3.91.5 \(\int \frac {1}{32} (8192 e^{4 x}+x-192 x^2+8192 x^3+e^{3 x} (8192+24576 x)+e^{2 x} (-64+24448 x+24576 x^2)+e^x (-256 x+24448 x^2+8192 x^3)) \, dx\)

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

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Rubi [B]  time = 0.17, antiderivative size = 82, normalized size of antiderivative = 3.28, number of steps used = 25, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 2194, 2176, 2196, 1594} \begin {gather*} 64 x^4+256 e^x x^3-2 x^3-4 e^x x^2+384 e^{2 x} x^2+\frac {x^2}{64}-2 e^{2 x} x-\frac {256 e^{3 x}}{3}+64 e^{4 x}+\frac {256}{3} e^{3 x} (3 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8192*E^(4*x) + x - 192*x^2 + 8192*x^3 + E^(3*x)*(8192 + 24576*x) + E^(2*x)*(-64 + 24448*x + 24576*x^2) +
E^x*(-256*x + 24448*x^2 + 8192*x^3))/32,x]

[Out]

(-256*E^(3*x))/3 + 64*E^(4*x) - 2*E^(2*x)*x + x^2/64 - 4*E^x*x^2 + 384*E^(2*x)*x^2 - 2*x^3 + 256*E^x*x^3 + 64*
x^4 + (256*E^(3*x)*(1 + 3*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 1594

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

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}{32} \int \left (8192 e^{4 x}+x-192 x^2+8192 x^3+e^{3 x} (8192+24576 x)+e^{2 x} \left (-64+24448 x+24576 x^2\right )+e^x \left (-256 x+24448 x^2+8192 x^3\right )\right ) \, dx\\ &=\frac {x^2}{64}-2 x^3+64 x^4+\frac {1}{32} \int e^{3 x} (8192+24576 x) \, dx+\frac {1}{32} \int e^{2 x} \left (-64+24448 x+24576 x^2\right ) \, dx+\frac {1}{32} \int e^x \left (-256 x+24448 x^2+8192 x^3\right ) \, dx+256 \int e^{4 x} \, dx\\ &=64 e^{4 x}+\frac {x^2}{64}-2 x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)+\frac {1}{32} \int e^x x \left (-256+24448 x+8192 x^2\right ) \, dx+\frac {1}{32} \int \left (-64 e^{2 x}+24448 e^{2 x} x+24576 e^{2 x} x^2\right ) \, dx-256 \int e^{3 x} \, dx\\ &=-\frac {256 e^{3 x}}{3}+64 e^{4 x}+\frac {x^2}{64}-2 x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)+\frac {1}{32} \int \left (-256 e^x x+24448 e^x x^2+8192 e^x x^3\right ) \, dx-2 \int e^{2 x} \, dx+764 \int e^{2 x} x \, dx+768 \int e^{2 x} x^2 \, dx\\ &=-e^{2 x}-\frac {256 e^{3 x}}{3}+64 e^{4 x}+382 e^{2 x} x+\frac {x^2}{64}+384 e^{2 x} x^2-2 x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)-8 \int e^x x \, dx+256 \int e^x x^3 \, dx-382 \int e^{2 x} \, dx+764 \int e^x x^2 \, dx-768 \int e^{2 x} x \, dx\\ &=-192 e^{2 x}-\frac {256 e^{3 x}}{3}+64 e^{4 x}-8 e^x x-2 e^{2 x} x+\frac {x^2}{64}+764 e^x x^2+384 e^{2 x} x^2-2 x^3+256 e^x x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)+8 \int e^x \, dx+384 \int e^{2 x} \, dx-768 \int e^x x^2 \, dx-1528 \int e^x x \, dx\\ &=8 e^x-\frac {256 e^{3 x}}{3}+64 e^{4 x}-1536 e^x x-2 e^{2 x} x+\frac {x^2}{64}-4 e^x x^2+384 e^{2 x} x^2-2 x^3+256 e^x x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)+1528 \int e^x \, dx+1536 \int e^x x \, dx\\ &=1536 e^x-\frac {256 e^{3 x}}{3}+64 e^{4 x}-2 e^{2 x} x+\frac {x^2}{64}-4 e^x x^2+384 e^{2 x} x^2-2 x^3+256 e^x x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)-1536 \int e^x \, dx\\ &=-\frac {256 e^{3 x}}{3}+64 e^{4 x}-2 e^{2 x} x+\frac {x^2}{64}-4 e^x x^2+384 e^{2 x} x^2-2 x^3+256 e^x x^3+64 x^4+\frac {256}{3} e^{3 x} (1+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 27, normalized size = 1.08 \begin {gather*} \frac {1}{64} \left (64 e^{2 x}+128 e^x x+x (-1+64 x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8192*E^(4*x) + x - 192*x^2 + 8192*x^3 + E^(3*x)*(8192 + 24576*x) + E^(2*x)*(-64 + 24448*x + 24576*x
^2) + E^x*(-256*x + 24448*x^2 + 8192*x^3))/32,x]

[Out]

(64*E^(2*x) + 128*E^x*x + x*(-1 + 64*x))^2/64

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fricas [B]  time = 0.47, size = 59, normalized size = 2.36 \begin {gather*} 64 \, x^{4} - 2 \, x^{3} + \frac {1}{64} \, x^{2} + 256 \, x e^{\left (3 \, x\right )} + 2 \, {\left (192 \, x^{2} - x\right )} e^{\left (2 \, x\right )} + 4 \, {\left (64 \, x^{3} - x^{2}\right )} e^{x} + 64 \, e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(256*exp(x)^4+1/32*(24576*x+8192)*exp(x)^3+1/32*(24576*x^2+24448*x-64)*exp(x)^2+1/32*(8192*x^3+24448*
x^2-256*x)*exp(x)+256*x^3-6*x^2+1/32*x,x, algorithm="fricas")

[Out]

64*x^4 - 2*x^3 + 1/64*x^2 + 256*x*e^(3*x) + 2*(192*x^2 - x)*e^(2*x) + 4*(64*x^3 - x^2)*e^x + 64*e^(4*x)

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giac [B]  time = 0.14, size = 59, normalized size = 2.36 \begin {gather*} 64 \, x^{4} - 2 \, x^{3} + \frac {1}{64} \, x^{2} + 256 \, x e^{\left (3 \, x\right )} + 2 \, {\left (192 \, x^{2} - x\right )} e^{\left (2 \, x\right )} + 4 \, {\left (64 \, x^{3} - x^{2}\right )} e^{x} + 64 \, e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(256*exp(x)^4+1/32*(24576*x+8192)*exp(x)^3+1/32*(24576*x^2+24448*x-64)*exp(x)^2+1/32*(8192*x^3+24448*
x^2-256*x)*exp(x)+256*x^3-6*x^2+1/32*x,x, algorithm="giac")

[Out]

64*x^4 - 2*x^3 + 1/64*x^2 + 256*x*e^(3*x) + 2*(192*x^2 - x)*e^(2*x) + 4*(64*x^3 - x^2)*e^x + 64*e^(4*x)

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maple [A]  time = 0.06, size = 60, normalized size = 2.40

method	result	size
default	\(\frac {x^{2}}{64}-2 x^{3}+64 x^{4}+64 \,{\mathrm e}^{4 x}+256 x \,{\mathrm e}^{3 x}-2 x \,{\mathrm e}^{2 x}+384 \,{\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{x} x^{2}+256 \,{\mathrm e}^{x} x^{3}\)	\(60\)
norman	\(\frac {x^{2}}{64}-2 x^{3}+64 x^{4}+64 \,{\mathrm e}^{4 x}+256 x \,{\mathrm e}^{3 x}-2 x \,{\mathrm e}^{2 x}+384 \,{\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{x} x^{2}+256 \,{\mathrm e}^{x} x^{3}\)	\(60\)
risch	\(\frac {x^{2}}{64}-2 x^{3}+64 x^{4}+64 \,{\mathrm e}^{4 x}+256 x \,{\mathrm e}^{3 x}-2 x \,{\mathrm e}^{2 x}+384 \,{\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{x} x^{2}+256 \,{\mathrm e}^{x} x^{3}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(256*exp(x)^4+1/32*(24576*x+8192)*exp(x)^3+1/32*(24576*x^2+24448*x-64)*exp(x)^2+1/32*(8192*x^3+24448*x^2-25
6*x)*exp(x)+256*x^3-6*x^2+1/32*x,x,method=_RETURNVERBOSE)

[Out]

1/64*x^2-2*x^3+64*x^4+64*exp(x)^4+256*x*exp(x)^3-2*x*exp(x)^2+384*exp(x)^2*x^2-4*exp(x)*x^2+256*exp(x)*x^3

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maxima [B]  time = 0.35, size = 59, normalized size = 2.36 \begin {gather*} 64 \, x^{4} - 2 \, x^{3} + \frac {1}{64} \, x^{2} + 256 \, x e^{\left (3 \, x\right )} + 2 \, {\left (192 \, x^{2} - x\right )} e^{\left (2 \, x\right )} + 4 \, {\left (64 \, x^{3} - x^{2}\right )} e^{x} + 64 \, e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(256*exp(x)^4+1/32*(24576*x+8192)*exp(x)^3+1/32*(24576*x^2+24448*x-64)*exp(x)^2+1/32*(8192*x^3+24448*
x^2-256*x)*exp(x)+256*x^3-6*x^2+1/32*x,x, algorithm="maxima")

[Out]

64*x^4 - 2*x^3 + 1/64*x^2 + 256*x*e^(3*x) + 2*(192*x^2 - x)*e^(2*x) + 4*(64*x^3 - x^2)*e^x + 64*e^(4*x)

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mupad [B]  time = 0.16, size = 24, normalized size = 0.96 \begin {gather*} \frac {{\left (64\,{\mathrm {e}}^{2\,x}-x+128\,x\,{\mathrm {e}}^x+64\,x^2\right )}^2}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/32 + 256*exp(4*x) + (exp(2*x)*(24448*x + 24576*x^2 - 64))/32 + (exp(3*x)*(24576*x + 8192))/32 - 6*x^2 +
256*x^3 + (exp(x)*(24448*x^2 - 256*x + 8192*x^3))/32,x)

[Out]

(64*exp(2*x) - x + 128*x*exp(x) + 64*x^2)^2/64

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sympy [B]  time = 0.16, size = 56, normalized size = 2.24 \begin {gather*} 64 x^{4} - 2 x^{3} + \frac {x^{2}}{64} + 256 x e^{3 x} + \left (384 x^{2} - 2 x\right ) e^{2 x} + \left (256 x^{3} - 4 x^{2}\right ) e^{x} + 64 e^{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(256*exp(x)**4+1/32*(24576*x+8192)*exp(x)**3+1/32*(24576*x**2+24448*x-64)*exp(x)**2+1/32*(8192*x**3+2
4448*x**2-256*x)*exp(x)+256*x**3-6*x**2+1/32*x,x)

[Out]

64*x**4 - 2*x**3 + x**2/64 + 256*x*exp(3*x) + (384*x**2 - 2*x)*exp(2*x) + (256*x**3 - 4*x**2)*exp(x) + 64*exp(
4*x)

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