3.104.17 \(\int \frac {1}{4} (12 e^{8+3 x}+e^{2 x} (e^4 (-32+32 x)+e^8 (16+32 x+48 x^2))+e^x (-12-16 x+16 x^2+e^4 (12-128 x+56 x^2+48 x^3)+e^8 (-3+68 x-32 x^2+120 x^3+36 x^4))) \, dx\)

Optimal. Leaf size=30 \[ e^x \left (-3+2 x+e^4 \left (\frac {3}{2}+e^x-x+3 x^2\right )\right )^2 \]

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Rubi [B]  time = 0.42, antiderivative size = 163, normalized size of antiderivative = 5.43, number of steps used = 52, number of rules used = 4, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {12, 2194, 2196, 2176} \begin {gather*} 9 e^{x+8} x^4+12 e^{x+4} x^3-6 e^{x+8} x^3+4 e^x x^2-22 e^{x+4} x^2+10 e^{x+8} x^2+6 e^{2 x+8} x^2-12 e^x x+12 e^{x+4} x-3 e^{x+8} x-2 e^{2 x+8} x+9 e^x-9 e^{x+4}+\frac {9 e^{x+8}}{4}-2 e^{2 x+4}+3 e^{2 x+8}+e^{3 x+8}-4 e^{2 x+4} (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*E^(8 + 3*x) + E^(2*x)*(E^4*(-32 + 32*x) + E^8*(16 + 32*x + 48*x^2)) + E^x*(-12 - 16*x + 16*x^2 + E^4*(
12 - 128*x + 56*x^2 + 48*x^3) + E^8*(-3 + 68*x - 32*x^2 + 120*x^3 + 36*x^4)))/4,x]

[Out]

9*E^x - 9*E^(4 + x) + (9*E^(8 + x))/4 - 2*E^(4 + 2*x) + 3*E^(8 + 2*x) + E^(8 + 3*x) - 4*E^(4 + 2*x)*(1 - x) -
12*E^x*x + 12*E^(4 + x)*x - 3*E^(8 + x)*x - 2*E^(8 + 2*x)*x + 4*E^x*x^2 - 22*E^(4 + x)*x^2 + 10*E^(8 + x)*x^2
+ 6*E^(8 + 2*x)*x^2 + 12*E^(4 + x)*x^3 - 6*E^(8 + x)*x^3 + 9*E^(8 + x)*x^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 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}{4} \int \left (12 e^{8+3 x}+e^{2 x} \left (e^4 (-32+32 x)+e^8 \left (16+32 x+48 x^2\right )\right )+e^x \left (-12-16 x+16 x^2+e^4 \left (12-128 x+56 x^2+48 x^3\right )+e^8 \left (-3+68 x-32 x^2+120 x^3+36 x^4\right )\right )\right ) \, dx\\ &=\frac {1}{4} \int e^{2 x} \left (e^4 (-32+32 x)+e^8 \left (16+32 x+48 x^2\right )\right ) \, dx+\frac {1}{4} \int e^x \left (-12-16 x+16 x^2+e^4 \left (12-128 x+56 x^2+48 x^3\right )+e^8 \left (-3+68 x-32 x^2+120 x^3+36 x^4\right )\right ) \, dx+3 \int e^{8+3 x} \, dx\\ &=e^{8+3 x}+\frac {1}{4} \int \left (32 e^{4+2 x} (-1+x)+16 e^{8+2 x} \left (1+2 x+3 x^2\right )\right ) \, dx+\frac {1}{4} \int \left (-12 e^x-16 e^x x+16 e^x x^2+4 e^{4+x} \left (3-32 x+14 x^2+12 x^3\right )+e^{8+x} \left (-3+68 x-32 x^2+120 x^3+36 x^4\right )\right ) \, dx\\ &=e^{8+3 x}+\frac {1}{4} \int e^{8+x} \left (-3+68 x-32 x^2+120 x^3+36 x^4\right ) \, dx-3 \int e^x \, dx-4 \int e^x x \, dx+4 \int e^x x^2 \, dx+4 \int e^{8+2 x} \left (1+2 x+3 x^2\right ) \, dx+8 \int e^{4+2 x} (-1+x) \, dx+\int e^{4+x} \left (3-32 x+14 x^2+12 x^3\right ) \, dx\\ &=-3 e^x+e^{8+3 x}-4 e^{4+2 x} (1-x)-4 e^x x+4 e^x x^2+\frac {1}{4} \int \left (-3 e^{8+x}+68 e^{8+x} x-32 e^{8+x} x^2+120 e^{8+x} x^3+36 e^{8+x} x^4\right ) \, dx+4 \int e^x \, dx-4 \int e^{4+2 x} \, dx+4 \int \left (e^{8+2 x}+2 e^{8+2 x} x+3 e^{8+2 x} x^2\right ) \, dx-8 \int e^x x \, dx+\int \left (3 e^{4+x}-32 e^{4+x} x+14 e^{4+x} x^2+12 e^{4+x} x^3\right ) \, dx\\ &=e^x-2 e^{4+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x+4 e^x x^2-\frac {3}{4} \int e^{8+x} \, dx+3 \int e^{4+x} \, dx+4 \int e^{8+2 x} \, dx+8 \int e^x \, dx+8 \int e^{8+2 x} x \, dx-8 \int e^{8+x} x^2 \, dx+9 \int e^{8+x} x^4 \, dx+12 \int e^{8+2 x} x^2 \, dx+12 \int e^{4+x} x^3 \, dx+14 \int e^{4+x} x^2 \, dx+17 \int e^{8+x} x \, dx+30 \int e^{8+x} x^3 \, dx-32 \int e^{4+x} x \, dx\\ &=9 e^x+3 e^{4+x}-\frac {3 e^{8+x}}{4}-2 e^{4+2 x}+2 e^{8+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x-32 e^{4+x} x+17 e^{8+x} x+4 e^{8+2 x} x+4 e^x x^2+14 e^{4+x} x^2-8 e^{8+x} x^2+6 e^{8+2 x} x^2+12 e^{4+x} x^3+30 e^{8+x} x^3+9 e^{8+x} x^4-4 \int e^{8+2 x} \, dx-12 \int e^{8+2 x} x \, dx+16 \int e^{8+x} x \, dx-17 \int e^{8+x} \, dx-28 \int e^{4+x} x \, dx+32 \int e^{4+x} \, dx-36 \int e^{4+x} x^2 \, dx-36 \int e^{8+x} x^3 \, dx-90 \int e^{8+x} x^2 \, dx\\ &=9 e^x+35 e^{4+x}-\frac {71 e^{8+x}}{4}-2 e^{4+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x-60 e^{4+x} x+33 e^{8+x} x-2 e^{8+2 x} x+4 e^x x^2-22 e^{4+x} x^2-98 e^{8+x} x^2+6 e^{8+2 x} x^2+12 e^{4+x} x^3-6 e^{8+x} x^3+9 e^{8+x} x^4+6 \int e^{8+2 x} \, dx-16 \int e^{8+x} \, dx+28 \int e^{4+x} \, dx+72 \int e^{4+x} x \, dx+108 \int e^{8+x} x^2 \, dx+180 \int e^{8+x} x \, dx\\ &=9 e^x+63 e^{4+x}-\frac {135 e^{8+x}}{4}-2 e^{4+2 x}+3 e^{8+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x+12 e^{4+x} x+213 e^{8+x} x-2 e^{8+2 x} x+4 e^x x^2-22 e^{4+x} x^2+10 e^{8+x} x^2+6 e^{8+2 x} x^2+12 e^{4+x} x^3-6 e^{8+x} x^3+9 e^{8+x} x^4-72 \int e^{4+x} \, dx-180 \int e^{8+x} \, dx-216 \int e^{8+x} x \, dx\\ &=9 e^x-9 e^{4+x}-\frac {855 e^{8+x}}{4}-2 e^{4+2 x}+3 e^{8+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x+12 e^{4+x} x-3 e^{8+x} x-2 e^{8+2 x} x+4 e^x x^2-22 e^{4+x} x^2+10 e^{8+x} x^2+6 e^{8+2 x} x^2+12 e^{4+x} x^3-6 e^{8+x} x^3+9 e^{8+x} x^4+216 \int e^{8+x} \, dx\\ &=9 e^x-9 e^{4+x}+\frac {9 e^{8+x}}{4}-2 e^{4+2 x}+3 e^{8+2 x}+e^{8+3 x}-4 e^{4+2 x} (1-x)-12 e^x x+12 e^{4+x} x-3 e^{8+x} x-2 e^{8+2 x} x+4 e^x x^2-22 e^{4+x} x^2+10 e^{8+x} x^2+6 e^{8+2 x} x^2+12 e^{4+x} x^3-6 e^{8+x} x^3+9 e^{8+x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 35, normalized size = 1.17 \begin {gather*} \frac {1}{4} e^x \left (-6+2 e^{4+x}+4 x+e^4 \left (3-2 x+6 x^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^(8 + 3*x) + E^(2*x)*(E^4*(-32 + 32*x) + E^8*(16 + 32*x + 48*x^2)) + E^x*(-12 - 16*x + 16*x^2 +
E^4*(12 - 128*x + 56*x^2 + 48*x^3) + E^8*(-3 + 68*x - 32*x^2 + 120*x^3 + 36*x^4)))/4,x]

[Out]

(E^x*(-6 + 2*E^(4 + x) + 4*x + E^4*(3 - 2*x + 6*x^2))^2)/4

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fricas [B]  time = 0.63, size = 91, normalized size = 3.03 \begin {gather*} {\left ({\left (6 \, x^{2} - 2 \, x + 3\right )} e^{8} + 2 \, {\left (2 \, x - 3\right )} e^{4}\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (16 \, x^{2} + {\left (36 \, x^{4} - 24 \, x^{3} + 40 \, x^{2} - 12 \, x + 9\right )} e^{8} + 4 \, {\left (12 \, x^{3} - 22 \, x^{2} + 12 \, x - 9\right )} e^{4} - 48 \, x + 36\right )} e^{x} + e^{\left (3 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*exp(4)^2*exp(x)^3+1/4*((48*x^2+32*x+16)*exp(4)^2+(32*x-32)*exp(4))*exp(x)^2+1/4*((36*x^4+120*x^3-3
2*x^2+68*x-3)*exp(4)^2+(48*x^3+56*x^2-128*x+12)*exp(4)+16*x^2-16*x-12)*exp(x),x, algorithm="fricas")

[Out]

((6*x^2 - 2*x + 3)*e^8 + 2*(2*x - 3)*e^4)*e^(2*x) + 1/4*(16*x^2 + (36*x^4 - 24*x^3 + 40*x^2 - 12*x + 9)*e^8 +
4*(12*x^3 - 22*x^2 + 12*x - 9)*e^4 - 48*x + 36)*e^x + e^(3*x + 8)

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giac [B]  time = 0.17, size = 96, normalized size = 3.20 \begin {gather*} {\left (6 \, x^{2} - 2 \, x + 3\right )} e^{\left (2 \, x + 8\right )} + 2 \, {\left (2 \, x - 3\right )} e^{\left (2 \, x + 4\right )} + \frac {1}{4} \, {\left (36 \, x^{4} - 24 \, x^{3} + 40 \, x^{2} - 12 \, x + 9\right )} e^{\left (x + 8\right )} + {\left (12 \, x^{3} - 22 \, x^{2} + 12 \, x - 9\right )} e^{\left (x + 4\right )} + {\left (4 \, x^{2} - 12 \, x + 9\right )} e^{x} + e^{\left (3 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*exp(4)^2*exp(x)^3+1/4*((48*x^2+32*x+16)*exp(4)^2+(32*x-32)*exp(4))*exp(x)^2+1/4*((36*x^4+120*x^3-3
2*x^2+68*x-3)*exp(4)^2+(48*x^3+56*x^2-128*x+12)*exp(4)+16*x^2-16*x-12)*exp(x),x, algorithm="giac")

[Out]

(6*x^2 - 2*x + 3)*e^(2*x + 8) + 2*(2*x - 3)*e^(2*x + 4) + 1/4*(36*x^4 - 24*x^3 + 40*x^2 - 12*x + 9)*e^(x + 8)
+ (12*x^3 - 22*x^2 + 12*x - 9)*e^(x + 4) + (4*x^2 - 12*x + 9)*e^x + e^(3*x + 8)

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maple [B]  time = 0.09, size = 107, normalized size = 3.57

method	result	size
risch	\({\mathrm e}^{3 x +8}+\frac {\left (24 x^{2} {\mathrm e}^{8}-8 x \,{\mathrm e}^{8}+12 \,{\mathrm e}^{8}+16 x \,{\mathrm e}^{4}-24 \,{\mathrm e}^{4}\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (36 x^{4} {\mathrm e}^{8}-24 x^{3} {\mathrm e}^{8}+40 x^{2} {\mathrm e}^{8}+48 x^{3} {\mathrm e}^{4}-12 x \,{\mathrm e}^{8}-88 x^{2} {\mathrm e}^{4}+9 \,{\mathrm e}^{8}+48 x \,{\mathrm e}^{4}+16 x^{2}-36 \,{\mathrm e}^{4}-48 x +36\right ) {\mathrm e}^{x}}{4}\)	\(107\)
norman	\(\left (3 \,{\mathrm e}^{8}-6 \,{\mathrm e}^{4}\right ) {\mathrm e}^{2 x}+\left (\frac {9 \,{\mathrm e}^{8}}{4}-9 \,{\mathrm e}^{4}+9\right ) {\mathrm e}^{x}+{\mathrm e}^{8} {\mathrm e}^{3 x}+\left (-6 \,{\mathrm e}^{8}+12 \,{\mathrm e}^{4}\right ) x^{3} {\mathrm e}^{x}+\left (-2 \,{\mathrm e}^{8}+4 \,{\mathrm e}^{4}\right ) x \,{\mathrm e}^{2 x}+\left (4+10 \,{\mathrm e}^{8}-22 \,{\mathrm e}^{4}\right ) x^{2} {\mathrm e}^{x}+\left (-3 \,{\mathrm e}^{8}+12 \,{\mathrm e}^{4}-12\right ) x \,{\mathrm e}^{x}+6 x^{2} {\mathrm e}^{8} {\mathrm e}^{2 x}+9 x^{4} {\mathrm e}^{8} {\mathrm e}^{x}\)	\(134\)
meijerg	\(-\frac {3 \,{\mathrm e}^{8} \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{2}+2 \,{\mathrm e}^{4} \left ({\mathrm e}^{4}+1\right ) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )-\left (-8 \,{\mathrm e}^{8}+14 \,{\mathrm e}^{4}+4\right ) \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )-\left (-17 \,{\mathrm e}^{8}+32 \,{\mathrm e}^{4}+4\right ) \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )-9 \,{\mathrm e}^{8} \left (24-\frac {\left (5 x^{4}-20 x^{3}+60 x^{2}-120 x +120\right ) {\mathrm e}^{x}}{5}\right )-\left (-30 \,{\mathrm e}^{8}-12 \,{\mathrm e}^{4}\right ) \left (6-\frac {\left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}}{4}\right )-{\mathrm e}^{8} \left (1-{\mathrm e}^{3 x}\right )-2 \,{\mathrm e}^{4} \left ({\mathrm e}^{4}-2\right ) \left (1-{\mathrm e}^{2 x}\right )-\left (-3-\frac {3 \,{\mathrm e}^{8}}{4}+3 \,{\mathrm e}^{4}\right ) \left (1-{\mathrm e}^{x}\right )\)	\(204\)
default	\(-4 \,{\mathrm e}^{4} {\mathrm e}^{2 x}+2 \,{\mathrm e}^{8} {\mathrm e}^{2 x}+8 \,{\mathrm e}^{8} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )+8 \,{\mathrm e}^{4} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )+12 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{2 x}}{4}\right )+4 \,{\mathrm e}^{x} x^{2}-12 \,{\mathrm e}^{x} x +9 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{4} {\mathrm e}^{x}-\frac {3 \,{\mathrm e}^{8} {\mathrm e}^{x}}{4}-32 \,{\mathrm e}^{4} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+17 \,{\mathrm e}^{8} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+12 \,{\mathrm e}^{4} \left ({\mathrm e}^{x} x^{3}-3 \,{\mathrm e}^{x} x^{2}+6 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}\right )+14 \,{\mathrm e}^{4} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )-8 \,{\mathrm e}^{8} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )+30 \,{\mathrm e}^{8} \left ({\mathrm e}^{x} x^{3}-3 \,{\mathrm e}^{x} x^{2}+6 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}\right )+9 \,{\mathrm e}^{8} \left ({\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{3}+12 \,{\mathrm e}^{x} x^{2}-24 \,{\mathrm e}^{x} x +24 \,{\mathrm e}^{x}\right )+{\mathrm e}^{8} {\mathrm e}^{3 x}\)	\(288\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*exp(4)^2*exp(x)^3+1/4*((48*x^2+32*x+16)*exp(4)^2+(32*x-32)*exp(4))*exp(x)^2+1/4*((36*x^4+120*x^3-32*x^2+
68*x-3)*exp(4)^2+(48*x^3+56*x^2-128*x+12)*exp(4)+16*x^2-16*x-12)*exp(x),x,method=_RETURNVERBOSE)

[Out]

exp(3*x+8)+1/4*(24*x^2*exp(8)-8*x*exp(8)+12*exp(8)+16*x*exp(4)-24*exp(4))*exp(2*x)+1/4*(36*x^4*exp(8)-24*x^3*e
xp(8)+40*x^2*exp(8)+48*x^3*exp(4)-12*x*exp(8)-88*x^2*exp(4)+9*exp(8)+48*x*exp(4)+16*x^2-36*exp(4)-48*x+36)*exp
(x)

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maxima [B]  time = 0.36, size = 227, normalized size = 7.57 \begin {gather*} {\left (6 \, x^{2} e^{8} - 2 \, x {\left (e^{8} - 2 \, e^{4}\right )} + 3 \, e^{8} - 6 \, e^{4}\right )} e^{\left (2 \, x\right )} + 9 \, {\left (x^{4} e^{8} - 4 \, x^{3} e^{8} + 12 \, x^{2} e^{8} - 24 \, x e^{8} + 24 \, e^{8}\right )} e^{x} + 30 \, {\left (x^{3} e^{8} - 3 \, x^{2} e^{8} + 6 \, x e^{8} - 6 \, e^{8}\right )} e^{x} + 12 \, {\left (x^{3} e^{4} - 3 \, x^{2} e^{4} + 6 \, x e^{4} - 6 \, e^{4}\right )} e^{x} - 8 \, {\left (x^{2} e^{8} - 2 \, x e^{8} + 2 \, e^{8}\right )} e^{x} + 14 \, {\left (x^{2} e^{4} - 2 \, x e^{4} + 2 \, e^{4}\right )} e^{x} + 4 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 17 \, {\left (x e^{8} - e^{8}\right )} e^{x} - 32 \, {\left (x e^{4} - e^{4}\right )} e^{x} - 4 \, {\left (x - 1\right )} e^{x} + e^{\left (3 \, x + 8\right )} - \frac {3}{4} \, e^{\left (x + 8\right )} + 3 \, e^{\left (x + 4\right )} - 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*exp(4)^2*exp(x)^3+1/4*((48*x^2+32*x+16)*exp(4)^2+(32*x-32)*exp(4))*exp(x)^2+1/4*((36*x^4+120*x^3-3
2*x^2+68*x-3)*exp(4)^2+(48*x^3+56*x^2-128*x+12)*exp(4)+16*x^2-16*x-12)*exp(x),x, algorithm="maxima")

[Out]

(6*x^2*e^8 - 2*x*(e^8 - 2*e^4) + 3*e^8 - 6*e^4)*e^(2*x) + 9*(x^4*e^8 - 4*x^3*e^8 + 12*x^2*e^8 - 24*x*e^8 + 24*
e^8)*e^x + 30*(x^3*e^8 - 3*x^2*e^8 + 6*x*e^8 - 6*e^8)*e^x + 12*(x^3*e^4 - 3*x^2*e^4 + 6*x*e^4 - 6*e^4)*e^x - 8
*(x^2*e^8 - 2*x*e^8 + 2*e^8)*e^x + 14*(x^2*e^4 - 2*x*e^4 + 2*e^4)*e^x + 4*(x^2 - 2*x + 2)*e^x + 17*(x*e^8 - e^
8)*e^x - 32*(x*e^4 - e^4)*e^x - 4*(x - 1)*e^x + e^(3*x + 8) - 3/4*e^(x + 8) + 3*e^(x + 4) - 3*e^x

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mupad [B]  time = 6.71, size = 33, normalized size = 1.10 \begin {gather*} \frac {{\mathrm {e}}^x\,{\left (4\,x+2\,{\mathrm {e}}^{x+4}+3\,{\mathrm {e}}^4-2\,x\,{\mathrm {e}}^4+6\,x^2\,{\mathrm {e}}^4-6\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*exp(3*x)*exp(8) + (exp(2*x)*(exp(8)*(32*x + 48*x^2 + 16) + exp(4)*(32*x - 32)))/4 + (exp(x)*(exp(4)*(56*
x^2 - 128*x + 48*x^3 + 12) - 16*x + exp(8)*(68*x - 32*x^2 + 120*x^3 + 36*x^4 - 3) + 16*x^2 - 12))/4,x)

[Out]

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

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sympy [B]  time = 0.29, size = 128, normalized size = 4.27 \begin {gather*} \frac {\left (24 x^{2} e^{8} - 8 x e^{8} + 16 x e^{4} - 24 e^{4} + 12 e^{8}\right ) e^{2 x}}{4} + \frac {\left (36 x^{4} e^{8} - 24 x^{3} e^{8} + 48 x^{3} e^{4} - 88 x^{2} e^{4} + 16 x^{2} + 40 x^{2} e^{8} - 12 x e^{8} - 48 x + 48 x e^{4} - 36 e^{4} + 36 + 9 e^{8}\right ) e^{x}}{4} + e^{8} e^{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*exp(4)**2*exp(x)**3+1/4*((48*x**2+32*x+16)*exp(4)**2+(32*x-32)*exp(4))*exp(x)**2+1/4*((36*x**4+120
*x**3-32*x**2+68*x-3)*exp(4)**2+(48*x**3+56*x**2-128*x+12)*exp(4)+16*x**2-16*x-12)*exp(x),x)

[Out]

(24*x**2*exp(8) - 8*x*exp(8) + 16*x*exp(4) - 24*exp(4) + 12*exp(8))*exp(2*x)/4 + (36*x**4*exp(8) - 24*x**3*exp
(8) + 48*x**3*exp(4) - 88*x**2*exp(4) + 16*x**2 + 40*x**2*exp(8) - 12*x*exp(8) - 48*x + 48*x*exp(4) - 36*exp(4
) + 36 + 9*exp(8))*exp(x)/4 + exp(8)*exp(3*x)

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