3.17.76 \(\int \frac {-e^8 x+(x^2)^{\frac {25 e^8-10 x^2}{e^8}} (50 e^8-20 x^2-20 x^2 \log (x^2))}{e^8 x} \, dx\)

Optimal. Leaf size=20 \[ -x+\left (x^2\right )^{5 \left (5-\frac {2 x^2}{e^8}\right )} \]

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Rubi [F]  time = 1.29, 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 {-e^8 x+\left (x^2\right )^{\frac {25 e^8-10 x^2}{e^8}} \left (50 e^8-20 x^2-20 x^2 \log \left (x^2\right )\right )}{e^8 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^8*x) + (x^2)^((25*E^8 - 10*x^2)/E^8)*(50*E^8 - 20*x^2 - 20*x^2*Log[x^2]))/(E^8*x),x]

[Out]

-x + 25*Defer[Subst][Defer[Int][x^(24 - (10*x)/E^8), x], x, x^2] - (10*Defer[Subst][Defer[Int][x^(25 - (10*x)/
E^8), x], x, x^2])/E^8 - (10*Log[x^2]*Defer[Subst][Defer[Int][x^(25 - (10*x)/E^8), x], x, x^2])/E^8 + (10*Defe
r[Subst][Defer[Int][Defer[Int][x^(25 - (10*x)/E^8), x]/x, x], x, x^2])/E^8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-e^8 x+\left (x^2\right )^{\frac {25 e^8-10 x^2}{e^8}} \left (50 e^8-20 x^2-20 x^2 \log \left (x^2\right )\right )}{x} \, dx}{e^8}\\ &=\frac {\int \left (-e^8-10 x^{49} \left (x^2\right )^{-\frac {10 x^2}{e^8}} \left (-5 e^8+2 x^2+2 x^2 \log \left (x^2\right )\right )\right ) \, dx}{e^8}\\ &=-x-\frac {10 \int x^{49} \left (x^2\right )^{-\frac {10 x^2}{e^8}} \left (-5 e^8+2 x^2+2 x^2 \log \left (x^2\right )\right ) \, dx}{e^8}\\ &=-x-\frac {5 \operatorname {Subst}\left (\int x^{24-\frac {10 x}{e^8}} \left (-5 e^8+2 x+2 x \log (x)\right ) \, dx,x,x^2\right )}{e^8}\\ &=-x-\frac {5 \operatorname {Subst}\left (\int \left (-5 e^8 x^{24-\frac {10 x}{e^8}}+2 x^{25-\frac {10 x}{e^8}}+2 x^{25-\frac {10 x}{e^8}} \log (x)\right ) \, dx,x,x^2\right )}{e^8}\\ &=-x+25 \operatorname {Subst}\left (\int x^{24-\frac {10 x}{e^8}} \, dx,x,x^2\right )-\frac {10 \operatorname {Subst}\left (\int x^{25-\frac {10 x}{e^8}} \, dx,x,x^2\right )}{e^8}-\frac {10 \operatorname {Subst}\left (\int x^{25-\frac {10 x}{e^8}} \log (x) \, dx,x,x^2\right )}{e^8}\\ &=-x+25 \operatorname {Subst}\left (\int x^{24-\frac {10 x}{e^8}} \, dx,x,x^2\right )-\frac {10 \operatorname {Subst}\left (\int x^{25-\frac {10 x}{e^8}} \, dx,x,x^2\right )}{e^8}+\frac {10 \operatorname {Subst}\left (\int \frac {\int x^{25-\frac {10 x}{e^8}} \, dx}{x} \, dx,x,x^2\right )}{e^8}-\frac {\left (10 \log \left (x^2\right )\right ) \operatorname {Subst}\left (\int x^{25-\frac {10 x}{e^8}} \, dx,x,x^2\right )}{e^8}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 18, normalized size = 0.90 \begin {gather*} -x+\left (x^2\right )^{25-\frac {10 x^2}{e^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x + (x^2)^(25 - (10*x^2)/E^8)

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fricas [B]  time = 0.68, size = 44, normalized size = 2.20 \begin {gather*} -\frac {{\left (x^{2}\right )}^{5 \, {\left (2 \, x^{2} - 5 \, e^{8}\right )} e^{\left (-8\right )}} x - 1}{{\left (x^{2}\right )}^{5 \, {\left (2 \, x^{2} - 5 \, e^{8}\right )} e^{\left (-8\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.14, size = 18, normalized size = 0.90

method	result	size
risch	\(\left (x^{2}\right )^{25-10 x^{2} {\mathrm e}^{-8}}-x\)	\(18\)
norman	\(\left ({\mathrm e}^{4} {\mathrm e}^{\left (25 \,{\mathrm e}^{8}-10 x^{2}\right ) \ln \left (x^{2}\right ) {\mathrm e}^{-8}}-x \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4}\)	\(37\)
default	\({\mathrm e}^{-8} \left ({\mathrm e}^{8} {\mathrm e}^{\left (25 \,{\mathrm e}^{8}-10 x^{2}\right ) \ln \left (x^{2}\right ) {\mathrm e}^{-8}}-x \,{\mathrm e}^{8}\right )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2)^(25-10*x^2*exp(-8))-x

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maxima [A]  time = 0.49, size = 25, normalized size = 1.25 \begin {gather*} {\left (x^{50} e^{\left (-20 \, x^{2} e^{\left (-8\right )} \log \relax (x) + 8\right )} - x e^{8}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2*log(x^2)+50*exp(4)^2-20*x^2)*exp((25*exp(4)^2-10*x^2)*log(x^2)/exp(4)^2)-x*exp(4)^2)/x/exp
(4)^2,x, algorithm="maxima")

[Out]

(x^50*e^(-20*x^2*e^(-8)*log(x) + 8) - x*e^8)*e^(-8)

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mupad [B]  time = 1.28, size = 21, normalized size = 1.05 \begin {gather*} \frac {x^{50}}{{\left (x^2\right )}^{10\,x^2\,{\mathrm {e}}^{-8}}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(x*exp(8) + exp(log(x^2)*exp(-8)*(25*exp(8) - 10*x^2))*(20*x^2*log(x^2) - 50*exp(8) + 20*x^2)))/
x,x)

[Out]

x^50/(x^2)^(10*x^2*exp(-8)) - x

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sympy [A]  time = 0.39, size = 20, normalized size = 1.00 \begin {gather*} - x + e^{\frac {\left (- 10 x^{2} + 25 e^{8}\right ) \log {\left (x^{2} \right )}}{e^{8}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + exp((-10*x**2 + 25*exp(8))*exp(-8)*log(x**2))

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