3.78.8 \(\int \frac {1}{5} e^x (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)) \, dx\)

Optimal. Leaf size=18 \[ e^{5+x} \left (2+e^{25+\frac {x}{5}}\right ) x \]

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Rubi [B]  time = 0.10, antiderivative size = 49, normalized size of antiderivative = 2.72, number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 6742, 2176, 2194} \begin {gather*} 2 e^{x+5} (x+1)-2 e^{x+5}-\frac {5}{6} e^{\frac {6 x}{5}+30}+\frac {1}{6} e^{\frac {6 x}{5}+30} (6 x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(E^(5 + (125 + x)/5)*(5 + 6*x) + E^5*(10 + 10*x)))/5,x]

[Out]

-2*E^(5 + x) - (5*E^(30 + (6*x)/5))/6 + 2*E^(5 + x)*(1 + x) + (E^(30 + (6*x)/5)*(5 + 6*x))/6

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^x \left (e^{5+\frac {125+x}{5}} (5+6 x)+e^5 (10+10 x)\right ) \, dx\\ &=\frac {1}{5} \int \left (10 e^{5+x} (1+x)+e^{30+\frac {6 x}{5}} (5+6 x)\right ) \, dx\\ &=\frac {1}{5} \int e^{30+\frac {6 x}{5}} (5+6 x) \, dx+2 \int e^{5+x} (1+x) \, dx\\ &=2 e^{5+x} (1+x)+\frac {1}{6} e^{30+\frac {6 x}{5}} (5+6 x)-2 \int e^{5+x} \, dx-\int e^{30+\frac {6 x}{5}} \, dx\\ &=-2 e^{5+x}-\frac {5}{6} e^{30+\frac {6 x}{5}}+2 e^{5+x} (1+x)+\frac {1}{6} e^{30+\frac {6 x}{5}} (5+6 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.11 \begin {gather*} 2 e^{5+x} x+e^{30+\frac {6 x}{5}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(E^(5 + (125 + x)/5)*(5 + 6*x) + E^5*(10 + 10*x)))/5,x]

[Out]

2*E^(5 + x)*x + E^(30 + (6*x)/5)*x

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fricas [A]  time = 0.54, size = 19, normalized size = 1.06 \begin {gather*} {\left (x e^{\left (\frac {6}{5} \, x + 180\right )} + 2 \, x e^{\left (x + 155\right )}\right )} e^{\left (-150\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((6*x+5)*exp(5)*exp(1/5*x+25)+(10*x+10)*exp(5))*exp(x+log(x))/x,x, algorithm="fricas")

[Out]

(x*e^(6/5*x + 180) + 2*x*e^(x + 155))*e^(-150)

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giac [A]  time = 0.14, size = 16, normalized size = 0.89 \begin {gather*} x e^{\left (\frac {6}{5} \, x + 30\right )} + 2 \, x e^{\left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((6*x+5)*exp(5)*exp(1/5*x+25)+(10*x+10)*exp(5))*exp(x+log(x))/x,x, algorithm="giac")

[Out]

x*e^(6/5*x + 30) + 2*x*e^(x + 5)

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maple [A]  time = 0.06, size = 17, normalized size = 0.94

method	result	size
risch	\(x \,{\mathrm e}^{\frac {6 x}{5}+30}+2 x \,{\mathrm e}^{5+x}\)	\(17\)
norman	\({\mathrm e}^{-125} {\mathrm e}^{5} x \,{\mathrm e}^{\frac {6 x}{5}+150}+2 \,{\mathrm e}^{-125} {\mathrm e}^{5} x \,{\mathrm e}^{x +125}\)	\(35\)
default	\(2 \,{\mathrm e}^{5} {\mathrm e}^{x}+2 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-25 \,{\mathrm e}^{\frac {6 x}{5}+30}+\frac {5 \,{\mathrm e}^{\frac {6 x}{5}+30} \left (\frac {6 x}{5}+30\right )}{6}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((6*x+5)*exp(5)*exp(1/5*x+25)+(10*x+10)*exp(5))*exp(x+ln(x))/x,x,method=_RETURNVERBOSE)

[Out]

x*exp(6/5*x+30)+2*x*exp(5+x)

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maxima [B]  time = 0.36, size = 44, normalized size = 2.44 \begin {gather*} \frac {1}{6} \, {\left (6 \, x e^{30} - 5 \, e^{30}\right )} e^{\left (\frac {6}{5} \, x\right )} + 2 \, {\left (x e^{5} - e^{5}\right )} e^{x} + \frac {5}{6} \, e^{\left (\frac {6}{5} \, x + 30\right )} + 2 \, e^{\left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((6*x+5)*exp(5)*exp(1/5*x+25)+(10*x+10)*exp(5))*exp(x+log(x))/x,x, algorithm="maxima")

[Out]

1/6*(6*x*e^30 - 5*e^30)*e^(6/5*x) + 2*(x*e^5 - e^5)*e^x + 5/6*e^(6/5*x + 30) + 2*e^(x + 5)

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mupad [B]  time = 0.09, size = 15, normalized size = 0.83 \begin {gather*} x\,\left (2\,{\mathrm {e}}^{x+5}+{\mathrm {e}}^{\frac {6\,x}{5}+30}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(x))*(exp(5)*(10*x + 10) + exp(5)*exp(x/5 + 25)*(6*x + 5)))/(5*x),x)

[Out]

x*(2*exp(x + 5) + exp((6*x)/5 + 30))

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sympy [A]  time = 0.19, size = 20, normalized size = 1.11 \begin {gather*} x e^{30} \left (e^{x}\right )^{\frac {6}{5}} + 2 x e^{5} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((6*x+5)*exp(5)*exp(1/5*x+25)+(10*x+10)*exp(5))*exp(x+ln(x))/x,x)

[Out]

x*exp(30)*exp(x)**(6/5) + 2*x*exp(5)*exp(x)

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