3.48.25 \(\int \frac {-10 e^6 \log (5)+e^x (10+e^3 (20-10 x)-5 x) \log (5)+e^{2 x} (-10+10 x) \log (5)}{6 x^3} \, dx\)

Optimal. Leaf size=26 \[ \frac {5 \left (-e^x+\left (e^3-e^x\right )^2\right ) \log (5)}{6 x^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 46, normalized size of antiderivative = 1.77, number of steps used = 5, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 2197} \begin {gather*} \frac {5 e^{2 x} \log (5)}{6 x^2}-\frac {5 \left (1+2 e^3\right ) e^x \log (5)}{6 x^2}+\frac {5 e^6 \log (5)}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*E^6*Log[5] + E^x*(10 + E^3*(20 - 10*x) - 5*x)*Log[5] + E^(2*x)*(-10 + 10*x)*Log[5])/(6*x^3),x]

[Out]

(5*E^6*Log[5])/(6*x^2) + (5*E^(2*x)*Log[5])/(6*x^2) - (5*E^x*(1 + 2*E^3)*Log[5])/(6*x^2)

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{6} \int \frac {-10 e^6 \log (5)+e^x \left (10+e^3 (20-10 x)-5 x\right ) \log (5)+e^{2 x} (-10+10 x) \log (5)}{x^3} \, dx\\ &=\frac {1}{6} \int \left (-\frac {10 e^6 \log (5)}{x^3}-\frac {5 e^x \left (1+2 e^3\right ) (-2+x) \log (5)}{x^3}+\frac {10 e^{2 x} (-1+x) \log (5)}{x^3}\right ) \, dx\\ &=\frac {5 e^6 \log (5)}{6 x^2}+\frac {1}{3} (5 \log (5)) \int \frac {e^{2 x} (-1+x)}{x^3} \, dx-\frac {1}{6} \left (5 \left (1+2 e^3\right ) \log (5)\right ) \int \frac {e^x (-2+x)}{x^3} \, dx\\ &=\frac {5 e^6 \log (5)}{6 x^2}+\frac {5 e^{2 x} \log (5)}{6 x^2}-\frac {5 e^x \left (1+2 e^3\right ) \log (5)}{6 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 30, normalized size = 1.15 \begin {gather*} \frac {5 \left (e^6-e^x+e^{2 x}-2 e^{3+x}\right ) \log (5)}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*E^6*Log[5] + E^x*(10 + E^3*(20 - 10*x) - 5*x)*Log[5] + E^(2*x)*(-10 + 10*x)*Log[5])/(6*x^3),x]

[Out]

(5*(E^6 - E^x + E^(2*x) - 2*E^(3 + x))*Log[5])/(6*x^2)

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fricas [A]  time = 0.55, size = 31, normalized size = 1.19 \begin {gather*} -\frac {5 \, {\left ({\left (2 \, e^{3} + 1\right )} e^{x} \log \relax (5) - e^{6} \log \relax (5) - e^{\left (2 \, x\right )} \log \relax (5)\right )}}{6 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((10*x-10)*log(5)*exp(x)^2+((20-10*x)*exp(3)-5*x+10)*log(5)*exp(x)-10*exp(3)^2*log(5))/x^3,x, al
gorithm="fricas")

[Out]

-5/6*((2*e^3 + 1)*e^x*log(5) - e^6*log(5) - e^(2*x)*log(5))/x^2

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giac [A]  time = 0.15, size = 32, normalized size = 1.23 \begin {gather*} \frac {5 \, {\left (e^{6} \log \relax (5) + e^{\left (2 \, x\right )} \log \relax (5) - 2 \, e^{\left (x + 3\right )} \log \relax (5) - e^{x} \log \relax (5)\right )}}{6 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((10*x-10)*log(5)*exp(x)^2+((20-10*x)*exp(3)-5*x+10)*log(5)*exp(x)-10*exp(3)^2*log(5))/x^3,x, al
gorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 36, normalized size = 1.38

method	result	size
norman	\(\frac {\left (-\frac {5 \,{\mathrm e}^{3} \ln \relax (5)}{3}-\frac {5 \ln \relax (5)}{6}\right ) {\mathrm e}^{x}+\frac {5 \,{\mathrm e}^{6} \ln \relax (5)}{6}+\frac {5 \ln \relax (5) {\mathrm e}^{2 x}}{6}}{x^{2}}\)	\(36\)
risch	\(\frac {5 \,{\mathrm e}^{6} \ln \relax (5)}{6 x^{2}}+\frac {5 \ln \relax (5) {\mathrm e}^{2 x}}{6 x^{2}}-\frac {5 \left (2 \,{\mathrm e}^{3}+1\right ) \ln \relax (5) {\mathrm e}^{x}}{6 x^{2}}\)	\(37\)
default	\(\frac {5 \,{\mathrm e}^{6} \ln \relax (5)}{6 x^{2}}-\frac {5 \,{\mathrm e}^{x} \ln \relax (5)}{6 x^{2}}+\frac {5 \ln \relax (5) {\mathrm e}^{2 x}}{6 x^{2}}-\frac {5 \,{\mathrm e}^{x} \ln \relax (5) {\mathrm e}^{3}}{3 x^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((10*x-10)*ln(5)*exp(x)^2+((20-10*x)*exp(3)-5*x+10)*ln(5)*exp(x)-10*exp(3)^2*ln(5))/x^3,x,method=_RETU
RNVERBOSE)

[Out]

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

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maxima [C]  time = 0.42, size = 68, normalized size = 2.62 \begin {gather*} -\frac {5}{3} \, e^{3} \Gamma \left (-1, -x\right ) \log \relax (5) - \frac {10}{3} \, e^{3} \Gamma \left (-2, -x\right ) \log \relax (5) - \frac {5}{6} \, \Gamma \left (-1, -x\right ) \log \relax (5) + \frac {10}{3} \, \Gamma \left (-1, -2 \, x\right ) \log \relax (5) - \frac {5}{3} \, \Gamma \left (-2, -x\right ) \log \relax (5) + \frac {20}{3} \, \Gamma \left (-2, -2 \, x\right ) \log \relax (5) + \frac {5 \, e^{6} \log \relax (5)}{6 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((10*x-10)*log(5)*exp(x)^2+((20-10*x)*exp(3)-5*x+10)*log(5)*exp(x)-10*exp(3)^2*log(5))/x^3,x, al
gorithm="maxima")

[Out]

-5/3*e^3*gamma(-1, -x)*log(5) - 10/3*e^3*gamma(-2, -x)*log(5) - 5/6*gamma(-1, -x)*log(5) + 10/3*gamma(-1, -2*x
)*log(5) - 5/3*gamma(-2, -x)*log(5) + 20/3*gamma(-2, -2*x)*log(5) + 5/6*e^6*log(5)/x^2

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mupad [B]  time = 0.09, size = 32, normalized size = 1.23 \begin {gather*} \frac {5\,{\mathrm {e}}^{2\,x}\,\ln \relax (5)+5\,{\mathrm {e}}^6\,\ln \relax (5)-5\,{\mathrm {e}}^x\,\ln \relax (5)\,\left (2\,{\mathrm {e}}^3+1\right )}{6\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*exp(6)*log(5))/3 - (exp(2*x)*log(5)*(10*x - 10))/6 + (exp(x)*log(5)*(5*x + exp(3)*(10*x - 20) - 10))/
6)/x^3,x)

[Out]

(5*exp(2*x)*log(5) + 5*exp(6)*log(5) - 5*exp(x)*log(5)*(2*exp(3) + 1))/(6*x^2)

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sympy [B]  time = 0.17, size = 56, normalized size = 2.15 \begin {gather*} \frac {5 e^{6} \log {\relax (5 )}}{6 x^{2}} + \frac {30 x^{2} e^{2 x} \log {\relax (5 )} + \left (- 60 x^{2} e^{3} \log {\relax (5 )} - 30 x^{2} \log {\relax (5 )}\right ) e^{x}}{36 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((10*x-10)*ln(5)*exp(x)**2+((20-10*x)*exp(3)-5*x+10)*ln(5)*exp(x)-10*exp(3)**2*ln(5))/x**3,x)

[Out]

5*exp(6)*log(5)/(6*x**2) + (30*x**2*exp(2*x)*log(5) + (-60*x**2*exp(3)*log(5) - 30*x**2*log(5))*exp(x))/(36*x*
*4)

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