3.38.55 \(\int \frac {e^{2 x} (200+100 x-50 x^2)+e^x (-4+18 x-6 x^2)+(e^{2 x} (-200 x+50 x^2)+e^x (4 x-9 x^2+2 x^3)) \log (-4 x+9 x^2-2 x^3+e^x (200 x-50 x^2))}{(4 x-9 x^2+2 x^3+e^x (-200 x+50 x^2)) \log ^2(-4 x+9 x^2-2 x^3+e^x (200 x-50 x^2))} \, dx\)

Optimal. Leaf size=27 \[ 1+\frac {e^x}{\log \left (2 (4-x) x \left (-\frac {1}{2}+25 e^x+x\right )\right )} \]

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Rubi [A]  time = 1.31, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, integrand size = 154, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6688, 2288} \begin {gather*} \frac {e^x}{\log \left (-\left (\left (-2 x-50 e^x+1\right ) (4-x) x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(200 + 100*x - 50*x^2) + E^x*(-4 + 18*x - 6*x^2) + (E^(2*x)*(-200*x + 50*x^2) + E^x*(4*x - 9*x^2
+ 2*x^3))*Log[-4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*x^2)])/((4*x - 9*x^2 + 2*x^3 + E^x*(-200*x + 50*x^2))*Log
[-4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*x^2)]^2),x]

[Out]

E^x/Log[-((1 - 50*E^x - 2*x)*(4 - x)*x)]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 {e^x \left (-4+18 x-6 x^2-50 e^x \left (-4-2 x+x^2\right )+(-4+x) x \left (-1+50 e^x+2 x\right ) \log \left (-\left ((-4+x) x \left (-1+50 e^x+2 x\right )\right )\right )\right )}{\left (1-50 e^x-2 x\right ) (4-x) x \log ^2\left (-\left ((-4+x) x \left (-1+50 e^x+2 x\right )\right )\right )} \, dx\\ &=\frac {e^x}{\log \left (-\left (\left (1-50 e^x-2 x\right ) (4-x) x\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 23, normalized size = 0.85 \begin {gather*} \frac {e^x}{\log \left (-\left ((-4+x) x \left (-1+50 e^x+2 x\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(200 + 100*x - 50*x^2) + E^x*(-4 + 18*x - 6*x^2) + (E^(2*x)*(-200*x + 50*x^2) + E^x*(4*x -
9*x^2 + 2*x^3))*Log[-4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*x^2)])/((4*x - 9*x^2 + 2*x^3 + E^x*(-200*x + 50*x^2
))*Log[-4*x + 9*x^2 - 2*x^3 + E^x*(200*x - 50*x^2)]^2),x]

[Out]

E^x/Log[-((-4 + x)*x*(-1 + 50*E^x + 2*x))]

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fricas [A]  time = 1.02, size = 31, normalized size = 1.15 \begin {gather*} \frac {e^{x}}{\log \left (-2 \, x^{3} + 9 \, x^{2} - 50 \, {\left (x^{2} - 4 \, x\right )} e^{x} - 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2+200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50
*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x-4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*e
xp(x)-2*x^3+9*x^2-4*x)^2,x, algorithm="fricas")

[Out]

e^x/log(-2*x^3 + 9*x^2 - 50*(x^2 - 4*x)*e^x - 4*x)

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giac [A]  time = 0.37, size = 32, normalized size = 1.19 \begin {gather*} \frac {e^{x}}{\log \left (-2 \, x^{3} - 50 \, x^{2} e^{x} + 9 \, x^{2} + 200 \, x e^{x} - 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2+200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50
*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x-4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*e
xp(x)-2*x^3+9*x^2-4*x)^2,x, algorithm="giac")

[Out]

e^x/log(-2*x^3 - 50*x^2*e^x + 9*x^2 + 200*x*e^x - 4*x)

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maple [C]  time = 0.14, size = 222, normalized size = 8.22

method	result	size
risch	\(\frac {2 i {\mathrm e}^{x}}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) \mathrm {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right ) \mathrm {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )^{2}-\pi \mathrm {csgn}\left (i x \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )\right )^{3}-2 \pi +2 i \ln \relax (x )+2 i \ln \left (x^{2}+\left (25 \,{\mathrm e}^{x}-\frac {9}{2}\right ) x -100 \,{\mathrm e}^{x}+2\right )}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*ln((-50*x^2+200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50*x^2+10
0*x+200)*exp(x)^2+(-6*x^2+18*x-4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/ln((-50*x^2+200*x)*exp(x)-2*
x^3+9*x^2-4*x)^2,x,method=_RETURNVERBOSE)

[Out]

2*I*exp(x)/(Pi*csgn(I*x)*csgn(I*(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))*csgn(I*x*(x^2+(25*exp(x)-9/2)*x-100*exp(
x)+2))-Pi*csgn(I*x)*csgn(I*x*(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))^2+2*Pi*csgn(I*x*(x^2+(25*exp(x)-9/2)*x-100*
exp(x)+2))^2-Pi*csgn(I*(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))*csgn(I*x*(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))^2-
Pi*csgn(I*x*(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))^3-2*Pi+2*I*ln(x)+2*I*ln(x^2+(25*exp(x)-9/2)*x-100*exp(x)+2))

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maxima [A]  time = 0.57, size = 22, normalized size = 0.81 \begin {gather*} \frac {e^{x}}{\log \left (x - 4\right ) + \log \relax (x) + \log \left (-2 \, x - 50 \, e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-200*x)*exp(x)^2+(2*x^3-9*x^2+4*x)*exp(x))*log((-50*x^2+200*x)*exp(x)-2*x^3+9*x^2-4*x)+(-50
*x^2+100*x+200)*exp(x)^2+(-6*x^2+18*x-4)*exp(x))/((50*x^2-200*x)*exp(x)+2*x^3-9*x^2+4*x)/log((-50*x^2+200*x)*e
xp(x)-2*x^3+9*x^2-4*x)^2,x, algorithm="maxima")

[Out]

e^x/(log(x - 4) + log(x) + log(-2*x - 50*e^x + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (200\,x-50\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^3-9\,x^2+4\,x\right )\right )-{\mathrm {e}}^{2\,x}\,\left (-50\,x^2+100\,x+200\right )+{\mathrm {e}}^x\,\left (6\,x^2-18\,x+4\right )}{{\ln \left ({\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-4\,x+9\,x^2-2\,x^3\right )}^2\,\left (4\,x-{\mathrm {e}}^x\,\left (200\,x-50\,x^2\right )-9\,x^2+2\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^2 - 2*x^3)*(exp(2*x)*(200*x - 50*x^2) - exp(x)*(4*x - 9*x^2 + 2*
x^3)) - exp(2*x)*(100*x - 50*x^2 + 200) + exp(x)*(6*x^2 - 18*x + 4))/(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^
2 - 2*x^3)^2*(4*x - exp(x)*(200*x - 50*x^2) - 9*x^2 + 2*x^3)),x)

[Out]

int(-(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^2 - 2*x^3)*(exp(2*x)*(200*x - 50*x^2) - exp(x)*(4*x - 9*x^2 + 2*
x^3)) - exp(2*x)*(100*x - 50*x^2 + 200) + exp(x)*(6*x^2 - 18*x + 4))/(log(exp(x)*(200*x - 50*x^2) - 4*x + 9*x^
2 - 2*x^3)^2*(4*x - exp(x)*(200*x - 50*x^2) - 9*x^2 + 2*x^3)), x)

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sympy [A]  time = 0.37, size = 29, normalized size = 1.07 \begin {gather*} \frac {e^{x}}{\log {\left (- 2 x^{3} + 9 x^{2} - 4 x + \left (- 50 x^{2} + 200 x\right ) e^{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x**2-200*x)*exp(x)**2+(2*x**3-9*x**2+4*x)*exp(x))*ln((-50*x**2+200*x)*exp(x)-2*x**3+9*x**2-4*x
)+(-50*x**2+100*x+200)*exp(x)**2+(-6*x**2+18*x-4)*exp(x))/((50*x**2-200*x)*exp(x)+2*x**3-9*x**2+4*x)/ln((-50*x
**2+200*x)*exp(x)-2*x**3+9*x**2-4*x)**2,x)

[Out]

exp(x)/log(-2*x**3 + 9*x**2 - 4*x + (-50*x**2 + 200*x)*exp(x))

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