3.65.39 \(\int \frac {21 x+4 x^2+e^{1+e^4+2 x} (10+22 x+4 x^2)+(5+x) \log (5+x)}{5+x} \, dx\)

Optimal. Leaf size=21 \[ x \left (2 \left (e^{1+e^4+2 x}+x\right )+\log (5+x)\right ) \]

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Rubi [B]  time = 0.26, antiderivative size = 48, normalized size of antiderivative = 2.29, number of steps used = 10, number of rules used = 7, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.159, Rules used = {6742, 2176, 2194, 6688, 77, 2389, 2295} \begin {gather*} 2 x^2-e^{2 x+e^4+1}+e^{2 x+e^4+1} (2 x+1)+(x+5) \log (x+5)-5 \log (x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(21*x + 4*x^2 + E^(1 + E^4 + 2*x)*(10 + 22*x + 4*x^2) + (5 + x)*Log[5 + x])/(5 + x),x]

[Out]

-E^(1 + E^4 + 2*x) + 2*x^2 + E^(1 + E^4 + 2*x)*(1 + 2*x) - 5*Log[5 + x] + (5 + x)*Log[5 + x]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{1+e^4+2 x} (1+2 x)+\frac {21 x+4 x^2+5 \log (5+x)+x \log (5+x)}{5+x}\right ) \, dx\\ &=2 \int e^{1+e^4+2 x} (1+2 x) \, dx+\int \frac {21 x+4 x^2+5 \log (5+x)+x \log (5+x)}{5+x} \, dx\\ &=e^{1+e^4+2 x} (1+2 x)-2 \int e^{1+e^4+2 x} \, dx+\int \left (\frac {x (21+4 x)}{5+x}+\log (5+x)\right ) \, dx\\ &=-e^{1+e^4+2 x}+e^{1+e^4+2 x} (1+2 x)+\int \frac {x (21+4 x)}{5+x} \, dx+\int \log (5+x) \, dx\\ &=-e^{1+e^4+2 x}+e^{1+e^4+2 x} (1+2 x)+\int \left (1+4 x-\frac {5}{5+x}\right ) \, dx+\operatorname {Subst}(\int \log (x) \, dx,x,5+x)\\ &=-e^{1+e^4+2 x}+2 x^2+e^{1+e^4+2 x} (1+2 x)-5 \log (5+x)+(5+x) \log (5+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 21, normalized size = 1.00 \begin {gather*} x \left (2 \left (e^{1+e^4+2 x}+x\right )+\log (5+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(21*x + 4*x^2 + E^(1 + E^4 + 2*x)*(10 + 22*x + 4*x^2) + (5 + x)*Log[5 + x])/(5 + x),x]

[Out]

x*(2*(E^(1 + E^4 + 2*x) + x) + Log[5 + x])

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fricas [A]  time = 0.94, size = 23, normalized size = 1.10 \begin {gather*} 2 \, x^{2} + 2 \, x e^{\left (2 \, x + e^{4} + 1\right )} + x \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)+(4*x^2+22*x+10)*exp(exp(2)^2+2*x+1)+4*x^2+21*x)/(5+x),x, algorithm="fricas")

[Out]

2*x^2 + 2*x*e^(2*x + e^4 + 1) + x*log(x + 5)

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giac [A]  time = 0.13, size = 23, normalized size = 1.10 \begin {gather*} 2 \, x^{2} + 2 \, x e^{\left (2 \, x + e^{4} + 1\right )} + x \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)+(4*x^2+22*x+10)*exp(exp(2)^2+2*x+1)+4*x^2+21*x)/(5+x),x, algorithm="giac")

[Out]

2*x^2 + 2*x*e^(2*x + e^4 + 1) + x*log(x + 5)

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maple [A]  time = 0.38, size = 24, normalized size = 1.14

method	result	size
risch	\(x \ln \left (5+x \right )+2 x^{2}+2 \,{\mathrm e}^{1+2 x +{\mathrm e}^{4}} x\)	\(24\)
norman	\(x \ln \left (5+x \right )+2 x^{2}+2 \,{\mathrm e}^{1+2 x +{\mathrm e}^{4}} x\)	\(26\)
default	\({\mathrm e}^{1+2 x +{\mathrm e}^{4}} \left (1+2 x +{\mathrm e}^{4}\right )-{\mathrm e}^{1+2 x +{\mathrm e}^{4}}-{\mathrm e}^{1+2 x +{\mathrm e}^{4}} {\mathrm e}^{4}+2 x^{2}-5 \ln \left (5+x \right )+\left (5+x \right ) \ln \left (5+x \right )-5\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5+x)*ln(5+x)+(4*x^2+22*x+10)*exp(exp(2)^2+2*x+1)+4*x^2+21*x)/(5+x),x,method=_RETURNVERBOSE)

[Out]

x*ln(5+x)+2*x^2+2*exp(1+2*x+exp(4))*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x^{2} + 2 \, x e^{\left (2 \, x + e^{4} + 1\right )} - 10 \, e^{\left (e^{4} - 9\right )} E_{1}\left (-2 \, x - 10\right ) + {\left (x + 5\right )} \log \left (x + 5\right ) - 10 \, \int \frac {e^{\left (2 \, x + e^{4} + 1\right )}}{x + 5}\,{d x} - 5 \, \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*log(5+x)+(4*x^2+22*x+10)*exp(exp(2)^2+2*x+1)+4*x^2+21*x)/(5+x),x, algorithm="maxima")

[Out]

2*x^2 + 2*x*e^(2*x + e^4 + 1) - 10*e^(e^4 - 9)*exp_integral_e(1, -2*x - 10) + (x + 5)*log(x + 5) - 10*integrat
e(e^(2*x + e^4 + 1)/(x + 5), x) - 5*log(x + 5)

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mupad [B]  time = 4.16, size = 24, normalized size = 1.14 \begin {gather*} x\,\ln \left (x+5\right )+2\,x^2+2\,x\,{\mathrm {e}}^{2\,x}\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((21*x + log(x + 5)*(x + 5) + exp(2*x + exp(4) + 1)*(22*x + 4*x^2 + 10) + 4*x^2)/(x + 5),x)

[Out]

x*log(x + 5) + 2*x^2 + 2*x*exp(2*x)*exp(1)*exp(exp(4))

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sympy [A]  time = 0.31, size = 24, normalized size = 1.14 \begin {gather*} 2 x^{2} + 2 x e^{2 x + 1 + e^{4}} + x \log {\left (x + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5+x)*ln(5+x)+(4*x**2+22*x+10)*exp(exp(2)**2+2*x+1)+4*x**2+21*x)/(5+x),x)

[Out]

2*x**2 + 2*x*exp(2*x + 1 + exp(4)) + x*log(x + 5)

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