3.13.54 \(\int \frac {28 e^4+e^{5+x^2} (2 x+e^4 (1-10 x+2 x^2))}{e^4} \, dx\)

Optimal. Leaf size=16 \[ \left (28+e^{5+x^2}\right ) \left (-5+\frac {1}{e^4}+x\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 28, normalized size of antiderivative = 1.75, number of steps used = 8, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 2226, 2204, 2209, 2212} \begin {gather*} e^{x^2+5} x+\left (1-5 e^4\right ) e^{x^2+1}+28 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(28*E^4 + E^(5 + x^2)*(2*x + E^4*(1 - 10*x + 2*x^2)))/E^4,x]

[Out]

E^(1 + x^2)*(1 - 5*E^4) + 28*x + E^(5 + x^2)*x

Rule 12

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (28 e^4+e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right )\right ) \, dx}{e^4}\\ &=28 x+\frac {\int e^{5+x^2} \left (2 x+e^4 \left (1-10 x+2 x^2\right )\right ) \, dx}{e^4}\\ &=28 x+\frac {\int \left (e^{9+x^2}-2 e^{5+x^2} \left (-1+5 e^4\right ) x+2 e^{9+x^2} x^2\right ) \, dx}{e^4}\\ &=28 x-\left (2 \left (5-\frac {1}{e^4}\right )\right ) \int e^{5+x^2} x \, dx+\frac {\int e^{9+x^2} \, dx}{e^4}+\frac {2 \int e^{9+x^2} x^2 \, dx}{e^4}\\ &=-\left (\left (5-\frac {1}{e^4}\right ) e^{5+x^2}\right )+28 x+e^{5+x^2} x+\frac {1}{2} e^5 \sqrt {\pi } \text {erfi}(x)-\frac {\int e^{9+x^2} \, dx}{e^4}\\ &=-\left (\left (5-\frac {1}{e^4}\right ) e^{5+x^2}\right )+28 x+e^{5+x^2} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.75 \begin {gather*} e^{1+x^2} \left (1-5 e^4\right )+28 x+e^{5+x^2} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(28*E^4 + E^(5 + x^2)*(2*x + E^4*(1 - 10*x + 2*x^2)))/E^4,x]

[Out]

E^(1 + x^2)*(1 - 5*E^4) + 28*x + E^(5 + x^2)*x

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fricas [A]  time = 0.91, size = 24, normalized size = 1.50 \begin {gather*} {\left (28 \, x e^{4} + {\left ({\left (x - 5\right )} e^{4} + 1\right )} e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x+1)*exp(4)+2*x)*exp(x^2+5)+28*exp(4))/exp(4),x, algorithm="fricas")

[Out]

(28*x*e^4 + ((x - 5)*e^4 + 1)*e^(x^2 + 5))*e^(-4)

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giac [A]  time = 0.31, size = 25, normalized size = 1.56 \begin {gather*} {\left (28 \, x e^{4} + {\left (x - 5\right )} e^{\left (x^{2} + 9\right )} + e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x+1)*exp(4)+2*x)*exp(x^2+5)+28*exp(4))/exp(4),x, algorithm="giac")

[Out]

(28*x*e^4 + (x - 5)*e^(x^2 + 9) + e^(x^2 + 5))*e^(-4)

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maple [A]  time = 0.03, size = 22, normalized size = 1.38

method	result	size
risch	\(28 x +\left (x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{4}+1\right ) {\mathrm e}^{x^{2}+1}\)	\(22\)
norman	\(x \,{\mathrm e}^{x^{2}+5}+28 x -\left (5 \,{\mathrm e}^{4}-1\right ) {\mathrm e}^{-4} {\mathrm e}^{x^{2}+5}\)	\(31\)
default	\({\mathrm e}^{-4} \left (\frac {{\mathrm e}^{4} {\mathrm e}^{5} \sqrt {\pi }\, \erfi \relax (x )}{2}+{\mathrm e}^{5} {\mathrm e}^{x^{2}}-5 \,{\mathrm e}^{x^{2}} {\mathrm e}^{5} {\mathrm e}^{4}+2 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \erfi \relax (x )}{4}\right )+28 x \,{\mathrm e}^{4}\right )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-10*x+1)*exp(4)+2*x)*exp(x^2+5)+28*exp(4))/exp(4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.41, size = 31, normalized size = 1.94 \begin {gather*} {\left (28 \, x e^{4} + x e^{\left (x^{2} + 9\right )} - 5 \, e^{\left (x^{2} + 9\right )} + e^{\left (x^{2} + 5\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-10*x+1)*exp(4)+2*x)*exp(x^2+5)+28*exp(4))/exp(4),x, algorithm="maxima")

[Out]

(28*x*e^4 + x*e^(x^2 + 9) - 5*e^(x^2 + 9) + e^(x^2 + 5))*e^(-4)

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mupad [B]  time = 0.92, size = 23, normalized size = 1.44 \begin {gather*} 28\,x+x\,{\mathrm {e}}^{x^2+5}+{\mathrm {e}}^{x^2+5}\,\left ({\mathrm {e}}^{-4}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4)*(28*exp(4) + exp(x^2 + 5)*(2*x + exp(4)*(2*x^2 - 10*x + 1))),x)

[Out]

28*x + x*exp(x^2 + 5) + exp(x^2 + 5)*(exp(-4) - 5)

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sympy [A]  time = 0.12, size = 24, normalized size = 1.50 \begin {gather*} 28 x + \frac {\left (x e^{4} - 5 e^{4} + 1\right ) e^{x^{2} + 5}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-10*x+1)*exp(4)+2*x)*exp(x**2+5)+28*exp(4))/exp(4),x)

[Out]

28*x + (x*exp(4) - 5*exp(4) + 1)*exp(-4)*exp(x**2 + 5)

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