3.40.64 \(\int \frac {e^{\frac {3+16 e^4 x-16 x^2+48 e^{3/4} x^2}{16 x}} (-3-16 x^2+48 e^{3/4} x^2)}{16 x^2} \, dx\) [3964]

Optimal. Leaf size=25 \[ e^{e^4-x+3 \left (e^{3/4}+\frac {1}{16 x^2}\right ) x} \]

[Out]

exp(3*(1/16/x^2+exp(3/4))*x+exp(4)-x)

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Rubi [A]
time = 0.37, antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 3, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6, 12, 6838} \begin {gather*} e^{\frac {48 e^{3/4} x^2-16 x^2+16 e^4 x+3}{16 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3 + 16*E^4*x - 16*x^2 + 48*E^(3/4)*x^2)/(16*x))*(-3 - 16*x^2 + 48*E^(3/4)*x^2))/(16*x^2),x]

[Out]

E^((3 + 16*E^4*x - 16*x^2 + 48*E^(3/4)*x^2)/(16*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 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3+16 e^4 x-16 x^2+48 e^{3/4} x^2}{16 x}} \left (-3+\left (-16+48 e^{3/4}\right ) x^2\right )}{16 x^2} \, dx\\ &=\frac {1}{16} \int \frac {e^{\frac {3+16 e^4 x-16 x^2+48 e^{3/4} x^2}{16 x}} \left (-3+\left (-16+48 e^{3/4}\right ) x^2\right )}{x^2} \, dx\\ &=e^{\frac {3+16 e^4 x-16 x^2+48 e^{3/4} x^2}{16 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 24, normalized size = 0.96 \begin {gather*} e^{e^4+\frac {3}{16 x}+\left (-1+3 e^{3/4}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 + 16*E^4*x - 16*x^2 + 48*E^(3/4)*x^2)/(16*x))*(-3 - 16*x^2 + 48*E^(3/4)*x^2))/(16*x^2),x]

[Out]

E^(E^4 + 3/(16*x) + (-1 + 3*E^(3/4))*x)

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Maple [A]
time = 0.30, size = 26, normalized size = 1.04

method	result	size
gosper	\({\mathrm e}^{\frac {16 x \,{\mathrm e}^{4}+48 x^{2} {\mathrm e}^{\frac {3}{4}}-16 x^{2}+3}{16 x}}\)	\(26\)
norman	\({\mathrm e}^{\frac {16 x \,{\mathrm e}^{4}+48 x^{2} {\mathrm e}^{\frac {3}{4}}-16 x^{2}+3}{16 x}}\)	\(26\)
risch	\({\mathrm e}^{\frac {16 x \,{\mathrm e}^{4}+48 x^{2} {\mathrm e}^{\frac {3}{4}}-16 x^{2}+3}{16 x}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(48*x^2*exp(3/4)-16*x^2-3)*exp(1/16*(16*x*exp(4)+48*x^2*exp(3/4)-16*x^2+3)/x)/x^2,x,method=_RETURNVER
BOSE)

[Out]

exp(1/16*(16*x*exp(4)+48*x^2*exp(3/4)-16*x^2+3)/x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
time = 0.56, size = 37, normalized size = 1.48 \begin {gather*} \frac {{\left (3 \, e^{\left (e^{4} + \frac {3}{4}\right )} - e^{\left (e^{4}\right )}\right )} e^{\left (3 \, x e^{\frac {3}{4}} - x + \frac {3}{16 \, x}\right )}}{3 \, e^{\frac {3}{4}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(48*x^2*exp(3/4)-16*x^2-3)*exp(1/16*(16*x*exp(4)+48*x^2*exp(3/4)-16*x^2+3)/x)/x^2,x, algorithm=
"maxima")

[Out]

(3*e^(e^4 + 3/4) - e^(e^4))*e^(3*x*e^(3/4) - x + 3/16/x)/(3*e^(3/4) - 1)

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Fricas [A]
time = 0.36, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (\frac {48 \, x^{2} e^{\frac {3}{4}} - 16 \, x^{2} + 16 \, x e^{4} + 3}{16 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(48*x^2*exp(3/4)-16*x^2-3)*exp(1/16*(16*x*exp(4)+48*x^2*exp(3/4)-16*x^2+3)/x)/x^2,x, algorithm=
"fricas")

[Out]

e^(1/16*(48*x^2*e^(3/4) - 16*x^2 + 16*x*e^4 + 3)/x)

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Sympy [A]
time = 0.07, size = 24, normalized size = 0.96 \begin {gather*} e^{\frac {- x^{2} + 3 x^{2} e^{\frac {3}{4}} + x e^{4} + \frac {3}{16}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(48*x**2*exp(3/4)-16*x**2-3)*exp(1/16*(16*x*exp(4)+48*x**2*exp(3/4)-16*x**2+3)/x)/x**2,x)

[Out]

exp((-x**2 + 3*x**2*exp(3/4) + x*exp(4) + 3/16)/x)

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Giac [A]
time = 0.42, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (\frac {48 \, x^{2} e^{\frac {3}{4}} - 16 \, x^{2} + 16 \, x e^{4} + 3}{16 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(48*x^2*exp(3/4)-16*x^2-3)*exp(1/16*(16*x*exp(4)+48*x^2*exp(3/4)-16*x^2+3)/x)/x^2,x, algorithm=
"giac")

[Out]

e^(1/16*(48*x^2*e^(3/4) - 16*x^2 + 16*x*e^4 + 3)/x)

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Mupad [B]
time = 3.03, size = 20, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {3}{16\,x}}\,{\mathrm {e}}^{3\,x\,{\mathrm {e}}^{3/4}}\,{\mathrm {e}}^{{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x*exp(4) + 3*x^2*exp(3/4) - x^2 + 3/16)/x)*(16*x^2 - 48*x^2*exp(3/4) + 3))/(16*x^2),x)

[Out]

exp(-x)*exp(3/(16*x))*exp(3*x*exp(3/4))*exp(exp(4))

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