3.39.56 \(\int \frac {4 e^{20+4 x}+96 x+16 e^{15+3 x} x-140 x^2-8 x^3+4 x^4+e^{10+2 x} (-48-104 x+24 x^2)+e^{5+x} (-192 x-112 x^2+16 x^3)+(-4 e^{20+4 x}-16 e^{15+3 x} x-4 x^2+8 x^3-4 x^4+e^{10+2 x} (8 x-24 x^2)+e^{5+x} (16 x^2-16 x^3)) \log (x)}{e^{20+4 x} x^2+4 e^{15+3 x} x^3+x^4-2 x^5+x^6+e^{10+2 x} (-2 x^3+6 x^4)+e^{5+x} (-4 x^4+4 x^5)} \, dx\)

Optimal. Leaf size=27 \[ 6+\frac {4 \left (-\frac {12}{x-\left (e^{5+x}+x\right )^2}+\log (x)\right )}{x} \]

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Rubi [F]  time = 3.78, 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 {4 e^{20+4 x}+96 x+16 e^{15+3 x} x-140 x^2-8 x^3+4 x^4+e^{10+2 x} \left (-48-104 x+24 x^2\right )+e^{5+x} \left (-192 x-112 x^2+16 x^3\right )+\left (-4 e^{20+4 x}-16 e^{15+3 x} x-4 x^2+8 x^3-4 x^4+e^{10+2 x} \left (8 x-24 x^2\right )+e^{5+x} \left (16 x^2-16 x^3\right )\right ) \log (x)}{e^{20+4 x} x^2+4 e^{15+3 x} x^3+x^4-2 x^5+x^6+e^{10+2 x} \left (-2 x^3+6 x^4\right )+e^{5+x} \left (-4 x^4+4 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*E^(20 + 4*x) + 96*x + 16*E^(15 + 3*x)*x - 140*x^2 - 8*x^3 + 4*x^4 + E^(10 + 2*x)*(-48 - 104*x + 24*x^2)
+ E^(5 + x)*(-192*x - 112*x^2 + 16*x^3) + (-4*E^(20 + 4*x) - 16*E^(15 + 3*x)*x - 4*x^2 + 8*x^3 - 4*x^4 + E^(1
0 + 2*x)*(8*x - 24*x^2) + E^(5 + x)*(16*x^2 - 16*x^3))*Log[x])/(E^(20 + 4*x)*x^2 + 4*E^(15 + 3*x)*x^3 + x^4 -
2*x^5 + x^6 + E^(10 + 2*x)*(-2*x^3 + 6*x^4) + E^(5 + x)*(-4*x^4 + 4*x^5)),x]

[Out]

(4*Log[x])/x - 192*Defer[Int][(E^(10 + 2*x) - x + 2*E^(5 + x)*x + x^2)^(-2), x] + 96*Defer[Int][E^(5 + x)/(E^(
10 + 2*x) - x + 2*E^(5 + x)*x + x^2)^2, x] + 48*Defer[Int][1/(x*(E^(10 + 2*x) - x + 2*E^(5 + x)*x + x^2)^2), x
] - 96*Defer[Int][E^(5 + x)/(x*(E^(10 + 2*x) - x + 2*E^(5 + x)*x + x^2)^2), x] + 96*Defer[Int][x/(E^(10 + 2*x)
- x + 2*E^(5 + x)*x + x^2)^2, x] - 48*Defer[Int][1/(x^2*(E^(10 + 2*x) - x + 2*E^(5 + x)*x + x^2)), x] - 96*De
fer[Int][1/(x*(E^(10 + 2*x) - x + 2*E^(5 + x)*x + x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^{4 (5+x)}+16 e^{3 (5+x)} x+16 e^{5+x} x \left (-12-7 x+x^2\right )+8 e^{2 (5+x)} \left (-6-13 x+3 x^2\right )+4 x \left (24-35 x-2 x^2+x^3\right )-4 \left (e^{2 (5+x)}+2 e^{5+x} x+(-1+x) x\right )^2 \log (x)}{x^2 \left (e^{2 (5+x)}+2 e^{5+x} x+(-1+x) x\right )^2} \, dx\\ &=\int \left (-\frac {48 (1+2 x)}{x^2 \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )}+\frac {48 \left (1-2 e^{5+x}-4 x+2 e^{5+x} x+2 x^2\right )}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}-\frac {4 (-1+\log (x))}{x^2}\right ) \, dx\\ &=-\left (4 \int \frac {-1+\log (x)}{x^2} \, dx\right )-48 \int \frac {1+2 x}{x^2 \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )} \, dx+48 \int \frac {1-2 e^{5+x}-4 x+2 e^{5+x} x+2 x^2}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx\\ &=\frac {4 \log (x)}{x}+48 \int \left (-\frac {4}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}+\frac {2 e^{5+x}}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}+\frac {1}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}-\frac {2 e^{5+x}}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}+\frac {2 x}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2}\right ) \, dx-48 \int \left (\frac {1}{x^2 \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )}+\frac {2}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )}\right ) \, dx\\ &=\frac {4 \log (x)}{x}+48 \int \frac {1}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx-48 \int \frac {1}{x^2 \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )} \, dx+96 \int \frac {e^{5+x}}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx-96 \int \frac {e^{5+x}}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx+96 \int \frac {x}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx-96 \int \frac {1}{x \left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )} \, dx-192 \int \frac {1}{\left (e^{10+2 x}-x+2 e^{5+x} x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \left (\frac {12}{e^{2 (5+x)}+2 e^{5+x} x+(-1+x) x}+\log (x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(20 + 4*x) + 96*x + 16*E^(15 + 3*x)*x - 140*x^2 - 8*x^3 + 4*x^4 + E^(10 + 2*x)*(-48 - 104*x + 2
4*x^2) + E^(5 + x)*(-192*x - 112*x^2 + 16*x^3) + (-4*E^(20 + 4*x) - 16*E^(15 + 3*x)*x - 4*x^2 + 8*x^3 - 4*x^4
+ E^(10 + 2*x)*(8*x - 24*x^2) + E^(5 + x)*(16*x^2 - 16*x^3))*Log[x])/(E^(20 + 4*x)*x^2 + 4*E^(15 + 3*x)*x^3 +
x^4 - 2*x^5 + x^6 + E^(10 + 2*x)*(-2*x^3 + 6*x^4) + E^(5 + x)*(-4*x^4 + 4*x^5)),x]

[Out]

(4*(12/(E^(2*(5 + x)) + 2*E^(5 + x)*x + (-1 + x)*x) + Log[x]))/x

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fricas [B]  time = 0.73, size = 55, normalized size = 2.04 \begin {gather*} \frac {4 \, {\left ({\left (x^{2} + 2 \, x e^{\left (x + 5\right )} - x + e^{\left (2 \, x + 10\right )}\right )} \log \relax (x) + 12\right )}}{x^{3} + 2 \, x^{2} e^{\left (x + 5\right )} - x^{2} + x e^{\left (2 \, x + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(5+x)^4-16*x*exp(5+x)^3+(-24*x^2+8*x)*exp(5+x)^2+(-16*x^3+16*x^2)*exp(5+x)-4*x^4+8*x^3-4*x^2
)*log(x)+4*exp(5+x)^4+16*x*exp(5+x)^3+(24*x^2-104*x-48)*exp(5+x)^2+(16*x^3-112*x^2-192*x)*exp(5+x)+4*x^4-8*x^3
-140*x^2+96*x)/(x^2*exp(5+x)^4+4*x^3*exp(5+x)^3+(6*x^4-2*x^3)*exp(5+x)^2+(4*x^5-4*x^4)*exp(5+x)+x^6-2*x^5+x^4)
,x, algorithm="fricas")

[Out]

4*((x^2 + 2*x*e^(x + 5) - x + e^(2*x + 10))*log(x) + 12)/(x^3 + 2*x^2*e^(x + 5) - x^2 + x*e^(2*x + 10))

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giac [B]  time = 0.27, size = 284, normalized size = 10.52 \begin {gather*} \frac {4 \, {\left (2 \, x^{5} \log \relax (x) + 6 \, x^{4} e^{\left (x + 5\right )} \log \relax (x) + 2 \, x^{5} + 2 \, x^{4} e^{\left (x + 5\right )} - 12 \, x^{4} \log \relax (x) + 4 \, x^{3} e^{\left (2 \, x + 10\right )} \log \relax (x) - 18 \, x^{3} e^{\left (x + 5\right )} \log \relax (x) - 4 \, x^{4} - 6 \, x^{3} e^{\left (x + 5\right )} + 19 \, x^{3} \log \relax (x) - 12 \, x^{2} e^{\left (2 \, x + 10\right )} \log \relax (x) + 10 \, x^{2} e^{\left (x + 5\right )} \log \relax (x) - 23 \, x^{3} - 66 \, x^{2} e^{\left (x + 5\right )} - 11 \, x^{2} \log \relax (x) + 8 \, x e^{\left (2 \, x + 10\right )} \log \relax (x) - 94 \, x^{2} + 118 \, x e^{\left (x + 5\right )} + 2 \, x \log \relax (x) - e^{\left (2 \, x + 10\right )} \log \relax (x) + 131 \, x - 48 \, e^{\left (x + 5\right )} - 36\right )}}{4 \, x^{6} + 8 \, x^{5} e^{\left (x + 5\right )} - 16 \, x^{5} + 4 \, x^{4} e^{\left (2 \, x + 10\right )} - 24 \, x^{4} e^{\left (x + 5\right )} + 20 \, x^{4} - 12 \, x^{3} e^{\left (2 \, x + 10\right )} + 16 \, x^{3} e^{\left (x + 5\right )} - 9 \, x^{3} + 8 \, x^{2} e^{\left (2 \, x + 10\right )} - 2 \, x^{2} e^{\left (x + 5\right )} + x^{2} - x e^{\left (2 \, x + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(5+x)^4-16*x*exp(5+x)^3+(-24*x^2+8*x)*exp(5+x)^2+(-16*x^3+16*x^2)*exp(5+x)-4*x^4+8*x^3-4*x^2
)*log(x)+4*exp(5+x)^4+16*x*exp(5+x)^3+(24*x^2-104*x-48)*exp(5+x)^2+(16*x^3-112*x^2-192*x)*exp(5+x)+4*x^4-8*x^3
-140*x^2+96*x)/(x^2*exp(5+x)^4+4*x^3*exp(5+x)^3+(6*x^4-2*x^3)*exp(5+x)^2+(4*x^5-4*x^4)*exp(5+x)+x^6-2*x^5+x^4)
,x, algorithm="giac")

[Out]

4*(2*x^5*log(x) + 6*x^4*e^(x + 5)*log(x) + 2*x^5 + 2*x^4*e^(x + 5) - 12*x^4*log(x) + 4*x^3*e^(2*x + 10)*log(x)
- 18*x^3*e^(x + 5)*log(x) - 4*x^4 - 6*x^3*e^(x + 5) + 19*x^3*log(x) - 12*x^2*e^(2*x + 10)*log(x) + 10*x^2*e^(
x + 5)*log(x) - 23*x^3 - 66*x^2*e^(x + 5) - 11*x^2*log(x) + 8*x*e^(2*x + 10)*log(x) - 94*x^2 + 118*x*e^(x + 5)
+ 2*x*log(x) - e^(2*x + 10)*log(x) + 131*x - 48*e^(x + 5) - 36)/(4*x^6 + 8*x^5*e^(x + 5) - 16*x^5 + 4*x^4*e^(
2*x + 10) - 24*x^4*e^(x + 5) + 20*x^4 - 12*x^3*e^(2*x + 10) + 16*x^3*e^(x + 5) - 9*x^3 + 8*x^2*e^(2*x + 10) -
2*x^2*e^(x + 5) + x^2 - x*e^(2*x + 10))

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maple [A]  time = 0.05, size = 36, normalized size = 1.33

method	result	size
risch	\(\frac {4 \ln \relax (x )}{x}+\frac {48}{x \left ({\mathrm e}^{2 x +10}+2 x \,{\mathrm e}^{5+x}+x^{2}-x \right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(5+x)^4-16*x*exp(5+x)^3+(-24*x^2+8*x)*exp(5+x)^2+(-16*x^3+16*x^2)*exp(5+x)-4*x^4+8*x^3-4*x^2)*ln(x
)+4*exp(5+x)^4+16*x*exp(5+x)^3+(24*x^2-104*x-48)*exp(5+x)^2+(16*x^3-112*x^2-192*x)*exp(5+x)+4*x^4-8*x^3-140*x^
2+96*x)/(x^2*exp(5+x)^4+4*x^3*exp(5+x)^3+(6*x^4-2*x^3)*exp(5+x)^2+(4*x^5-4*x^4)*exp(5+x)+x^6-2*x^5+x^4),x,meth
od=_RETURNVERBOSE)

[Out]

4*ln(x)/x+48/x/(exp(2*x+10)+2*x*exp(5+x)+x^2-x)

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maxima [B]  time = 0.54, size = 60, normalized size = 2.22 \begin {gather*} \frac {4 \, {\left (2 \, x e^{\left (x + 5\right )} \log \relax (x) + {\left (x^{2} - x\right )} \log \relax (x) + e^{\left (2 \, x + 10\right )} \log \relax (x) + 12\right )}}{x^{3} + 2 \, x^{2} e^{\left (x + 5\right )} - x^{2} + x e^{\left (2 \, x + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(5+x)^4-16*x*exp(5+x)^3+(-24*x^2+8*x)*exp(5+x)^2+(-16*x^3+16*x^2)*exp(5+x)-4*x^4+8*x^3-4*x^2
)*log(x)+4*exp(5+x)^4+16*x*exp(5+x)^3+(24*x^2-104*x-48)*exp(5+x)^2+(16*x^3-112*x^2-192*x)*exp(5+x)+4*x^4-8*x^3
-140*x^2+96*x)/(x^2*exp(5+x)^4+4*x^3*exp(5+x)^3+(6*x^4-2*x^3)*exp(5+x)^2+(4*x^5-4*x^4)*exp(5+x)+x^6-2*x^5+x^4)
,x, algorithm="maxima")

[Out]

4*(2*x*e^(x + 5)*log(x) + (x^2 - x)*log(x) + e^(2*x + 10)*log(x) + 12)/(x^3 + 2*x^2*e^(x + 5) - x^2 + x*e^(2*x
+ 10))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{x+5}\,\left (-16\,x^3+112\,x^2+192\,x\right )-4\,{\mathrm {e}}^{4\,x+20}-96\,x+{\mathrm {e}}^{2\,x+10}\,\left (-24\,x^2+104\,x+48\right )-16\,x\,{\mathrm {e}}^{3\,x+15}+\ln \relax (x)\,\left (4\,{\mathrm {e}}^{4\,x+20}-{\mathrm {e}}^{2\,x+10}\,\left (8\,x-24\,x^2\right )-{\mathrm {e}}^{x+5}\,\left (16\,x^2-16\,x^3\right )+16\,x\,{\mathrm {e}}^{3\,x+15}+4\,x^2-8\,x^3+4\,x^4\right )+140\,x^2+8\,x^3-4\,x^4}{4\,x^3\,{\mathrm {e}}^{3\,x+15}-{\mathrm {e}}^{2\,x+10}\,\left (2\,x^3-6\,x^4\right )-{\mathrm {e}}^{x+5}\,\left (4\,x^4-4\,x^5\right )+x^2\,{\mathrm {e}}^{4\,x+20}+x^4-2\,x^5+x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + 5)*(192*x + 112*x^2 - 16*x^3) - 4*exp(4*x + 20) - 96*x + exp(2*x + 10)*(104*x - 24*x^2 + 48) - 1
6*x*exp(3*x + 15) + log(x)*(4*exp(4*x + 20) - exp(2*x + 10)*(8*x - 24*x^2) - exp(x + 5)*(16*x^2 - 16*x^3) + 16
*x*exp(3*x + 15) + 4*x^2 - 8*x^3 + 4*x^4) + 140*x^2 + 8*x^3 - 4*x^4)/(4*x^3*exp(3*x + 15) - exp(2*x + 10)*(2*x
^3 - 6*x^4) - exp(x + 5)*(4*x^4 - 4*x^5) + x^2*exp(4*x + 20) + x^4 - 2*x^5 + x^6),x)

[Out]

int(-(exp(x + 5)*(192*x + 112*x^2 - 16*x^3) - 4*exp(4*x + 20) - 96*x + exp(2*x + 10)*(104*x - 24*x^2 + 48) - 1
6*x*exp(3*x + 15) + log(x)*(4*exp(4*x + 20) - exp(2*x + 10)*(8*x - 24*x^2) - exp(x + 5)*(16*x^2 - 16*x^3) + 16
*x*exp(3*x + 15) + 4*x^2 - 8*x^3 + 4*x^4) + 140*x^2 + 8*x^3 - 4*x^4)/(4*x^3*exp(3*x + 15) - exp(2*x + 10)*(2*x
^3 - 6*x^4) - exp(x + 5)*(4*x^4 - 4*x^5) + x^2*exp(4*x + 20) + x^4 - 2*x^5 + x^6), x)

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sympy [A]  time = 0.39, size = 32, normalized size = 1.19 \begin {gather*} \frac {48}{x^{3} + 2 x^{2} e^{x + 5} - x^{2} + x e^{2 x + 10}} + \frac {4 \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(5+x)**4-16*x*exp(5+x)**3+(-24*x**2+8*x)*exp(5+x)**2+(-16*x**3+16*x**2)*exp(5+x)-4*x**4+8*x*
*3-4*x**2)*ln(x)+4*exp(5+x)**4+16*x*exp(5+x)**3+(24*x**2-104*x-48)*exp(5+x)**2+(16*x**3-112*x**2-192*x)*exp(5+
x)+4*x**4-8*x**3-140*x**2+96*x)/(x**2*exp(5+x)**4+4*x**3*exp(5+x)**3+(6*x**4-2*x**3)*exp(5+x)**2+(4*x**5-4*x**
4)*exp(5+x)+x**6-2*x**5+x**4),x)

[Out]

48/(x**3 + 2*x**2*exp(x + 5) - x**2 + x*exp(2*x + 10)) + 4*log(x)/x

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