3.33.35 \(\int \frac {10+20 x+75 x^2-40 x^3+e^2 (5+10 x+75 x^2-20 x^3)+e^x (2+6 x+17 x^2-3 x^3-2 x^4+e^2 (1+3 x+16 x^2+x^3-x^4))}{2+e^2} \, dx\)

Optimal. Leaf size=29 \[ \left (5+e^x\right ) x \left (1+x \left (1+\left (5-\frac {5}{2+e^2}-x\right ) x\right )\right ) \]

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Rubi [B]  time = 0.40, antiderivative size = 227, normalized size of antiderivative = 7.83, number of steps used = 37, number of rules used = 4, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -\frac {2 e^x x^4}{2+e^2}-\frac {e^{x+2} x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {10 x^4}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {5 e^{x+2} x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {e^{x+2} x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {2 e^x x}{2+e^2}+\frac {e^{x+2} x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {10 x}{2+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + 20*x + 75*x^2 - 40*x^3 + E^2*(5 + 10*x + 75*x^2 - 20*x^3) + E^x*(2 + 6*x + 17*x^2 - 3*x^3 - 2*x^4 +
E^2*(1 + 3*x + 16*x^2 + x^3 - x^4)))/(2 + E^2),x]

[Out]

(10*x)/(2 + E^2) + (5*E^2*x)/(2 + E^2) + (2*E^x*x)/(2 + E^2) + (E^(2 + x)*x)/(2 + E^2) + (10*x^2)/(2 + E^2) +
(5*E^2*x^2)/(2 + E^2) + (2*E^x*x^2)/(2 + E^2) + (E^(2 + x)*x^2)/(2 + E^2) + (25*x^3)/(2 + E^2) + (25*E^2*x^3)/
(2 + E^2) + (5*E^x*x^3)/(2 + E^2) + (5*E^(2 + x)*x^3)/(2 + E^2) - (10*x^4)/(2 + E^2) - (5*E^2*x^4)/(2 + E^2) -
(2*E^x*x^4)/(2 + E^2) - (E^(2 + x)*x^4)/(2 + E^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {\int \left (10+20 x+75 x^2-40 x^3+e^2 \left (5+10 x+75 x^2-20 x^3\right )+e^x \left (2+6 x+17 x^2-3 x^3-2 x^4+e^2 \left (1+3 x+16 x^2+x^3-x^4\right )\right )\right ) \, dx}{2+e^2}\\ &=\frac {10 x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {25 x^3}{2+e^2}-\frac {10 x^4}{2+e^2}+\frac {\int e^x \left (2+6 x+17 x^2-3 x^3-2 x^4+e^2 \left (1+3 x+16 x^2+x^3-x^4\right )\right ) \, dx}{2+e^2}+\frac {e^2 \int \left (5+10 x+75 x^2-20 x^3\right ) \, dx}{2+e^2}\\ &=\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}+\frac {\int \left (2 e^x+6 e^x x+17 e^x x^2-3 e^x x^3-2 e^x x^4-e^{2+x} \left (-1-3 x-16 x^2-x^3+x^4\right )\right ) \, dx}{2+e^2}\\ &=\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {\int e^{2+x} \left (-1-3 x-16 x^2-x^3+x^4\right ) \, dx}{2+e^2}+\frac {2 \int e^x \, dx}{2+e^2}-\frac {2 \int e^x x^4 \, dx}{2+e^2}-\frac {3 \int e^x x^3 \, dx}{2+e^2}+\frac {6 \int e^x x \, dx}{2+e^2}+\frac {17 \int e^x x^2 \, dx}{2+e^2}\\ &=\frac {2 e^x}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {6 e^x x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {17 e^x x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}-\frac {3 e^x x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {\int \left (-e^{2+x}-3 e^{2+x} x-16 e^{2+x} x^2-e^{2+x} x^3+e^{2+x} x^4\right ) \, dx}{2+e^2}-\frac {6 \int e^x \, dx}{2+e^2}+\frac {8 \int e^x x^3 \, dx}{2+e^2}+\frac {9 \int e^x x^2 \, dx}{2+e^2}-\frac {34 \int e^x x \, dx}{2+e^2}\\ &=-\frac {4 e^x}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}-\frac {28 e^x x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {26 e^x x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}+\frac {\int e^{2+x} \, dx}{2+e^2}+\frac {\int e^{2+x} x^3 \, dx}{2+e^2}-\frac {\int e^{2+x} x^4 \, dx}{2+e^2}+\frac {3 \int e^{2+x} x \, dx}{2+e^2}+\frac {16 \int e^{2+x} x^2 \, dx}{2+e^2}-\frac {18 \int e^x x \, dx}{2+e^2}-\frac {24 \int e^x x^2 \, dx}{2+e^2}+\frac {34 \int e^x \, dx}{2+e^2}\\ &=\frac {30 e^x}{2+e^2}+\frac {e^{2+x}}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}-\frac {46 e^x x}{2+e^2}+\frac {3 e^{2+x} x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {16 e^{2+x} x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {e^{2+x} x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {e^{2+x} x^4}{2+e^2}-\frac {3 \int e^{2+x} \, dx}{2+e^2}-\frac {3 \int e^{2+x} x^2 \, dx}{2+e^2}+\frac {4 \int e^{2+x} x^3 \, dx}{2+e^2}+\frac {18 \int e^x \, dx}{2+e^2}-\frac {32 \int e^{2+x} x \, dx}{2+e^2}+\frac {48 \int e^x x \, dx}{2+e^2}\\ &=\frac {48 e^x}{2+e^2}-\frac {2 e^{2+x}}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {2 e^x x}{2+e^2}-\frac {29 e^{2+x} x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {13 e^{2+x} x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {5 e^{2+x} x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {e^{2+x} x^4}{2+e^2}+\frac {6 \int e^{2+x} x \, dx}{2+e^2}-\frac {12 \int e^{2+x} x^2 \, dx}{2+e^2}+\frac {32 \int e^{2+x} \, dx}{2+e^2}-\frac {48 \int e^x \, dx}{2+e^2}\\ &=\frac {30 e^{2+x}}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {2 e^x x}{2+e^2}-\frac {23 e^{2+x} x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {e^{2+x} x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {5 e^{2+x} x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {e^{2+x} x^4}{2+e^2}-\frac {6 \int e^{2+x} \, dx}{2+e^2}+\frac {24 \int e^{2+x} x \, dx}{2+e^2}\\ &=\frac {24 e^{2+x}}{2+e^2}+\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {2 e^x x}{2+e^2}+\frac {e^{2+x} x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {e^{2+x} x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {5 e^{2+x} x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {e^{2+x} x^4}{2+e^2}-\frac {24 \int e^{2+x} \, dx}{2+e^2}\\ &=\frac {10 x}{2+e^2}+\frac {5 e^2 x}{2+e^2}+\frac {2 e^x x}{2+e^2}+\frac {e^{2+x} x}{2+e^2}+\frac {10 x^2}{2+e^2}+\frac {5 e^2 x^2}{2+e^2}+\frac {2 e^x x^2}{2+e^2}+\frac {e^{2+x} x^2}{2+e^2}+\frac {25 x^3}{2+e^2}+\frac {25 e^2 x^3}{2+e^2}+\frac {5 e^x x^3}{2+e^2}+\frac {5 e^{2+x} x^3}{2+e^2}-\frac {10 x^4}{2+e^2}-\frac {5 e^2 x^4}{2+e^2}-\frac {2 e^x x^4}{2+e^2}-\frac {e^{2+x} x^4}{2+e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 47, normalized size = 1.62 \begin {gather*} -\frac {\left (5+e^x\right ) x \left (-2-2 x-5 x^2+2 x^3+e^2 \left (-1-x-5 x^2+x^3\right )\right )}{2+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 20*x + 75*x^2 - 40*x^3 + E^2*(5 + 10*x + 75*x^2 - 20*x^3) + E^x*(2 + 6*x + 17*x^2 - 3*x^3 - 2*
x^4 + E^2*(1 + 3*x + 16*x^2 + x^3 - x^4)))/(2 + E^2),x]

[Out]

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

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fricas [B]  time = 0.66, size = 90, normalized size = 3.10 \begin {gather*} -\frac {10 \, x^{4} - 25 \, x^{3} - 10 \, x^{2} + 5 \, {\left (x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{2} + {\left (2 \, x^{4} - 5 \, x^{3} - 2 \, x^{2} + {\left (x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{2} - 2 \, x\right )} e^{x} - 10 \, x}{e^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="fricas")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (2*x^4 - 5*x^3 - 2*x^2 + (x^4 - 5*x^3 - x^2 - x)*
e^2 - 2*x)*e^x - 10*x)/(e^2 + 2)

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giac [B]  time = 0.14, size = 92, normalized size = 3.17 \begin {gather*} -\frac {10 \, x^{4} - 25 \, x^{3} - 10 \, x^{2} + 5 \, {\left (x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{2} + {\left (x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{\left (x + 2\right )} + {\left (2 \, x^{4} - 5 \, x^{3} - 2 \, x^{2} - 2 \, x\right )} e^{x} - 10 \, x}{e^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="giac")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (x^4 - 5*x^3 - x^2 - x)*e^(x + 2) + (2*x^4 - 5*x^
3 - 2*x^2 - 2*x)*e^x - 10*x)/(e^2 + 2)

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maple [B]  time = 0.04, size = 64, normalized size = 2.21

method	result	size
norman	\({\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}+5 x +5 x^{2}-5 x^{4}-{\mathrm e}^{x} x^{4}+\frac {25 \left ({\mathrm e}^{2}+1\right ) x^{3}}{{\mathrm e}^{2}+2}+\frac {5 \left ({\mathrm e}^{2}+1\right ) x^{3} {\mathrm e}^{x}}{{\mathrm e}^{2}+2}\)	\(64\)
risch	\(-\frac {5 x^{4} {\mathrm e}^{2}}{{\mathrm e}^{2}+2}+\frac {25 x^{3} {\mathrm e}^{2}}{{\mathrm e}^{2}+2}-\frac {10 x^{4}}{{\mathrm e}^{2}+2}+\frac {5 x^{2} {\mathrm e}^{2}}{{\mathrm e}^{2}+2}+\frac {25 x^{3}}{{\mathrm e}^{2}+2}+\frac {5 \,{\mathrm e}^{2} x}{{\mathrm e}^{2}+2}+\frac {10 x^{2}}{{\mathrm e}^{2}+2}+\frac {10 x}{{\mathrm e}^{2}+2}+\frac {\left (-x^{4} {\mathrm e}^{2}+5 x^{3} {\mathrm e}^{2}-2 x^{4}+x^{2} {\mathrm e}^{2}+5 x^{3}+{\mathrm e}^{2} x +2 x^{2}+2 x \right ) {\mathrm e}^{x}}{{\mathrm e}^{2}+2}\)	\(146\)
default	\(\frac {10 x +{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} \left ({\mathrm e}^{x} x^{3}-3 \,{\mathrm e}^{x} x^{2}+6 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}\right )+2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} x^{3}-2 \,{\mathrm e}^{x} x^{4}+3 \,{\mathrm e}^{2} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+16 \,{\mathrm e}^{2} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )-{\mathrm e}^{2} \left ({\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{3}+12 \,{\mathrm e}^{x} x^{2}-24 \,{\mathrm e}^{x} x +24 \,{\mathrm e}^{x}\right )+{\mathrm e}^{2} \left (-5 x^{4}+25 x^{3}+5 x^{2}+5 x \right )+10 x^{2}+25 x^{3}-10 x^{4}}{{\mathrm e}^{2}+2}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*x^3+75
*x^2+20*x+10)/(exp(2)+2),x,method=_RETURNVERBOSE)

[Out]

exp(x)*x+exp(x)*x^2+5*x+5*x^2-5*x^4-exp(x)*x^4+25*(exp(2)+1)/(exp(2)+2)*x^3+5*(exp(2)+1)/(exp(2)+2)*x^3*exp(x)

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maxima [B]  time = 0.38, size = 85, normalized size = 2.93 \begin {gather*} -\frac {10 \, x^{4} - 25 \, x^{3} - 10 \, x^{2} + 5 \, {\left (x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{2} + {\left (x^{4} {\left (e^{2} + 2\right )} - 5 \, x^{3} {\left (e^{2} + 1\right )} - x^{2} {\left (e^{2} + 2\right )} - x {\left (e^{2} + 2\right )}\right )} e^{x} - 10 \, x}{e^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="maxima")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (x^4*(e^2 + 2) - 5*x^3*(e^2 + 1) - x^2*(e^2 + 2)
- x*(e^2 + 2))*e^x - 10*x)/(e^2 + 2)

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mupad [B]  time = 2.04, size = 47, normalized size = 1.62 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^x+5\right )\,\left (2\,x+{\mathrm {e}}^2+x\,{\mathrm {e}}^2+5\,x^2\,{\mathrm {e}}^2-x^3\,{\mathrm {e}}^2+5\,x^2-2\,x^3+2\right )}{{\mathrm {e}}^2+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + exp(x)*(6*x + exp(2)*(3*x + 16*x^2 + x^3 - x^4 + 1) + 17*x^2 - 3*x^3 - 2*x^4 + 2) + exp(2)*(10*x +
75*x^2 - 20*x^3 + 5) + 75*x^2 - 40*x^3 + 10)/(exp(2) + 2),x)

[Out]

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

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sympy [B]  time = 0.23, size = 82, normalized size = 2.83 \begin {gather*} - 5 x^{4} + \frac {x^{3} \left (25 + 25 e^{2}\right )}{2 + e^{2}} + 5 x^{2} + 5 x + \frac {\left (- x^{4} e^{2} - 2 x^{4} + 5 x^{3} + 5 x^{3} e^{2} + 2 x^{2} + x^{2} e^{2} + 2 x + x e^{2}\right ) e^{x}}{2 + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**4+x**3+16*x**2+3*x+1)*exp(2)-2*x**4-3*x**3+17*x**2+6*x+2)*exp(x)+(-20*x**3+75*x**2+10*x+5)*ex
p(2)-40*x**3+75*x**2+20*x+10)/(exp(2)+2),x)

[Out]

-5*x**4 + x**3*(25 + 25*exp(2))/(2 + exp(2)) + 5*x**2 + 5*x + (-x**4*exp(2) - 2*x**4 + 5*x**3 + 5*x**3*exp(2)
+ 2*x**2 + x**2*exp(2) + 2*x + x*exp(2))*exp(x)/(2 + exp(2))

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