3.43.42 \(\int \frac {-580+600 x-150 x^2+(100-100 x+25 x^2) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+(100-100 x+25 x^2) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx\)

Optimal. Leaf size=27 \[ 3+x+\frac {x}{\left (5-x-\frac {3}{2} \left (e^x+x\right )\right ) (-6+\log (2))} \]

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Rubi [F]  time = 0.81, 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 {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-580 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E^x*(354 - 174*x + (-
60 + 30*x)*Log[2]))/(-600 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E^x*(
360 - 180*x + (-60 + 30*x)*Log[2])),x]

[Out]

x - (30*Defer[Int][x/(-10 + 3*E^x + 5*x)^2, x])/(6 - Log[2]) + (10*Defer[Int][x^2/(-10 + 3*E^x + 5*x)^2, x])/(
6 - Log[2]) + (2*Defer[Int][(-10 + 3*E^x + 5*x)^(-1), x])/(6 - Log[2]) - (2*Defer[Int][x/(-10 + 3*E^x + 5*x),
x])/(6 - Log[2])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2 (6-\log (2))} \, dx\\ &=\frac {\int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2} \, dx}{6-\log (2)}\\ &=\frac {\int \left (\frac {10 (-3+x) x}{\left (-10+3 e^x+5 x\right )^2}-\frac {2 (-1+x)}{-10+3 e^x+5 x}+6 \left (1-\frac {\log (2)}{6}\right )\right ) \, dx}{6-\log (2)}\\ &=x-\frac {2 \int \frac {-1+x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {(-3+x) x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}\\ &=x-\frac {2 \int \left (-\frac {1}{-10+3 e^x+5 x}+\frac {x}{-10+3 e^x+5 x}\right ) \, dx}{6-\log (2)}+\frac {10 \int \left (-\frac {3 x}{\left (-10+3 e^x+5 x\right )^2}+\frac {x^2}{\left (-10+3 e^x+5 x\right )^2}\right ) \, dx}{6-\log (2)}\\ &=x+\frac {2 \int \frac {1}{-10+3 e^x+5 x} \, dx}{6-\log (2)}-\frac {2 \int \frac {x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {x^2}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}-\frac {30 \int \frac {x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} \frac {x \left (-6+\frac {2}{10-3 e^x-5 x}+\log (2)\right )}{-6+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-580 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E^x*(354 - 174*
x + (-60 + 30*x)*Log[2]))/(-600 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) +
E^x*(360 - 180*x + (-60 + 30*x)*Log[2])),x]

[Out]

(x*(-6 + 2/(10 - 3*E^x - 5*x) + Log[2]))/(-6 + Log[2])

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fricas [B]  time = 1.06, size = 56, normalized size = 2.07 \begin {gather*} -\frac {30 \, x^{2} - 3 \, {\left (x \log \relax (2) - 6 \, x\right )} e^{x} - 5 \, {\left (x^{2} - 2 \, x\right )} \log \relax (2) - 58 \, x}{3 \, {\left (\log \relax (2) - 6\right )} e^{x} + 5 \, {\left (x - 2\right )} \log \relax (2) - 30 \, x + 60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-
580)/((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),
x, algorithm="fricas")

[Out]

-(30*x^2 - 3*(x*log(2) - 6*x)*e^x - 5*(x^2 - 2*x)*log(2) - 58*x)/(3*(log(2) - 6)*e^x + 5*(x - 2)*log(2) - 30*x
+ 60)

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giac [B]  time = 0.14, size = 60, normalized size = 2.22 \begin {gather*} \frac {5 \, x^{2} \log \relax (2) + 3 \, x e^{x} \log \relax (2) - 30 \, x^{2} - 18 \, x e^{x} - 10 \, x \log \relax (2) + 58 \, x}{5 \, x \log \relax (2) + 3 \, e^{x} \log \relax (2) - 30 \, x - 18 \, e^{x} - 10 \, \log \relax (2) + 60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-
580)/((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),
x, algorithm="giac")

[Out]

(5*x^2*log(2) + 3*x*e^x*log(2) - 30*x^2 - 18*x*e^x - 10*x*log(2) + 58*x)/(5*x*log(2) + 3*e^x*log(2) - 30*x - 1
8*e^x - 10*log(2) + 60)

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maple [A]  time = 0.11, size = 23, normalized size = 0.85

method	result	size
risch	\(x -\frac {2 x}{\left (\ln \relax (2)-6\right ) \left (-10+5 x +3 \,{\mathrm e}^{x}\right )}\)	\(23\)
norman	\(\frac {\frac {6 \left (5 \ln \relax (2)-29\right ) {\mathrm e}^{x}}{5 \left (\ln \relax (2)-6\right )}+5 x^{2}+3 \,{\mathrm e}^{x} x -\frac {4 \left (5 \ln \relax (2)-29\right )}{\ln \relax (2)-6}}{-10+5 x +3 \,{\mathrm e}^{x}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*ln(2)-54)*exp(x)^2+((30*x-60)*ln(2)-174*x+354)*exp(x)+(25*x^2-100*x+100)*ln(2)-150*x^2+600*x-580)/((9*
ln(2)-54)*exp(x)^2+((30*x-60)*ln(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*ln(2)-150*x^2+600*x-600),x,method=_RE
TURNVERBOSE)

[Out]

x-2*x/(ln(2)-6)/(-10+5*x+3*exp(x))

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maxima [B]  time = 0.47, size = 52, normalized size = 1.93 \begin {gather*} \frac {5 \, x^{2} {\left (\log \relax (2) - 6\right )} + 3 \, x {\left (\log \relax (2) - 6\right )} e^{x} - 2 \, x {\left (5 \, \log \relax (2) - 29\right )}}{5 \, x {\left (\log \relax (2) - 6\right )} + 3 \, {\left (\log \relax (2) - 6\right )} e^{x} - 10 \, \log \relax (2) + 60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-
580)/((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),
x, algorithm="maxima")

[Out]

(5*x^2*(log(2) - 6) + 3*x*(log(2) - 6)*e^x - 2*x*(5*log(2) - 29))/(5*x*(log(2) - 6) + 3*(log(2) - 6)*e^x - 10*
log(2) + 60)

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mupad [B]  time = 0.55, size = 54, normalized size = 2.00 \begin {gather*} -\frac {x^2\,\left (\ln \left (32\right )-30\right )-x\,\left (10\,\ln \relax (2)-58\right )+x\,{\mathrm {e}}^x\,\left (\ln \relax (8)-18\right )}{30\,x+10\,\ln \relax (2)+18\,{\mathrm {e}}^x-5\,x\,\ln \relax (2)-3\,{\mathrm {e}}^x\,\ln \relax (2)-60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((600*x + log(2)*(25*x^2 - 100*x + 100) + exp(x)*(log(2)*(30*x - 60) - 174*x + 354) + exp(2*x)*(9*log(2) -
54) - 150*x^2 - 580)/(600*x + log(2)*(25*x^2 - 100*x + 100) + exp(x)*(log(2)*(30*x - 60) - 180*x + 360) + exp(
2*x)*(9*log(2) - 54) - 150*x^2 - 600),x)

[Out]

-(x^2*(log(32) - 30) - x*(10*log(2) - 58) + x*exp(x)*(log(8) - 18))/(30*x + 10*log(2) + 18*exp(x) - 5*x*log(2)
- 3*exp(x)*log(2) - 60)

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sympy [A]  time = 0.17, size = 31, normalized size = 1.15 \begin {gather*} x - \frac {2 x}{- 30 x + 5 x \log {\relax (2 )} + \left (-18 + 3 \log {\relax (2 )}\right ) e^{x} - 10 \log {\relax (2 )} + 60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*ln(2)-54)*exp(x)**2+((30*x-60)*ln(2)-174*x+354)*exp(x)+(25*x**2-100*x+100)*ln(2)-150*x**2+600*x-
580)/((9*ln(2)-54)*exp(x)**2+((30*x-60)*ln(2)-180*x+360)*exp(x)+(25*x**2-100*x+100)*ln(2)-150*x**2+600*x-600),
x)

[Out]

x - 2*x/(-30*x + 5*x*log(2) + (-18 + 3*log(2))*exp(x) - 10*log(2) + 60)

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