3.100.48 \(\int \frac {-12+e^x (-3+x)-3 x^4 \log (100)}{x^4 \log (100)} \, dx\)

Optimal. Leaf size=18 \[ -3-3 x+\frac {4+e^x}{x^3 \log (100)} \]

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^x}{x^3 \log (100)}+\frac {4}{x^3 \log (100)}-3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 + E^x*(-3 + x) - 3*x^4*Log[100])/(x^4*Log[100]),x]

[Out]

-3*x + 4/(x^3*Log[100]) + E^x/(x^3*Log[100])

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 {\int \frac {-12+e^x (-3+x)-3 x^4 \log (100)}{x^4} \, dx}{\log (100)}\\ &=\frac {\int \left (\frac {e^x (-3+x)}{x^4}-\frac {3 \left (4+x^4 \log (100)\right )}{x^4}\right ) \, dx}{\log (100)}\\ &=\frac {\int \frac {e^x (-3+x)}{x^4} \, dx}{\log (100)}-\frac {3 \int \frac {4+x^4 \log (100)}{x^4} \, dx}{\log (100)}\\ &=\frac {e^x}{x^3 \log (100)}-\frac {3 \int \left (\frac {4}{x^4}+\log (100)\right ) \, dx}{\log (100)}\\ &=-3 x+\frac {4}{x^3 \log (100)}+\frac {e^x}{x^3 \log (100)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 1.28 \begin {gather*} \frac {\frac {4}{x^3}+\frac {e^x}{x^3}-3 x \log (100)}{\log (100)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 + E^x*(-3 + x) - 3*x^4*Log[100])/(x^4*Log[100]),x]

[Out]

(4/x^3 + E^x/x^3 - 3*x*Log[100])/Log[100]

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fricas [A]  time = 0.54, size = 22, normalized size = 1.22 \begin {gather*} -\frac {6 \, x^{4} \log \left (10\right ) - e^{x} - 4}{2 \, x^{3} \log \left (10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-3)*exp(x)-6*x^4*log(10)-12)/x^4/log(10),x, algorithm="fricas")

[Out]

-1/2*(6*x^4*log(10) - e^x - 4)/(x^3*log(10))

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giac [A]  time = 0.14, size = 22, normalized size = 1.22 \begin {gather*} -\frac {6 \, x^{4} \log \left (10\right ) - e^{x} - 4}{2 \, x^{3} \log \left (10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-3)*exp(x)-6*x^4*log(10)-12)/x^4/log(10),x, algorithm="giac")

[Out]

-1/2*(6*x^4*log(10) - e^x - 4)/(x^3*log(10))

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maple [A]  time = 0.05, size = 24, normalized size = 1.33

method	result	size
default	\(\frac {\frac {4}{x^{3}}+\frac {{\mathrm e}^{x}}{x^{3}}-6 \ln \left (10\right ) x}{2 \ln \left (10\right )}\)	\(24\)
norman	\(\frac {-3 x^{4}+\frac {2}{\ln \left (10\right )}+\frac {{\mathrm e}^{x}}{2 \ln \left (10\right )}}{x^{3}}\)	\(25\)
risch	\(-\frac {3 x \ln \relax (2)}{\ln \relax (2)+\ln \relax (5)}-\frac {3 x \ln \relax (5)}{\ln \relax (2)+\ln \relax (5)}+\frac {2}{\left (\ln \relax (2)+\ln \relax (5)\right ) x^{3}}+\frac {{\mathrm e}^{x}}{2 \left (\ln \relax (2)+\ln \relax (5)\right ) x^{3}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((x-3)*exp(x)-6*x^4*ln(10)-12)/x^4/ln(10),x,method=_RETURNVERBOSE)

[Out]

1/2/ln(10)*(4/x^3+exp(x)/x^3-6*ln(10)*x)

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maxima [C]  time = 0.38, size = 29, normalized size = 1.61 \begin {gather*} -\frac {6 \, x \log \left (10\right ) - \frac {4}{x^{3}} + \Gamma \left (-2, -x\right ) + 3 \, \Gamma \left (-3, -x\right )}{2 \, \log \left (10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-3)*exp(x)-6*x^4*log(10)-12)/x^4/log(10),x, algorithm="maxima")

[Out]

-1/2*(6*x*log(10) - 4/x^3 + gamma(-2, -x) + 3*gamma(-3, -x))/log(10)

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mupad [B]  time = 6.86, size = 18, normalized size = 1.00 \begin {gather*} \frac {\frac {{\mathrm {e}}^x}{2}+2}{x^3\,\ln \left (10\right )}-3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^4*log(10) - (exp(x)*(x - 3))/2 + 6)/(x^4*log(10)),x)

[Out]

(exp(x)/2 + 2)/(x^3*log(10)) - 3*x

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sympy [A]  time = 0.16, size = 26, normalized size = 1.44 \begin {gather*} \frac {- 3 x \log {\left (10 \right )} + \frac {2}{x^{3}}}{\log {\left (10 \right )}} + \frac {e^{x}}{2 x^{3} \log {\left (10 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-3)*exp(x)-6*x**4*ln(10)-12)/x**4/ln(10),x)

[Out]

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

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