\(\int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx\) [2410]

Optimal result

Integrand size = 29, antiderivative size = 18 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \left (2-\frac {e^{100}}{2}\right )^4}{64 x^2} \]

[Out]

9/64*(2-1/2*exp(100))^4/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 30} \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \left (4-e^{100}\right )^4}{1024 x^2} \]

[In]

Int[(-2304 + 2304*E^100 - 864*E^200 + 144*E^300 - 9*E^400)/(512*x^3),x]

[Out]

(9*(4 - E^100)^4)/(1024*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{512} \left (9 \left (4-e^{100}\right )^4\right ) \int \frac {1}{x^3} \, dx\right ) \\ & = \frac {9 \left (4-e^{100}\right )^4}{1024 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \left (-4+e^{100}\right )^4}{1024 x^2} \]

[In]

Integrate[(-2304 + 2304*E^100 - 864*E^200 + 144*E^300 - 9*E^400)/(512*x^3),x]

[Out]

(9*(-4 + E^100)^4)/(1024*x^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(27\) vs. \(2(13)=26\).

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56

method	result	size
gosper	\(\frac {\frac {9 \,{\mathrm e}^{400}}{1024}-\frac {9 \,{\mathrm e}^{300}}{64}+\frac {27 \,{\mathrm e}^{200}}{32}-\frac {9 \,{\mathrm e}^{100}}{4}+\frac {9}{4}}{x^{2}}\)	\(28\)
norman	\(\frac {\frac {9 \,{\mathrm e}^{400}}{1024}-\frac {9 \,{\mathrm e}^{300}}{64}+\frac {27 \,{\mathrm e}^{200}}{32}-\frac {9 \,{\mathrm e}^{100}}{4}+\frac {9}{4}}{x^{2}}\)	\(29\)
default	\(-\frac {-\frac {9 \,{\mathrm e}^{400}}{512}+\frac {9 \,{\mathrm e}^{300}}{32}-\frac {27 \,{\mathrm e}^{200}}{16}+\frac {9 \,{\mathrm e}^{100}}{2}-\frac {9}{2}}{2 x^{2}}\)	\(30\)
parallelrisch	\(-\frac {-\frac {9 \,{\mathrm e}^{400}}{512}+\frac {9 \,{\mathrm e}^{300}}{32}-\frac {27 \,{\mathrm e}^{200}}{16}+\frac {9 \,{\mathrm e}^{100}}{2}-\frac {9}{2}}{2 x^{2}}\)	\(30\)
risch	\(\frac {9 \,{\mathrm e}^{400}}{1024 x^{2}}-\frac {9 \,{\mathrm e}^{300}}{64 x^{2}}+\frac {27 \,{\mathrm e}^{200}}{32 x^{2}}-\frac {9 \,{\mathrm e}^{100}}{4 x^{2}}+\frac {9}{4 x^{2}}\)	\(35\)

[In]

int(1/512*(-9*exp(100)^4+144*exp(100)^3-864*exp(100)^2+2304*exp(100)-2304)/x^3,x,method=_RETURNVERBOSE)

[Out]

9/1024*(exp(100)^4-16*exp(100)^3+96*exp(100)^2-256*exp(100)+256)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \, {\left (e^{400} - 16 \, e^{300} + 96 \, e^{200} - 256 \, e^{100} + 256\right )}}{1024 \, x^{2}} \]

[In]

integrate(1/512*(-9*exp(100)^4+144*exp(100)^3-864*exp(100)^2+2304*exp(100)-2304)/x^3,x, algorithm="fricas")

[Out]

9/1024*(e^400 - 16*e^300 + 96*e^200 - 256*e^100 + 256)/x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (14) = 28\).

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.00 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=- \frac {- \frac {9 e^{400}}{512} - \frac {27 e^{200}}{16} - \frac {9}{2} + \frac {9 e^{100}}{2} + \frac {9 e^{300}}{32}}{2 x^{2}} \]

[In]

integrate(1/512*(-9*exp(100)**4+144*exp(100)**3-864*exp(100)**2+2304*exp(100)-2304)/x**3,x)

[Out]

-(-9*exp(400)/512 - 27*exp(200)/16 - 9/2 + 9*exp(100)/2 + 9*exp(300)/32)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \, {\left (e^{400} - 16 \, e^{300} + 96 \, e^{200} - 256 \, e^{100} + 256\right )}}{1024 \, x^{2}} \]

[In]

integrate(1/512*(-9*exp(100)^4+144*exp(100)^3-864*exp(100)^2+2304*exp(100)-2304)/x^3,x, algorithm="maxima")

[Out]

9/1024*(e^400 - 16*e^300 + 96*e^200 - 256*e^100 + 256)/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9 \, {\left (e^{400} - 16 \, e^{300} + 96 \, e^{200} - 256 \, e^{100} + 256\right )}}{1024 \, x^{2}} \]

[In]

integrate(1/512*(-9*exp(100)^4+144*exp(100)^3-864*exp(100)^2+2304*exp(100)-2304)/x^3,x, algorithm="giac")

[Out]

9/1024*(e^400 - 16*e^300 + 96*e^200 - 256*e^100 + 256)/x^2

Mupad [B] (verification not implemented)

Time = 9.73 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {-2304+2304 e^{100}-864 e^{200}+144 e^{300}-9 e^{400}}{512 x^3} \, dx=\frac {9\,{\left ({\mathrm {e}}^{100}-4\right )}^4}{1024\,x^2} \]

[In]

int(-((27*exp(200))/16 - (9*exp(100))/2 - (9*exp(300))/32 + (9*exp(400))/512 + 9/2)/x^3,x)

[Out]

(9*(exp(100) - 4)^4)/(1024*x^2)