3.45.54 \(\int \frac {-20-10 x+e^x (-15 x-5 x^2)-5 e^x x \log (20 x^2)}{27 x+9 x^2+e^x (-18 x-6 x^2)+e^{2 x} (3 x+x^2)+(9 x-6 e^x x+e^{2 x} x) \log (20 x^2)+(-18 x-6 x^2+e^x (6 x+2 x^2)+(-6 x+2 e^x x) \log (20 x^2)) \log (9+6 x+x^2+(6+2 x) \log (20 x^2)+\log ^2(20 x^2))+(3 x+x^2+x \log (20 x^2)) \log ^2(9+6 x+x^2+(6+2 x) \log (20 x^2)+\log ^2(20 x^2))} \, dx\)

Optimal. Leaf size=21 \[ \frac {5}{-3+e^x+\log \left (\left (3+x+\log \left (20 x^2\right )\right )^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, integrand size = 208, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {5}{-\log \left (\left (\log \left (20 x^2\right )+x+3\right )^2\right )-e^x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 10*x + E^x*(-15*x - 5*x^2) - 5*E^x*x*Log[20*x^2])/(27*x + 9*x^2 + E^x*(-18*x - 6*x^2) + E^(2*x)*(3*
x + x^2) + (9*x - 6*E^x*x + E^(2*x)*x)*Log[20*x^2] + (-18*x - 6*x^2 + E^x*(6*x + 2*x^2) + (-6*x + 2*E^x*x)*Log
[20*x^2])*Log[9 + 6*x + x^2 + (6 + 2*x)*Log[20*x^2] + Log[20*x^2]^2] + (3*x + x^2 + x*Log[20*x^2])*Log[9 + 6*x
+ x^2 + (6 + 2*x)*Log[20*x^2] + Log[20*x^2]^2]^2),x]

[Out]

-5/(3 - E^x - Log[(3 + x + Log[20*x^2])^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (-4-\left (2+3 e^x\right ) x-e^x x^2-e^x x \log \left (20 x^2\right )\right )}{x \left (3+x+\log \left (20 x^2\right )\right ) \left (3-e^x-\log \left (\left (3+x+\log \left (20 x^2\right )\right )^2\right )\right )^2} \, dx\\ &=5 \int \frac {-4-\left (2+3 e^x\right ) x-e^x x^2-e^x x \log \left (20 x^2\right )}{x \left (3+x+\log \left (20 x^2\right )\right ) \left (3-e^x-\log \left (\left (3+x+\log \left (20 x^2\right )\right )^2\right )\right )^2} \, dx\\ &=-\frac {5}{3-e^x-\log \left (\left (3+x+\log \left (20 x^2\right )\right )^2\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 21, normalized size = 1.00 \begin {gather*} \frac {5}{-3+e^x+\log \left (\left (3+x+\log \left (20 x^2\right )\right )^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 10*x + E^x*(-15*x - 5*x^2) - 5*E^x*x*Log[20*x^2])/(27*x + 9*x^2 + E^x*(-18*x - 6*x^2) + E^(2*
x)*(3*x + x^2) + (9*x - 6*E^x*x + E^(2*x)*x)*Log[20*x^2] + (-18*x - 6*x^2 + E^x*(6*x + 2*x^2) + (-6*x + 2*E^x*
x)*Log[20*x^2])*Log[9 + 6*x + x^2 + (6 + 2*x)*Log[20*x^2] + Log[20*x^2]^2] + (3*x + x^2 + x*Log[20*x^2])*Log[9
+ 6*x + x^2 + (6 + 2*x)*Log[20*x^2] + Log[20*x^2]^2]^2),x]

[Out]

5/(-3 + E^x + Log[(3 + x + Log[20*x^2])^2])

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{e^{x} + \log \left (x^{2} + 2 \, {\left (x + 3\right )} \log \left (20 \, x^{2}\right ) + \log \left (20 \, x^{2}\right )^{2} + 6 \, x + 9\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*exp(x)*log(20*x^2)+(-5*x^2-15*x)*exp(x)-10*x-20)/((x*log(20*x^2)+x^2+3*x)*log(log(20*x^2)^2+(2
*x+6)*log(20*x^2)+x^2+6*x+9)^2+((2*exp(x)*x-6*x)*log(20*x^2)+(2*x^2+6*x)*exp(x)-6*x^2-18*x)*log(log(20*x^2)^2+
(2*x+6)*log(20*x^2)+x^2+6*x+9)+(x*exp(x)^2-6*exp(x)*x+9*x)*log(20*x^2)+(x^2+3*x)*exp(x)^2+(-6*x^2-18*x)*exp(x)
+9*x^2+27*x),x, algorithm="fricas")

[Out]

5/(e^x + log(x^2 + 2*(x + 3)*log(20*x^2) + log(20*x^2)^2 + 6*x + 9) - 3)

________________________________________________________________________________________

giac [B]  time = 1.85, size = 42, normalized size = 2.00 \begin {gather*} \frac {5}{e^{x} + \log \left (x^{2} + 2 \, x \log \left (20 \, x^{2}\right ) + \log \left (20 \, x^{2}\right )^{2} + 6 \, x + 6 \, \log \left (20 \, x^{2}\right ) + 9\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*exp(x)*log(20*x^2)+(-5*x^2-15*x)*exp(x)-10*x-20)/((x*log(20*x^2)+x^2+3*x)*log(log(20*x^2)^2+(2
*x+6)*log(20*x^2)+x^2+6*x+9)^2+((2*exp(x)*x-6*x)*log(20*x^2)+(2*x^2+6*x)*exp(x)-6*x^2-18*x)*log(log(20*x^2)^2+
(2*x+6)*log(20*x^2)+x^2+6*x+9)+(x*exp(x)^2-6*exp(x)*x+9*x)*log(20*x^2)+(x^2+3*x)*exp(x)^2+(-6*x^2-18*x)*exp(x)
+9*x^2+27*x),x, algorithm="giac")

[Out]

5/(e^x + log(x^2 + 2*x*log(20*x^2) + log(20*x^2)^2 + 6*x + 6*log(20*x^2) + 9) - 3)

________________________________________________________________________________________

maple [C]  time = 0.88, size = 509, normalized size = 24.24

method	result	size
risch	\(\frac {10 i}{2 \pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )\right )^{2} \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )\right ) \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \ln \left (20\right )+2 i x +4 i \ln \relax (x )+6 i\right )^{2}\right )^{3}-2 \pi -4 i \ln \relax (2)+2 i {\mathrm e}^{x}+4 i \ln \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \left (2 \ln \relax (2)+\ln \relax (5)\right )+2 i x +4 i \ln \relax (x )+6 i\right )-6 i}\)	\(509\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*x*exp(x)*ln(20*x^2)+(-5*x^2-15*x)*exp(x)-10*x-20)/((x*ln(20*x^2)+x^2+3*x)*ln(ln(20*x^2)^2+(2*x+6)*ln(2
0*x^2)+x^2+6*x+9)^2+((2*exp(x)*x-6*x)*ln(20*x^2)+(2*x^2+6*x)*exp(x)-6*x^2-18*x)*ln(ln(20*x^2)^2+(2*x+6)*ln(20*
x^2)+x^2+6*x+9)+(x*exp(x)^2-6*exp(x)*x+9*x)*ln(20*x^2)+(x^2+3*x)*exp(x)^2+(-6*x^2-18*x)*exp(x)+9*x^2+27*x),x,m
ethod=_RETURNVERBOSE)

[Out]

10*I/(2*Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+2*I*ln(20)+2*I*x+4
*I*ln(x)+6*I)^2)^2+Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+2*I*ln(
20)+2*I*x+4*I*ln(x)+6*I))^2*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+2
*I*ln(20)+2*I*x+4*I*ln(x)+6*I)^2)-2*Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn
(I*x^2)^3+2*I*ln(20)+2*I*x+4*I*ln(x)+6*I))*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*
csgn(I*x^2)^3+2*I*ln(20)+2*I*x+4*I*ln(x)+6*I)^2)^2-Pi*csgn(I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I
*x^2)^2+Pi*csgn(I*x^2)^3+2*I*ln(20)+2*I*x+4*I*ln(x)+6*I)^2)^3-2*Pi-4*I*ln(2)+2*I*exp(x)+4*I*ln(Pi*csgn(I*x)^2*
csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+2*I*(2*ln(2)+ln(5))+2*I*x+4*I*ln(x)+6*I)-6*I)

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 24, normalized size = 1.14 \begin {gather*} \frac {5}{e^{x} + 2 \, \log \left (x + \log \relax (5) + 2 \, \log \relax (2) + 2 \, \log \relax (x) + 3\right ) - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*exp(x)*log(20*x^2)+(-5*x^2-15*x)*exp(x)-10*x-20)/((x*log(20*x^2)+x^2+3*x)*log(log(20*x^2)^2+(2
*x+6)*log(20*x^2)+x^2+6*x+9)^2+((2*exp(x)*x-6*x)*log(20*x^2)+(2*x^2+6*x)*exp(x)-6*x^2-18*x)*log(log(20*x^2)^2+
(2*x+6)*log(20*x^2)+x^2+6*x+9)+(x*exp(x)^2-6*exp(x)*x+9*x)*log(20*x^2)+(x^2+3*x)*exp(x)^2+(-6*x^2-18*x)*exp(x)
+9*x^2+27*x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.38, size = 37, normalized size = 1.76 \begin {gather*} \frac {5}{\ln \left (6\,x+{\ln \left (20\,x^2\right )}^2+\ln \left (20\,x^2\right )\,\left (2\,x+6\right )+x^2+9\right )+{\mathrm {e}}^x-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + exp(x)*(15*x + 5*x^2) + 5*x*exp(x)*log(20*x^2) + 20)/(27*x + log(20*x^2)*(9*x + x*exp(2*x) - 6*x*
exp(x)) - log(6*x + log(20*x^2)^2 + log(20*x^2)*(2*x + 6) + x^2 + 9)*(18*x - exp(x)*(6*x + 2*x^2) + 6*x^2 + lo
g(20*x^2)*(6*x - 2*x*exp(x))) + exp(2*x)*(3*x + x^2) - exp(x)*(18*x + 6*x^2) + 9*x^2 + log(6*x + log(20*x^2)^2
+ log(20*x^2)*(2*x + 6) + x^2 + 9)^2*(3*x + x*log(20*x^2) + x^2)),x)

[Out]

5/(log(6*x + log(20*x^2)^2 + log(20*x^2)*(2*x + 6) + x^2 + 9) + exp(x) - 3)

________________________________________________________________________________________

sympy [A]  time = 0.60, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{e^{x} + \log {\left (x^{2} + 6 x + \left (2 x + 6\right ) \log {\left (20 x^{2} \right )} + \log {\left (20 x^{2} \right )}^{2} + 9 \right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*exp(x)*ln(20*x**2)+(-5*x**2-15*x)*exp(x)-10*x-20)/((x*ln(20*x**2)+x**2+3*x)*ln(ln(20*x**2)**2+
(2*x+6)*ln(20*x**2)+x**2+6*x+9)**2+((2*exp(x)*x-6*x)*ln(20*x**2)+(2*x**2+6*x)*exp(x)-6*x**2-18*x)*ln(ln(20*x**
2)**2+(2*x+6)*ln(20*x**2)+x**2+6*x+9)+(x*exp(x)**2-6*exp(x)*x+9*x)*ln(20*x**2)+(x**2+3*x)*exp(x)**2+(-6*x**2-1
8*x)*exp(x)+9*x**2+27*x),x)

[Out]

5/(exp(x) + log(x**2 + 6*x + (2*x + 6)*log(20*x**2) + log(20*x**2)**2 + 9) - 3)

________________________________________________________________________________________