3.52.97 \(\int (-1+15 x^2+e^{2 x} (1+2 x)) \, dx\)

Optimal. Leaf size=17 \[ 2-x+e^{2 x} x+5 x^3 \]

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.88, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2176, 2194} \begin {gather*} 5 x^3-x-\frac {e^{2 x}}{2}+\frac {1}{2} e^{2 x} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + 15*x^2 + E^(2*x)*(1 + 2*x),x]

[Out]

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

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x+5 x^3+\int e^{2 x} (1+2 x) \, dx\\ &=-x+5 x^3+\frac {1}{2} e^{2 x} (1+2 x)-\int e^{2 x} \, dx\\ &=-\frac {e^{2 x}}{2}-x+5 x^3+\frac {1}{2} e^{2 x} (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.94 \begin {gather*} -x+e^{2 x} x+5 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + 15*x^2 + E^(2*x)*(1 + 2*x),x]

[Out]

-x + E^(2*x)*x + 5*x^3

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fricas [A]  time = 0.70, size = 15, normalized size = 0.88 \begin {gather*} 5 \, x^{3} + x e^{\left (2 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(2*x)+15*x^2-1,x, algorithm="fricas")

[Out]

5*x^3 + x*e^(2*x) - x

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giac [A]  time = 1.10, size = 15, normalized size = 0.88 \begin {gather*} 5 \, x^{3} + x e^{\left (2 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(2*x)+15*x^2-1,x, algorithm="giac")

[Out]

5*x^3 + x*e^(2*x) - x

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maple [A]  time = 0.04, size = 16, normalized size = 0.94

method	result	size
derivativedivides	\(-x +5 x^{3}+x \,{\mathrm e}^{2 x}\)	\(16\)
default	\(-x +5 x^{3}+x \,{\mathrm e}^{2 x}\)	\(16\)
norman	\(-x +5 x^{3}+x \,{\mathrm e}^{2 x}\)	\(16\)
risch	\(-x +5 x^{3}+x \,{\mathrm e}^{2 x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x+1)*exp(2*x)+15*x^2-1,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 15, normalized size = 0.88 \begin {gather*} 5 \, x^{3} + x e^{\left (2 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(2*x)+15*x^2-1,x, algorithm="maxima")

[Out]

5*x^3 + x*e^(2*x) - x

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mupad [B]  time = 0.04, size = 13, normalized size = 0.76 \begin {gather*} x\,\left ({\mathrm {e}}^{2\,x}+5\,x^2-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*(2*x + 1) + 15*x^2 - 1,x)

[Out]

x*(exp(2*x) + 5*x^2 - 1)

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sympy [A]  time = 0.08, size = 12, normalized size = 0.71 \begin {gather*} 5 x^{3} + x e^{2 x} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(2*x)+15*x**2-1,x)

[Out]

5*x**3 + x*exp(2*x) - x

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