3.74.74 \(\int \frac {540 x-315 x^2+50 x^3-25 x^4+e^x (-1125 x^2+1125 x^3-125 x^4+125 x^5)+(-90 x+45 x^2-10 x^3+5 x^4+e^x (450 x^2-450 x^3+50 x^4-50 x^5)) \log (\frac {9+x^2}{x^2})+e^x (-45 x^2+45 x^3-5 x^4+5 x^5) \log ^2(\frac {9+x^2}{x^2})}{27-54 x+30 x^2-6 x^3+3 x^4+e^x (-270 x+270 x^2-30 x^3+30 x^4)+e^{2 x} (675 x^2+75 x^4)+(e^{2 x} (-270 x^2-30 x^4)+e^x (54 x-54 x^2+6 x^3-6 x^4)) \log (\frac {9+x^2}{x^2})+e^{2 x} (27 x^2+3 x^4) \log ^2(\frac {9+x^2}{x^2})} \, dx\)

Optimal. Leaf size=32 \[ \frac {5 x}{3 \left (-e^x+\frac {-1+x}{x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}\right )} \]

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Rubi [F]  time = 24.20, 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 {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(540*x - 315*x^2 + 50*x^3 - 25*x^4 + E^x*(-1125*x^2 + 1125*x^3 - 125*x^4 + 125*x^5) + (-90*x + 45*x^2 - 10
*x^3 + 5*x^4 + E^x*(450*x^2 - 450*x^3 + 50*x^4 - 50*x^5))*Log[(9 + x^2)/x^2] + E^x*(-45*x^2 + 45*x^3 - 5*x^4 +
5*x^5)*Log[(9 + x^2)/x^2]^2)/(27 - 54*x + 30*x^2 - 6*x^3 + 3*x^4 + E^x*(-270*x + 270*x^2 - 30*x^3 + 30*x^4) +
E^(2*x)*(675*x^2 + 75*x^4) + (E^(2*x)*(-270*x^2 - 30*x^4) + E^x*(54*x - 54*x^2 + 6*x^3 - 6*x^4))*Log[(9 + x^2
)/x^2] + E^(2*x)*(27*x^2 + 3*x^4)*Log[(9 + x^2)/x^2]^2),x]

[Out]

-30*Defer[Int][(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^(-2), x] - (15 - 45*I)*Defer[Int][1/((3*I - x)*(1 - x
- 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2), x] + (25*Defer[Int][x/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2, x])/3
+ (25*Defer[Int][x^2/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2, x])/3 - (25*Defer[Int][x^3/(1 - x - 5*E^x*x +
E^x*x*Log[1 + 9/x^2])^2, x])/3 + (15 + 45*I)*Defer[Int][1/((3*I + x)*(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])
^2), x] - (5*Defer[Int][(x*Log[1 + 9/x^2])/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2, x])/3 - (5*Defer[Int][(
x^2*Log[1 + 9/x^2])/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2, x])/3 + (5*Defer[Int][(x^3*Log[1 + 9/x^2])/(1
- x - 5*E^x*x + E^x*x*Log[1 + 9/x^2])^2, x])/3 + (25*Defer[Int][x/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2]), x]
)/3 - (25*Defer[Int][x^2/(1 - x - 5*E^x*x + E^x*x*Log[1 + 9/x^2]), x])/3 - (5*Defer[Int][(x*Log[1 + 9/x^2])/(1
- x - 5*E^x*x + E^x*x*Log[1 + 9/x^2]), x])/3 + (5*Defer[Int][(x^2*Log[1 + 9/x^2])/(1 - x - 5*E^x*x + E^x*x*Lo
g[1 + 9/x^2]), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{3 \left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \frac {x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \left (\frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx\\ &=\frac {5}{3} \int \frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4+\left (-9-9 x+8 x^2-x^3+x^4\right ) \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {5}{3} \int \left (-\frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx\\ &=-\left (\frac {5}{3} \int \frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx\right )+\frac {5}{3} \int \frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \left (\frac {63 x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {27 x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {40 x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {5 x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {5 x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {8 x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {5}{3} \int \frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\right )+\frac {5}{3} \int \frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {5}{3} \int \left (-\frac {5 x}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {5}{3} \int \left (-\frac {5 x^2}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {25}{3} \int \frac {x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {25}{3} \int \frac {x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {40}{3} \int \frac {x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+45 \int \frac {x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {200}{3} \int \frac {x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+105 \int \frac {x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 44, normalized size = 1.38 \begin {gather*} -\frac {5 x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{3 \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(540*x - 315*x^2 + 50*x^3 - 25*x^4 + E^x*(-1125*x^2 + 1125*x^3 - 125*x^4 + 125*x^5) + (-90*x + 45*x^
2 - 10*x^3 + 5*x^4 + E^x*(450*x^2 - 450*x^3 + 50*x^4 - 50*x^5))*Log[(9 + x^2)/x^2] + E^x*(-45*x^2 + 45*x^3 - 5
*x^4 + 5*x^5)*Log[(9 + x^2)/x^2]^2)/(27 - 54*x + 30*x^2 - 6*x^3 + 3*x^4 + E^x*(-270*x + 270*x^2 - 30*x^3 + 30*
x^4) + E^(2*x)*(675*x^2 + 75*x^4) + (E^(2*x)*(-270*x^2 - 30*x^4) + E^x*(54*x - 54*x^2 + 6*x^3 - 6*x^4))*Log[(9
+ x^2)/x^2] + E^(2*x)*(27*x^2 + 3*x^4)*Log[(9 + x^2)/x^2]^2),x]

[Out]

(-5*x^2*(-5 + Log[1 + 9/x^2]))/(3*(1 - (1 + 5*E^x)*x + E^x*x*Log[1 + 9/x^2]))

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fricas [A]  time = 1.00, size = 48, normalized size = 1.50 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^
4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2)+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)
/((3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6*x^3-54*x^2+54*x)*exp(x))*lo
g((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, a
lgorithm="fricas")

[Out]

-5/3*(x^2*log((x^2 + 9)/x^2) - 5*x^2)/(x*e^x*log((x^2 + 9)/x^2) - 5*x*e^x - x + 1)

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giac [A]  time = 0.95, size = 48, normalized size = 1.50 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^
4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2)+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)
/((3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6*x^3-54*x^2+54*x)*exp(x))*lo
g((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, a
lgorithm="giac")

[Out]

-5/3*(x^2*log((x^2 + 9)/x^2) - 5*x^2)/(x*e^x*log((x^2 + 9)/x^2) - 5*x*e^x - x + 1)

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maple [C]  time = 0.20, size = 216, normalized size = 6.75

method	result	size
risch	\(-\frac {5 x \,{\mathrm e}^{-x}}{3}+\frac {10 i x \left (x -1\right ) {\mathrm e}^{-x}}{3 \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}-\pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+9\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right ) {\mathrm e}^{x}+\pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{2} {\mathrm e}^{x}+\pi x \,\mathrm {csgn}\left (i \left (x^{2}+9\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{2} {\mathrm e}^{x}-\pi x \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{3} {\mathrm e}^{x}+4 i x \,{\mathrm e}^{x} \ln \relax (x )-2 i x \,{\mathrm e}^{x} \ln \left (x^{2}+9\right )+10 i x \,{\mathrm e}^{x}+2 i x -2 i\right )}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*ln((x^2+9)/x^2)^2+((-50*x^5+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^4-10*x^
3+45*x^2-90*x)*ln((x^2+9)/x^2)+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)/((3*x^4
+27*x^2)*exp(x)^2*ln((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6*x^3-54*x^2+54*x)*exp(x))*ln((x^2+9)/
x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x,method=_RETU
RNVERBOSE)

[Out]

-5/3*x*exp(-x)+10/3*I*x*(x-1)/(Pi*x*csgn(I*x)^2*csgn(I*x^2)*exp(x)-2*Pi*x*csgn(I*x)*csgn(I*x^2)^2*exp(x)+Pi*x*
csgn(I*x^2)^3*exp(x)-Pi*x*csgn(I/x^2)*csgn(I*(x^2+9))*csgn(I/x^2*(x^2+9))*exp(x)+Pi*x*csgn(I/x^2)*csgn(I/x^2*(
x^2+9))^2*exp(x)+Pi*x*csgn(I*(x^2+9))*csgn(I/x^2*(x^2+9))^2*exp(x)-Pi*x*csgn(I/x^2*(x^2+9))^3*exp(x)+4*I*x*exp
(x)*ln(x)-2*I*x*exp(x)*ln(x^2+9)+10*I*x*exp(x)+2*I*x-2*I)*exp(-x)

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maxima [B]  time = 0.63, size = 69, normalized size = 2.16 \begin {gather*} -\frac {5 \, {\left (x^{2} e^{x} \log \left (x^{2} + 9\right ) - {\left (2 \, x^{2} \log \relax (x) + 5 \, x^{2}\right )} e^{x}\right )}}{3 \, {\left (x e^{\left (2 \, x\right )} \log \left (x^{2} + 9\right ) - {\left (2 \, x \log \relax (x) + 5 \, x\right )} e^{\left (2 \, x\right )} - {\left (x - 1\right )} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^5-5*x^4+45*x^3-45*x^2)*exp(x)*log((x^2+9)/x^2)^2+((-50*x^5+50*x^4-450*x^3+450*x^2)*exp(x)+5*x^
4-10*x^3+45*x^2-90*x)*log((x^2+9)/x^2)+(125*x^5-125*x^4+1125*x^3-1125*x^2)*exp(x)-25*x^4+50*x^3-315*x^2+540*x)
/((3*x^4+27*x^2)*exp(x)^2*log((x^2+9)/x^2)^2+((-30*x^4-270*x^2)*exp(x)^2+(-6*x^4+6*x^3-54*x^2+54*x)*exp(x))*lo
g((x^2+9)/x^2)+(75*x^4+675*x^2)*exp(x)^2+(30*x^4-30*x^3+270*x^2-270*x)*exp(x)+3*x^4-6*x^3+30*x^2-54*x+27),x, a
lgorithm="maxima")

[Out]

-5/3*(x^2*e^x*log(x^2 + 9) - (2*x^2*log(x) + 5*x^2)*e^x)/(x*e^(2*x)*log(x^2 + 9) - (2*x*log(x) + 5*x)*e^(2*x)
- (x - 1)*e^x)

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mupad [B]  time = 5.35, size = 44, normalized size = 1.38 \begin {gather*} \frac {5\,x^2\,\left (\ln \left (\frac {x^2+9}{x^2}\right )-5\right )}{3\,\left (x+5\,x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\ln \left (\frac {x^2+9}{x^2}\right )-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(1125*x^2 - 1125*x^3 + 125*x^4 - 125*x^5) - 540*x - log((x^2 + 9)/x^2)*(exp(x)*(450*x^2 - 450*x^3
+ 50*x^4 - 50*x^5) - 90*x + 45*x^2 - 10*x^3 + 5*x^4) + 315*x^2 - 50*x^3 + 25*x^4 + exp(x)*log((x^2 + 9)/x^2)^
2*(45*x^2 - 45*x^3 + 5*x^4 - 5*x^5))/(log((x^2 + 9)/x^2)*(exp(x)*(54*x - 54*x^2 + 6*x^3 - 6*x^4) - exp(2*x)*(2
70*x^2 + 30*x^4)) - exp(x)*(270*x - 270*x^2 + 30*x^3 - 30*x^4) - 54*x + exp(2*x)*(675*x^2 + 75*x^4) + 30*x^2 -
6*x^3 + 3*x^4 + exp(2*x)*log((x^2 + 9)/x^2)^2*(27*x^2 + 3*x^4) + 27),x)

[Out]

(5*x^2*(log((x^2 + 9)/x^2) - 5))/(3*(x + 5*x*exp(x) - x*exp(x)*log((x^2 + 9)/x^2) - 1))

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sympy [A]  time = 0.63, size = 44, normalized size = 1.38 \begin {gather*} \frac {- 5 x^{2} \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} + 25 x^{2}}{- 3 x + \left (3 x \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} - 15 x\right ) e^{x} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**5-5*x**4+45*x**3-45*x**2)*exp(x)*ln((x**2+9)/x**2)**2+((-50*x**5+50*x**4-450*x**3+450*x**2)*e
xp(x)+5*x**4-10*x**3+45*x**2-90*x)*ln((x**2+9)/x**2)+(125*x**5-125*x**4+1125*x**3-1125*x**2)*exp(x)-25*x**4+50
*x**3-315*x**2+540*x)/((3*x**4+27*x**2)*exp(x)**2*ln((x**2+9)/x**2)**2+((-30*x**4-270*x**2)*exp(x)**2+(-6*x**4
+6*x**3-54*x**2+54*x)*exp(x))*ln((x**2+9)/x**2)+(75*x**4+675*x**2)*exp(x)**2+(30*x**4-30*x**3+270*x**2-270*x)*
exp(x)+3*x**4-6*x**3+30*x**2-54*x+27),x)

[Out]

(-5*x**2*log((x**2 + 9)/x**2) + 25*x**2)/(-3*x + (3*x*log((x**2 + 9)/x**2) - 15*x)*exp(x) + 3)

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