3.20.88 \(\int \frac {1}{2} (72+e^{\frac {1}{2} (8 e^{12}-3 x)} (2-3 x)+192 x+96 x^2+e^{\frac {1}{4} (8 e^{12}-3 x)} (-24-14 x+12 x^2)) \, dx\)

Optimal. Leaf size=26 \[ x \left (-e^{2 e^{12}-\frac {3 x}{4}}+x+3 (2+x)\right )^2 \]

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Rubi [B]  time = 0.15, antiderivative size = 95, normalized size of antiderivative = 3.65, number of steps used = 12, number of rules used = 4, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 2176, 2194, 2196} \begin {gather*} 16 x^3-8 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2+48 x^2-12 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x+36 x+\frac {2}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )}-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(72 + E^((8*E^12 - 3*x)/2)*(2 - 3*x) + 192*x + 96*x^2 + E^((8*E^12 - 3*x)/4)*(-24 - 14*x + 12*x^2))/2,x]

[Out]

(2*E^((8*E^12 - 3*x)/2))/3 - (E^((8*E^12 - 3*x)/2)*(2 - 3*x))/3 + 36*x - 12*E^((8*E^12 - 3*x)/4)*x + 48*x^2 -
8*E^((8*E^12 - 3*x)/4)*x^2 + 16*x^3

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (72+e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+192 x+96 x^2+e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} \left (-24-14 x+12 x^2\right )\right ) \, dx\\ &=36 x+48 x^2+16 x^3+\frac {1}{2} \int e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x) \, dx+\frac {1}{2} \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} \left (-24-14 x+12 x^2\right ) \, dx\\ &=-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+36 x+48 x^2+16 x^3+\frac {1}{2} \int \left (-24 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )}-14 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x+12 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2\right ) \, dx-\int e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} \, dx\\ &=\frac {2}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )}-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+36 x+48 x^2+16 x^3+6 \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2 \, dx-7 \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x \, dx-12 \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} \, dx\\ &=16 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )}+\frac {2}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )}-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+36 x+\frac {28}{3} e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x+48 x^2-8 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2+16 x^3-\frac {28}{3} \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} \, dx+16 \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x \, dx\\ &=\frac {256}{9} e^{\frac {1}{4} \left (8 e^{12}-3 x\right )}+\frac {2}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )}-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+36 x-12 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x+48 x^2-8 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2+16 x^3+\frac {64}{3} \int e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} \, dx\\ &=\frac {2}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )}-\frac {1}{3} e^{\frac {1}{2} \left (8 e^{12}-3 x\right )} (2-3 x)+36 x-12 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x+48 x^2-8 e^{\frac {1}{4} \left (8 e^{12}-3 x\right )} x^2+16 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 33, normalized size = 1.27 \begin {gather*} e^{-3 x/2} x \left (e^{2 e^{12}}-2 e^{3 x/4} (3+2 x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(72 + E^((8*E^12 - 3*x)/2)*(2 - 3*x) + 192*x + 96*x^2 + E^((8*E^12 - 3*x)/4)*(-24 - 14*x + 12*x^2))/
2,x]

[Out]

(x*(E^(2*E^12) - 2*E^((3*x)/4)*(3 + 2*x))^2)/E^((3*x)/2)

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fricas [B]  time = 0.62, size = 45, normalized size = 1.73 \begin {gather*} 16 \, x^{3} + 48 \, x^{2} - 4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{\left (-\frac {3}{4} \, x + 2 \, e^{12}\right )} + x e^{\left (-\frac {3}{2} \, x + 4 \, e^{12}\right )} + 36 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x+2)*exp(2*exp(12)-3/4*x)^2+1/2*(12*x^2-14*x-24)*exp(2*exp(12)-3/4*x)+48*x^2+96*x+36,x, algo
rithm="fricas")

[Out]

16*x^3 + 48*x^2 - 4*(2*x^2 + 3*x)*e^(-3/4*x + 2*e^12) + x*e^(-3/2*x + 4*e^12) + 36*x

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giac [B]  time = 0.20, size = 45, normalized size = 1.73 \begin {gather*} 16 \, x^{3} + 48 \, x^{2} - 4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{\left (-\frac {3}{4} \, x + 2 \, e^{12}\right )} + x e^{\left (-\frac {3}{2} \, x + 4 \, e^{12}\right )} + 36 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x+2)*exp(2*exp(12)-3/4*x)^2+1/2*(12*x^2-14*x-24)*exp(2*exp(12)-3/4*x)+48*x^2+96*x+36,x, algo
rithm="giac")

[Out]

16*x^3 + 48*x^2 - 4*(2*x^2 + 3*x)*e^(-3/4*x + 2*e^12) + x*e^(-3/2*x + 4*e^12) + 36*x

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maple [B]  time = 0.07, size = 46, normalized size = 1.77

method	result	size
risch	\({\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} x +\frac {\left (-16 x^{2}-24 x \right ) {\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}}}{2}+16 x^{3}+48 x^{2}+36 x\)	\(46\)
norman	\({\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} x +36 x +48 x^{2}+16 x^{3}-12 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} x -8 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} x^{2}\)	\(54\)
default	\(16 x^{3}+48 x^{2}+36 x -\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )}{3}+\frac {8 \,{\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} {\mathrm e}^{12}}{3}+16 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )-\frac {128 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )^{2}}{9}+\frac {224 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} {\mathrm e}^{12}}{9}-\frac {512 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} {\mathrm e}^{24}}{9}+\frac {512 \,{\mathrm e}^{12} \left ({\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )-{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}}\right )}{9}\)	\(153\)
derivativedivides	\(-96 \,{\mathrm e}^{12}+36 x +\frac {256 \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )^{2}}{3}+16 x^{3}-\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )}{3}+\frac {8 \,{\mathrm e}^{4 \,{\mathrm e}^{12}-\frac {3 x}{2}} {\mathrm e}^{12}}{3}+16 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )-\frac {128 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )^{2}}{9}+\frac {224 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} {\mathrm e}^{12}}{9}-\frac {512 \,{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} {\mathrm e}^{24}}{9}+\frac {512 \,{\mathrm e}^{12} \left ({\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}} \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right )-{\mathrm e}^{2 \,{\mathrm e}^{12}-\frac {3 x}{4}}\right )}{9}-\frac {1024 \left (2 \,{\mathrm e}^{12}-\frac {3 x}{4}\right ) {\mathrm e}^{12}}{3}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-3*x+2)*exp(2*exp(12)-3/4*x)^2+1/2*(12*x^2-14*x-24)*exp(2*exp(12)-3/4*x)+48*x^2+96*x+36,x,method=_RET
URNVERBOSE)

[Out]

exp(4*exp(12)-3/2*x)*x+1/2*(-16*x^2-24*x)*exp(2*exp(12)-3/4*x)+16*x^3+48*x^2+36*x

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maxima [B]  time = 0.66, size = 50, normalized size = 1.92 \begin {gather*} 16 \, x^{3} + 48 \, x^{2} - 4 \, {\left (2 \, x^{2} e^{\left (2 \, e^{12}\right )} + 3 \, x e^{\left (2 \, e^{12}\right )}\right )} e^{\left (-\frac {3}{4} \, x\right )} + x e^{\left (-\frac {3}{2} \, x + 4 \, e^{12}\right )} + 36 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x+2)*exp(2*exp(12)-3/4*x)^2+1/2*(12*x^2-14*x-24)*exp(2*exp(12)-3/4*x)+48*x^2+96*x+36,x, algo
rithm="maxima")

[Out]

16*x^3 + 48*x^2 - 4*(2*x^2*e^(2*e^12) + 3*x*e^(2*e^12))*e^(-3/4*x) + x*e^(-3/2*x + 4*e^12) + 36*x

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mupad [B]  time = 1.17, size = 38, normalized size = 1.46 \begin {gather*} x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{12}-\frac {3\,x}{2}}\,{\left (6\,{\mathrm {e}}^{\frac {3\,x}{4}-2\,{\mathrm {e}}^{12}}+4\,x\,{\mathrm {e}}^{\frac {3\,x}{4}-2\,{\mathrm {e}}^{12}}-1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(96*x - (exp(2*exp(12) - (3*x)/4)*(14*x - 12*x^2 + 24))/2 + 48*x^2 - (exp(4*exp(12) - (3*x)/2)*(3*x - 2))/2
+ 36,x)

[Out]

x*exp(4*exp(12) - (3*x)/2)*(6*exp((3*x)/4 - 2*exp(12)) + 4*x*exp((3*x)/4 - 2*exp(12)) - 1)^2

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sympy [B]  time = 0.13, size = 48, normalized size = 1.85 \begin {gather*} 16 x^{3} + 48 x^{2} + x e^{- \frac {3 x}{2} + 4 e^{12}} + 36 x + \left (- 8 x^{2} - 12 x\right ) e^{- \frac {3 x}{4} + 2 e^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x+2)*exp(2*exp(12)-3/4*x)**2+1/2*(12*x**2-14*x-24)*exp(2*exp(12)-3/4*x)+48*x**2+96*x+36,x)

[Out]

16*x**3 + 48*x**2 + x*exp(-3*x/2 + 4*exp(12)) + 36*x + (-8*x**2 - 12*x)*exp(-3*x/4 + 2*exp(12))

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