3.95.21 \(\int (26+54 x+6 x^2+4 x^3+e^4 (2+4 x)+e^{2 x} (2 x+2 x^2)+e^x (26+30 x+8 x^2+2 x^3+e^4 (2+2 x))) \, dx\)

Optimal. Leaf size=20 \[ \left (-3+e^4+x+x \left (e^x+\frac {16}{x}+x\right )\right )^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 79, normalized size of antiderivative = 3.95, number of steps used = 23, number of rules used = 4, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} x^4+2 e^x x^3+2 x^3+2 e^x x^2+e^{2 x} x^2+27 x^2+26 e^x x+26 x-2 e^{x+4}+\frac {1}{2} e^4 (2 x+1)^2+2 e^{x+4} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[26 + 54*x + 6*x^2 + 4*x^3 + E^4*(2 + 4*x) + E^(2*x)*(2*x + 2*x^2) + E^x*(26 + 30*x + 8*x^2 + 2*x^3 + E^4*(
2 + 2*x)),x]

[Out]

-2*E^(4 + x) + 26*x + 26*E^x*x + 27*x^2 + 2*E^x*x^2 + E^(2*x)*x^2 + 2*x^3 + 2*E^x*x^3 + x^4 + 2*E^(4 + x)*(1 +
x) + (E^4*(1 + 2*x)^2)/2

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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} &=26 x+27 x^2+2 x^3+x^4+\frac {1}{2} e^4 (1+2 x)^2+\int e^{2 x} \left (2 x+2 x^2\right ) \, dx+\int e^x \left (26+30 x+8 x^2+2 x^3+e^4 (2+2 x)\right ) \, dx\\ &=26 x+27 x^2+2 x^3+x^4+\frac {1}{2} e^4 (1+2 x)^2+\int e^{2 x} x (2+2 x) \, dx+\int \left (26 e^x+30 e^x x+8 e^x x^2+2 e^x x^3+2 e^{4+x} (1+x)\right ) \, dx\\ &=26 x+27 x^2+2 x^3+x^4+\frac {1}{2} e^4 (1+2 x)^2+2 \int e^x x^3 \, dx+2 \int e^{4+x} (1+x) \, dx+8 \int e^x x^2 \, dx+26 \int e^x \, dx+30 \int e^x x \, dx+\int \left (2 e^{2 x} x+2 e^{2 x} x^2\right ) \, dx\\ &=26 e^x+26 x+30 e^x x+27 x^2+8 e^x x^2+2 x^3+2 e^x x^3+x^4+2 e^{4+x} (1+x)+\frac {1}{2} e^4 (1+2 x)^2-2 \int e^{4+x} \, dx+2 \int e^{2 x} x \, dx+2 \int e^{2 x} x^2 \, dx-6 \int e^x x^2 \, dx-16 \int e^x x \, dx-30 \int e^x \, dx\\ &=-4 e^x-2 e^{4+x}+26 x+14 e^x x+e^{2 x} x+27 x^2+2 e^x x^2+e^{2 x} x^2+2 x^3+2 e^x x^3+x^4+2 e^{4+x} (1+x)+\frac {1}{2} e^4 (1+2 x)^2-2 \int e^{2 x} x \, dx+12 \int e^x x \, dx+16 \int e^x \, dx-\int e^{2 x} \, dx\\ &=12 e^x-\frac {e^{2 x}}{2}-2 e^{4+x}+26 x+26 e^x x+27 x^2+2 e^x x^2+e^{2 x} x^2+2 x^3+2 e^x x^3+x^4+2 e^{4+x} (1+x)+\frac {1}{2} e^4 (1+2 x)^2-12 \int e^x \, dx+\int e^{2 x} \, dx\\ &=-2 e^{4+x}+26 x+26 e^x x+27 x^2+2 e^x x^2+e^{2 x} x^2+2 x^3+2 e^x x^3+x^4+2 e^{4+x} (1+x)+\frac {1}{2} e^4 (1+2 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 59, normalized size = 2.95 \begin {gather*} 26 x+2 e^4 x+27 x^2+2 e^4 x^2+e^{2 x} x^2+2 x^3+x^4+2 e^x \left (\left (13+e^4\right ) x+x^2+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[26 + 54*x + 6*x^2 + 4*x^3 + E^4*(2 + 4*x) + E^(2*x)*(2*x + 2*x^2) + E^x*(26 + 30*x + 8*x^2 + 2*x^3 +
E^4*(2 + 2*x)),x]

[Out]

26*x + 2*E^4*x + 27*x^2 + 2*E^4*x^2 + E^(2*x)*x^2 + 2*x^3 + x^4 + 2*E^x*((13 + E^4)*x + x^2 + x^3)

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fricas [B]  time = 0.93, size = 52, normalized size = 2.60 \begin {gather*} x^{4} + 2 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 27 \, x^{2} + 2 \, {\left (x^{2} + x\right )} e^{4} + 2 \, {\left (x^{3} + x^{2} + x e^{4} + 13 \, x\right )} e^{x} + 26 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*exp(x)^2+((2*x+2)*exp(4)+2*x^3+8*x^2+30*x+26)*exp(x)+(4*x+2)*exp(4)+4*x^3+6*x^2+54*x+26,
x, algorithm="fricas")

[Out]

x^4 + 2*x^3 + x^2*e^(2*x) + 27*x^2 + 2*(x^2 + x)*e^4 + 2*(x^3 + x^2 + x*e^4 + 13*x)*e^x + 26*x

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giac [B]  time = 0.16, size = 55, normalized size = 2.75 \begin {gather*} x^{4} + 2 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 27 \, x^{2} + 2 \, {\left (x^{2} + x\right )} e^{4} + 2 \, x e^{\left (x + 4\right )} + 2 \, {\left (x^{3} + x^{2} + 13 \, x\right )} e^{x} + 26 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*exp(x)^2+((2*x+2)*exp(4)+2*x^3+8*x^2+30*x+26)*exp(x)+(4*x+2)*exp(4)+4*x^3+6*x^2+54*x+26,
x, algorithm="giac")

[Out]

x^4 + 2*x^3 + x^2*e^(2*x) + 27*x^2 + 2*(x^2 + x)*e^4 + 2*x*e^(x + 4) + 2*(x^3 + x^2 + 13*x)*e^x + 26*x

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maple [B]  time = 0.05, size = 60, normalized size = 3.00

method	result	size
norman	\(x^{4}+\left (2 \,{\mathrm e}^{4}+26\right ) x +\left (2 \,{\mathrm e}^{4}+27\right ) x^{2}+{\mathrm e}^{2 x} x^{2}+\left (2 \,{\mathrm e}^{4}+26\right ) x \,{\mathrm e}^{x}+2 x^{3}+2 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{x} x^{3}\)	\(60\)
risch	\({\mathrm e}^{2 x} x^{2}+\left (2 x^{3}+2 x \,{\mathrm e}^{4}+2 x^{2}+26 x \right ) {\mathrm e}^{x}+2 x^{2} {\mathrm e}^{4}+2 x \,{\mathrm e}^{4}+x^{4}+2 x^{3}+27 x^{2}+26 x\)	\(60\)
default	\(26 x +\left (2 x^{2}+2 x \right ) {\mathrm e}^{4}+26 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{x} x^{3}+2 \,{\mathrm e}^{4} {\mathrm e}^{x}+2 \,{\mathrm e}^{4} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+{\mathrm e}^{2 x} x^{2}+27 x^{2}+2 x^{3}+x^{4}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+2*x)*exp(x)^2+((2*x+2)*exp(4)+2*x^3+8*x^2+30*x+26)*exp(x)+(4*x+2)*exp(4)+4*x^3+6*x^2+54*x+26,x,meth
od=_RETURNVERBOSE)

[Out]

x^4+(2*exp(4)+26)*x+(2*exp(4)+27)*x^2+exp(x)^2*x^2+(2*exp(4)+26)*x*exp(x)+2*x^3+2*exp(x)*x^2+2*exp(x)*x^3

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maxima [B]  time = 0.35, size = 51, normalized size = 2.55 \begin {gather*} x^{4} + 2 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 27 \, x^{2} + 2 \, {\left (x^{2} + x\right )} e^{4} + 2 \, {\left (x^{3} + x^{2} + x {\left (e^{4} + 13\right )}\right )} e^{x} + 26 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*exp(x)^2+((2*x+2)*exp(4)+2*x^3+8*x^2+30*x+26)*exp(x)+(4*x+2)*exp(4)+4*x^3+6*x^2+54*x+26,
x, algorithm="maxima")

[Out]

x^4 + 2*x^3 + x^2*e^(2*x) + 27*x^2 + 2*(x^2 + x)*e^4 + 2*(x^3 + x^2 + x*(e^4 + 13))*e^x + 26*x

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mupad [B]  time = 7.45, size = 21, normalized size = 1.05 \begin {gather*} x\,\left (x+{\mathrm {e}}^x+1\right )\,\left (x+2\,{\mathrm {e}}^4+x\,{\mathrm {e}}^x+x^2+26\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(54*x + exp(2*x)*(2*x + 2*x^2) + 6*x^2 + 4*x^3 + exp(x)*(30*x + 8*x^2 + 2*x^3 + exp(4)*(2*x + 2) + 26) + ex
p(4)*(4*x + 2) + 26,x)

[Out]

x*(x + exp(x) + 1)*(x + 2*exp(4) + x*exp(x) + x^2 + 26)

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sympy [B]  time = 0.15, size = 58, normalized size = 2.90 \begin {gather*} x^{4} + 2 x^{3} + x^{2} e^{2 x} + x^{2} \left (27 + 2 e^{4}\right ) + x \left (26 + 2 e^{4}\right ) + \left (2 x^{3} + 2 x^{2} + 26 x + 2 x e^{4}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+2*x)*exp(x)**2+((2*x+2)*exp(4)+2*x**3+8*x**2+30*x+26)*exp(x)+(4*x+2)*exp(4)+4*x**3+6*x**2+54
*x+26,x)

[Out]

x**4 + 2*x**3 + x**2*exp(2*x) + x**2*(27 + 2*exp(4)) + x*(26 + 2*exp(4)) + (2*x**3 + 2*x**2 + 26*x + 2*x*exp(4
))*exp(x)

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