\(\int \frac {1}{4} e^{2 x} (155-32 x-19 x^2-2 x^3-2 x^4+e^5 (-5-10 x+3 x^2+2 x^3)) \, dx\) [7268]

Optimal result

Integrand size = 48, antiderivative size = 31 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=e^{2 x} \left (4-x-\frac {1}{4} \left (-3+e^5-x\right ) x\right ) \left (5-x^2\right ) \]

[Out]

(4-x-1/4*(exp(5)-x-3)*x)*exp(x)^2*(-x^2+5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(80\) vs. \(2(31)=62\).

Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.58, number of steps used = 30, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} e^{2 x} x^4+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{2 x+5} x^3-\frac {11}{4} e^{2 x} x^2-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{2 x+5} x+20 e^{2 x} \]

[In]

Int[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]

[Out]

20*E^(2*x) - (5*E^(2*x)*x)/4 - (5*E^(5 + 2*x)*x)/4 - (11*E^(2*x)*x^2)/4 + (E^(2*x)*x^3)/4 + (E^(5 + 2*x)*x^3)/
4 - (E^(2*x)*x^4)/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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2227

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx \\ & = \frac {1}{4} \int \left (155 e^{2 x}-32 e^{2 x} x-19 e^{2 x} x^2-2 e^{2 x} x^3-2 e^{2 x} x^4+e^{5+2 x} \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx \\ & = \frac {1}{4} \int e^{5+2 x} \left (-5-10 x+3 x^2+2 x^3\right ) \, dx-\frac {1}{2} \int e^{2 x} x^3 \, dx-\frac {1}{2} \int e^{2 x} x^4 \, dx-\frac {19}{4} \int e^{2 x} x^2 \, dx-8 \int e^{2 x} x \, dx+\frac {155}{4} \int e^{2 x} \, dx \\ & = \frac {155 e^{2 x}}{8}-4 e^{2 x} x-\frac {19}{8} e^{2 x} x^2-\frac {1}{4} e^{2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {1}{4} \int \left (-5 e^{5+2 x}-10 e^{5+2 x} x+3 e^{5+2 x} x^2+2 e^{5+2 x} x^3\right ) \, dx+\frac {3}{4} \int e^{2 x} x^2 \, dx+4 \int e^{2 x} \, dx+\frac {19}{4} \int e^{2 x} x \, dx+\int e^{2 x} x^3 \, dx \\ & = \frac {171 e^{2 x}}{8}-\frac {13}{8} e^{2 x} x-2 e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {1}{2} \int e^{5+2 x} x^3 \, dx-\frac {3}{4} \int e^{2 x} x \, dx+\frac {3}{4} \int e^{5+2 x} x^2 \, dx-\frac {5}{4} \int e^{5+2 x} \, dx-\frac {3}{2} \int e^{2 x} x^2 \, dx-\frac {19}{8} \int e^{2 x} \, dx-\frac {5}{2} \int e^{5+2 x} x \, dx \\ & = \frac {323 e^{2 x}}{16}-\frac {5}{8} e^{5+2 x}-2 e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {3}{8} e^{5+2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {3}{8} \int e^{2 x} \, dx-\frac {3}{4} \int e^{5+2 x} x \, dx-\frac {3}{4} \int e^{5+2 x} x^2 \, dx+\frac {5}{4} \int e^{5+2 x} \, dx+\frac {3}{2} \int e^{2 x} x \, dx \\ & = \frac {163 e^{2 x}}{8}-\frac {5}{4} e^{2 x} x-\frac {13}{8} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4+\frac {3}{8} \int e^{5+2 x} \, dx-\frac {3}{4} \int e^{2 x} \, dx+\frac {3}{4} \int e^{5+2 x} x \, dx \\ & = 20 e^{2 x}+\frac {3}{16} e^{5+2 x}-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4-\frac {3}{8} \int e^{5+2 x} \, dx \\ & = 20 e^{2 x}-\frac {5}{4} e^{2 x} x-\frac {5}{4} e^{5+2 x} x-\frac {11}{4} e^{2 x} x^2+\frac {1}{4} e^{2 x} x^3+\frac {1}{4} e^{5+2 x} x^3-\frac {1}{4} e^{2 x} x^4 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} e^{2 x} \left (-5+x^2\right ) \left (16-\left (1+e^5\right ) x+x^2\right ) \]

[In]

Integrate[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]

[Out]

-1/4*(E^(2*x)*(-5 + x^2)*(16 - (1 + E^5)*x + x^2))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16

method	result	size
gosper	\(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\)	\(36\)
risch	\(\frac {{\mathrm e}^{2 x} \left (x^{3} {\mathrm e}^{5}-x^{4}+x^{3}-5 x \,{\mathrm e}^{5}-11 x^{2}-5 x +80\right )}{4}\)	\(36\)
norman	\(\left (-\frac {5}{4}-\frac {5 \,{\mathrm e}^{5}}{4}\right ) x \,{\mathrm e}^{2 x}+\left (\frac {1}{4}+\frac {{\mathrm e}^{5}}{4}\right ) x^{3} {\mathrm e}^{2 x}+20 \,{\mathrm e}^{2 x}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}\)	\(52\)
parallelrisch	\(\frac {{\mathrm e}^{2 x} x^{3} {\mathrm e}^{5}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}+\frac {{\mathrm e}^{2 x} x^{3}}{4}-\frac {5 x \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{4}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}-\frac {5 x \,{\mathrm e}^{2 x}}{4}+20 \,{\mathrm e}^{2 x}\)	\(62\)
meijerg	\(-19+\frac {155 \,{\mathrm e}^{2 x}}{8}-\frac {\left (80 x^{4}-160 x^{3}+240 x^{2}-240 x +120\right ) {\mathrm e}^{2 x}}{320}+\frac {\left (\frac {{\mathrm e}^{5}}{2}-\frac {1}{2}\right ) \left (6-\frac {\left (-32 x^{3}+48 x^{2}-48 x +24\right ) {\mathrm e}^{2 x}}{4}\right )}{16}-\frac {\left (\frac {3 \,{\mathrm e}^{5}}{4}-\frac {19}{4}\right ) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{8}+\frac {\left (-\frac {5 \,{\mathrm e}^{5}}{2}-8\right ) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{4}+\frac {5 \,{\mathrm e}^{5} \left (1-{\mathrm e}^{2 x}\right )}{8}\)	\(125\)
default	\(20 \,{\mathrm e}^{2 x}-\frac {5 x \,{\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{8}-\frac {11 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {{\mathrm e}^{2 x} x^{3}}{4}-\frac {{\mathrm e}^{2 x} x^{4}}{4}-\frac {5 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )}{2}+\frac {3 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-\frac {x \,{\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{2 x}}{4}\right )}{4}+\frac {{\mathrm e}^{5} \left (\frac {{\mathrm e}^{2 x} x^{3}}{2}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{4}+\frac {3 x \,{\mathrm e}^{2 x}}{4}-\frac {3 \,{\mathrm e}^{2 x}}{8}\right )}{2}\)	\(131\)

[In]

int(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*exp(x)^2*(x^3*exp(5)-x^4+x^3-5*x*exp(5)-11*x^2-5*x+80)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} - {\left (x^{3} - 5 \, x\right )} e^{5} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} \]

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="fricas")

[Out]

-1/4*(x^4 - x^3 + 11*x^2 - (x^3 - 5*x)*e^5 + 5*x - 80)*e^(2*x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=\frac {\left (- x^{4} + x^{3} + x^{3} e^{5} - 11 x^{2} - 5 x e^{5} - 5 x + 80\right ) e^{2 x}}{4} \]

[In]

integrate(1/4*((2*x**3+3*x**2-10*x-5)*exp(5)-2*x**4-2*x**3-19*x**2-32*x+155)*exp(x)**2,x)

[Out]

(-x**4 + x**3 + x**3*exp(5) - 11*x**2 - 5*x*exp(5) - 5*x + 80)*exp(2*x)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (25) = 50\).

Time = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 5.03 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{8} \, {\left (2 \, x^{4} - 4 \, x^{3} + 6 \, x^{2} - 6 \, x + 3\right )} e^{\left (2 \, x\right )} + \frac {1}{16} \, {\left (4 \, x^{3} e^{5} - 6 \, x^{2} e^{5} + 6 \, x e^{5} - 3 \, e^{5}\right )} e^{\left (2 \, x\right )} - \frac {1}{16} \, {\left (4 \, x^{3} - 6 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {3}{16} \, {\left (2 \, x^{2} e^{5} - 2 \, x e^{5} + e^{5}\right )} e^{\left (2 \, x\right )} - \frac {19}{16} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (2 \, x e^{5} - e^{5}\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {155}{8} \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{\left (2 \, x + 5\right )} \]

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="maxima")

[Out]

-1/8*(2*x^4 - 4*x^3 + 6*x^2 - 6*x + 3)*e^(2*x) + 1/16*(4*x^3*e^5 - 6*x^2*e^5 + 6*x*e^5 - 3*e^5)*e^(2*x) - 1/16
*(4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 3/16*(2*x^2*e^5 - 2*x*e^5 + e^5)*e^(2*x) - 19/16*(2*x^2 - 2*x + 1)*e^(2*x
) - 5/8*(2*x*e^5 - e^5)*e^(2*x) - 2*(2*x - 1)*e^(2*x) + 155/8*e^(2*x) - 5/8*e^(2*x + 5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=-\frac {1}{4} \, {\left (x^{4} - x^{3} + 11 \, x^{2} + 5 \, x - 80\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{3} - 5 \, x\right )} e^{\left (2 \, x + 5\right )} \]

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="giac")

[Out]

-1/4*(x^4 - x^3 + 11*x^2 + 5*x - 80)*e^(2*x) + 1/4*(x^3 - 5*x)*e^(2*x + 5)

Mupad [B] (verification not implemented)

Time = 10.71 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{4} e^{2 x} \left (155-32 x-19 x^2-2 x^3-2 x^4+e^5 \left (-5-10 x+3 x^2+2 x^3\right )\right ) \, dx=\frac {{\mathrm {e}}^{2\,x}\,\left (x^2-5\right )\,\left (x+x\,{\mathrm {e}}^5-x^2-16\right )}{4} \]

[In]

int(-(exp(2*x)*(32*x + exp(5)*(10*x - 3*x^2 - 2*x^3 + 5) + 19*x^2 + 2*x^3 + 2*x^4 - 155))/4,x)

[Out]

(exp(2*x)*(x^2 - 5)*(x + x*exp(5) - x^2 - 16))/4