3.35.29 \(\int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+(e^x (-220-240 x-20 x^2+(-44 x-48 x^2-4 x^3) \log (2))+e^x (44+48 x+4 x^2) \log (2) \log (11+x)) \log (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)})}{-55-60 x-5 x^2+(-11 x-12 x^2-x^3) \log (2)+(11+12 x+x^2) \log (2) \log (11+x)} \, dx\)

Optimal. Leaf size=26 \[ 4 e^x \log \left (\frac {1+x}{-x-\frac {5}{\log (2)}+\log (11+x)}\right ) \]

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Rubi [A]  time = 1.10, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6741, 6688, 12, 2288} \begin {gather*} 4 e^x \log \left (-\frac {(x+1) \log (2)}{x \log (2)-\log (2) \log (x+11)+5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + x] + (E^x*(-220 - 240*x - 20*x^2 + (-44*x
- 48*x^2 - 4*x^3)*Log[2]) + E^x*(44 + 48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log[2] +
Log[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log[2] + (11 + 12*x + x^2)*Log[2]*Log[11 +
x]),x]

[Out]

4*E^x*Log[-(((1 + x)*Log[2])/(5 + x*Log[2] - Log[2]*Log[11 + x]))]

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x (-220-20 x+40 \log (2))-e^x (44+4 x) \log (2) \log (11+x)-\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=\int \frac {4 e^x \left (5 (11+x-\log (4))-(11+x) \log (2) \log (11+x)+\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x)) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=4 \int \frac {e^x \left (5 (11+x-\log (4))-(11+x) \log (2) \log (11+x)+\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x)) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )\right )}{\left (11+12 x+x^2\right ) (5+x \log (2)-\log (2) \log (11+x))} \, dx\\ &=4 e^x \log \left (-\frac {(1+x) \log (2)}{5+x \log (2)-\log (2) \log (11+x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 28, normalized size = 1.08 \begin {gather*} 4 e^x \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + x] + (E^x*(-220 - 240*x - 20*x^2 + (
-44*x - 48*x^2 - 4*x^3)*Log[2]) + E^x*(44 + 48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log
[2] + Log[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log[2] + (11 + 12*x + x^2)*Log[2]*Log
[11 + x]),x]

[Out]

4*E^x*Log[((1 + x)*Log[2])/(-5 - x*Log[2] + Log[2]*Log[11 + x])]

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fricas [A]  time = 0.62, size = 28, normalized size = 1.08 \begin {gather*} 4 \, e^{x} \log \left (-\frac {{\left (x + 1\right )} \log \relax (2)}{x \log \relax (2) - \log \relax (2) \log \left (x + 11\right ) + 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)*log(2)-20*x^2-240*x-220)*exp(x))*log
((x+1)*log(2)/(log(2)*log(11+x)-x*log(2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x
^2+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, algorithm="fricas")

[Out]

4*e^x*log(-(x + 1)*log(2)/(x*log(2) - log(2)*log(x + 11) + 5))

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giac [A]  time = 0.75, size = 35, normalized size = 1.35 \begin {gather*} -4 \, e^{x} \log \left (x \log \relax (2) - \log \relax (2) \log \left (x + 11\right ) + 5\right ) + 4 \, e^{x} \log \left (-x \log \relax (2) - \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)*log(2)-20*x^2-240*x-220)*exp(x))*log
((x+1)*log(2)/(log(2)*log(11+x)-x*log(2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x
^2+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, algorithm="giac")

[Out]

-4*e^x*log(x*log(2) - log(2)*log(x + 11) + 5) + 4*e^x*log(-x*log(2) - log(2))

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maple [C]  time = 0.18, size = 241, normalized size = 9.27

method	result	size
risch	\(-4 \,{\mathrm e}^{x} \ln \left (5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)\right )-2 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) {\mathrm e}^{x}+2 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}-4 i \pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}+2 i \pi \,\mathrm {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{2} {\mathrm e}^{x}+2 i \pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \relax (2)}\right )^{3} {\mathrm e}^{x}+4 i \pi \,{\mathrm e}^{x}+4 \ln \left (\ln \relax (2)\right ) {\mathrm e}^{x}+4 \ln \left (x +1\right ) {\mathrm e}^{x}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x^3-48*x^2-44*x)*ln(2)-20*x^2-240*x-220)*exp(x))*ln((x+1)*ln(
2)/(ln(2)*ln(11+x)-x*ln(2)-5))+(4*x+44)*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x^2+12*x+11)*ln(2)
*ln(11+x)+(-x^3-12*x^2-11*x)*ln(2)-5*x^2-60*x-55),x,method=_RETURNVERBOSE)

[Out]

-4*exp(x)*ln(5+(x-ln(11+x))*ln(2))-2*I*Pi*csgn(I*(x+1))*csgn(I/(5+(x-ln(11+x))*ln(2)))*csgn(I/(5+(x-ln(11+x))*
ln(2))*(x+1))*exp(x)+2*I*Pi*csgn(I*(x+1))*csgn(I/(5+(x-ln(11+x))*ln(2))*(x+1))^2*exp(x)-4*I*Pi*csgn(I/(5+(x-ln
(11+x))*ln(2))*(x+1))^2*exp(x)+2*I*Pi*csgn(I/(5+(x-ln(11+x))*ln(2)))*csgn(I/(5+(x-ln(11+x))*ln(2))*(x+1))^2*ex
p(x)+2*I*Pi*csgn(I/(5+(x-ln(11+x))*ln(2))*(x+1))^3*exp(x)+4*I*Pi*exp(x)+4*ln(ln(2))*exp(x)+4*ln(x+1)*exp(x)

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maxima [A]  time = 0.76, size = 35, normalized size = 1.35 \begin {gather*} -4 \, e^{x} \log \left (-x \log \relax (2) + \log \relax (2) \log \left (x + 11\right ) - 5\right ) + 4 \, e^{x} \log \left (x + 1\right ) + 4 \, e^{x} \log \left (\log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)*log(2)-20*x^2-240*x-220)*exp(x))*log
((x+1)*log(2)/(log(2)*log(11+x)-x*log(2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x
^2+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, algorithm="maxima")

[Out]

-4*e^x*log(-x*log(2) + log(2)*log(x + 11) - 5) + 4*e^x*log(x + 1) + 4*e^x*log(log(2))

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mupad [B]  time = 0.87, size = 28, normalized size = 1.08 \begin {gather*} 4\,\ln \left (-\frac {\ln \relax (2)\,\left (x+1\right )}{x\,\ln \relax (2)-\ln \left (x+11\right )\,\ln \relax (2)+5}\right )\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(20*x - 40*log(2) + 220) + log(-(log(2)*(x + 1))/(x*log(2) - log(x + 11)*log(2) + 5))*(exp(x)*(240
*x + log(2)*(44*x + 48*x^2 + 4*x^3) + 20*x^2 + 220) - log(x + 11)*exp(x)*log(2)*(48*x + 4*x^2 + 44)) - log(x +
11)*exp(x)*log(2)*(4*x + 44))/(60*x + log(2)*(11*x + 12*x^2 + x^3) + 5*x^2 - log(x + 11)*log(2)*(12*x + x^2 +
11) + 55),x)

[Out]

4*log(-(log(2)*(x + 1))/(x*log(2) - log(x + 11)*log(2) + 5))*exp(x)

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sympy [A]  time = 27.47, size = 27, normalized size = 1.04 \begin {gather*} 4 e^{x} \log {\left (\frac {\left (x + 1\right ) \log {\relax (2 )}}{- x \log {\relax (2 )} + \log {\relax (2 )} \log {\left (x + 11 \right )} - 5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x**3-48*x**2-44*x)*ln(2)-20*x**2-240*x-220)*exp(x))*ln
((x+1)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5))+(4*x+44)*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x**2+12*
x+11)*ln(2)*ln(11+x)+(-x**3-12*x**2-11*x)*ln(2)-5*x**2-60*x-55),x)

[Out]

4*exp(x)*log((x + 1)*log(2)/(-x*log(2) + log(2)*log(x + 11) - 5))

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