3.82.23 \(\int \frac {e^{4 x} (4-16 x)-10 x}{5 x^2} \, dx\)

Optimal. Leaf size=25 \[ 2 \left (\frac {2}{5} \left (3-\frac {e^{4 x}}{x}\right )-\log (5 x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2197} \begin {gather*} -\frac {4 e^{4 x}}{5 x}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4*x)*(4 - 16*x) - 10*x)/(5*x^2),x]

[Out]

(-4*E^(4*x))/(5*x) - 2*Log[x]

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}{5} \int \frac {e^{4 x} (4-16 x)-10 x}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {10}{x}-\frac {4 e^{4 x} (-1+4 x)}{x^2}\right ) \, dx\\ &=-2 \log (x)-\frac {4}{5} \int \frac {e^{4 x} (-1+4 x)}{x^2} \, dx\\ &=-\frac {4 e^{4 x}}{5 x}-2 \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 19, normalized size = 0.76 \begin {gather*} -\frac {2}{5} \left (\frac {2 e^{4 x}}{x}+5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x)*(4 - 16*x) - 10*x)/(5*x^2),x]

[Out]

(-2*((2*E^(4*x))/x + 5*Log[x]))/5

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 17, normalized size = 0.68 \begin {gather*} -\frac {2 \, {\left (5 \, x \log \relax (x) + 2 \, e^{\left (4 \, x\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-16*x+4)*exp(4*x)-10*x)/x^2,x, algorithm="fricas")

[Out]

-2/5*(5*x*log(x) + 2*e^(4*x))/x

________________________________________________________________________________________

giac [A]  time = 0.31, size = 17, normalized size = 0.68 \begin {gather*} -\frac {2 \, {\left (5 \, x \log \relax (x) + 2 \, e^{\left (4 \, x\right )}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-16*x+4)*exp(4*x)-10*x)/x^2,x, algorithm="giac")

[Out]

-2/5*(5*x*log(x) + 2*e^(4*x))/x

________________________________________________________________________________________

maple [A]  time = 0.09, size = 15, normalized size = 0.60

method	result	size
norman	\(-\frac {4 \,{\mathrm e}^{4 x}}{5 x}-2 \ln \relax (x )\)	\(15\)
risch	\(-\frac {4 \,{\mathrm e}^{4 x}}{5 x}-2 \ln \relax (x )\)	\(15\)
derivativedivides	\(-2 \ln \left (4 x \right )-\frac {4 \,{\mathrm e}^{4 x}}{5 x}\)	\(17\)
default	\(-2 \ln \left (4 x \right )-\frac {4 \,{\mathrm e}^{4 x}}{5 x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-16*x+4)*exp(4*x)-10*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-4/5*exp(4*x)/x-2*ln(x)

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 18, normalized size = 0.72 \begin {gather*} -\frac {16}{5} \, {\rm Ei}\left (4 \, x\right ) + \frac {16}{5} \, \Gamma \left (-1, -4 \, x\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-16*x+4)*exp(4*x)-10*x)/x^2,x, algorithm="maxima")

[Out]

-16/5*Ei(4*x) + 16/5*gamma(-1, -4*x) - 2*log(x)

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 14, normalized size = 0.56 \begin {gather*} -2\,\ln \relax (x)-\frac {4\,{\mathrm {e}}^{4\,x}}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + (exp(4*x)*(16*x - 4))/5)/x^2,x)

[Out]

- 2*log(x) - (4*exp(4*x))/(5*x)

________________________________________________________________________________________

sympy [A]  time = 0.09, size = 15, normalized size = 0.60 \begin {gather*} - 2 \log {\relax (x )} - \frac {4 e^{4 x}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-16*x+4)*exp(4*x)-10*x)/x**2,x)

[Out]

-2*log(x) - 4*exp(4*x)/(5*x)

________________________________________________________________________________________