3.98.91 \(\int \frac {25+8 x+160 e^4 x^3+(5+2 x+32 e^4 x^3) \log (2)}{160 e^4 x^3+32 e^4 x^3 \log (2)} \, dx\)

Optimal. Leaf size=26 \[ x-\frac {\frac {5}{4}+x-\frac {x}{5+\log (2)}}{16 e^4 x^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6, 12, 14} \begin {gather*} -\frac {5}{64 e^4 x^2}+x-\frac {8+\log (4)}{32 e^4 x (5+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 8*x + 160*E^4*x^3 + (5 + 2*x + 32*E^4*x^3)*Log[2])/(160*E^4*x^3 + 32*E^4*x^3*Log[2]),x]

[Out]

-5/(64*E^4*x^2) + x - (8 + Log[4])/(32*E^4*x*(5 + Log[2]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{e^4 x^3 (160+32 \log (2))} \, dx\\ &=\frac {\int \frac {25+8 x+160 e^4 x^3+\left (5+2 x+32 e^4 x^3\right ) \log (2)}{x^3} \, dx}{32 e^4 (5+\log (2))}\\ &=\frac {\int \left (32 e^4 (5+\log (2))+\frac {5 (5+\log (2))}{x^3}+\frac {8+\log (4)}{x^2}\right ) \, dx}{32 e^4 (5+\log (2))}\\ &=-\frac {5}{64 e^4 x^2}+x-\frac {8+\log (4)}{32 e^4 x (5+\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.50 \begin {gather*} -\frac {25-64 e^4 x^3 (5+\log (2))+2 x (8+\log (4))+\log (32)}{64 e^4 x^2 (5+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 8*x + 160*E^4*x^3 + (5 + 2*x + 32*E^4*x^3)*Log[2])/(160*E^4*x^3 + 32*E^4*x^3*Log[2]),x]

[Out]

-1/64*(25 - 64*E^4*x^3*(5 + Log[2]) + 2*x*(8 + Log[4]) + Log[32])/(E^4*x^2*(5 + Log[2]))

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fricas [B]  time = 0.92, size = 47, normalized size = 1.81 \begin {gather*} \frac {320 \, x^{3} e^{4} + {\left (64 \, x^{3} e^{4} - 4 \, x - 5\right )} \log \relax (2) - 16 \, x - 25}{64 \, {\left (x^{2} e^{4} \log \relax (2) + 5 \, x^{2} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*exp(2)^2+5+2*x)*log(2)+160*x^3*exp(2)^2+8*x+25)/(32*x^3*exp(2)^2*log(2)+160*x^3*exp(2)^2),x
, algorithm="fricas")

[Out]

1/64*(320*x^3*e^4 + (64*x^3*e^4 - 4*x - 5)*log(2) - 16*x - 25)/(x^2*e^4*log(2) + 5*x^2*e^4)

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giac [B]  time = 0.25, size = 57, normalized size = 2.19 \begin {gather*} \frac {x e^{4} \log \relax (2) + 5 \, x e^{4}}{e^{4} \log \relax (2) + 5 \, e^{4}} - \frac {4 \, x \log \relax (2) + 16 \, x + 5 \, \log \relax (2) + 25}{64 \, {\left (e^{4} \log \relax (2) + 5 \, e^{4}\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*exp(2)^2+5+2*x)*log(2)+160*x^3*exp(2)^2+8*x+25)/(32*x^3*exp(2)^2*log(2)+160*x^3*exp(2)^2),x
, algorithm="giac")

[Out]

(x*e^4*log(2) + 5*x*e^4)/(e^4*log(2) + 5*e^4) - 1/64*(4*x*log(2) + 16*x + 5*log(2) + 25)/((e^4*log(2) + 5*e^4)
*x^2)

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maple [A]  time = 0.08, size = 29, normalized size = 1.12

method	result	size
risch	\(x +\frac {\left (\left (-\frac {\ln \relax (2)}{16}-\frac {1}{4}\right ) x -\frac {5 \ln \relax (2)}{64}-\frac {25}{64}\right ) {\mathrm e}^{-4}}{x^{2} \left (\ln \relax (2)+5\right )}\)	\(29\)
norman	\(\frac {\left (x^{3} {\mathrm e}^{2}-\frac {5 \,{\mathrm e}^{-2}}{64}-\frac {\left (4+\ln \relax (2)\right ) {\mathrm e}^{-2} x}{16 \left (\ln \relax (2)+5\right )}\right ) {\mathrm e}^{-2}}{x^{2}}\)	\(39\)
default	\(\frac {{\mathrm e}^{-4} \left (32 x \,{\mathrm e}^{4} \ln \relax (2)+160 x \,{\mathrm e}^{4}-\frac {5 \ln \relax (2)+25}{2 x^{2}}-\frac {8+2 \ln \relax (2)}{x}\right )}{32 \ln \relax (2)+160}\)	\(48\)
gosper	\(\frac {\left (64 x^{3} {\mathrm e}^{4} \ln \relax (2)+320 x^{3} {\mathrm e}^{4}-4 x \ln \relax (2)-5 \ln \relax (2)-16 x -25\right ) {\mathrm e}^{-4}}{64 x^{2} \left (\ln \relax (2)+5\right )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^3*exp(2)^2+5+2*x)*ln(2)+160*x^3*exp(2)^2+8*x+25)/(32*x^3*exp(2)^2*ln(2)+160*x^3*exp(2)^2),x,method=
_RETURNVERBOSE)

[Out]

x+((-1/16*ln(2)-1/4)*x-5/64*ln(2)-25/64)/x^2*exp(-4)/(ln(2)+5)

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maxima [A]  time = 0.38, size = 32, normalized size = 1.23 \begin {gather*} x - \frac {4 \, x {\left (\log \relax (2) + 4\right )} + 5 \, \log \relax (2) + 25}{64 \, {\left (e^{4} \log \relax (2) + 5 \, e^{4}\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*exp(2)^2+5+2*x)*log(2)+160*x^3*exp(2)^2+8*x+25)/(32*x^3*exp(2)^2*log(2)+160*x^3*exp(2)^2),x
, algorithm="maxima")

[Out]

x - 1/64*(4*x*(log(2) + 4) + 5*log(2) + 25)/((e^4*log(2) + 5*e^4)*x^2)

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mupad [B]  time = 0.13, size = 32, normalized size = 1.23 \begin {gather*} x-\frac {\frac {\ln \left (32\right )}{2}+x\,\left (\ln \relax (4)+8\right )+\frac {25}{2}}{x^2\,\left (160\,{\mathrm {e}}^4+32\,{\mathrm {e}}^4\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 160*x^3*exp(4) + log(2)*(2*x + 32*x^3*exp(4) + 5) + 25)/(160*x^3*exp(4) + 32*x^3*exp(4)*log(2)),x)

[Out]

x - (log(32)/2 + x*(log(4) + 8) + 25/2)/(x^2*(160*exp(4) + 32*exp(4)*log(2)))

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sympy [B]  time = 0.20, size = 49, normalized size = 1.88 \begin {gather*} \frac {x \left (32 e^{4} \log {\relax (2 )} + 160 e^{4}\right ) + \frac {x \left (-16 - 4 \log {\relax (2 )}\right ) - 25 - 5 \log {\relax (2 )}}{2 x^{2}}}{32 e^{4} \log {\relax (2 )} + 160 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**3*exp(2)**2+5+2*x)*ln(2)+160*x**3*exp(2)**2+8*x+25)/(32*x**3*exp(2)**2*ln(2)+160*x**3*exp(2)
**2),x)

[Out]

(x*(32*exp(4)*log(2) + 160*exp(4)) + (x*(-16 - 4*log(2)) - 25 - 5*log(2))/(2*x**2))/(32*exp(4)*log(2) + 160*ex
p(4))

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