3.55.95 \(\int \frac {32-226 x^2-64 x^3+(-16+65 x^2) \log (4)+e^x (-32 x^3+16 x^3 \log (4))}{-2 x^2+x^2 \log (4)} \, dx\)

Optimal. Leaf size=28 \[ x-16 (-1+x) \left (-4-e^x+\frac {1}{x}+\frac {2 (4+x)}{-2+\log (4)}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 47, normalized size of antiderivative = 1.68, number of steps used = 8, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6, 12, 14, 2176, 2194} \begin {gather*} \frac {32 x^2}{2-\log (4)}+16 e^x x-16 e^x+\frac {16}{x}+\frac {2 x (113-65 \log (2))}{2-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 226*x^2 - 64*x^3 + (-16 + 65*x^2)*Log[4] + E^x*(-32*x^3 + 16*x^3*Log[4]))/(-2*x^2 + x^2*Log[4]),x]

[Out]

-16*E^x + 16/x + 16*E^x*x + (32*x^2)/(2 - Log[4]) + (2*x*(113 - 65*Log[2]))/(2 - Log[4])

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]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32-226 x^2-64 x^3+\left (-16+65 x^2\right ) \log (4)+e^x \left (-32 x^3+16 x^3 \log (4)\right )}{x^2 (-2+\log (4))} \, dx\\ &=\frac {\int \frac {32-226 x^2-64 x^3+\left (-16+65 x^2\right ) \log (4)+e^x \left (-32 x^3+16 x^3 \log (4)\right )}{x^2} \, dx}{-2+\log (4)}\\ &=\frac {\int \left (\frac {-64 x^3-x^2 (226-65 \log (4))+16 (2-\log (4))}{x^2}+16 e^x x (-2+\log (4))\right ) \, dx}{-2+\log (4)}\\ &=16 \int e^x x \, dx+\frac {\int \frac {-64 x^3-x^2 (226-65 \log (4))+16 (2-\log (4))}{x^2} \, dx}{-2+\log (4)}\\ &=16 e^x x-16 \int e^x \, dx+\frac {\int \left (-64 x-226 \left (1-\frac {65 \log (2)}{113}\right )-\frac {16 (-2+\log (4))}{x^2}\right ) \, dx}{-2+\log (4)}\\ &=-16 e^x+\frac {16}{x}+16 e^x x+\frac {32 x^2}{2-\log (4)}+\frac {2 x (113-65 \log (2))}{2-\log (4)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 39, normalized size = 1.39 \begin {gather*} 16 e^x (-1+x)+\frac {16}{x}-\frac {32 x^2}{-2+\log (4)}+\frac {x (-226+65 \log (4))}{-2+\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 226*x^2 - 64*x^3 + (-16 + 65*x^2)*Log[4] + E^x*(-32*x^3 + 16*x^3*Log[4]))/(-2*x^2 + x^2*Log[4]
),x]

[Out]

16*E^x*(-1 + x) + 16/x - (32*x^2)/(-2 + Log[4]) + (x*(-226 + 65*Log[4]))/(-2 + Log[4])

________________________________________________________________________________________

fricas [B]  time = 3.08, size = 57, normalized size = 2.04 \begin {gather*} -\frac {16 \, x^{3} + 113 \, x^{2} + 16 \, {\left (x^{2} - {\left (x^{2} - x\right )} \log \relax (2) - x\right )} e^{x} - {\left (65 \, x^{2} + 16\right )} \log \relax (2) + 16}{x \log \relax (2) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*log(2)-32*x^3)*exp(x)+2*(65*x^2-16)*log(2)-64*x^3-226*x^2+32)/(2*x^2*log(2)-2*x^2),x, algor
ithm="fricas")

[Out]

-(16*x^3 + 113*x^2 + 16*(x^2 - (x^2 - x)*log(2) - x)*e^x - (65*x^2 + 16)*log(2) + 16)/(x*log(2) - x)

________________________________________________________________________________________

giac [B]  time = 0.18, size = 62, normalized size = 2.21 \begin {gather*} \frac {16 \, x^{2} e^{x} \log \relax (2) - 16 \, x^{3} - 16 \, x^{2} e^{x} + 65 \, x^{2} \log \relax (2) - 16 \, x e^{x} \log \relax (2) - 113 \, x^{2} + 16 \, x e^{x} + 16 \, \log \relax (2) - 16}{x \log \relax (2) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*log(2)-32*x^3)*exp(x)+2*(65*x^2-16)*log(2)-64*x^3-226*x^2+32)/(2*x^2*log(2)-2*x^2),x, algor
ithm="giac")

[Out]

(16*x^2*e^x*log(2) - 16*x^3 - 16*x^2*e^x + 65*x^2*log(2) - 16*x*e^x*log(2) - 113*x^2 + 16*x*e^x + 16*log(2) -
16)/(x*log(2) - x)

________________________________________________________________________________________

maple [A]  time = 0.24, size = 46, normalized size = 1.64

method	result	size
norman	\(\frac {16+\frac {\left (65 \ln \relax (2)-113\right ) x^{2}}{\ln \relax (2)-1}-\frac {16 x^{3}}{\ln \relax (2)-1}-16 \,{\mathrm e}^{x} x +16 \,{\mathrm e}^{x} x^{2}}{x}\)	\(46\)
risch	\(\frac {65 \ln \relax (2) x}{\ln \relax (2)-1}-\frac {16 x^{2}}{\ln \relax (2)-1}-\frac {113 x}{\ln \relax (2)-1}+\frac {16 \ln \relax (2)^{2}}{\left (\ln \relax (2)-1\right )^{2} x}-\frac {32 \ln \relax (2)}{\left (\ln \relax (2)-1\right )^{2} x}+\frac {16}{\left (\ln \relax (2)-1\right )^{2} x}+\left (16 x -16\right ) {\mathrm e}^{x}\)	\(80\)
default	\(-\frac {16}{\left (\ln \relax (2)-1\right ) x}-\frac {113 x}{\ln \relax (2)-1}-\frac {16 x^{2}}{\ln \relax (2)-1}+\frac {16 \ln \relax (2)}{\left (\ln \relax (2)-1\right ) x}+\frac {65 \ln \relax (2) x}{\ln \relax (2)-1}-\frac {16 \,{\mathrm e}^{x} x}{\ln \relax (2)-1}+\frac {16 \,{\mathrm e}^{x}}{\ln \relax (2)-1}+\frac {16 \ln \relax (2) {\mathrm e}^{x} x}{\ln \relax (2)-1}-\frac {16 \ln \relax (2) {\mathrm e}^{x}}{\ln \relax (2)-1}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^3*ln(2)-32*x^3)*exp(x)+2*(65*x^2-16)*ln(2)-64*x^3-226*x^2+32)/(2*x^2*ln(2)-2*x^2),x,method=_RETURNV
ERBOSE)

[Out]

(16+(65*ln(2)-113)/(ln(2)-1)*x^2-16/(ln(2)-1)*x^3-16*exp(x)*x+16*exp(x)*x^2)/x

________________________________________________________________________________________

maxima [B]  time = 0.47, size = 84, normalized size = 3.00 \begin {gather*} \frac {16 \, {\left (x - 1\right )} e^{x} \log \relax (2)}{\log \relax (2) - 1} - \frac {16 \, x^{2}}{\log \relax (2) - 1} - \frac {16 \, {\left (x - 1\right )} e^{x}}{\log \relax (2) - 1} + \frac {65 \, x \log \relax (2)}{\log \relax (2) - 1} - \frac {113 \, x}{\log \relax (2) - 1} + \frac {16 \, \log \relax (2)}{x {\left (\log \relax (2) - 1\right )}} - \frac {16}{x {\left (\log \relax (2) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^3*log(2)-32*x^3)*exp(x)+2*(65*x^2-16)*log(2)-64*x^3-226*x^2+32)/(2*x^2*log(2)-2*x^2),x, algor
ithm="maxima")

[Out]

16*(x - 1)*e^x*log(2)/(log(2) - 1) - 16*x^2/(log(2) - 1) - 16*(x - 1)*e^x/(log(2) - 1) + 65*x*log(2)/(log(2) -
1) - 113*x/(log(2) - 1) + 16*log(2)/(x*(log(2) - 1)) - 16/(x*(log(2) - 1))

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 54, normalized size = 1.93 \begin {gather*} \frac {16\,\ln \relax (2)+x^2\,\left (65\,\ln \relax (2)-113\right )-16\,x^3+x^2\,{\mathrm {e}}^x\,\left (16\,\ln \relax (2)-16\right )-x\,{\mathrm {e}}^x\,\left (16\,\ln \relax (2)-16\right )-16}{x\,\left (\ln \relax (2)-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2)*(65*x^2 - 16) + exp(x)*(32*x^3*log(2) - 32*x^3) - 226*x^2 - 64*x^3 + 32)/(2*x^2*log(2) - 2*x^2),
x)

[Out]

(16*log(2) + x^2*(65*log(2) - 113) - 16*x^3 + x^2*exp(x)*(16*log(2) - 16) - x*exp(x)*(16*log(2) - 16) - 16)/(x
*(log(2) - 1))

________________________________________________________________________________________

sympy [A]  time = 0.17, size = 36, normalized size = 1.29 \begin {gather*} \left (16 x - 16\right ) e^{x} + \frac {- 16 x^{2} - x \left (113 - 65 \log {\relax (2 )}\right ) - \frac {16 - 16 \log {\relax (2 )}}{x}}{-1 + \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**3*ln(2)-32*x**3)*exp(x)+2*(65*x**2-16)*ln(2)-64*x**3-226*x**2+32)/(2*x**2*ln(2)-2*x**2),x)

[Out]

(16*x - 16)*exp(x) + (-16*x**2 - x*(113 - 65*log(2)) - (16 - 16*log(2))/x)/(-1 + log(2))

________________________________________________________________________________________