3.98.49 \(\int \frac {-50 x-20 e^4 x-2 e^8 x+(-20 x-4 e^4 x) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} (25+e^8+14 x^3+e^4 (10+3 x^3)+(9 x^2+2 e^4 x^2) \log (4)+(10+2 e^4+3 x^3+2 x^2 \log (4)) \log (x)+\log ^2(x))}{25+10 e^4+e^8+(10+2 e^4) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \left (e^{\frac {x^2 (x+\log (4))}{5+e^4+\log (x)}}-x\right ) x \]

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Rubi [B]  time = 2.98, antiderivative size = 99, normalized size of antiderivative = 3.96, number of steps used = 6, number of rules used = 4, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6, 6741, 6742, 2288} \begin {gather*} -x^2-\frac {4^{\frac {x^2}{\log (x)+e^4+5}} e^{\frac {x^3}{\log (x)+e^4+5}} \left (x^2 \log (16) \log (x)+\left (9+2 e^4\right ) x^2 \log (4)\right )}{\log (4) \left (\log (x)+e^4+5\right )^2 \left (\frac {x}{\left (\log (x)+e^4+5\right )^2}-\frac {2 x}{\log (x)+e^4+5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log[x]^2 + E^((x^3 + x^2*Log[4])/(5 + E^4 + L
og[x]))*(25 + E^8 + 14*x^3 + E^4*(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2*Log[4
])*Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[x]^2),x]

[Out]

-x^2 - (4^(x^2/(5 + E^4 + Log[x]))*E^(x^3/(5 + E^4 + Log[x]))*((9 + 2*E^4)*x^2*Log[4] + x^2*Log[16]*Log[x]))/(
Log[4]*(5 + E^4 + Log[x])^2*(x/(5 + E^4 + Log[x])^2 - (2*x)/(5 + E^4 + Log[x])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 6741

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

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 {-2 e^8 x+\left (-50-20 e^4\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx\\ &=\int \frac {\left (-50-20 e^4-2 e^8\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx\\ &=\int \frac {\left (-50-20 e^4-2 e^8\right ) x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2} \, dx\\ &=\int \left (-2 x+\frac {4^{\frac {x^2}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} e^{\frac {x^3}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} \left (25 \left (1+\frac {1}{25} e^4 \left (10+e^4\right )\right )+14 \left (1+\frac {3 e^4}{14}\right ) x^3+9 x^2 \log (4) \left (1+\frac {e^4 \log (16)}{9 \log (4)}\right )+10 \left (1+\frac {e^4}{5}\right ) \log (x)+3 x^3 \log (x)+x^2 \log (16) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2}\right ) \, dx\\ &=-x^2+\int \frac {4^{\frac {x^2}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} e^{\frac {x^3}{5 \left (1+\frac {e^4}{5}\right )+\log (x)}} \left (25 \left (1+\frac {1}{25} e^4 \left (10+e^4\right )\right )+14 \left (1+\frac {3 e^4}{14}\right ) x^3+9 x^2 \log (4) \left (1+\frac {e^4 \log (16)}{9 \log (4)}\right )+10 \left (1+\frac {e^4}{5}\right ) \log (x)+3 x^3 \log (x)+x^2 \log (16) \log (x)+\log ^2(x)\right )}{\left (5 \left (1+\frac {e^4}{5}\right )+\log (x)\right )^2} \, dx\\ &=-x^2-\frac {4^{\frac {x^2}{5+e^4+\log (x)}} e^{\frac {x^3}{5+e^4+\log (x)}} \left (x^2 \left (9 \log (4)+e^4 \log (16)\right )+x^2 \log (16) \log (x)\right )}{\log (4) \left (5+e^4+\log (x)\right )^2 \left (\frac {x}{\left (5+e^4+\log (x)\right )^2}-\frac {2 x}{5+e^4+\log (x)}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log[x]^2 + E^((x^3 + x^2*Log[4])/(5 + E
^4 + Log[x]))*(25 + E^8 + 14*x^3 + E^4*(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2
*Log[4])*Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[x]^2),x]

[Out]

Integrate[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log[x]^2 + E^((x^3 + x^2*Log[4])/(5 + E
^4 + Log[x]))*(25 + E^8 + 14*x^3 + E^4*(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2
*Log[4])*Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[x]^2), x]

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fricas [A]  time = 0.87, size = 29, normalized size = 1.16 \begin {gather*} -x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \relax (2)}{e^{4} + \log \relax (x) + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp(4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)
*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)
^2-20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),x, algorithm="fricas")

[Out]

-x^2 + x*e^((x^3 + 2*x^2*log(2))/(e^4 + log(x) + 5))

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giac [A]  time = 1.06, size = 29, normalized size = 1.16 \begin {gather*} -x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \relax (2)}{e^{4} + \log \relax (x) + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp(4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)
*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)
^2-20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),x, algorithm="giac")

[Out]

-x^2 + x*e^((x^3 + 2*x^2*log(2))/(e^4 + log(x) + 5))

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maple [A]  time = 0.24, size = 28, normalized size = 1.12

method	result	size
risch	\(-x^{2}+x \,{\mathrm e}^{\frac {\left (x +2 \ln \relax (2)\right ) x^{2}}{\ln \relax (x )+5+{\mathrm e}^{4}}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)^2+(4*x^2*ln(2)+2*exp(4)+3*x^3+10)*ln(x)+2*(2*x^2*exp(4)+9*x^2)*ln(2)+exp(4)^2+(3*x^3+10)*exp(4)+14
*x^3+25)*exp((2*x^2*ln(2)+x^3)/(ln(x)+5+exp(4)))-2*x*ln(x)^2+(-4*x*exp(4)-20*x)*ln(x)-2*x*exp(4)^2-20*x*exp(4)
-50*x)/(ln(x)^2+(2*exp(4)+10)*ln(x)+exp(4)^2+10*exp(4)+25),x,method=_RETURNVERBOSE)

[Out]

-x^2+x*exp((x+2*ln(2))*x^2/(ln(x)+5+exp(4)))

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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(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp(4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)
*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)
^2-20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),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 [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {50\,x+2\,x\,{\ln \relax (x)}^2+20\,x\,{\mathrm {e}}^4+2\,x\,{\mathrm {e}}^8+\ln \relax (x)\,\left (20\,x+4\,x\,{\mathrm {e}}^4\right )-{\mathrm {e}}^{\frac {x^3+2\,\ln \relax (2)\,x^2}{{\mathrm {e}}^4+\ln \relax (x)+5}}\,\left ({\mathrm {e}}^8+\ln \relax (x)\,\left (3\,x^3+4\,\ln \relax (2)\,x^2+2\,{\mathrm {e}}^4+10\right )+{\ln \relax (x)}^2+{\mathrm {e}}^4\,\left (3\,x^3+10\right )+14\,x^3+2\,\ln \relax (2)\,\left (2\,x^2\,{\mathrm {e}}^4+9\,x^2\right )+25\right )}{{\ln \relax (x)}^2+\left (2\,{\mathrm {e}}^4+10\right )\,\ln \relax (x)+10\,{\mathrm {e}}^4+{\mathrm {e}}^8+25} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x + 2*x*log(x)^2 + 20*x*exp(4) + 2*x*exp(8) + log(x)*(20*x + 4*x*exp(4)) - exp((2*x^2*log(2) + x^3)/(
exp(4) + log(x) + 5))*(exp(8) + log(x)*(2*exp(4) + 4*x^2*log(2) + 3*x^3 + 10) + log(x)^2 + exp(4)*(3*x^3 + 10)
+ 14*x^3 + 2*log(2)*(2*x^2*exp(4) + 9*x^2) + 25))/(10*exp(4) + exp(8) + log(x)^2 + log(x)*(2*exp(4) + 10) + 2
5),x)

[Out]

-int((50*x + 2*x*log(x)^2 + 20*x*exp(4) + 2*x*exp(8) + log(x)*(20*x + 4*x*exp(4)) - exp((2*x^2*log(2) + x^3)/(
exp(4) + log(x) + 5))*(exp(8) + log(x)*(2*exp(4) + 4*x^2*log(2) + 3*x^3 + 10) + log(x)^2 + exp(4)*(3*x^3 + 10)
+ 14*x^3 + 2*log(2)*(2*x^2*exp(4) + 9*x^2) + 25))/(10*exp(4) + exp(8) + log(x)^2 + log(x)*(2*exp(4) + 10) + 2
5), x)

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sympy [A]  time = 5.09, size = 26, normalized size = 1.04 \begin {gather*} - x^{2} + x e^{\frac {x^{3} + 2 x^{2} \log {\relax (2 )}}{\log {\relax (x )} + 5 + e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)**2+(4*x**2*ln(2)+2*exp(4)+3*x**3+10)*ln(x)+2*(2*x**2*exp(4)+9*x**2)*ln(2)+exp(4)**2+(3*x**3+
10)*exp(4)+14*x**3+25)*exp((2*x**2*ln(2)+x**3)/(ln(x)+5+exp(4)))-2*x*ln(x)**2+(-4*x*exp(4)-20*x)*ln(x)-2*x*exp
(4)**2-20*x*exp(4)-50*x)/(ln(x)**2+(2*exp(4)+10)*ln(x)+exp(4)**2+10*exp(4)+25),x)

[Out]

-x**2 + x*exp((x**3 + 2*x**2*log(2))/(log(x) + 5 + exp(4)))

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