3.82.91 \(\int \frac {e^x (-5-5 x)+15 x^2 \log (5)+e^{5-3 x+x^2} (e^x (-10+10 x)+(-10 x+15 x^2-10 x^3) \log (5))}{e^{10} \log (5)} \, dx\)

Optimal. Leaf size=32 \[ \frac {5 \left (e^{5+(-3+x) x}-x\right ) \left (-x^2+\frac {e^x}{\log (5)}\right )}{e^{10}} \]

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Rubi [B]  time = 0.19, antiderivative size = 79, normalized size of antiderivative = 2.47, number of steps used = 8, number of rules used = 6, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2176, 2194, 6742, 2236, 2288} \begin {gather*} \frac {5 x^3}{e^{10}}-\frac {5 e^{x^2-3 x-5} \left (3 x-2 x^2\right ) x}{3-2 x}+\frac {5 e^{x^2-2 x-5}}{\log (5)}+\frac {5 e^{x-10}}{\log (5)}-\frac {5 e^{x-10} (x+1)}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-5 - 5*x) + 15*x^2*Log[5] + E^(5 - 3*x + x^2)*(E^x*(-10 + 10*x) + (-10*x + 15*x^2 - 10*x^3)*Log[5]))
/(E^10*Log[5]),x]

[Out]

(5*x^3)/E^10 - (5*E^(-5 - 3*x + x^2)*x*(3*x - 2*x^2))/(3 - 2*x) + (5*E^(-10 + x))/Log[5] + (5*E^(-5 - 2*x + x^
2))/Log[5] - (5*E^(-10 + x)*(1 + x))/Log[5]

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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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 {\int \left (e^x (-5-5 x)+15 x^2 \log (5)+e^{5-3 x+x^2} \left (e^x (-10+10 x)+\left (-10 x+15 x^2-10 x^3\right ) \log (5)\right )\right ) \, dx}{e^{10} \log (5)}\\ &=\frac {5 x^3}{e^{10}}+\frac {\int e^x (-5-5 x) \, dx}{e^{10} \log (5)}+\frac {\int e^{5-3 x+x^2} \left (e^x (-10+10 x)+\left (-10 x+15 x^2-10 x^3\right ) \log (5)\right ) \, dx}{e^{10} \log (5)}\\ &=\frac {5 x^3}{e^{10}}-\frac {5 e^{-10+x} (1+x)}{\log (5)}+\frac {\int \left (10 e^{5-2 x+x^2} (-1+x)-5 e^{5-3 x+x^2} x \left (2-3 x+2 x^2\right ) \log (5)\right ) \, dx}{e^{10} \log (5)}+\frac {5 \int e^x \, dx}{e^{10} \log (5)}\\ &=\frac {5 x^3}{e^{10}}+\frac {5 e^{-10+x}}{\log (5)}-\frac {5 e^{-10+x} (1+x)}{\log (5)}-\frac {5 \int e^{5-3 x+x^2} x \left (2-3 x+2 x^2\right ) \, dx}{e^{10}}+\frac {10 \int e^{5-2 x+x^2} (-1+x) \, dx}{e^{10} \log (5)}\\ &=\frac {5 x^3}{e^{10}}-\frac {5 e^{-5-3 x+x^2} x \left (3 x-2 x^2\right )}{3-2 x}+\frac {5 e^{-10+x}}{\log (5)}+\frac {5 e^{-5-2 x+x^2}}{\log (5)}-\frac {5 e^{-10+x} (1+x)}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 49, normalized size = 1.53 \begin {gather*} \frac {-5 e^x x+5 x^3 \log (5)+5 e^{-3 x+x^2} \left (e^{5+x}-e^5 x^2 \log (5)\right )}{e^{10} \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-5 - 5*x) + 15*x^2*Log[5] + E^(5 - 3*x + x^2)*(E^x*(-10 + 10*x) + (-10*x + 15*x^2 - 10*x^3)*Lo
g[5]))/(E^10*Log[5]),x]

[Out]

(-5*E^x*x + 5*x^3*Log[5] + 5*E^(-3*x + x^2)*(E^(5 + x) - E^5*x^2*Log[5]))/(E^10*Log[5])

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fricas [A]  time = 0.61, size = 42, normalized size = 1.31 \begin {gather*} \frac {5 \, {\left (x^{3} \log \relax (5) - {\left (x^{2} \log \relax (5) - e^{x}\right )} e^{\left (x^{2} - 3 \, x + 5\right )} - x e^{x}\right )} e^{\left (-10\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x-10)*exp(x)+(-10*x^3+15*x^2-10*x)*log(5))*exp(x^2-3*x+5)+(-5*x-5)*exp(x)+15*x^2*log(5))/exp(1
0)/log(5),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.19, size = 67, normalized size = 2.09 \begin {gather*} \frac {5 \, {\left (4 \, x^{3} \log \relax (5) - {\left ({\left (2 \, x - 3\right )}^{2} \log \relax (5) + 6 \, {\left (2 \, x - 3\right )} \log \relax (5) + 9 \, \log \relax (5)\right )} e^{\left (x^{2} - 3 \, x + 5\right )} - 4 \, x e^{x} + 4 \, e^{\left (x^{2} - 2 \, x + 5\right )}\right )} e^{\left (-10\right )}}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x-10)*exp(x)+(-10*x^3+15*x^2-10*x)*log(5))*exp(x^2-3*x+5)+(-5*x-5)*exp(x)+15*x^2*log(5))/exp(1
0)/log(5),x, algorithm="giac")

[Out]

5/4*(4*x^3*log(5) - ((2*x - 3)^2*log(5) + 6*(2*x - 3)*log(5) + 9*log(5))*e^(x^2 - 3*x + 5) - 4*x*e^x + 4*e^(x^
2 - 2*x + 5))*e^(-10)/log(5)

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maple [A]  time = 0.12, size = 46, normalized size = 1.44

method	result	size
risch	\(5 \,{\mathrm e}^{-10} x^{3}-\frac {5 x \,{\mathrm e}^{x -10}}{\ln \relax (5)}+\frac {\left (-5 x^{2} \ln \relax (5)+5 \,{\mathrm e}^{x}\right ) {\mathrm e}^{x^{2}-3 x -5}}{\ln \relax (5)}\)	\(46\)
default	\(\frac {{\mathrm e}^{-10} \left (-5 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x^{2}-2 x +5}-5 \ln \relax (5) {\mathrm e}^{x^{2}-3 x +5} x^{2}+5 x^{3} \ln \relax (5)\right )}{\ln \relax (5)}\)	\(50\)
norman	\(5 \,{\mathrm e}^{-10} x^{3}-5 \,{\mathrm e}^{-10} x^{2} {\mathrm e}^{x^{2}-3 x +5}-\frac {5 \,{\mathrm e}^{-10} x \,{\mathrm e}^{x}}{\ln \relax (5)}+\frac {5 \,{\mathrm e}^{-10} {\mathrm e}^{x} {\mathrm e}^{x^{2}-3 x +5}}{\ln \relax (5)}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x-10)*exp(x)+(-10*x^3+15*x^2-10*x)*ln(5))*exp(x^2-3*x+5)+(-5*x-5)*exp(x)+15*x^2*ln(5))/exp(10)/ln(5)
,x,method=_RETURNVERBOSE)

[Out]

5*exp(-10)*x^3-5*x/ln(5)*exp(x-10)+(-5*x^2*ln(5)+5*exp(x))/ln(5)*exp(x^2-3*x-5)

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maxima [A]  time = 0.46, size = 51, normalized size = 1.59 \begin {gather*} \frac {5 \, {\left (x^{3} \log \relax (5) - {\left (x^{2} e^{5} \log \relax (5) - e^{\left (x + 5\right )}\right )} e^{\left (x^{2} - 3 \, x\right )} - {\left (x - 1\right )} e^{x} - e^{x}\right )} e^{\left (-10\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x-10)*exp(x)+(-10*x^3+15*x^2-10*x)*log(5))*exp(x^2-3*x+5)+(-5*x-5)*exp(x)+15*x^2*log(5))/exp(1
0)/log(5),x, algorithm="maxima")

[Out]

5*(x^3*log(5) - (x^2*e^5*log(5) - e^(x + 5))*e^(x^2 - 3*x) - (x - 1)*e^x - e^x)*e^(-10)/log(5)

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mupad [B]  time = 0.24, size = 31, normalized size = 0.97 \begin {gather*} -\frac {5\,{\mathrm {e}}^{-10}\,\left (x-{\mathrm {e}}^{x^2-3\,x+5}\right )\,\left ({\mathrm {e}}^x-x^2\,\ln \relax (5)\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-10)*(exp(x^2 - 3*x + 5)*(log(5)*(10*x - 15*x^2 + 10*x^3) - exp(x)*(10*x - 10)) + exp(x)*(5*x + 5) -
15*x^2*log(5)))/log(5),x)

[Out]

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

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sympy [A]  time = 0.31, size = 51, normalized size = 1.59 \begin {gather*} \frac {5 x^{3}}{e^{10}} - \frac {5 x e^{x}}{e^{10} \log {\relax (5 )}} + \frac {\left (- 5 x^{2} \log {\relax (5 )} + 5 e^{x}\right ) e^{x^{2} - 3 x + 5}}{e^{10} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x-10)*exp(x)+(-10*x**3+15*x**2-10*x)*ln(5))*exp(x**2-3*x+5)+(-5*x-5)*exp(x)+15*x**2*ln(5))/exp
(10)/ln(5),x)

[Out]

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

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