3.97.39 \(\int \frac {1}{5} e^{\frac {1}{5} (5 x+x \log (3))} (-155-75 x+(-16-15 x) \log (3)) \, dx\)

Optimal. Leaf size=21 \[ 5+e^{x+\frac {1}{5} x \log (3)} (x-16 (1+x)) \]

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Rubi [B]  time = 0.05, antiderivative size = 52, normalized size of antiderivative = 2.48, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 2187, 2176, 2194} \begin {gather*} \frac {25\ 3^{\frac {x}{5}+1} e^x}{5+\log (3)}-\frac {3^{x/5} e^x (15 x (5+\log (3))+155+16 \log (3))}{5+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((5*x + x*Log[3])/5)*(-155 - 75*x + (-16 - 15*x)*Log[3]))/5,x]

[Out]

(25*3^(1 + x/5)*E^x)/(5 + Log[3]) - (3^(x/5)*E^x*(155 + 16*Log[3] + 15*x*(5 + Log[3])))/(5 + Log[3])

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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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} &=\frac {1}{5} \int e^{\frac {1}{5} (5 x+x \log (3))} (-155-75 x+(-16-15 x) \log (3)) \, dx\\ &=\frac {1}{5} \int e^{\frac {1}{5} x (5+\log (3))} (-155-16 \log (3)-15 x (5+\log (3))) \, dx\\ &=-\frac {3^{x/5} e^x (155+16 \log (3)+15 x (5+\log (3)))}{5+\log (3)}+15 \int e^{\frac {1}{5} x (5+\log (3))} \, dx\\ &=\frac {25\ 3^{1+\frac {x}{5}} e^x}{5+\log (3)}-\frac {3^{x/5} e^x (155+16 \log (3)+15 x (5+\log (3)))}{5+\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{5} 3^{x/5} e^x (80+75 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5*x + x*Log[3])/5)*(-155 - 75*x + (-16 - 15*x)*Log[3]))/5,x]

[Out]

-1/5*(3^(x/5)*E^x*(80 + 75*x))

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fricas [A]  time = 0.53, size = 15, normalized size = 0.71 \begin {gather*} -{\left (15 \, x + 16\right )} e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-15*x-16)*log(3)-75*x-155)*exp(1/5*x*log(3)+x),x, algorithm="fricas")

[Out]

-(15*x + 16)*e^(1/5*x*log(3) + x)

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giac [B]  time = 0.24, size = 49, normalized size = 2.33 \begin {gather*} -\frac {{\left (15 \, x \log \relax (3)^{2} + 150 \, x \log \relax (3) + 16 \, \log \relax (3)^{2} + 375 \, x + 160 \, \log \relax (3) + 400\right )} e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )}}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-15*x-16)*log(3)-75*x-155)*exp(1/5*x*log(3)+x),x, algorithm="giac")

[Out]

-(15*x*log(3)^2 + 150*x*log(3) + 16*log(3)^2 + 375*x + 160*log(3) + 400)*e^(1/5*x*log(3) + x)/(log(3)^2 + 10*l
og(3) + 25)

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maple [A]  time = 0.06, size = 15, normalized size = 0.71

method	result	size
risch	\(\frac {\left (-75 x -80\right ) 3^{\frac {x}{5}} {\mathrm e}^{x}}{5}\)	\(15\)
gosper	\(-{\mathrm e}^{\frac {x \ln \relax (3)}{5}+x} \left (15 x +16\right )\)	\(16\)
norman	\(-15 x \,{\mathrm e}^{\frac {x \ln \relax (3)}{5}+x}-16 \,{\mathrm e}^{\frac {x \ln \relax (3)}{5}+x}\)	\(23\)
meijerg	\(-\frac {155 \left (1-{\mathrm e}^{-\frac {x \left (-\ln \relax (3)-5\right )}{5}}\right )}{-\ln \relax (3)-5}+\frac {5 \left (-15 \ln \relax (3)-75\right ) \left (1-\frac {\left (2+\frac {2 x \left (-\ln \relax (3)-5\right )}{5}\right ) {\mathrm e}^{-\frac {x \left (-\ln \relax (3)-5\right )}{5}}}{2}\right )}{\left (-\ln \relax (3)-5\right )^{2}}-\frac {16 \ln \relax (3) \left (1-{\mathrm e}^{-\frac {x \left (-\ln \relax (3)-5\right )}{5}}\right )}{-\ln \relax (3)-5}\)	\(93\)
derivativedivides	\(\frac {-16 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \ln \relax (3)-\frac {375 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \left (\frac {\ln \relax (3)}{5}+1\right ) x}{5+\ln \relax (3)}+\frac {375 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}-\frac {75 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \ln \relax (3) \left (\frac {\ln \relax (3)}{5}+1\right ) x}{5+\ln \relax (3)}+\frac {75 \ln \relax (3) {\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}-155 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}\)	\(119\)
default	\(\frac {-16 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \ln \relax (3)-\frac {375 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \left (\frac {\ln \relax (3)}{5}+1\right ) x}{5+\ln \relax (3)}+\frac {375 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}-\frac {75 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x} \ln \relax (3) \left (\frac {\ln \relax (3)}{5}+1\right ) x}{5+\ln \relax (3)}+\frac {75 \ln \relax (3) {\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}-155 \,{\mathrm e}^{\left (\frac {\ln \relax (3)}{5}+1\right ) x}}{5+\ln \relax (3)}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-15*x-16)*ln(3)-75*x-155)*exp(1/5*x*ln(3)+x),x,method=_RETURNVERBOSE)

[Out]

1/5*(-75*x-80)*3^(1/5*x)*exp(x)

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maxima [B]  time = 0.44, size = 97, normalized size = 4.62 \begin {gather*} -\frac {15 \, {\left (x {\left (\log \relax (3) + 5\right )} - 5\right )} e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )} \log \relax (3)}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} - \frac {75 \, {\left (x {\left (\log \relax (3) + 5\right )} - 5\right )} e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )}}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} - \frac {16 \, e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )} \log \relax (3)}{\log \relax (3) + 5} - \frac {155 \, e^{\left (\frac {1}{5} \, x \log \relax (3) + x\right )}}{\log \relax (3) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-15*x-16)*log(3)-75*x-155)*exp(1/5*x*log(3)+x),x, algorithm="maxima")

[Out]

-15*(x*(log(3) + 5) - 5)*e^(1/5*x*log(3) + x)*log(3)/(log(3)^2 + 10*log(3) + 25) - 75*(x*(log(3) + 5) - 5)*e^(
1/5*x*log(3) + x)/(log(3)^2 + 10*log(3) + 25) - 16*e^(1/5*x*log(3) + x)*log(3)/(log(3) + 5) - 155*e^(1/5*x*log
(3) + x)/(log(3) + 5)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.67 \begin {gather*} -3^{x/5}\,{\mathrm {e}}^x\,\left (15\,x+16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + (x*log(3))/5)*(75*x + log(3)*(15*x + 16) + 155))/5,x)

[Out]

-3^(x/5)*exp(x)*(15*x + 16)

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sympy [A]  time = 0.12, size = 15, normalized size = 0.71 \begin {gather*} \left (- 15 x - 16\right ) e^{\frac {x \log {\relax (3 )}}{5} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-15*x-16)*ln(3)-75*x-155)*exp(1/5*x*ln(3)+x),x)

[Out]

(-15*x - 16)*exp(x*log(3)/5 + x)

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