3.56.19 \(\int (200 x+120 x^2+16 x^3+e^{2 x} (-2 x+2 x^2)+e^x (-20+12 x+28 x^2+4 x^3)) \, dx\)

Optimal. Leaf size=20 \[ 1+\left (-e^x+x \left (10+e^x+2 x\right )\right )^2 \]

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Rubi [B]  time = 0.13, antiderivative size = 60, normalized size of antiderivative = 3.00, number of steps used = 21, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} 4 x^4+4 e^x x^3+40 x^3+16 e^x x^2+e^{2 x} x^2+100 x^2-20 e^x x-2 e^{2 x} x+e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[200*x + 120*x^2 + 16*x^3 + E^(2*x)*(-2*x + 2*x^2) + E^x*(-20 + 12*x + 28*x^2 + 4*x^3),x]

[Out]

E^(2*x) - 20*E^x*x - 2*E^(2*x)*x + 100*x^2 + 16*E^x*x^2 + E^(2*x)*x^2 + 40*x^3 + 4*E^x*x^3 + 4*x^4

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} &=100 x^2+40 x^3+4 x^4+\int e^{2 x} \left (-2 x+2 x^2\right ) \, dx+\int e^x \left (-20+12 x+28 x^2+4 x^3\right ) \, dx\\ &=100 x^2+40 x^3+4 x^4+\int e^{2 x} x (-2+2 x) \, dx+\int \left (-20 e^x+12 e^x x+28 e^x x^2+4 e^x x^3\right ) \, dx\\ &=100 x^2+40 x^3+4 x^4+4 \int e^x x^3 \, dx+12 \int e^x x \, dx-20 \int e^x \, dx+28 \int e^x x^2 \, dx+\int \left (-2 e^{2 x} x+2 e^{2 x} x^2\right ) \, dx\\ &=-20 e^x+12 e^x x+100 x^2+28 e^x x^2+40 x^3+4 e^x x^3+4 x^4-2 \int e^{2 x} x \, dx+2 \int e^{2 x} x^2 \, dx-12 \int e^x \, dx-12 \int e^x x^2 \, dx-56 \int e^x x \, dx\\ &=-32 e^x-44 e^x x-e^{2 x} x+100 x^2+16 e^x x^2+e^{2 x} x^2+40 x^3+4 e^x x^3+4 x^4-2 \int e^{2 x} x \, dx+24 \int e^x x \, dx+56 \int e^x \, dx+\int e^{2 x} \, dx\\ &=24 e^x+\frac {e^{2 x}}{2}-20 e^x x-2 e^{2 x} x+100 x^2+16 e^x x^2+e^{2 x} x^2+40 x^3+4 e^x x^3+4 x^4-24 \int e^x \, dx+\int e^{2 x} \, dx\\ &=e^{2 x}-20 e^x x-2 e^{2 x} x+100 x^2+16 e^x x^2+e^{2 x} x^2+40 x^3+4 e^x x^3+4 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 16, normalized size = 0.80 \begin {gather*} \left (e^x (-1+x)+2 x (5+x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[200*x + 120*x^2 + 16*x^3 + E^(2*x)*(-2*x + 2*x^2) + E^x*(-20 + 12*x + 28*x^2 + 4*x^3),x]

[Out]

(E^x*(-1 + x) + 2*x*(5 + x))^2

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fricas [B]  time = 0.85, size = 45, normalized size = 2.25 \begin {gather*} 4 \, x^{4} + 40 \, x^{3} + 100 \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{3} + 4 \, x^{2} - 5 \, x\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+(4*x^3+28*x^2+12*x-20)*exp(x)+16*x^3+120*x^2+200*x,x, algorithm="fricas")

[Out]

4*x^4 + 40*x^3 + 100*x^2 + (x^2 - 2*x + 1)*e^(2*x) + 4*(x^3 + 4*x^2 - 5*x)*e^x

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giac [B]  time = 0.14, size = 45, normalized size = 2.25 \begin {gather*} 4 \, x^{4} + 40 \, x^{3} + 100 \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{3} + 4 \, x^{2} - 5 \, x\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+(4*x^3+28*x^2+12*x-20)*exp(x)+16*x^3+120*x^2+200*x,x, algorithm="giac")

[Out]

4*x^4 + 40*x^3 + 100*x^2 + (x^2 - 2*x + 1)*e^(2*x) + 4*(x^3 + 4*x^2 - 5*x)*e^x

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maple [B]  time = 0.04, size = 47, normalized size = 2.35

method	result	size
risch	\(\left (x^{2}-2 x +1\right ) {\mathrm e}^{2 x}+\left (4 x^{3}+16 x^{2}-20 x \right ) {\mathrm e}^{x}+4 x^{4}+40 x^{3}+100 x^{2}\)	\(47\)
default	\({\mathrm e}^{2 x} x^{2}-2 x \,{\mathrm e}^{2 x}+{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x^{3}+16 \,{\mathrm e}^{x} x^{2}-20 \,{\mathrm e}^{x} x +100 x^{2}+40 x^{3}+4 x^{4}\)	\(55\)
norman	\({\mathrm e}^{2 x} x^{2}-2 x \,{\mathrm e}^{2 x}+{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x^{3}+16 \,{\mathrm e}^{x} x^{2}-20 \,{\mathrm e}^{x} x +100 x^{2}+40 x^{3}+4 x^{4}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2-2*x)*exp(x)^2+(4*x^3+28*x^2+12*x-20)*exp(x)+16*x^3+120*x^2+200*x,x,method=_RETURNVERBOSE)

[Out]

(x^2-2*x+1)*exp(2*x)+(4*x^3+16*x^2-20*x)*exp(x)+4*x^4+40*x^3+100*x^2

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maxima [B]  time = 0.36, size = 45, normalized size = 2.25 \begin {gather*} 4 \, x^{4} + 40 \, x^{3} + 100 \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{3} + 4 \, x^{2} - 5 \, x\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+(4*x^3+28*x^2+12*x-20)*exp(x)+16*x^3+120*x^2+200*x,x, algorithm="maxima")

[Out]

4*x^4 + 40*x^3 + 100*x^2 + (x^2 - 2*x + 1)*e^(2*x) + 4*(x^3 + 4*x^2 - 5*x)*e^x

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mupad [B]  time = 0.11, size = 19, normalized size = 0.95 \begin {gather*} {\left (10\,x-{\mathrm {e}}^x+x\,{\mathrm {e}}^x+2\,x^2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(200*x - exp(2*x)*(2*x - 2*x^2) + 120*x^2 + 16*x^3 + exp(x)*(12*x + 28*x^2 + 4*x^3 - 20),x)

[Out]

(10*x - exp(x) + x*exp(x) + 2*x^2)^2

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sympy [B]  time = 0.12, size = 44, normalized size = 2.20 \begin {gather*} 4 x^{4} + 40 x^{3} + 100 x^{2} + \left (x^{2} - 2 x + 1\right ) e^{2 x} + \left (4 x^{3} + 16 x^{2} - 20 x\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+(4*x**3+28*x**2+12*x-20)*exp(x)+16*x**3+120*x**2+200*x,x)

[Out]

4*x**4 + 40*x**3 + 100*x**2 + (x**2 - 2*x + 1)*exp(2*x) + (4*x**3 + 16*x**2 - 20*x)*exp(x)

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