3.84.25 \(\int \frac {(600-160 x-136 x^2-16 x^3) \log ^2(2 x)+e^{\frac {x}{(40+8 x) \log (2 x)}} (-5 x-x^2+5 x \log (2 x)+(200+80 x+8 x^2) \log ^2(2 x))}{(200+80 x+8 x^2) \log ^2(2 x)} \, dx\)

Optimal. Leaf size=25 \[ \left (3+e^{\frac {x}{8 (5+x) \log (2 x)}}-x\right ) x \]

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Rubi [B]  time = 1.17, antiderivative size = 97, normalized size of antiderivative = 3.88, number of steps used = 5, number of rules used = 4, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 12, 6742, 2288} \begin {gather*} \frac {e^{\frac {x}{8 (x+5) \log (2 x)}} \left (x^2+5 x-5 x \log (2 x)\right )}{(x+5)^2 \left (\frac {1}{(x+5) \log ^2(2 x)}+\frac {x}{(x+5)^2 \log (2 x)}-\frac {1}{(x+5) \log (2 x)}\right ) \log ^2(2 x)}-\frac {1}{4} (3-2 x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((600 - 160*x - 136*x^2 - 16*x^3)*Log[2*x]^2 + E^(x/((40 + 8*x)*Log[2*x]))*(-5*x - x^2 + 5*x*Log[2*x] + (2
00 + 80*x + 8*x^2)*Log[2*x]^2))/((200 + 80*x + 8*x^2)*Log[2*x]^2),x]

[Out]

-1/4*(3 - 2*x)^2 + (E^(x/(8*(5 + x)*Log[2*x]))*(5*x + x^2 - 5*x*Log[2*x]))/((5 + x)^2*(1/((5 + x)*Log[2*x]^2)
+ x/((5 + x)^2*Log[2*x]) - 1/((5 + x)*Log[2*x]))*Log[2*x]^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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (600-160 x-136 x^2-16 x^3\right ) \log ^2(2 x)+e^{\frac {x}{(40+8 x) \log (2 x)}} \left (-5 x-x^2+5 x \log (2 x)+\left (200+80 x+8 x^2\right ) \log ^2(2 x)\right )}{8 (5+x)^2 \log ^2(2 x)} \, dx\\ &=\frac {1}{8} \int \frac {\left (600-160 x-136 x^2-16 x^3\right ) \log ^2(2 x)+e^{\frac {x}{(40+8 x) \log (2 x)}} \left (-5 x-x^2+5 x \log (2 x)+\left (200+80 x+8 x^2\right ) \log ^2(2 x)\right )}{(5+x)^2 \log ^2(2 x)} \, dx\\ &=\frac {1}{8} \int \left (-8 (-3+2 x)+\frac {e^{\frac {x}{8 (5+x) \log (2 x)}} \left (-5 x-x^2+5 x \log (2 x)+200 \log ^2(2 x)+80 x \log ^2(2 x)+8 x^2 \log ^2(2 x)\right )}{(5+x)^2 \log ^2(2 x)}\right ) \, dx\\ &=-\frac {1}{4} (3-2 x)^2+\frac {1}{8} \int \frac {e^{\frac {x}{8 (5+x) \log (2 x)}} \left (-5 x-x^2+5 x \log (2 x)+200 \log ^2(2 x)+80 x \log ^2(2 x)+8 x^2 \log ^2(2 x)\right )}{(5+x)^2 \log ^2(2 x)} \, dx\\ &=-\frac {1}{4} (3-2 x)^2+\frac {e^{\frac {x}{8 (5+x) \log (2 x)}} \left (5 x+x^2-5 x \log (2 x)\right )}{(5+x)^2 \left (\frac {1}{(5+x) \log ^2(2 x)}+\frac {x}{(5+x)^2 \log (2 x)}-\frac {1}{(5+x) \log (2 x)}\right ) \log ^2(2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 26, normalized size = 1.04 \begin {gather*} -x \left (-3-e^{\frac {x}{8 (5+x) \log (2 x)}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((600 - 160*x - 136*x^2 - 16*x^3)*Log[2*x]^2 + E^(x/((40 + 8*x)*Log[2*x]))*(-5*x - x^2 + 5*x*Log[2*x
] + (200 + 80*x + 8*x^2)*Log[2*x]^2))/((200 + 80*x + 8*x^2)*Log[2*x]^2),x]

[Out]

-(x*(-3 - E^(x/(8*(5 + x)*Log[2*x])) + x))

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fricas [A]  time = 0.67, size = 26, normalized size = 1.04 \begin {gather*} -x^{2} + x e^{\left (\frac {x}{8 \, {\left (x + 5\right )} \log \left (2 \, x\right )}\right )} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+80*x+200)*log(2*x)^2+5*x*log(2*x)-x^2-5*x)*exp(x/(8*x+40)/log(2*x))+(-16*x^3-136*x^2-160*x+
600)*log(2*x)^2)/(8*x^2+80*x+200)/log(2*x)^2,x, algorithm="fricas")

[Out]

-x^2 + x*e^(1/8*x/((x + 5)*log(2*x))) + 3*x

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giac [A]  time = 0.38, size = 30, normalized size = 1.20 \begin {gather*} -x^{2} + x e^{\left (\frac {x}{8 \, {\left (x \log \left (2 \, x\right ) + 5 \, \log \left (2 \, x\right )\right )}}\right )} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+80*x+200)*log(2*x)^2+5*x*log(2*x)-x^2-5*x)*exp(x/(8*x+40)/log(2*x))+(-16*x^3-136*x^2-160*x+
600)*log(2*x)^2)/(8*x^2+80*x+200)/log(2*x)^2,x, algorithm="giac")

[Out]

-x^2 + x*e^(1/8*x/(x*log(2*x) + 5*log(2*x))) + 3*x

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maple [A]  time = 0.71, size = 27, normalized size = 1.08

method	result	size
risch	\(-x^{2}+{\mathrm e}^{\frac {x}{8 \left (5+x \right ) \ln \left (2 x \right )}} x +3 x\)	\(27\)
default	\(3 x +\frac {40 \ln \relax (2) {\mathrm e}^{\frac {x}{8 \left (5+x \right ) \left (\ln \relax (2)+\ln \relax (x )\right )}} x +40 \ln \relax (x ) {\mathrm e}^{\frac {x}{8 \left (5+x \right ) \left (\ln \relax (2)+\ln \relax (x )\right )}} x +8 x^{2} \ln \relax (2) {\mathrm e}^{\frac {x}{8 \left (5+x \right ) \left (\ln \relax (2)+\ln \relax (x )\right )}}+8 \ln \relax (x ) x^{2} {\mathrm e}^{\frac {x}{8 \left (5+x \right ) \left (\ln \relax (2)+\ln \relax (x )\right )}}}{8 \left (5+x \right ) \left (\ln \relax (2)+\ln \relax (x )\right )}-x^{2}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2+80*x+200)*ln(2*x)^2+5*x*ln(2*x)-x^2-5*x)*exp(x/(8*x+40)/ln(2*x))+(-16*x^3-136*x^2-160*x+600)*ln(2
*x)^2)/(8*x^2+80*x+200)/ln(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

-x^2+exp(1/8*x/(5+x)/ln(2*x))*x+3*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+80*x+200)*log(2*x)^2+5*x*log(2*x)-x^2-5*x)*exp(x/(8*x+40)/log(2*x))+(-16*x^3-136*x^2-160*x+
600)*log(2*x)^2)/(8*x^2+80*x+200)/log(2*x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 5.35, size = 31, normalized size = 1.24 \begin {gather*} x\,\left ({\mathrm {e}}^{\frac {x}{40\,\ln \relax (2)+40\,\ln \relax (x)+8\,x\,\ln \relax (2)+8\,x\,\ln \relax (x)}}-x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x/(log(2*x)*(8*x + 40)))*(5*x - 5*x*log(2*x) - log(2*x)^2*(80*x + 8*x^2 + 200) + x^2) + log(2*x)^2*(
160*x + 136*x^2 + 16*x^3 - 600))/(log(2*x)^2*(80*x + 8*x^2 + 200)),x)

[Out]

x*(exp(x/(40*log(2) + 40*log(x) + 8*x*log(2) + 8*x*log(x))) - x + 3)

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sympy [A]  time = 51.16, size = 20, normalized size = 0.80 \begin {gather*} - x^{2} + x e^{\frac {x}{\left (8 x + 40\right ) \log {\left (2 x \right )}}} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2+80*x+200)*ln(2*x)**2+5*x*ln(2*x)-x**2-5*x)*exp(x/(8*x+40)/ln(2*x))+(-16*x**3-136*x**2-160*
x+600)*ln(2*x)**2)/(8*x**2+80*x+200)/ln(2*x)**2,x)

[Out]

-x**2 + x*exp(x/((8*x + 40)*log(2*x))) + 3*x

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