3.59.2 \(\int \frac {e^{-\frac {2 x^4}{625+1050 x+641 x^2+168 x^3+16 x^4}} (-750 x^3-315 x^4)}{15625+39375 x+40575 x^2+21861 x^3+6492 x^4+1008 x^5+64 x^6} \, dx\)

Optimal. Leaf size=22 \[ \frac {15}{4} e^{-\frac {2 x^4}{\left (x+(5+2 x)^2\right )^2}} \]

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Rubi [A]  time = 0.69, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {1593, 6688, 12, 6706} \begin {gather*} \frac {15}{4} e^{-\frac {2 x^4}{\left (4 x^2+21 x+25\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-750*x^3 - 315*x^4)/(E^((2*x^4)/(625 + 1050*x + 641*x^2 + 168*x^3 + 16*x^4))*(15625 + 39375*x + 40575*x^2
+ 21861*x^3 + 6492*x^4 + 1008*x^5 + 64*x^6)),x]

[Out]

15/(4*E^((2*x^4)/(25 + 21*x + 4*x^2)^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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6688

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

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {2 x^4}{625+1050 x+641 x^2+168 x^3+16 x^4}} (-750-315 x) x^3}{15625+39375 x+40575 x^2+21861 x^3+6492 x^4+1008 x^5+64 x^6} \, dx\\ &=\int \frac {15 e^{-\frac {2 x^4}{\left (25+21 x+4 x^2\right )^2}} (-50-21 x) x^3}{\left (25+21 x+4 x^2\right )^3} \, dx\\ &=15 \int \frac {e^{-\frac {2 x^4}{\left (25+21 x+4 x^2\right )^2}} (-50-21 x) x^3}{\left (25+21 x+4 x^2\right )^3} \, dx\\ &=\frac {15}{4} e^{-\frac {2 x^4}{\left (25+21 x+4 x^2\right )^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.52, size = 23, normalized size = 1.05 \begin {gather*} \frac {15}{4} e^{-\frac {2 x^4}{\left (25+21 x+4 x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-750*x^3 - 315*x^4)/(E^((2*x^4)/(625 + 1050*x + 641*x^2 + 168*x^3 + 16*x^4))*(15625 + 39375*x + 405
75*x^2 + 21861*x^3 + 6492*x^4 + 1008*x^5 + 64*x^6)),x]

[Out]

15/(4*E^((2*x^4)/(25 + 21*x + 4*x^2)^2))

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fricas [A]  time = 0.56, size = 30, normalized size = 1.36 \begin {gather*} \frac {15}{4} \, e^{\left (-\frac {2 \, x^{4}}{16 \, x^{4} + 168 \, x^{3} + 641 \, x^{2} + 1050 \, x + 625}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-315*x^4-750*x^3)/(64*x^6+1008*x^5+6492*x^4+21861*x^3+40575*x^2+39375*x+15625)/exp(x^4/(16*x^4+168*
x^3+641*x^2+1050*x+625))^2,x, algorithm="fricas")

[Out]

15/4*e^(-2*x^4/(16*x^4 + 168*x^3 + 641*x^2 + 1050*x + 625))

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giac [A]  time = 0.20, size = 30, normalized size = 1.36 \begin {gather*} \frac {15}{4} \, e^{\left (-\frac {2 \, x^{4}}{16 \, x^{4} + 168 \, x^{3} + 641 \, x^{2} + 1050 \, x + 625}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-315*x^4-750*x^3)/(64*x^6+1008*x^5+6492*x^4+21861*x^3+40575*x^2+39375*x+15625)/exp(x^4/(16*x^4+168*
x^3+641*x^2+1050*x+625))^2,x, algorithm="giac")

[Out]

15/4*e^(-2*x^4/(16*x^4 + 168*x^3 + 641*x^2 + 1050*x + 625))

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maple [A]  time = 0.13, size = 21, normalized size = 0.95

method	result	size
risch	\(\frac {15 \,{\mathrm e}^{-\frac {2 x^{4}}{\left (4 x^{2}+21 x +25\right )^{2}}}}{4}\)	\(21\)
gosper	\(\frac {15 \,{\mathrm e}^{-\frac {2 x^{4}}{16 x^{4}+168 x^{3}+641 x^{2}+1050 x +625}}}{4}\)	\(32\)
norman	\(\frac {\left (\frac {9375}{4}+\frac {7875}{2} x +\frac {9615}{4} x^{2}+630 x^{3}+60 x^{4}\right ) {\mathrm e}^{-\frac {2 x^{4}}{16 x^{4}+168 x^{3}+641 x^{2}+1050 x +625}}}{\left (4 x^{2}+21 x +25\right )^{2}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-315*x^4-750*x^3)/(64*x^6+1008*x^5+6492*x^4+21861*x^3+40575*x^2+39375*x+15625)/exp(x^4/(16*x^4+168*x^3+64
1*x^2+1050*x+625))^2,x,method=_RETURNVERBOSE)

[Out]

15/4*exp(-2*x^4/(4*x^2+21*x+25)^2)

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maxima [B]  time = 4.99, size = 83, normalized size = 3.77 \begin {gather*} \frac {15}{4} \, e^{\left (\frac {5061 \, x}{32 \, {\left (16 \, x^{4} + 168 \, x^{3} + 641 \, x^{2} + 1050 \, x + 625\right )}} + \frac {21 \, x}{4 \, {\left (4 \, x^{2} + 21 \, x + 25\right )}} + \frac {8525}{32 \, {\left (16 \, x^{4} + 168 \, x^{3} + 641 \, x^{2} + 1050 \, x + 625\right )}} - \frac {241}{32 \, {\left (4 \, x^{2} + 21 \, x + 25\right )}} - \frac {1}{8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-315*x^4-750*x^3)/(64*x^6+1008*x^5+6492*x^4+21861*x^3+40575*x^2+39375*x+15625)/exp(x^4/(16*x^4+168*
x^3+641*x^2+1050*x+625))^2,x, algorithm="maxima")

[Out]

15/4*e^(5061/32*x/(16*x^4 + 168*x^3 + 641*x^2 + 1050*x + 625) + 21/4*x/(4*x^2 + 21*x + 25) + 8525/32/(16*x^4 +
168*x^3 + 641*x^2 + 1050*x + 625) - 241/32/(4*x^2 + 21*x + 25) - 1/8)

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mupad [B]  time = 4.36, size = 30, normalized size = 1.36 \begin {gather*} \frac {15\,{\mathrm {e}}^{-\frac {2\,x^4}{16\,x^4+168\,x^3+641\,x^2+1050\,x+625}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*x^4)/(1050*x + 641*x^2 + 168*x^3 + 16*x^4 + 625))*(750*x^3 + 315*x^4))/(39375*x + 40575*x^2 + 21
861*x^3 + 6492*x^4 + 1008*x^5 + 64*x^6 + 15625),x)

[Out]

(15*exp(-(2*x^4)/(1050*x + 641*x^2 + 168*x^3 + 16*x^4 + 625)))/4

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sympy [A]  time = 0.25, size = 29, normalized size = 1.32 \begin {gather*} \frac {15 e^{- \frac {2 x^{4}}{16 x^{4} + 168 x^{3} + 641 x^{2} + 1050 x + 625}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-315*x**4-750*x**3)/(64*x**6+1008*x**5+6492*x**4+21861*x**3+40575*x**2+39375*x+15625)/exp(x**4/(16*
x**4+168*x**3+641*x**2+1050*x+625))**2,x)

[Out]

15*exp(-2*x**4/(16*x**4 + 168*x**3 + 641*x**2 + 1050*x + 625))/4

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