3.27.43 \(\int \frac {-20-8 x-x^2+e^x (3+x)+(-1+e^5)^{e^{8 x}} (-1+e^{8 x} (32+8 x) \log (-1+e^5))}{16+8 x+x^2} \, dx\)

Optimal. Leaf size=26 \[ -x+\frac {4+e^x+\left (-1+e^5\right )^{e^{8 x}}}{4+x} \]

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Rubi [B]  time = 0.48, antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 9, number of rules used = 5, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {27, 6742, 43, 2197, 2288} \begin {gather*} -x+\frac {e^{-8 x} \left (e^5-1\right )^{e^{8 x}} \left (e^{8 x} x+4 e^{8 x}\right )}{(x+4)^2}+\frac {e^x}{x+4}+\frac {4}{x+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 8*x - x^2 + E^x*(3 + x) + (-1 + E^5)^E^(8*x)*(-1 + E^(8*x)*(32 + 8*x)*Log[-1 + E^5]))/(16 + 8*x + x
^2),x]

[Out]

-x + 4/(4 + x) + E^x/(4 + x) + ((-1 + E^5)^E^(8*x)*(4*E^(8*x) + E^(8*x)*x))/(E^(8*x)*(4 + x)^2)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 {-20-8 x-x^2+e^x (3+x)+\left (-1+e^5\right )^{e^{8 x}} \left (-1+e^{8 x} (32+8 x) \log \left (-1+e^5\right )\right )}{(4+x)^2} \, dx\\ &=\int \left (-\frac {20}{(4+x)^2}-\frac {8 x}{(4+x)^2}-\frac {x^2}{(4+x)^2}+\frac {e^x (3+x)}{(4+x)^2}+\frac {\left (-1+e^5\right )^{e^{8 x}} \left (-1+32 e^{8 x} \log \left (-1+e^5\right )+8 e^{8 x} x \log \left (-1+e^5\right )\right )}{(4+x)^2}\right ) \, dx\\ &=\frac {20}{4+x}-8 \int \frac {x}{(4+x)^2} \, dx-\int \frac {x^2}{(4+x)^2} \, dx+\int \frac {e^x (3+x)}{(4+x)^2} \, dx+\int \frac {\left (-1+e^5\right )^{e^{8 x}} \left (-1+32 e^{8 x} \log \left (-1+e^5\right )+8 e^{8 x} x \log \left (-1+e^5\right )\right )}{(4+x)^2} \, dx\\ &=\frac {20}{4+x}+\frac {e^x}{4+x}+\frac {e^{-8 x} \left (-1+e^5\right )^{e^{8 x}} \left (4 e^{8 x}+e^{8 x} x\right )}{(4+x)^2}-8 \int \left (-\frac {4}{(4+x)^2}+\frac {1}{4+x}\right ) \, dx-\int \left (1+\frac {16}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx\\ &=-x+\frac {4}{4+x}+\frac {e^x}{4+x}+\frac {e^{-8 x} \left (-1+e^5\right )^{e^{8 x}} \left (4 e^{8 x}+e^{8 x} x\right )}{(4+x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 30, normalized size = 1.15 \begin {gather*} \frac {4+e^x+\left (-1+e^5\right )^{e^{8 x}}-4 x-x^2}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 8*x - x^2 + E^x*(3 + x) + (-1 + E^5)^E^(8*x)*(-1 + E^(8*x)*(32 + 8*x)*Log[-1 + E^5]))/(16 + 8
*x + x^2),x]

[Out]

(4 + E^x + (-1 + E^5)^E^(8*x) - 4*x - x^2)/(4 + x)

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fricas [A]  time = 0.51, size = 30, normalized size = 1.15 \begin {gather*} -\frac {x^{2} + 4 \, x - {\left (e^{5} - 1\right )}^{e^{\left (8 \, x\right )}} - e^{x} - 4}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+32)*exp(4*x)^2*log(exp(5)-1)-1)*exp(exp(4*x)^2*log(exp(5)-1))+(3+x)*exp(x)-x^2-8*x-20)/(x^2+8
*x+16),x, algorithm="fricas")

[Out]

-(x^2 + 4*x - (e^5 - 1)^e^(8*x) - e^x - 4)/(x + 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} - {\left (8 \, {\left (x + 4\right )} e^{\left (8 \, x\right )} \log \left (e^{5} - 1\right ) - 1\right )} {\left (e^{5} - 1\right )}^{e^{\left (8 \, x\right )}} - {\left (x + 3\right )} e^{x} + 8 \, x + 20}{x^{2} + 8 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+32)*exp(4*x)^2*log(exp(5)-1)-1)*exp(exp(4*x)^2*log(exp(5)-1))+(3+x)*exp(x)-x^2-8*x-20)/(x^2+8
*x+16),x, algorithm="giac")

[Out]

integrate(-(x^2 - (8*(x + 4)*e^(8*x)*log(e^5 - 1) - 1)*(e^5 - 1)^e^(8*x) - (x + 3)*e^x + 8*x + 20)/(x^2 + 8*x
+ 16), x)

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maple [A]  time = 0.42, size = 35, normalized size = 1.35

method	result	size
risch	\(-x +\frac {4}{4+x}+\frac {{\mathrm e}^{x}}{4+x}+\frac {\left ({\mathrm e}^{5}-1\right )^{{\mathrm e}^{8 x}}}{4+x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x+32)*exp(4*x)^2*ln(exp(5)-1)-1)*exp(exp(4*x)^2*ln(exp(5)-1))+(3+x)*exp(x)-x^2-8*x-20)/(x^2+8*x+16),x
,method=_RETURNVERBOSE)

[Out]

-x+4/(4+x)+exp(x)/(4+x)+1/(4+x)*(exp(5)-1)^exp(8*x)

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maxima [B]  time = 0.57, size = 57, normalized size = 2.19 \begin {gather*} -\frac {x^{2} + 4 \, x - e^{\left (e^{\left (8 \, x\right )} \log \left (e^{4} + e^{3} + e^{2} + e + 1\right ) + e^{\left (8 \, x\right )} \log \left (e - 1\right )\right )} - e^{x} + 16}{x + 4} + \frac {20}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+32)*exp(4*x)^2*log(exp(5)-1)-1)*exp(exp(4*x)^2*log(exp(5)-1))+(3+x)*exp(x)-x^2-8*x-20)/(x^2+8
*x+16),x, algorithm="maxima")

[Out]

-(x^2 + 4*x - e^(e^(8*x)*log(e^4 + e^3 + e^2 + e + 1) + e^(8*x)*log(e - 1)) - e^x + 16)/(x + 4) + 20/(x + 4)

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mupad [B]  time = 1.71, size = 34, normalized size = 1.31 \begin {gather*} \frac {4}{x+4}-x+\frac {{\mathrm {e}}^x}{x+4}+\frac {{\left ({\mathrm {e}}^5-1\right )}^{{\mathrm {e}}^{8\,x}}}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x - exp(x)*(x + 3) + x^2 - exp(log(exp(5) - 1)*exp(8*x))*(log(exp(5) - 1)*exp(8*x)*(8*x + 32) - 1) + 2
0)/(8*x + x^2 + 16),x)

[Out]

4/(x + 4) - x + exp(x)/(x + 4) + (exp(5) - 1)^exp(8*x)/(x + 4)

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sympy [A]  time = 0.21, size = 29, normalized size = 1.12 \begin {gather*} - x + \frac {e^{x}}{x + 4} + \frac {e^{e^{8 x} \log {\left (-1 + e^{5} \right )}}}{x + 4} + \frac {4}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+32)*exp(4*x)**2*ln(exp(5)-1)-1)*exp(exp(4*x)**2*ln(exp(5)-1))+(3+x)*exp(x)-x**2-8*x-20)/(x**2
+8*x+16),x)

[Out]

-x + exp(x)/(x + 4) + exp(exp(8*x)*log(-1 + exp(5)))/(x + 4) + 4/(x + 4)

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