3.64.46 \(\int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+(32 e^{x/4}+32 e^{x/2}) \log (x^4)+(4 e^{x/2}+e^{x/4} (4+x)) \log ^2(x^4)}{1+2 e^{x/4}+e^{x/2}} \, dx\) [6346]

Optimal. Leaf size=26 \[ 4 \left (x-\frac {x}{1+e^{x/4}}\right ) \left (25+\log ^2\left (x^4\right )\right ) \]

[Out]

4*(x-x/(exp(1/4*x)+1))*(ln(x^4)^2+25)

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Rubi [F]
time = 1.20, 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 {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100*E^(x/2) + E^(x/4)*(100 + 25*x) + (32*E^(x/4) + 32*E^(x/2))*Log[x^4] + (4*E^(x/2) + E^(x/4)*(4 + x))*L
og[x^4]^2)/(1 + 2*E^(x/4) + E^(x/2)),x]

[Out]

100*x - (100*x)/(1 + E^(x/4)) + 128*Defer[Subst][Defer[Int][(E^x*Log[256*x^4])/(1 + E^x), x], x, x/4] + 16*Def
er[Subst][Defer[Int][(E^x*Log[256*x^4]^2)/(1 + E^x), x], x, x/4] + 16*Defer[Subst][Defer[Int][(E^x*x*Log[256*x
^4]^2)/(1 + E^x)^2, x], x, x/4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{x/4} \left (25 \left (4+4 e^{x/4}+x\right )+32 \left (1+e^{x/4}\right ) \log \left (x^4\right )+\left (4+4 e^{x/4}+x\right ) \log ^2\left (x^4\right )\right )}{\left (1+e^{x/4}\right )^2} \, dx\\ &=4 \text {Subst}\left (\int \frac {e^x \left (25 \left (4+4 e^x+4 x\right )+32 \left (1+e^x\right ) \log \left (256 x^4\right )+\left (4+4 e^x+4 x\right ) \log ^2\left (256 x^4\right )\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )\\ &=4 \text {Subst}\left (\int \left (\frac {4 e^x x \left (25+\log ^2\left (256 x^4\right )\right )}{\left (1+e^x\right )^2}+\frac {4 e^x \left (25+8 \log \left (256 x^4\right )+\log ^2\left (256 x^4\right )\right )}{1+e^x}\right ) \, dx,x,\frac {x}{4}\right )\\ &=16 \text {Subst}\left (\int \frac {e^x x \left (25+\log ^2\left (256 x^4\right )\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x \left (25+8 \log \left (256 x^4\right )+\log ^2\left (256 x^4\right )\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )\\ &=16 \text {Subst}\left (\int \left (\frac {25 e^x}{1+e^x}+\frac {8 e^x \log \left (256 x^4\right )}{1+e^x}+\frac {e^x \log ^2\left (256 x^4\right )}{1+e^x}\right ) \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \left (\frac {25 e^x x}{\left (1+e^x\right )^2}+\frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2}\right ) \, dx,x,\frac {x}{4}\right )\\ &=16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {e^x}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {e^x x}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )\\ &=-\frac {100 x}{1+e^{x/4}}+16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {1}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^{x/4}\right )\\ &=-\frac {100 x}{1+e^{x/4}}+400 \log \left (1+e^{x/4}\right )+16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,e^{x/4}\right )\\ &=-\frac {100 x}{1+e^{x/4}}+400 \log \left (1+e^{x/4}\right )+16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+400 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{x/4}\right )-400 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^{x/4}\right )\\ &=100 x-\frac {100 x}{1+e^{x/4}}+16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )+16 \text {Subst}\left (\int \frac {e^x x \log ^2\left (256 x^4\right )}{\left (1+e^x\right )^2} \, dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x} \, dx,x,\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.24, size = 29, normalized size = 1.12 \begin {gather*} \frac {4 e^{x/4} x \left (25+\log ^2\left (x^4\right )\right )}{1+e^{x/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100*E^(x/2) + E^(x/4)*(100 + 25*x) + (32*E^(x/4) + 32*E^(x/2))*Log[x^4] + (4*E^(x/2) + E^(x/4)*(4 +
x))*Log[x^4]^2)/(1 + 2*E^(x/4) + E^(x/2)),x]

[Out]

(4*E^(x/4)*x*(25 + Log[x^4]^2))/(1 + E^(x/4))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(23)=46\).
time = 1.10, size = 61, normalized size = 2.35

method	result	size
norman	\(\frac {100 x \,{\mathrm e}^{\frac {x}{4}}+4 \ln \left (x^{4}\right )^{2} x \,{\mathrm e}^{\frac {x}{4}}}{{\mathrm e}^{\frac {x}{4}}+1}\)	\(31\)
default	\(\frac {\left (4 \left (\ln \left (x^{4}\right )-4 \ln \left (x \right )\right )^{2}+100\right ) x \,{\mathrm e}^{\frac {x}{4}}+64 \,{\mathrm e}^{\frac {x}{4}} \ln \left (x \right )^{2} x +32 \,{\mathrm e}^{\frac {x}{4}} \ln \left (x \right ) x \left (\ln \left (x^{4}\right )-4 \ln \left (x \right )\right )}{{\mathrm e}^{\frac {x}{4}}+1}\)	\(61\)
risch	\(\text {Expression too large to display}\)	\(1515\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*ln(x^4)^2+(32*exp(1/4*x)^2+32*exp(1/4*x))*ln(x^4)+100*exp(1/4*x)^2+(25*
x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*exp(1/4*x)+1),x,method=_RETURNVERBOSE)

[Out]

((4*(ln(x^4)-4*ln(x))^2+100)*x*exp(1/4*x)+64*exp(1/4*x)*ln(x)^2*x+32*exp(1/4*x)*ln(x)*x*(ln(x^4)-4*ln(x)))/(ex
p(1/4*x)+1)

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Maxima [A]
time = 0.33, size = 35, normalized size = 1.35 \begin {gather*} \frac {64 \, x e^{\left (\frac {1}{4} \, x\right )} \log \left (x\right )^{2}}{e^{\left (\frac {1}{4} \, x\right )} + 1} + \frac {100 \, x e^{\left (\frac {1}{4} \, x\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+32*exp(1/4*x))*log(x^4)+100*exp(1/4*x
)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*exp(1/4*x)+1),x, algorithm="maxima")

[Out]

64*x*e^(1/4*x)*log(x)^2/(e^(1/4*x) + 1) + 100*x*e^(1/4*x)/(e^(1/4*x) + 1)

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Fricas [A]
time = 0.36, size = 30, normalized size = 1.15 \begin {gather*} \frac {4 \, {\left (x e^{\left (\frac {1}{4} \, x\right )} \log \left (x^{4}\right )^{2} + 25 \, x e^{\left (\frac {1}{4} \, x\right )}\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+32*exp(1/4*x))*log(x^4)+100*exp(1/4*x
)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*exp(1/4*x)+1),x, algorithm="fricas")

[Out]

4*(x*e^(1/4*x)*log(x^4)^2 + 25*x*e^(1/4*x))/(e^(1/4*x) + 1)

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Sympy [A]
time = 0.07, size = 34, normalized size = 1.31 \begin {gather*} 4 x \log {\left (x^{4} \right )}^{2} + 100 x + \frac {- 4 x \log {\left (x^{4} \right )}^{2} - 100 x}{e^{\frac {x}{4}} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1/4*x)**2+(4+x)*exp(1/4*x))*ln(x**4)**2+(32*exp(1/4*x)**2+32*exp(1/4*x))*ln(x**4)+100*exp(1/
4*x)**2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)**2+2*exp(1/4*x)+1),x)

[Out]

4*x*log(x**4)**2 + 100*x + (-4*x*log(x**4)**2 - 100*x)/(exp(x/4) + 1)

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Giac [A]
time = 0.41, size = 30, normalized size = 1.15 \begin {gather*} \frac {4 \, {\left (x e^{\left (\frac {1}{4} \, x\right )} \log \left (x^{4}\right )^{2} + 25 \, x e^{\left (\frac {1}{4} \, x\right )}\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+32*exp(1/4*x))*log(x^4)+100*exp(1/4*x
)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*exp(1/4*x)+1),x, algorithm="giac")

[Out]

4*(x*e^(1/4*x)*log(x^4)^2 + 25*x*e^(1/4*x))/(e^(1/4*x) + 1)

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Mupad [B]
time = 4.57, size = 23, normalized size = 0.88 \begin {gather*} \frac {4\,x\,{\mathrm {e}}^{x/4}\,\left ({\ln \left (x^4\right )}^2+25\right )}{{\mathrm {e}}^{x/4}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*exp(x/2) + log(x^4)*(32*exp(x/2) + 32*exp(x/4)) + log(x^4)^2*(4*exp(x/2) + exp(x/4)*(x + 4)) + exp(x/
4)*(25*x + 100))/(exp(x/2) + 2*exp(x/4) + 1),x)

[Out]

(4*x*exp(x/4)*(log(x^4)^2 + 25))/(exp(x/4) + 1)

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