3.54.86 \(\int \frac {-160+100 x+e^5 (-6+4 x)+(-50-2 e^5) \log (3)+e^x (50+50 x+e^5 (2+2 x)+(-50-2 e^5) \log (3))}{25+e^5} \, dx\)

Optimal. Leaf size=27 \[ -2 \left (3-e^x+\frac {5}{25+e^5}-x\right ) (x-\log (3)) \]

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Rubi [B]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 2.44, number of steps used = 5, number of rules used = 4, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 2187, 2176, 2194} \begin {gather*} \frac {50 x^2}{25+e^5}+\frac {e^5 (3-2 x)^2}{2 \left (25+e^5\right )}-2 e^x+2 e^x (x+1-\log (3))-2 x \left (\frac {80}{25+e^5}+\log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-160 + 100*x + E^5*(-6 + 4*x) + (-50 - 2*E^5)*Log[3] + E^x*(50 + 50*x + E^5*(2 + 2*x) + (-50 - 2*E^5)*Log
[3]))/(25 + E^5),x]

[Out]

-2*E^x + (E^5*(3 - 2*x)^2)/(2*(25 + E^5)) + (50*x^2)/(25 + E^5) + 2*E^x*(1 + x - Log[3]) - 2*x*(80/(25 + E^5)
+ Log[3])

Rule 12

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

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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-160+100 x+e^5 (-6+4 x)+\left (-50-2 e^5\right ) \log (3)+e^x \left (50+50 x+e^5 (2+2 x)+\left (-50-2 e^5\right ) \log (3)\right )\right ) \, dx}{25+e^5}\\ &=\frac {e^5 (3-2 x)^2}{2 \left (25+e^5\right )}+\frac {50 x^2}{25+e^5}-2 x \left (\frac {80}{25+e^5}+\log (3)\right )+\frac {\int e^x \left (50+50 x+e^5 (2+2 x)+\left (-50-2 e^5\right ) \log (3)\right ) \, dx}{25+e^5}\\ &=\frac {e^5 (3-2 x)^2}{2 \left (25+e^5\right )}+\frac {50 x^2}{25+e^5}-2 x \left (\frac {80}{25+e^5}+\log (3)\right )+\frac {\int e^x \left (2 \left (25+e^5\right ) x+2 \left (25+e^5\right ) (1-\log (3))\right ) \, dx}{25+e^5}\\ &=\frac {e^5 (3-2 x)^2}{2 \left (25+e^5\right )}+\frac {50 x^2}{25+e^5}+2 e^x (1+x-\log (3))-2 x \left (\frac {80}{25+e^5}+\log (3)\right )-2 \int e^x \, dx\\ &=-2 e^x+\frac {e^5 (3-2 x)^2}{2 \left (25+e^5\right )}+\frac {50 x^2}{25+e^5}+2 e^x (1+x-\log (3))-2 x \left (\frac {80}{25+e^5}+\log (3)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 64, normalized size = 2.37 \begin {gather*} \frac {2 \left (-80 x+25 x^2+e^5 x^2+e^5 x (-3-\log (3))+25 e^x (x-\log (3))+e^{5+x} (x-\log (3))-25 x \log (3)\right )}{25+e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-160 + 100*x + E^5*(-6 + 4*x) + (-50 - 2*E^5)*Log[3] + E^x*(50 + 50*x + E^5*(2 + 2*x) + (-50 - 2*E^
5)*Log[3]))/(25 + E^5),x]

[Out]

(2*(-80*x + 25*x^2 + E^5*x^2 + E^5*x*(-3 - Log[3]) + 25*E^x*(x - Log[3]) + E^(5 + x)*(x - Log[3]) - 25*x*Log[3
]))/(25 + E^5)

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fricas [B]  time = 0.54, size = 58, normalized size = 2.15 \begin {gather*} \frac {2 \, {\left (25 \, x^{2} + {\left (x^{2} - 3 \, x\right )} e^{5} + {\left (x e^{5} - {\left (e^{5} + 25\right )} \log \relax (3) + 25 \, x\right )} e^{x} - {\left (x e^{5} + 25 \, x\right )} \log \relax (3) - 80 \, x\right )}}{e^{5} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(5)-50)*log(3)+(2*x+2)*exp(5)+50*x+50)*exp(x)+(-2*exp(5)-50)*log(3)+(4*x-6)*exp(5)+100*x-16
0)/(exp(5)+25),x, algorithm="fricas")

[Out]

2*(25*x^2 + (x^2 - 3*x)*e^5 + (x*e^5 - (e^5 + 25)*log(3) + 25*x)*e^x - (x*e^5 + 25*x)*log(3) - 80*x)/(e^5 + 25
)

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giac [B]  time = 0.15, size = 58, normalized size = 2.15 \begin {gather*} -\frac {2 \, {\left (x {\left (e^{5} + 25\right )} \log \relax (3) - 25 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{5} - {\left (x - \log \relax (3)\right )} e^{\left (x + 5\right )} - 25 \, {\left (x - \log \relax (3)\right )} e^{x} + 80 \, x\right )}}{e^{5} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(5)-50)*log(3)+(2*x+2)*exp(5)+50*x+50)*exp(x)+(-2*exp(5)-50)*log(3)+(4*x-6)*exp(5)+100*x-16
0)/(exp(5)+25),x, algorithm="giac")

[Out]

-2*(x*(e^5 + 25)*log(3) - 25*x^2 - (x^2 - 3*x)*e^5 - (x - log(3))*e^(x + 5) - 25*(x - log(3))*e^x + 80*x)/(e^5
+ 25)

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maple [A]  time = 0.04, size = 42, normalized size = 1.56

method	result	size
norman	\(2 x^{2}+2 \,{\mathrm e}^{x} x -2 \ln \relax (3) {\mathrm e}^{x}-\frac {2 \left ({\mathrm e}^{5} \ln \relax (3)+3 \,{\mathrm e}^{5}+25 \ln \relax (3)+80\right ) x}{{\mathrm e}^{5}+25}\)	\(42\)
default	\(\frac {-160 x +{\mathrm e}^{5} \left (2 x^{2}-6 x \right )-2 x \,{\mathrm e}^{5} \ln \relax (3)-50 x \ln \relax (3)+50 \,{\mathrm e}^{x} x -50 \ln \relax (3) {\mathrm e}^{x}+2 x \,{\mathrm e}^{5} {\mathrm e}^{x}-2 \,{\mathrm e}^{5} {\mathrm e}^{x} \ln \relax (3)+50 x^{2}}{{\mathrm e}^{5}+25}\)	\(67\)
risch	\(-\frac {2 x \,{\mathrm e}^{5} \ln \relax (3)}{{\mathrm e}^{5}+25}+\frac {2 x^{2} {\mathrm e}^{5}}{{\mathrm e}^{5}+25}-\frac {6 x \,{\mathrm e}^{5}}{{\mathrm e}^{5}+25}-\frac {50 x \ln \relax (3)}{{\mathrm e}^{5}+25}+\frac {50 x^{2}}{{\mathrm e}^{5}+25}-\frac {160 x}{{\mathrm e}^{5}+25}+\frac {\left (-2 \,{\mathrm e}^{5} \ln \relax (3)+2 x \,{\mathrm e}^{5}-50 \ln \relax (3)+50 x \right ) {\mathrm e}^{x}}{{\mathrm e}^{5}+25}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*exp(5)-50)*ln(3)+(2*x+2)*exp(5)+50*x+50)*exp(x)+(-2*exp(5)-50)*ln(3)+(4*x-6)*exp(5)+100*x-160)/(exp(
5)+25),x,method=_RETURNVERBOSE)

[Out]

2*x^2+2*exp(x)*x-2*ln(3)*exp(x)-2*(exp(5)*ln(3)+3*exp(5)+25*ln(3)+80)/(exp(5)+25)*x

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maxima [B]  time = 0.44, size = 57, normalized size = 2.11 \begin {gather*} -\frac {2 \, {\left (x {\left (e^{5} + 25\right )} \log \relax (3) - 25 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{5} - {\left (x {\left (e^{5} + 25\right )} - e^{5} \log \relax (3) - 25 \, \log \relax (3)\right )} e^{x} + 80 \, x\right )}}{e^{5} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(5)-50)*log(3)+(2*x+2)*exp(5)+50*x+50)*exp(x)+(-2*exp(5)-50)*log(3)+(4*x-6)*exp(5)+100*x-16
0)/(exp(5)+25),x, algorithm="maxima")

[Out]

-2*(x*(e^5 + 25)*log(3) - 25*x^2 - (x^2 - 3*x)*e^5 - (x*(e^5 + 25) - e^5*log(3) - 25*log(3))*e^x + 80*x)/(e^5
+ 25)

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mupad [B]  time = 0.10, size = 57, normalized size = 2.11 \begin {gather*} \frac {x^2\,\left (2\,{\mathrm {e}}^5+50\right )-x\,\left (6\,{\mathrm {e}}^5+50\,\ln \relax (3)+2\,{\mathrm {e}}^5\,\ln \relax (3)+160\right )-2\,{\mathrm {e}}^x\,\ln \relax (3)\,\left ({\mathrm {e}}^5+25\right )+x\,{\mathrm {e}}^x\,\left (2\,{\mathrm {e}}^5+50\right )}{{\mathrm {e}}^5+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*x + exp(x)*(50*x - log(3)*(2*exp(5) + 50) + exp(5)*(2*x + 2) + 50) - log(3)*(2*exp(5) + 50) + exp(5)*
(4*x - 6) - 160)/(exp(5) + 25),x)

[Out]

(x^2*(2*exp(5) + 50) - x*(6*exp(5) + 50*log(3) + 2*exp(5)*log(3) + 160) - 2*exp(x)*log(3)*(exp(5) + 25) + x*ex
p(x)*(2*exp(5) + 50))/(exp(5) + 25)

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sympy [B]  time = 0.16, size = 44, normalized size = 1.63 \begin {gather*} 2 x^{2} + \frac {x \left (- 6 e^{5} - 2 e^{5} \log {\relax (3 )} - 160 - 50 \log {\relax (3 )}\right )}{25 + e^{5}} + \left (2 x - 2 \log {\relax (3 )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*exp(5)-50)*ln(3)+(2*x+2)*exp(5)+50*x+50)*exp(x)+(-2*exp(5)-50)*ln(3)+(4*x-6)*exp(5)+100*x-160)
/(exp(5)+25),x)

[Out]

2*x**2 + x*(-6*exp(5) - 2*exp(5)*log(3) - 160 - 50*log(3))/(25 + exp(5)) + (2*x - 2*log(3))*exp(x)

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