3.25.51 \(\int \frac {-32 x^3+e^x (3-3 x-80 x^2) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2))}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx\)

Optimal. Leaf size=33 \[ 4-e^{4 x}-x^2-\frac {3}{-5+\frac {1}{1+\frac {e^x \log (2)}{x}}} \]

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Rubi [F]  time = 1.67, 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 {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{16 x^2+40 e^x x \log (2)+25 e^{2 x} \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-32*x^3 + E^x*(3 - 3*x - 80*x^2)*Log[2] - 50*E^(2*x)*x*Log[2]^2 + E^(4*x)*(-64*x^2 - 160*E^x*x*Log[2] - 1
00*E^(2*x)*Log[2]^2))/(16*x^2 + 40*E^x*x*Log[2] + 25*E^(2*x)*Log[2]^2),x]

[Out]

(-25*E^(4*x)*Log[2]^2)/Log[32]^2 - (25*x^2*Log[2]^2)/Log[32]^2 - (4*Log[8]*Defer[Int][x/(4*x + E^x*Log[32])^2,
x])/Log[32] + (4*Log[8]*Defer[Int][x^2/(4*x + E^x*Log[32])^2, x])/Log[32] + (Log[8]*Defer[Int][(4*x + E^x*Log
[32])^(-1), x])/Log[32] - (Log[8]*Defer[Int][x/(4*x + E^x*Log[32]), x])/Log[32] + (8192*(75*Log[2]^2 - Log[32]
*Log[32768])*Defer[Int][x^5/(4*x + E^x*Log[32]), x])/Log[32]^6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32 x^3+e^x \left (3-3 x-80 x^2\right ) \log (2)-50 e^{2 x} x \log ^2(2)+e^{4 x} \left (-64 x^2-160 e^x x \log (2)-100 e^{2 x} \log ^2(2)\right )}{\left (4 x+e^x \log (32)\right )^2} \, dx\\ &=\int \left (-\frac {100 e^{4 x} \log ^2(2)}{\log ^2(32)}+\frac {4 (-1+x) x \log (8)}{\log (32) \left (4 x+e^x \log (32)\right )^2}-\frac {64 e^{2 x} x^2 \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )}{\log ^4(32)}+\frac {512 e^x x^3 \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )}{\log ^5(32)}+\frac {2 x \left (-512 x^3 (25 \log (2)-3 \log (32)) (5 \log (2)-\log (32))-25 \log ^2(2) \log ^4(32)\right )}{\log ^6(32)}+\frac {\log (8) \log ^5(32)-x \log (8) \log ^5(32)+614400 x^5 \log ^2(2) \left (1-\frac {\log (32) \log (32768)}{75 \log ^2(2)}\right )}{\log ^6(32) \left (4 x+e^x \log (32)\right )}\right ) \, dx\\ &=\frac {\int \frac {\log (8) \log ^5(32)-x \log (8) \log ^5(32)+614400 x^5 \log ^2(2) \left (1-\frac {\log (32) \log (32768)}{75 \log ^2(2)}\right )}{4 x+e^x \log (32)} \, dx}{\log ^6(32)}+\frac {2 \int x \left (-512 x^3 (25 \log (2)-3 \log (32)) (5 \log (2)-\log (32))-25 \log ^2(2) \log ^4(32)\right ) \, dx}{\log ^6(32)}-\frac {\left (100 \log ^2(2)\right ) \int e^{4 x} \, dx}{\log ^2(32)}+\frac {(4 \log (8)) \int \frac {(-1+x) x}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}-\frac {\left (64 \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^{2 x} x^2 \, dx}{\log ^4(32)}+\frac {\left (512 \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^x x^3 \, dx}{\log ^5(32)}\\ &=\frac {512 e^x x^3 (5 \log (2)-\log (32))}{\log ^4(32)}-\frac {32 e^{2 x} x^2 (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {25 e^{4 x} \log ^2(2)}{\log ^2(32)}+\frac {\int \left (\frac {\log (8) \log ^5(32)}{4 x+e^x \log (32)}-\frac {x \log (8) \log ^5(32)}{4 x+e^x \log (32)}-\frac {8192 x^5 \left (-75 \log ^2(2)+\log (32) \log (32768)\right )}{4 x+e^x \log (32)}\right ) \, dx}{\log ^6(32)}+\frac {2 \int -25 x \log ^2(2) \log ^4(32) \, dx}{\log ^6(32)}+\frac {(4 \log (8)) \int \left (-\frac {x}{\left (4 x+e^x \log (32)\right )^2}+\frac {x^2}{\left (4 x+e^x \log (32)\right )^2}\right ) \, dx}{\log (32)}+\frac {\left (64 \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^{2 x} x \, dx}{\log ^4(32)}-\frac {\left (1536 \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^x x^2 \, dx}{\log ^5(32)}\\ &=-\frac {1536 e^x x^2 (5 \log (2)-\log (32))}{\log ^4(32)}+\frac {512 e^x x^3 (5 \log (2)-\log (32))}{\log ^4(32)}+\frac {32 e^{2 x} x (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {32 e^{2 x} x^2 (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {25 e^{4 x} \log ^2(2)}{\log ^2(32)}-\frac {\left (50 \log ^2(2)\right ) \int x \, dx}{\log ^2(32)}+\frac {\log (8) \int \frac {1}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {\log (8) \int \frac {x}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {(4 \log (8)) \int \frac {x}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}+\frac {(4 \log (8)) \int \frac {x^2}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}-\frac {\left (32 \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^{2 x} \, dx}{\log ^4(32)}+\frac {\left (3072 \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^x x \, dx}{\log ^5(32)}+\frac {\left (8192 \left (75 \log ^2(2)-\log (32) \log (32768)\right )\right ) \int \frac {x^5}{4 x+e^x \log (32)} \, dx}{\log ^6(32)}\\ &=\frac {3072 e^x x (5 \log (2)-\log (32))}{\log ^4(32)}-\frac {1536 e^x x^2 (5 \log (2)-\log (32))}{\log ^4(32)}+\frac {512 e^x x^3 (5 \log (2)-\log (32))}{\log ^4(32)}-\frac {16 e^{2 x} (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}+\frac {32 e^{2 x} x (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {32 e^{2 x} x^2 (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {25 e^{4 x} \log ^2(2)}{\log ^2(32)}-\frac {25 x^2 \log ^2(2)}{\log ^2(32)}+\frac {\log (8) \int \frac {1}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {\log (8) \int \frac {x}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {(4 \log (8)) \int \frac {x}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}+\frac {(4 \log (8)) \int \frac {x^2}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}-\frac {\left (3072 \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )\right ) \int e^x \, dx}{\log ^5(32)}+\frac {\left (8192 \left (75 \log ^2(2)-\log (32) \log (32768)\right )\right ) \int \frac {x^5}{4 x+e^x \log (32)} \, dx}{\log ^6(32)}\\ &=-\frac {3072 e^x (5 \log (2)-\log (32))}{\log ^4(32)}+\frac {3072 e^x x (5 \log (2)-\log (32))}{\log ^4(32)}-\frac {1536 e^x x^2 (5 \log (2)-\log (32))}{\log ^4(32)}+\frac {512 e^x x^3 (5 \log (2)-\log (32))}{\log ^4(32)}-\frac {16 e^{2 x} (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}+\frac {32 e^{2 x} x (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {32 e^{2 x} x^2 (5 \log (2)-\log (32)) (15 \log (2)-\log (32))}{\log ^4(32)}-\frac {25 e^{4 x} \log ^2(2)}{\log ^2(32)}-\frac {25 x^2 \log ^2(2)}{\log ^2(32)}+\frac {\log (8) \int \frac {1}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {\log (8) \int \frac {x}{4 x+e^x \log (32)} \, dx}{\log (32)}-\frac {(4 \log (8)) \int \frac {x}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}+\frac {(4 \log (8)) \int \frac {x^2}{\left (4 x+e^x \log (32)\right )^2} \, dx}{\log (32)}+\frac {\left (8192 \left (75 \log ^2(2)-\log (32) \log (32768)\right )\right ) \int \frac {x^5}{4 x+e^x \log (32)} \, dx}{\log ^6(32)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.17, size = 157, normalized size = 4.76 \begin {gather*} -\frac {125 e^{4 x} \log ^2(2) \log ^4(32)+125 x^2 \log ^2(2) \log ^4(32)-\frac {5 x \log (8) \log ^5(32)}{4 x+e^x \log (32)}+80 e^{2 x} \left (1-2 x+2 x^2\right ) \log ^2(32) \left (75 \log ^2(2)-20 \log (2) \log (32)+\log ^2(32)\right )-2560 e^x \left (-6+6 x-3 x^2+x^3\right ) \log (32) \left (50 \log ^2(2)-15 \log (2) \log (32)+\log ^2(32)\right )+1024 x^5 \left (125 \log ^2(2)-40 \log (2) \log (32)+3 \log ^2(32)\right )}{5 \log ^6(32)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x^3 + E^x*(3 - 3*x - 80*x^2)*Log[2] - 50*E^(2*x)*x*Log[2]^2 + E^(4*x)*(-64*x^2 - 160*E^x*x*Log[
2] - 100*E^(2*x)*Log[2]^2))/(16*x^2 + 40*E^x*x*Log[2] + 25*E^(2*x)*Log[2]^2),x]

[Out]

-1/5*(125*E^(4*x)*Log[2]^2*Log[32]^4 + 125*x^2*Log[2]^2*Log[32]^4 - (5*x*Log[8]*Log[32]^5)/(4*x + E^x*Log[32])
+ 80*E^(2*x)*(1 - 2*x + 2*x^2)*Log[32]^2*(75*Log[2]^2 - 20*Log[2]*Log[32] + Log[32]^2) - 2560*E^x*(-6 + 6*x -
3*x^2 + x^3)*Log[32]*(50*Log[2]^2 - 15*Log[2]*Log[32] + Log[32]^2) + 1024*x^5*(125*Log[2]^2 - 40*Log[2]*Log[3
2] + 3*Log[32]^2))/Log[32]^6

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fricas [A]  time = 0.54, size = 47, normalized size = 1.42 \begin {gather*} -\frac {25 \, x^{2} e^{x} \log \relax (2) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \relax (2) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \relax (2) + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50*x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)
*log(2)*exp(x)-32*x^3)/(25*log(2)^2*exp(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="fricas")

[Out]

-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x)/(5*e^x*log(2) + 4*x)

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giac [A]  time = 0.34, size = 47, normalized size = 1.42 \begin {gather*} -\frac {25 \, x^{2} e^{x} \log \relax (2) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \relax (2) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \relax (2) + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50*x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)
*log(2)*exp(x)-32*x^3)/(25*log(2)^2*exp(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="giac")

[Out]

-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x)/(5*e^x*log(2) + 4*x)

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maple [A]  time = 0.14, size = 28, normalized size = 0.85

method	result	size
risch	\(-{\mathrm e}^{4 x}-x^{2}+\frac {3 x}{5 \left (5 \,{\mathrm e}^{x} \ln \relax (2)+4 x \right )}\)	\(28\)
norman	\(\frac {-\frac {3 \,{\mathrm e}^{x} \ln \relax (2)}{4}-4 x^{3}-4 x \,{\mathrm e}^{4 x}-5 \ln \relax (2) {\mathrm e}^{5 x}-5 x^{2} \ln \relax (2) {\mathrm e}^{x}}{5 \,{\mathrm e}^{x} \ln \relax (2)+4 x}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-100*ln(2)^2*exp(x)^2-160*x*ln(2)*exp(x)-64*x^2)*exp(4*x)-50*x*ln(2)^2*exp(x)^2+(-80*x^2-3*x+3)*ln(2)*ex
p(x)-32*x^3)/(25*ln(2)^2*exp(x)^2+40*x*ln(2)*exp(x)+16*x^2),x,method=_RETURNVERBOSE)

[Out]

-exp(4*x)-x^2+3/5*x/(5*exp(x)*ln(2)+4*x)

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maxima [A]  time = 0.59, size = 47, normalized size = 1.42 \begin {gather*} -\frac {25 \, x^{2} e^{x} \log \relax (2) + 20 \, x^{3} + 20 \, x e^{\left (4 \, x\right )} + 25 \, e^{\left (5 \, x\right )} \log \relax (2) - 3 \, x}{5 \, {\left (5 \, e^{x} \log \relax (2) + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*log(2)^2*exp(x)^2-160*x*log(2)*exp(x)-64*x^2)*exp(4*x)-50*x*log(2)^2*exp(x)^2+(-80*x^2-3*x+3)
*log(2)*exp(x)-32*x^3)/(25*log(2)^2*exp(x)^2+40*x*log(2)*exp(x)+16*x^2),x, algorithm="maxima")

[Out]

-1/5*(25*x^2*e^x*log(2) + 20*x^3 + 20*x*e^(4*x) + 25*e^(5*x)*log(2) - 3*x)/(5*e^x*log(2) + 4*x)

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mupad [B]  time = 1.46, size = 47, normalized size = 1.42 \begin {gather*} -\frac {5\,{\mathrm {e}}^{5\,x}\,\ln \left (32\right )-3\,x+20\,x\,{\mathrm {e}}^{4\,x}+20\,x^3+5\,x^2\,{\mathrm {e}}^x\,\ln \left (32\right )}{5\,\left (4\,x+5\,{\mathrm {e}}^x\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4*x)*(100*exp(2*x)*log(2)^2 + 64*x^2 + 160*x*exp(x)*log(2)) + 32*x^3 + exp(x)*log(2)*(3*x + 80*x^2 -
3) + 50*x*exp(2*x)*log(2)^2)/(25*exp(2*x)*log(2)^2 + 16*x^2 + 40*x*exp(x)*log(2)),x)

[Out]

-(5*exp(5*x)*log(32) - 3*x + 20*x*exp(4*x) + 20*x^3 + 5*x^2*exp(x)*log(32))/(5*(4*x + 5*exp(x)*log(2)))

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sympy [A]  time = 0.16, size = 22, normalized size = 0.67 \begin {gather*} - x^{2} + \frac {3 x}{20 x + 25 e^{x} \log {\relax (2 )}} - e^{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*ln(2)**2*exp(x)**2-160*x*ln(2)*exp(x)-64*x**2)*exp(4*x)-50*x*ln(2)**2*exp(x)**2+(-80*x**2-3*x
+3)*ln(2)*exp(x)-32*x**3)/(25*ln(2)**2*exp(x)**2+40*x*ln(2)*exp(x)+16*x**2),x)

[Out]

-x**2 + 3*x/(20*x + 25*exp(x)*log(2)) - exp(4*x)

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