3.87.86 \(\int \frac {-5 \log (5)+e^{\frac {3 x+x^2 \log (5)+x \log ^2(5)}{\log (5)}} (-15-10 x \log (5)-5 \log ^2(5))}{4 \log (5)} \, dx\)

Optimal. Leaf size=28 \[ 3+\frac {1}{4} \left (4-5 \left (e^{\frac {3 x}{\log (5)}+x (x+\log (5))}+x\right )\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 2244, 2236} \begin {gather*} -\frac {5}{4} e^{x^2+\frac {x \left (3+\log ^2(5)\right )}{\log (5)}}-\frac {5 x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*Log[5] + E^((3*x + x^2*Log[5] + x*Log[5]^2)/Log[5])*(-15 - 10*x*Log[5] - 5*Log[5]^2))/(4*Log[5]),x]

[Out]

(-5*E^(x^2 + (x*(3 + Log[5]^2))/Log[5]))/4 - (5*x)/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 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 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-5 \log (5)+e^{\frac {3 x+x^2 \log (5)+x \log ^2(5)}{\log (5)}} \left (-15-10 x \log (5)-5 \log ^2(5)\right )\right ) \, dx}{4 \log (5)}\\ &=-\frac {5 x}{4}+\frac {\int e^{\frac {3 x+x^2 \log (5)+x \log ^2(5)}{\log (5)}} \left (-15-10 x \log (5)-5 \log ^2(5)\right ) \, dx}{4 \log (5)}\\ &=-\frac {5 x}{4}+\frac {\int e^{x^2+\frac {x \left (3+\log ^2(5)\right )}{\log (5)}} \left (-10 x \log (5)-5 \left (3+\log ^2(5)\right )\right ) \, dx}{4 \log (5)}\\ &=-\frac {5}{4} e^{x^2+\frac {x \left (3+\log ^2(5)\right )}{\log (5)}}-\frac {5 x}{4}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.48, size = 140, normalized size = 5.00 \begin {gather*} -\frac {5 \left (x \log (5)+\frac {1}{2} e^{-\frac {\left (3+\log ^2(5)\right )^2}{4 \log ^2(5)}} \sqrt {\pi } \text {erfi}\left (\frac {3+\log ^2(5)+x \log (25)}{\log (25)}\right ) \left (3+\log ^2(5)\right )\right )}{\log (625)}-\frac {e^{-\frac {\left (3+\log ^2(5)\right )^2}{4 \log ^2(5)}} \left (-5 \sqrt {\pi } \text {erfi}\left (\frac {3+\log ^2(5)+x \log (25)}{\log (25)}\right ) \left (3+\log ^2(5)\right )+5 e^{\frac {\left (3+\log ^2(5)+x \log (25)\right )^2}{\log ^2(25)}} \log (25)\right )}{2 \log (625)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*Log[5] + E^((3*x + x^2*Log[5] + x*Log[5]^2)/Log[5])*(-15 - 10*x*Log[5] - 5*Log[5]^2))/(4*Log[5])
,x]

[Out]

(-5*(x*Log[5] + (Sqrt[Pi]*Erfi[(3 + Log[5]^2 + x*Log[25])/Log[25]]*(3 + Log[5]^2))/(2*E^((3 + Log[5]^2)^2/(4*L
og[5]^2)))))/Log[625] - (-5*Sqrt[Pi]*Erfi[(3 + Log[5]^2 + x*Log[25])/Log[25]]*(3 + Log[5]^2) + 5*E^((3 + Log[5
]^2 + x*Log[25])^2/Log[25]^2)*Log[25])/(2*E^((3 + Log[5]^2)^2/(4*Log[5]^2))*Log[625])

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fricas [A]  time = 1.01, size = 28, normalized size = 1.00 \begin {gather*} -\frac {5}{4} \, x - \frac {5}{4} \, e^{\left (\frac {x^{2} \log \relax (5) + x \log \relax (5)^{2} + 3 \, x}{\log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*log(5)^2-10*x*log(5)-15)*exp((x*log(5)^2+x^2*log(5)+3*x)/log(5))-5*log(5))/log(5),x, algori
thm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 30, normalized size = 1.07 \begin {gather*} -\frac {5 \, {\left (x \log \relax (5) + e^{\left (x^{2} + x \log \relax (5) + \frac {3 \, x}{\log \relax (5)}\right )} \log \relax (5)\right )}}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*log(5)^2-10*x*log(5)-15)*exp((x*log(5)^2+x^2*log(5)+3*x)/log(5))-5*log(5))/log(5),x, algori
thm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 23, normalized size = 0.82

method	result	size
risch	\(-\frac {5 x}{4}-\frac {5 \,5^{x} {\mathrm e}^{\frac {x \left (x \ln \relax (5)+3\right )}{\ln \relax (5)}}}{4}\)	\(23\)
norman	\(-\frac {5 x}{4}-\frac {5 \,{\mathrm e}^{\frac {x \ln \relax (5)^{2}+x^{2} \ln \relax (5)+3 x}{\ln \relax (5)}}}{4}\)	\(29\)
default	\(\frac {-5 \ln \relax (5) {\mathrm e}^{x^{2}+\frac {\left (\ln \relax (5)^{2}+3\right ) x}{\ln \relax (5)}}-5 x \ln \relax (5)}{4 \ln \relax (5)}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-5*ln(5)^2-10*x*ln(5)-15)*exp((x*ln(5)^2+x^2*ln(5)+3*x)/ln(5))-5*ln(5))/ln(5),x,method=_RETURNVERBOS
E)

[Out]

-5/4*x-5/4*5^x*exp(x*(x*ln(5)+3)/ln(5))

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maxima [A]  time = 0.36, size = 36, normalized size = 1.29 \begin {gather*} -\frac {5 \, {\left (x \log \relax (5) + e^{\left (\frac {x^{2} \log \relax (5) + x \log \relax (5)^{2} + 3 \, x}{\log \relax (5)}\right )} \log \relax (5)\right )}}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*log(5)^2-10*x*log(5)-15)*exp((x*log(5)^2+x^2*log(5)+3*x)/log(5))-5*log(5))/log(5),x, algori
thm="maxima")

[Out]

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

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mupad [B]  time = 0.16, size = 21, normalized size = 0.75 \begin {gather*} -\frac {5\,x}{4}-\frac {5\,5^x\,{\mathrm {e}}^{\frac {3\,x}{\ln \relax (5)}}\,{\mathrm {e}}^{x^2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*log(5))/4 + (exp((3*x + x*log(5)^2 + x^2*log(5))/log(5))*(10*x*log(5) + 5*log(5)^2 + 15))/4)/log(5),x
)

[Out]

- (5*x)/4 - (5*5^x*exp((3*x)/log(5))*exp(x^2))/4

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sympy [A]  time = 0.13, size = 31, normalized size = 1.11 \begin {gather*} - \frac {5 x}{4} - \frac {5 e^{\frac {x^{2} \log {\relax (5 )} + x \log {\relax (5 )}^{2} + 3 x}{\log {\relax (5 )}}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*ln(5)**2-10*x*ln(5)-15)*exp((x*ln(5)**2+x**2*ln(5)+3*x)/ln(5))-5*ln(5))/ln(5),x)

[Out]

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

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