)*log(2)^2+2*(-500*x^2-4992*x+
80)*log(2)+250*x^3+4996*x^2+24920*x-400)/(4*log(2)^2+2*(-2*x-20)*log(2)+x^2+20*x+100),x, algorithm="maxima")

[Out]

-1000*(2*(log(2) - 5)/(x - 2*log(2) + 10) - log(x - 2*log(2) + 10))*log(2)^2 + 125*x^2 + 1000*x*(log(2) - 5) -
2*(log(2) - 5)*integrate(e^x/(x^2 - 4*x*(log(2) - 5) + 4*log(2)^2 - 40*log(2) + 100), x) - 1000*(4*(log(2) -
5)*log(x - 2*log(2) + 10) + x - 4*(log(2)^2 - 10*log(2) + 25)/(x - 2*log(2) + 10))*log(2) + 9984*(2*(log(2) -
5)/(x - 2*log(2) + 10) - log(x - 2*log(2) + 10))*log(2) - 8*e^(-10)*exp_integral_e(2, -x + 2*log(2) - 10)*log(
2)/(x - 2*log(2) + 10) + 3000*(log(2)^2 - 10*log(2) + 25)*log(x - 2*log(2) + 10) + 19984*(log(2) - 5)*log(x -
2*log(2) + 10) + 4996*x + (25*x^2 - x)*e^x/(x - 2*log(2) + 10) + 40*e^(-10)*exp_integral_e(2, -x + 2*log(2) -
10)/(x - 2*log(2) + 10) + 16*log(2)^2/(x - 2*log(2) + 10) - 2000*(log(2)^3 - 15*log(2)^2 + 75*log(2) - 125)/(x
- 2*log(2) + 10) - 19984*(log(2)^2 - 10*log(2) + 25)/(x - 2*log(2) + 10) - 49840*(log(2) - 5)/(x - 2*log(2) +
10) - 160*log(2)/(x - 2*log(2) + 10) + 400/(x - 2*log(2) + 10) + 24920*log(x - 2*log(2) + 10)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {24920\,x+{\mathrm {e}}^x\,\left (490\,x-2\,\ln \relax (2)\,\left (25\,x^2+49\,x-1\right )+274\,x^2+25\,x^3-10\right )-2\,\ln \relax (2)\,\left (500\,x^2+4992\,x-80\right )+4\,{\ln \relax (2)}^2\,\left (250\,x-4\right )+4996\,x^2+250\,x^3-400}{20\,x-2\,\ln \relax (2)\,\left (2\,x+20\right )+4\,{\ln \relax (2)}^2+x^2+100} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24920*x + exp(x)*(490*x - 2*log(2)*(49*x + 25*x^2 - 1) + 274*x^2 + 25*x^3 - 10) - 2*log(2)*(4992*x + 500*
x^2 - 80) + 4*log(2)^2*(250*x - 4) + 4996*x^2 + 250*x^3 - 400)/(20*x - 2*log(2)*(2*x + 20) + 4*log(2)^2 + x^2
+ 100),x)

[Out]

int((24920*x + exp(x)*(490*x - 2*log(2)*(49*x + 25*x^2 - 1) + 274*x^2 + 25*x^3 - 10) - 2*log(2)*(4992*x + 500*
x^2 - 80) + 4*log(2)^2*(250*x - 4) + 4996*x^2 + 250*x^3 - 400)/(20*x - 2*log(2)*(2*x + 20) + 4*log(2)^2 + x^2
+ 100), x)

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sympy [A]  time = 0.18, size = 26, normalized size = 0.81 \begin {gather*} 125 x^{2} - 4 x + \frac {\left (25 x^{2} - x\right ) e^{x}}{x - 2 \log {\relax (2 )} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-25*x**2-49*x+1)*ln(2)+25*x**3+274*x**2+490*x-10)*exp(x)+4*(250*x-4)*ln(2)**2+2*(-500*x**2-4992
*x+80)*ln(2)+250*x**3+4996*x**2+24920*x-400)/(4*ln(2)**2+2*(-2*x-20)*ln(2)+x**2+20*x+100),x)

[Out]

125*x**2 - 4*x + (25*x**2 - x)*exp(x)/(x - 2*log(2) + 10)

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