\(\int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} (10 x-60 x^{11}+65 x^{12}+e^x (-60 x^{11}-5 x^{12})) \, dx\) [5887]

Optimal result

Integrand size = 55, antiderivative size = 23 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} e^{x^2-\left (1+e^x-x\right ) x^{12}} \]

[Out]

5/4*exp(x^2-x^12*(1+exp(x)-x))

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 6838} \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} e^{x^{13}-e^x x^{12}-x^{12}+x^2} \]

[In]

Int[(E^(x^2 - x^12 - E^x*x^12 + x^13)*(10*x - 60*x^11 + 65*x^12 + E^x*(-60*x^11 - 5*x^12)))/4,x]

[Out]

(5*E^(x^2 - x^12 - E^x*x^12 + x^13))/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx \\ & = \frac {5}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} e^{x^2-\left (1+e^x\right ) x^{12}+x^{13}} \]

[In]

Integrate[(E^(x^2 - x^12 - E^x*x^12 + x^13)*(10*x - 60*x^11 + 65*x^12 + E^x*(-60*x^11 - 5*x^12)))/4,x]

[Out]

(5*E^(x^2 - (1 + E^x)*x^12 + x^13))/4

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {5 \,{\mathrm e}^{-x^{12} {\mathrm e}^{x}+x^{13}-x^{12}+x^{2}}}{4}\)	\(23\)
risch	\(\frac {5 \,{\mathrm e}^{-x^{2} \left ({\mathrm e}^{x} x^{10}-x^{11}+x^{10}-1\right )}}{4}\)	\(25\)

[In]

int(1/4*((-5*x^12-60*x^11)*exp(x)+65*x^12-60*x^11+10*x)*exp(-x^12*exp(x)+x^13-x^12+x^2),x,method=_RETURNVERBOS
E)

[Out]

5/4*exp(-x^12*exp(x)+x^13-x^12+x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} \, e^{\left (x^{13} - x^{12} e^{x} - x^{12} + x^{2}\right )} \]

[In]

integrate(1/4*((-5*x^12-60*x^11)*exp(x)+65*x^12-60*x^11+10*x)*exp(-x^12*exp(x)+x^13-x^12+x^2),x, algorithm="fr
icas")

[Out]

5/4*e^(x^13 - x^12*e^x - x^12 + x^2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5 e^{x^{13} - x^{12} e^{x} - x^{12} + x^{2}}}{4} \]

[In]

integrate(1/4*((-5*x**12-60*x**11)*exp(x)+65*x**12-60*x**11+10*x)*exp(-x**12*exp(x)+x**13-x**12+x**2),x)

[Out]

5*exp(x**13 - x**12*exp(x) - x**12 + x**2)/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} \, e^{\left (x^{13} - x^{12} e^{x} - x^{12} + x^{2}\right )} \]

[In]

integrate(1/4*((-5*x^12-60*x^11)*exp(x)+65*x^12-60*x^11+10*x)*exp(-x^12*exp(x)+x^13-x^12+x^2),x, algorithm="ma
xima")

[Out]

5/4*e^(x^13 - x^12*e^x - x^12 + x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5}{4} \, e^{\left (x^{13} - x^{12} e^{x} - x^{12} + x^{2}\right )} \]

[In]

integrate(1/4*((-5*x^12-60*x^11)*exp(x)+65*x^12-60*x^11+10*x)*exp(-x^12*exp(x)+x^13-x^12+x^2),x, algorithm="gi
ac")

[Out]

5/4*e^(x^13 - x^12*e^x - x^12 + x^2)

Mupad [B] (verification not implemented)

Time = 12.59 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{4} e^{x^2-x^{12}-e^x x^{12}+x^{13}} \left (10 x-60 x^{11}+65 x^{12}+e^x \left (-60 x^{11}-5 x^{12}\right )\right ) \, dx=\frac {5\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{x^{13}}\,{\mathrm {e}}^{-x^{12}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^{12}}}{4} \]

[In]

int((exp(x^2 - x^12*exp(x) - x^12 + x^13)*(10*x - exp(x)*(60*x^11 + 5*x^12) - 60*x^11 + 65*x^12))/4,x)

[Out]

(5*exp(x^2)*exp(x^13)*exp(-x^12*exp(x))*exp(-x^12))/4