\(\int \frac {e^x (21-3 x-3 \log ^2(5))}{-125+75 x-15 x^2+x^3+(75-30 x+3 x^2) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx\) [9057]

Optimal result

Integrand size = 59, antiderivative size = 18 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=3 \left (2-\frac {e^x}{\left (-5+x+\log ^2(5)\right )^2}\right ) \]

[Out]

6-3/(ln(5)^2-5+x)^2*exp(x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2230, 2208, 2209} \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=-\frac {3 e^x}{\left (-x+5-\log ^2(5)\right )^2} \]

[In]

Int[(E^x*(21 - 3*x - 3*Log[5]^2))/(-125 + 75*x - 15*x^2 + x^3 + (75 - 30*x + 3*x^2)*Log[5]^2 + (-15 + 3*x)*Log
[5]^4 + Log[5]^6),x]

[Out]

(-3*E^x)/(5 - x - Log[5]^2)^2

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {6 e^x}{\left (-5+x+\log ^2(5)\right )^3}-\frac {3 e^x}{\left (-5+x+\log ^2(5)\right )^2}\right ) \, dx \\ & = -\left (3 \int \frac {e^x}{\left (-5+x+\log ^2(5)\right )^2} \, dx\right )+6 \int \frac {e^x}{\left (-5+x+\log ^2(5)\right )^3} \, dx \\ & = -\frac {3 e^x}{\left (5-x-\log ^2(5)\right )^2}-\frac {3 e^x}{5-x-\log ^2(5)}+3 \int \frac {e^x}{\left (-5+x+\log ^2(5)\right )^2} \, dx-3 \int \frac {e^x}{-5+x+\log ^2(5)} \, dx \\ & = -3 e^{5-\log ^2(5)} \operatorname {ExpIntegralEi}\left (-5+x+\log ^2(5)\right )-\frac {3 e^x}{\left (5-x-\log ^2(5)\right )^2}+3 \int \frac {e^x}{-5+x+\log ^2(5)} \, dx \\ & = -\frac {3 e^x}{\left (5-x-\log ^2(5)\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=-\frac {3 e^x}{\left (-5+x+\log ^2(5)\right )^2} \]

[In]

Integrate[(E^x*(21 - 3*x - 3*Log[5]^2))/(-125 + 75*x - 15*x^2 + x^3 + (75 - 30*x + 3*x^2)*Log[5]^2 + (-15 + 3*
x)*Log[5]^4 + Log[5]^6),x]

[Out]

(-3*E^x)/(-5 + x + Log[5]^2)^2

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78

method	result	size
norman	\(-\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (5\right )^{2}-5+x \right )^{2}}\)	\(14\)
risch	\(-\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (5\right )^{2}-5+x \right )^{2}}\)	\(14\)
gosper	\(-\frac {3 \,{\mathrm e}^{x}}{\ln \left (5\right )^{4}+2 x \ln \left (5\right )^{2}-10 \ln \left (5\right )^{2}+x^{2}-10 x +25}\)	\(32\)
parallelrisch	\(-\frac {3 \,{\mathrm e}^{x}}{\ln \left (5\right )^{4}+2 x \ln \left (5\right )^{2}-10 \ln \left (5\right )^{2}+x^{2}-10 x +25}\)	\(32\)
default	\(-\frac {21 \,{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )^{2}}-\frac {15 \,{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )}-\frac {15 \,{\mathrm e}^{-\ln \left (5\right )^{2}+5} \operatorname {Ei}_{1}\left (-x -\ln \left (5\right )^{2}+5\right )}{2}-3 \left (-\ln \left (5\right )^{2}+5\right ) \left (-\frac {{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )^{2}}-\frac {{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )}-\frac {{\mathrm e}^{-\ln \left (5\right )^{2}+5} \operatorname {Ei}_{1}\left (-x -\ln \left (5\right )^{2}+5\right )}{2}\right )-3 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )^{2}}-\frac {{\mathrm e}^{x}}{2 \left (\ln \left (5\right )^{2}-5+x \right )}-\frac {{\mathrm e}^{-\ln \left (5\right )^{2}+5} \operatorname {Ei}_{1}\left (-x -\ln \left (5\right )^{2}+5\right )}{2}\right )\)	\(170\)

[In]

int((-3*ln(5)^2+21-3*x)*exp(x)/(ln(5)^6+(3*x-15)*ln(5)^4+(3*x^2-30*x+75)*ln(5)^2+x^3-15*x^2+75*x-125),x,method
=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=-\frac {3 \, e^{x}}{\log \left (5\right )^{4} + 2 \, {\left (x - 5\right )} \log \left (5\right )^{2} + x^{2} - 10 \, x + 25} \]

[In]

integrate((-3*log(5)^2+21-3*x)*exp(x)/(log(5)^6+(3*x-15)*log(5)^4+(3*x^2-30*x+75)*log(5)^2+x^3-15*x^2+75*x-125
),x, algorithm="fricas")

[Out]

-3*e^x/(log(5)^4 + 2*(x - 5)*log(5)^2 + x^2 - 10*x + 25)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=- \frac {3 e^{x}}{x^{2} - 10 x + 2 x \log {\left (5 \right )}^{2} - 10 \log {\left (5 \right )}^{2} + \log {\left (5 \right )}^{4} + 25} \]

[In]

integrate((-3*ln(5)**2+21-3*x)*exp(x)/(ln(5)**6+(3*x-15)*ln(5)**4+(3*x**2-30*x+75)*ln(5)**2+x**3-15*x**2+75*x-
125),x)

[Out]

-3*exp(x)/(x**2 - 10*x + 2*x*log(5)**2 - 10*log(5)**2 + log(5)**4 + 25)

Maxima [F]

\[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=\int { -\frac {3 \, {\left (\log \left (5\right )^{2} + x - 7\right )} e^{x}}{\log \left (5\right )^{6} + 3 \, {\left (x - 5\right )} \log \left (5\right )^{4} + x^{3} + 3 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (5\right )^{2} - 15 \, x^{2} + 75 \, x - 125} \,d x } \]

[In]

integrate((-3*log(5)^2+21-3*x)*exp(x)/(log(5)^6+(3*x-15)*log(5)^4+(3*x^2-30*x+75)*log(5)^2+x^3-15*x^2+75*x-125
),x, algorithm="maxima")

[Out]

3*e^(-log(5)^2 + 5)*exp_integral_e(3, -log(5)^2 - x + 5)*log(5)^2/(log(5)^2 + x - 5)^2 - 3*x*e^x/(log(5)^6 - 1
5*log(5)^4 + 3*(log(5)^2 - 5)*x^2 + x^3 + 3*(log(5)^4 - 10*log(5)^2 + 25)*x + 75*log(5)^2 - 125) - 21*e^(-log(
5)^2 + 5)*exp_integral_e(3, -log(5)^2 - x + 5)/(log(5)^2 + x - 5)^2 - 3*integrate(-(log(5)^2 - 2*x - 5)*e^x/(l
og(5)^8 - 20*log(5)^6 + 4*(log(5)^2 - 5)*x^3 + x^4 + 150*log(5)^4 + 6*(log(5)^4 - 10*log(5)^2 + 25)*x^2 + 4*(l
og(5)^6 - 15*log(5)^4 + 75*log(5)^2 - 125)*x - 500*log(5)^2 + 625), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=-\frac {3 \, e^{x}}{\log \left (5\right )^{4} + 2 \, x \log \left (5\right )^{2} + x^{2} - 10 \, \log \left (5\right )^{2} - 10 \, x + 25} \]

[In]

integrate((-3*log(5)^2+21-3*x)*exp(x)/(log(5)^6+(3*x-15)*log(5)^4+(3*x^2-30*x+75)*log(5)^2+x^3-15*x^2+75*x-125
),x, algorithm="giac")

[Out]

-3*e^x/(log(5)^4 + 2*x*log(5)^2 + x^2 - 10*log(5)^2 - 10*x + 25)

Mupad [B] (verification not implemented)

Time = 13.78 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {e^x \left (21-3 x-3 \log ^2(5)\right )}{-125+75 x-15 x^2+x^3+\left (75-30 x+3 x^2\right ) \log ^2(5)+(-15+3 x) \log ^4(5)+\log ^6(5)} \, dx=-\frac {3\,{\mathrm {e}}^x}{{\left (x+{\ln \left (5\right )}^2-5\right )}^2} \]

[In]

int(-(exp(x)*(3*x + 3*log(5)^2 - 21))/(75*x + log(5)^4*(3*x - 15) + log(5)^2*(3*x^2 - 30*x + 75) + log(5)^6 -
15*x^2 + x^3 - 125),x)

[Out]

-(3*exp(x))/(x + log(5)^2 - 5)^2