3.103.43 \(\int \frac {-24576+1152 x^2+96 x^3+e^{2 x} (8192 x+2560 x^2-2560 x^3+96 x^4+64 x^5)}{64+16 x+x^2} \, dx\)

Optimal. Leaf size=26 \[ 16 (4-x)^2 \left (3+\frac {2 e^{2 x} x^2}{8+x}\right ) \]

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Rubi [B]  time = 0.35, antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 18, number of rules used = 7, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {27, 6742, 2199, 2194, 2176, 2177, 2178} \begin {gather*} 32 e^{2 x} x^3-512 e^{2 x} x^2+4608 e^{2 x} x-36864 e^{2 x}+48 (4-x)^2+\frac {294912 e^{2 x}}{x+8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24576 + 1152*x^2 + 96*x^3 + E^(2*x)*(8192*x + 2560*x^2 - 2560*x^3 + 96*x^4 + 64*x^5))/(64 + 16*x + x^2),
x]

[Out]

-36864*E^(2*x) + 48*(4 - x)^2 + 4608*E^(2*x)*x - 512*E^(2*x)*x^2 + 32*E^(2*x)*x^3 + (294912*E^(2*x))/(8 + x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[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 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24576+1152 x^2+96 x^3+e^{2 x} \left (8192 x+2560 x^2-2560 x^3+96 x^4+64 x^5\right )}{(8+x)^2} \, dx\\ &=\int \left (96 (-4+x)+\frac {32 e^{2 x} (-4+x) x \left (-64-36 x+11 x^2+2 x^3\right )}{(8+x)^2}\right ) \, dx\\ &=48 (4-x)^2+32 \int \frac {e^{2 x} (-4+x) x \left (-64-36 x+11 x^2+2 x^3\right )}{(8+x)^2} \, dx\\ &=48 (4-x)^2+32 \int \left (-2160 e^{2 x}+256 e^{2 x} x-29 e^{2 x} x^2+2 e^{2 x} x^3-\frac {9216 e^{2 x}}{(8+x)^2}+\frac {18432 e^{2 x}}{8+x}\right ) \, dx\\ &=48 (4-x)^2+64 \int e^{2 x} x^3 \, dx-928 \int e^{2 x} x^2 \, dx+8192 \int e^{2 x} x \, dx-69120 \int e^{2 x} \, dx-294912 \int \frac {e^{2 x}}{(8+x)^2} \, dx+589824 \int \frac {e^{2 x}}{8+x} \, dx\\ &=-34560 e^{2 x}+48 (4-x)^2+4096 e^{2 x} x-464 e^{2 x} x^2+32 e^{2 x} x^3+\frac {294912 e^{2 x}}{8+x}+\frac {589824 \text {Ei}(2 (8+x))}{e^{16}}-96 \int e^{2 x} x^2 \, dx+928 \int e^{2 x} x \, dx-4096 \int e^{2 x} \, dx-589824 \int \frac {e^{2 x}}{8+x} \, dx\\ &=-36608 e^{2 x}+48 (4-x)^2+4560 e^{2 x} x-512 e^{2 x} x^2+32 e^{2 x} x^3+\frac {294912 e^{2 x}}{8+x}+96 \int e^{2 x} x \, dx-464 \int e^{2 x} \, dx\\ &=-36840 e^{2 x}+48 (4-x)^2+4608 e^{2 x} x-512 e^{2 x} x^2+32 e^{2 x} x^3+\frac {294912 e^{2 x}}{8+x}-48 \int e^{2 x} \, dx\\ &=-36864 e^{2 x}+48 (4-x)^2+4608 e^{2 x} x-512 e^{2 x} x^2+32 e^{2 x} x^3+\frac {294912 e^{2 x}}{8+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 29, normalized size = 1.12 \begin {gather*} \frac {16 x \left (2 e^{2 x} (-4+x)^2 x+3 \left (-64+x^2\right )\right )}{8+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24576 + 1152*x^2 + 96*x^3 + E^(2*x)*(8192*x + 2560*x^2 - 2560*x^3 + 96*x^4 + 64*x^5))/(64 + 16*x +
x^2),x]

[Out]

(16*x*(2*E^(2*x)*(-4 + x)^2*x + 3*(-64 + x^2)))/(8 + x)

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fricas [A]  time = 0.83, size = 36, normalized size = 1.38 \begin {gather*} \frac {16 \, {\left (3 \, x^{3} + 2 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} e^{\left (2 \, x\right )} - 192 \, x\right )}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^5+96*x^4-2560*x^3+2560*x^2+8192*x)*exp(2*x)+96*x^3+1152*x^2-24576)/(x^2+16*x+64),x, algorithm
="fricas")

[Out]

16*(3*x^3 + 2*(x^4 - 8*x^3 + 16*x^2)*e^(2*x) - 192*x)/(x + 8)

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giac [A]  time = 0.14, size = 43, normalized size = 1.65 \begin {gather*} \frac {16 \, {\left (2 \, x^{4} e^{\left (2 \, x\right )} - 16 \, x^{3} e^{\left (2 \, x\right )} + 3 \, x^{3} + 32 \, x^{2} e^{\left (2 \, x\right )} - 192 \, x\right )}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^5+96*x^4-2560*x^3+2560*x^2+8192*x)*exp(2*x)+96*x^3+1152*x^2-24576)/(x^2+16*x+64),x, algorithm
="giac")

[Out]

16*(2*x^4*e^(2*x) - 16*x^3*e^(2*x) + 3*x^3 + 32*x^2*e^(2*x) - 192*x)/(x + 8)

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maple [A]  time = 0.10, size = 32, normalized size = 1.23

method	result	size
risch	\(48 x^{2}-384 x +\frac {32 x^{2} \left (x^{2}-8 x +16\right ) {\mathrm e}^{2 x}}{x +8}\)	\(32\)
norman	\(\frac {48 x^{3}+512 \,{\mathrm e}^{2 x} x^{2}-256 \,{\mathrm e}^{2 x} x^{3}+32 \,{\mathrm e}^{2 x} x^{4}+24576}{x +8}\)	\(41\)
derivativedivides	\(48 x^{2}-384 x +\frac {589824 \,{\mathrm e}^{2 x}}{2 x +16}-36864 \,{\mathrm e}^{2 x}+4608 x \,{\mathrm e}^{2 x}-512 \,{\mathrm e}^{2 x} x^{2}+32 \,{\mathrm e}^{2 x} x^{3}\)	\(54\)
default	\(48 x^{2}-384 x +\frac {589824 \,{\mathrm e}^{2 x}}{2 x +16}-36864 \,{\mathrm e}^{2 x}+4608 x \,{\mathrm e}^{2 x}-512 \,{\mathrm e}^{2 x} x^{2}+32 \,{\mathrm e}^{2 x} x^{3}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x^5+96*x^4-2560*x^3+2560*x^2+8192*x)*exp(2*x)+96*x^3+1152*x^2-24576)/(x^2+16*x+64),x,method=_RETURNVE
RBOSE)

[Out]

48*x^2-384*x+32*x^2*(x^2-8*x+16)/(x+8)*exp(2*x)

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maxima [A]  time = 0.40, size = 34, normalized size = 1.31 \begin {gather*} 48 \, x^{2} - 384 \, x + \frac {32 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} e^{\left (2 \, x\right )}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^5+96*x^4-2560*x^3+2560*x^2+8192*x)*exp(2*x)+96*x^3+1152*x^2-24576)/(x^2+16*x+64),x, algorithm
="maxima")

[Out]

48*x^2 - 384*x + 32*(x^4 - 8*x^3 + 16*x^2)*e^(2*x)/(x + 8)

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mupad [B]  time = 6.12, size = 40, normalized size = 1.54 \begin {gather*} \frac {16\,x\,\left (32\,x\,{\mathrm {e}}^{2\,x}-16\,x^2\,{\mathrm {e}}^{2\,x}+2\,x^3\,{\mathrm {e}}^{2\,x}+3\,x^2-192\right )}{x+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(8192*x + 2560*x^2 - 2560*x^3 + 96*x^4 + 64*x^5) + 1152*x^2 + 96*x^3 - 24576)/(16*x + x^2 + 64),
x)

[Out]

(16*x*(32*x*exp(2*x) - 16*x^2*exp(2*x) + 2*x^3*exp(2*x) + 3*x^2 - 192))/(x + 8)

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sympy [A]  time = 0.13, size = 31, normalized size = 1.19 \begin {gather*} 48 x^{2} - 384 x + \frac {\left (32 x^{4} - 256 x^{3} + 512 x^{2}\right ) e^{2 x}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x**5+96*x**4-2560*x**3+2560*x**2+8192*x)*exp(2*x)+96*x**3+1152*x**2-24576)/(x**2+16*x+64),x)

[Out]

48*x**2 - 384*x + (32*x**4 - 256*x**3 + 512*x**2)*exp(2*x)/(x + 8)

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