3.1.99 \(\int \frac {100-105 x+31 x^2+3 e^{2 x} x^2+e^x (-30 x+24 x^2-x^3)+(30 x-19 x^2-6 e^x x^2) \log (x)+3 x^2 \log ^2(x)}{75+10 x-83 x^2+30 x^3+e^{2 x} (3 x^2+3 x^3)+e^x (-30 x-17 x^2+19 x^3)+(30 x+17 x^2-19 x^3+e^x (-6 x^2-6 x^3)) \log (x)+(3 x^2+3 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (x+\frac {1}{3} \left (3-\frac {-5+x}{-3-e^x+\frac {5}{x}+\log (x)}\right )\right ) \]

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Rubi [F]  time = 23.27, 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 {100-105 x+31 x^2+3 e^{2 x} x^2+e^x \left (-30 x+24 x^2-x^3\right )+\left (30 x-19 x^2-6 e^x x^2\right ) \log (x)+3 x^2 \log ^2(x)}{75+10 x-83 x^2+30 x^3+e^{2 x} \left (3 x^2+3 x^3\right )+e^x \left (-30 x-17 x^2+19 x^3\right )+\left (30 x+17 x^2-19 x^3+e^x \left (-6 x^2-6 x^3\right )\right ) \log (x)+\left (3 x^2+3 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100 - 105*x + 31*x^2 + 3*E^(2*x)*x^2 + E^x*(-30*x + 24*x^2 - x^3) + (30*x - 19*x^2 - 6*E^x*x^2)*Log[x] +
3*x^2*Log[x]^2)/(75 + 10*x - 83*x^2 + 30*x^3 + E^(2*x)*(3*x^2 + 3*x^3) + E^x*(-30*x - 17*x^2 + 19*x^3) + (30*x
+ 17*x^2 - 19*x^3 + E^x*(-6*x^2 - 6*x^3))*Log[x] + (3*x^2 + 3*x^3)*Log[x]^2),x]

[Out]

Log[1 + x] - 4*Defer[Int][(-5 + 3*x + E^x*x - x*Log[x])^(-1), x] - 5*Defer[Int][1/(x*(-5 + 3*x + E^x*x - x*Log
[x])), x] + 3*Defer[Int][x/(-5 + 3*x + E^x*x - x*Log[x]), x] + Defer[Int][(x*Log[x])/(5 - 3*x - E^x*x + x*Log[
x]), x] + 33*Defer[Int][(-15 - 11*x + 3*E^x*x + 10*x^2 + 3*E^x*x^2 - 3*x*Log[x] - 3*x^2*Log[x])^(-1), x] + 15*
Defer[Int][1/(x*(-15 - 11*x + 3*E^x*x + 10*x^2 + 3*E^x*x^2 - 3*x*Log[x] - 3*x^2*Log[x])), x] + 8*Defer[Int][x/
(-15 - 11*x + 3*E^x*x + 10*x^2 + 3*E^x*x^2 - 3*x*Log[x] - 3*x^2*Log[x]), x] - 10*Defer[Int][x^2/(-15 - 11*x +
3*E^x*x + 10*x^2 + 3*E^x*x^2 - 3*x*Log[x] - 3*x^2*Log[x]), x] - 6*Defer[Int][1/((1 + x)*(-15 - 11*x + 3*E^x*x
+ 10*x^2 + 3*E^x*x^2 - 3*x*Log[x] - 3*x^2*Log[x])), x] - 3*Defer[Int][(x*Log[x])/(15 + 11*x - 3*E^x*x - 10*x^2
- 3*E^x*x^2 + 3*x*Log[x] + 3*x^2*Log[x]), x] - 3*Defer[Int][(x^2*Log[x])/(15 + 11*x - 3*E^x*x - 10*x^2 - 3*E^
x*x^2 + 3*x*Log[x] + 3*x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{1+x}-\frac {5+4 x-3 x^2+x^2 \log (x)}{x \left (-5+3 x+e^x x-x \log (x)\right )}+\frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{x (1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}\right ) \, dx\\ &=\log (1+x)-\int \frac {5+4 x-3 x^2+x^2 \log (x)}{x \left (-5+3 x+e^x x-x \log (x)\right )} \, dx+\int \frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{x (1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx\\ &=\log (1+x)-\int \left (\frac {4}{-5+3 x+e^x x-x \log (x)}+\frac {5}{x \left (-5+3 x+e^x x-x \log (x)\right )}-\frac {3 x}{-5+3 x+e^x x-x \log (x)}-\frac {x \log (x)}{5-3 x-e^x x+x \log (x)}\right ) \, dx+\int \left (\frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{x \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}-\frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}\right ) \, dx\\ &=\log (1+x)+3 \int \frac {x}{-5+3 x+e^x x-x \log (x)} \, dx-4 \int \frac {1}{-5+3 x+e^x x-x \log (x)} \, dx-5 \int \frac {1}{x \left (-5+3 x+e^x x-x \log (x)\right )} \, dx+\int \frac {x \log (x)}{5-3 x-e^x x+x \log (x)} \, dx+\int \frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{x \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx-\int \frac {15+42 x+41 x^2-2 x^3-10 x^4+3 x^2 \log (x)+6 x^3 \log (x)+3 x^4 \log (x)}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx\\ &=\log (1+x)+3 \int \frac {x}{-5+3 x+e^x x-x \log (x)} \, dx-4 \int \frac {1}{-5+3 x+e^x x-x \log (x)} \, dx-5 \int \frac {1}{x \left (-5+3 x+e^x x-x \log (x)\right )} \, dx+\int \frac {x \log (x)}{5-3 x-e^x x+x \log (x)} \, dx+\int \left (\frac {42}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)}+\frac {15}{x \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}+\frac {41 x}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)}-\frac {2 x^2}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)}-\frac {10 x^3}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)}-\frac {3 x \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)}-\frac {6 x^2 \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)}-\frac {3 x^3 \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)}\right ) \, dx-\int \left (\frac {15}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}+\frac {42 x}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}+\frac {41 x^2}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}-\frac {2 x^3}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}-\frac {10 x^4}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )}-\frac {3 x^2 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )}-\frac {6 x^3 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )}-\frac {3 x^4 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )}\right ) \, dx\\ &=\log (1+x)-2 \int \frac {x^2}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)} \, dx+2 \int \frac {x^3}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx+3 \int \frac {x}{-5+3 x+e^x x-x \log (x)} \, dx-3 \int \frac {x \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)} \, dx-3 \int \frac {x^3 \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)} \, dx+3 \int \frac {x^2 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )} \, dx+3 \int \frac {x^4 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )} \, dx-4 \int \frac {1}{-5+3 x+e^x x-x \log (x)} \, dx-5 \int \frac {1}{x \left (-5+3 x+e^x x-x \log (x)\right )} \, dx-6 \int \frac {x^2 \log (x)}{15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)} \, dx+6 \int \frac {x^3 \log (x)}{(1+x) \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right )} \, dx-10 \int \frac {x^3}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)} \, dx+10 \int \frac {x^4}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx+15 \int \frac {1}{x \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx-15 \int \frac {1}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx+41 \int \frac {x}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)} \, dx-41 \int \frac {x^2}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx+42 \int \frac {1}{-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)} \, dx-42 \int \frac {x}{(1+x) \left (-15-11 x+3 e^x x+10 x^2+3 e^x x^2-3 x \log (x)-3 x^2 \log (x)\right )} \, dx+\int \frac {x \log (x)}{5-3 x-e^x x+x \log (x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.35, size = 70, normalized size = 2.33 \begin {gather*} \log (x)+\log (1+x)-\log (x (1+x))-\log \left (5-3 x-e^x x+x \log (x)\right )+\log \left (15+11 x-3 e^x x-10 x^2-3 e^x x^2+3 x \log (x)+3 x^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 105*x + 31*x^2 + 3*E^(2*x)*x^2 + E^x*(-30*x + 24*x^2 - x^3) + (30*x - 19*x^2 - 6*E^x*x^2)*Log
[x] + 3*x^2*Log[x]^2)/(75 + 10*x - 83*x^2 + 30*x^3 + E^(2*x)*(3*x^2 + 3*x^3) + E^x*(-30*x - 17*x^2 + 19*x^3) +
(30*x + 17*x^2 - 19*x^3 + E^x*(-6*x^2 - 6*x^3))*Log[x] + (3*x^2 + 3*x^3)*Log[x]^2),x]

[Out]

Log[x] + Log[1 + x] - Log[x*(1 + x)] - Log[5 - 3*x - E^x*x + x*Log[x]] + Log[15 + 11*x - 3*E^x*x - 10*x^2 - 3*
E^x*x^2 + 3*x*Log[x] + 3*x^2*Log[x]]

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fricas [B]  time = 0.87, size = 65, normalized size = 2.17 \begin {gather*} \log \left (x + 1\right ) + \log \left (-\frac {10 \, x^{2} + 3 \, {\left (x^{2} + x\right )} e^{x} - 3 \, {\left (x^{2} + x\right )} \log \relax (x) - 11 \, x - 15}{x^{2} + x}\right ) - \log \left (-\frac {x e^{x} - x \log \relax (x) + 3 \, x - 5}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)^2+(-6*exp(x)*x^2-19*x^2+30*x)*log(x)+3*exp(x)^2*x^2+(-x^3+24*x^2-30*x)*exp(x)+31*x^2-1
05*x+100)/((3*x^3+3*x^2)*log(x)^2+((-6*x^3-6*x^2)*exp(x)-19*x^3+17*x^2+30*x)*log(x)+(3*x^3+3*x^2)*exp(x)^2+(19
*x^3-17*x^2-30*x)*exp(x)+30*x^3-83*x^2+10*x+75),x, algorithm="fricas")

[Out]

log(x + 1) + log(-(10*x^2 + 3*(x^2 + x)*e^x - 3*(x^2 + x)*log(x) - 11*x - 15)/(x^2 + x)) - log(-(x*e^x - x*log
(x) + 3*x - 5)/x)

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giac [B]  time = 1.98, size = 53, normalized size = 1.77 \begin {gather*} \log \left (-3 \, x^{2} e^{x} + 3 \, x^{2} \log \relax (x) - 10 \, x^{2} - 3 \, x e^{x} + 3 \, x \log \relax (x) + 11 \, x + 15\right ) - \log \left (-x e^{x} + x \log \relax (x) - 3 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)^2+(-6*exp(x)*x^2-19*x^2+30*x)*log(x)+3*exp(x)^2*x^2+(-x^3+24*x^2-30*x)*exp(x)+31*x^2-1
05*x+100)/((3*x^3+3*x^2)*log(x)^2+((-6*x^3-6*x^2)*exp(x)-19*x^3+17*x^2+30*x)*log(x)+(3*x^3+3*x^2)*exp(x)^2+(19
*x^3-17*x^2-30*x)*exp(x)+30*x^3-83*x^2+10*x+75),x, algorithm="giac")

[Out]

log(-3*x^2*e^x + 3*x^2*log(x) - 10*x^2 - 3*x*e^x + 3*x*log(x) + 11*x + 15) - log(-x*e^x + x*log(x) - 3*x + 5)

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maple [B]  time = 0.07, size = 62, normalized size = 2.07

method	result	size
risch	\(\ln \left (x +1\right )+\ln \left (\ln \relax (x )-\frac {3 \,{\mathrm e}^{x} x^{2}+10 x^{2}+3 \,{\mathrm e}^{x} x -11 x -15}{3 \left (x +1\right ) x}\right )-\ln \left (\ln \relax (x )-\frac {{\mathrm e}^{x} x +3 x -5}{x}\right )\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*ln(x)^2+(-6*exp(x)*x^2-19*x^2+30*x)*ln(x)+3*exp(x)^2*x^2+(-x^3+24*x^2-30*x)*exp(x)+31*x^2-105*x+100
)/((3*x^3+3*x^2)*ln(x)^2+((-6*x^3-6*x^2)*exp(x)-19*x^3+17*x^2+30*x)*ln(x)+(3*x^3+3*x^2)*exp(x)^2+(19*x^3-17*x^
2-30*x)*exp(x)+30*x^3-83*x^2+10*x+75),x,method=_RETURNVERBOSE)

[Out]

ln(x+1)+ln(ln(x)-1/3*(3*exp(x)*x^2+10*x^2+3*exp(x)*x-11*x-15)/(x+1)/x)-ln(ln(x)-(exp(x)*x+3*x-5)/x)

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maxima [B]  time = 0.55, size = 64, normalized size = 2.13 \begin {gather*} \log \left (x + 1\right ) + \log \left (\frac {10 \, x^{2} + 3 \, {\left (x^{2} + x\right )} e^{x} - 3 \, {\left (x^{2} + x\right )} \log \relax (x) - 11 \, x - 15}{3 \, {\left (x^{2} + x\right )}}\right ) - \log \left (\frac {x e^{x} - x \log \relax (x) + 3 \, x - 5}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)^2+(-6*exp(x)*x^2-19*x^2+30*x)*log(x)+3*exp(x)^2*x^2+(-x^3+24*x^2-30*x)*exp(x)+31*x^2-1
05*x+100)/((3*x^3+3*x^2)*log(x)^2+((-6*x^3-6*x^2)*exp(x)-19*x^3+17*x^2+30*x)*log(x)+(3*x^3+3*x^2)*exp(x)^2+(19
*x^3-17*x^2-30*x)*exp(x)+30*x^3-83*x^2+10*x+75),x, algorithm="maxima")

[Out]

log(x + 1) + log(1/3*(10*x^2 + 3*(x^2 + x)*e^x - 3*(x^2 + x)*log(x) - 11*x - 15)/(x^2 + x)) - log((x*e^x - x*l
og(x) + 3*x - 5)/x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {3\,x^2\,{\mathrm {e}}^{2\,x}-105\,x+3\,x^2\,{\ln \relax (x)}^2-{\mathrm {e}}^x\,\left (x^3-24\,x^2+30\,x\right )-\ln \relax (x)\,\left (6\,x^2\,{\mathrm {e}}^x-30\,x+19\,x^2\right )+31\,x^2+100}{10\,x+{\mathrm {e}}^{2\,x}\,\left (3\,x^3+3\,x^2\right )+{\ln \relax (x)}^2\,\left (3\,x^3+3\,x^2\right )-83\,x^2+30\,x^3-{\mathrm {e}}^x\,\left (-19\,x^3+17\,x^2+30\,x\right )+\ln \relax (x)\,\left (30\,x-{\mathrm {e}}^x\,\left (6\,x^3+6\,x^2\right )+17\,x^2-19\,x^3\right )+75} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*exp(2*x) - 105*x + 3*x^2*log(x)^2 - exp(x)*(30*x - 24*x^2 + x^3) - log(x)*(6*x^2*exp(x) - 30*x + 19
*x^2) + 31*x^2 + 100)/(10*x + exp(2*x)*(3*x^2 + 3*x^3) + log(x)^2*(3*x^2 + 3*x^3) - 83*x^2 + 30*x^3 - exp(x)*(
30*x + 17*x^2 - 19*x^3) + log(x)*(30*x - exp(x)*(6*x^2 + 6*x^3) + 17*x^2 - 19*x^3) + 75),x)

[Out]

int((3*x^2*exp(2*x) - 105*x + 3*x^2*log(x)^2 - exp(x)*(30*x - 24*x^2 + x^3) - log(x)*(6*x^2*exp(x) - 30*x + 19
*x^2) + 31*x^2 + 100)/(10*x + exp(2*x)*(3*x^2 + 3*x^3) + log(x)^2*(3*x^2 + 3*x^3) - 83*x^2 + 30*x^3 - exp(x)*(
30*x + 17*x^2 - 19*x^3) + log(x)*(30*x - exp(x)*(6*x^2 + 6*x^3) + 17*x^2 - 19*x^3) + 75), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2*ln(x)**2+(-6*exp(x)*x**2-19*x**2+30*x)*ln(x)+3*exp(x)**2*x**2+(-x**3+24*x**2-30*x)*exp(x)+31
*x**2-105*x+100)/((3*x**3+3*x**2)*ln(x)**2+((-6*x**3-6*x**2)*exp(x)-19*x**3+17*x**2+30*x)*ln(x)+(3*x**3+3*x**2
)*exp(x)**2+(19*x**3-17*x**2-30*x)*exp(x)+30*x**3-83*x**2+10*x+75),x)

[Out]

Exception raised: PolynomialError

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