\(\int \frac {1}{3} (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)) \, dx\) [7145]

Optimal result

Integrand size = 27, antiderivative size = 19 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=-x+x \left (9+e^{3+\frac {5 x}{3}}+\log (4)\right ) \]

[Out]

x*(exp(5/3*x+3)+9+2*ln(2))-x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(42\) vs. \(2(19)=38\).

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2207, 2225} \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=-\frac {3}{5} e^{\frac {1}{3} (5 x+9)}+\frac {1}{5} e^{\frac {1}{3} (5 x+9)} (5 x+3)+x (8+\log (4)) \]

[In]

Int[(24 + E^((9 + 5*x)/3)*(3 + 5*x) + 3*Log[4])/3,x]

[Out]

(-3*E^((9 + 5*x)/3))/5 + (E^((9 + 5*x)/3)*(3 + 5*x))/5 + x*(8 + Log[4])

Rule 12

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

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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{align*} \text {integral}& = \frac {1}{3} \int \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx \\ & = x (8+\log (4))+\frac {1}{3} \int e^{\frac {1}{3} (9+5 x)} (3+5 x) \, dx \\ & = \frac {1}{5} e^{\frac {1}{3} (9+5 x)} (3+5 x)+x (8+\log (4))-\int e^{\frac {1}{3} (9+5 x)} \, dx \\ & = -\frac {3}{5} e^{\frac {1}{3} (9+5 x)}+\frac {1}{5} e^{\frac {1}{3} (9+5 x)} (3+5 x)+x (8+\log (4)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=\frac {1}{3} \left (24 x+3 e^{3+\frac {5 x}{3}} x+x \log (64)\right ) \]

[In]

Integrate[(24 + E^((9 + 5*x)/3)*(3 + 5*x) + 3*Log[4])/3,x]

[Out]

(24*x + 3*E^(3 + (5*x)/3)*x + x*Log[64])/3

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
norman	\(\left (8+2 \ln \left (2\right )\right ) x +{\mathrm e}^{\frac {5 x}{3}+3} x\)	\(18\)
risch	\({\mathrm e}^{\frac {5 x}{3}+3} x +2 x \ln \left (2\right )+8 x\)	\(18\)
parallelrisch	\(\left (8+2 \ln \left (2\right )\right ) x +{\mathrm e}^{\frac {5 x}{3}+3} x\)	\(18\)
default	\(8 x +\frac {3 \,{\mathrm e}^{\frac {5 x}{3}+3} \left (\frac {5 x}{3}+3\right )}{5}-\frac {9 \,{\mathrm e}^{\frac {5 x}{3}+3}}{5}+2 x \ln \left (2\right )\)	\(31\)
parts	\(8 x +\frac {3 \,{\mathrm e}^{\frac {5 x}{3}+3} \left (\frac {5 x}{3}+3\right )}{5}-\frac {9 \,{\mathrm e}^{\frac {5 x}{3}+3}}{5}+2 x \ln \left (2\right )\)	\(31\)
derivativedivides	\(8 x +\frac {72}{5}+\frac {3 \,{\mathrm e}^{\frac {5 x}{3}+3} \left (\frac {5 x}{3}+3\right )}{5}-\frac {9 \,{\mathrm e}^{\frac {5 x}{3}+3}}{5}+\frac {6 \left (\frac {5 x}{3}+3\right ) \ln \left (2\right )}{5}\)	\(36\)

[In]

int(1/3*(5*x+3)*exp(5/3*x+3)+2*ln(2)+8,x,method=_RETURNVERBOSE)

[Out]

(8+2*ln(2))*x+exp(5/3*x+3)*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=x e^{\left (\frac {5}{3} \, x + 3\right )} + 2 \, x \log \left (2\right ) + 8 \, x \]

[In]

integrate(1/3*(5*x+3)*exp(5/3*x+3)+2*log(2)+8,x, algorithm="fricas")

[Out]

x*e^(5/3*x + 3) + 2*x*log(2) + 8*x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=x e^{\frac {5 x}{3} + 3} + x \left (2 \log {\left (2 \right )} + 8\right ) \]

[In]

integrate(1/3*(5*x+3)*exp(5/3*x+3)+2*ln(2)+8,x)

[Out]

x*exp(5*x/3 + 3) + x*(2*log(2) + 8)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=x e^{\left (\frac {5}{3} \, x + 3\right )} + 2 \, x \log \left (2\right ) + 8 \, x \]

[In]

integrate(1/3*(5*x+3)*exp(5/3*x+3)+2*log(2)+8,x, algorithm="maxima")

[Out]

x*e^(5/3*x + 3) + 2*x*log(2) + 8*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=x e^{\left (\frac {5}{3} \, x + 3\right )} + 2 \, x \log \left (2\right ) + 8 \, x \]

[In]

integrate(1/3*(5*x+3)*exp(5/3*x+3)+2*log(2)+8,x, algorithm="giac")

[Out]

x*e^(5/3*x + 3) + 2*x*log(2) + 8*x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{3} \left (24+e^{\frac {1}{3} (9+5 x)} (3+5 x)+3 \log (4)\right ) \, dx=x\,\left (\frac {\ln \left (64\right )}{3}+8\right )+x\,{\mathrm {e}}^{\frac {5\,x}{3}+3} \]

[In]

int(2*log(2) + (exp((5*x)/3 + 3)*(5*x + 3))/3 + 8,x)

[Out]

x*(log(64)/3 + 8) + x*exp((5*x)/3 + 3)