3.36.78 \(\int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+(2 x^2+2^{5+5 x} (4+(-20 x-5 x^2) \log (2))) \log (x)}{2 x^2} \, dx\)

Optimal. Leaf size=25 \[ 1-\frac {\left (2^{5 (1+x)}-2 x\right ) (4+x) \log (x)}{2 x} \]

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Rubi [A]  time = 0.12, antiderivative size = 41, normalized size of antiderivative = 1.64, number of steps used = 9, number of rules used = 5, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 43, 2295, 2288} \begin {gather*} -\frac {2^{5 x+4} \left (x^2 \log (32) \log (x)+20 x \log (2) \log (x)\right )}{x^2 \log (32)}+x \log (x)+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2^(5 + 5*x)*(-4 - x) + 8*x + 2*x^2 + (2*x^2 + 2^(5 + 5*x)*(4 + (-20*x - 5*x^2)*Log[2]))*Log[x])/(2*x^2),x
]

[Out]

4*Log[x] + x*Log[x] - (2^(4 + 5*x)*(20*x*Log[2]*Log[x] + x^2*Log[32]*Log[x]))/(x^2*Log[32])

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 (4+x+x \log (x))}{x}-\frac {32^{1+x} \left (4+x-4 \log (x)+20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {32^{1+x} \left (4+x-4 \log (x)+20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2} \, dx\right )+\int \frac {4+x+x \log (x)}{x} \, dx\\ &=-\frac {2^{4+5 x} \left (20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2 \log (32)}+\int \left (\frac {4+x}{x}+\log (x)\right ) \, dx\\ &=-\frac {2^{4+5 x} \left (20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2 \log (32)}+\int \frac {4+x}{x} \, dx+\int \log (x) \, dx\\ &=-x+x \log (x)-\frac {2^{4+5 x} \left (20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2 \log (32)}+\int \left (1+\frac {4}{x}\right ) \, dx\\ &=4 \log (x)+x \log (x)-\frac {2^{4+5 x} \left (20 x \log (2) \log (x)+x^2 \log (32) \log (x)\right )}{x^2 \log (32)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 20, normalized size = 0.80 \begin {gather*} \frac {(4+x) \left (-2^{4+5 x}+x\right ) \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2^(5 + 5*x)*(-4 - x) + 8*x + 2*x^2 + (2*x^2 + 2^(5 + 5*x)*(4 + (-20*x - 5*x^2)*Log[2]))*Log[x])/(2*
x^2),x]

[Out]

((4 + x)*(-2^(4 + 5*x) + x)*Log[x])/x

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fricas [A]  time = 0.58, size = 27, normalized size = 1.08 \begin {gather*} -\frac {{\left (2^{5 \, x + 5} {\left (x + 4\right )} - 2 \, x^{2} - 8 \, x\right )} \log \relax (x)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x)+(-x-4)*exp((5*x+5)*log(2))+2*x^2+8*
x)/x^2,x, algorithm="fricas")

[Out]

-1/2*(2^(5*x + 5)*(x + 4) - 2*x^2 - 8*x)*log(x)/x

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giac [A]  time = 0.21, size = 30, normalized size = 1.20 \begin {gather*} -16 \cdot 2^{5 \, x} \log \relax (x) + x \log \relax (x) - \frac {64 \cdot 2^{5 \, x} \log \relax (x)}{x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x)+(-x-4)*exp((5*x+5)*log(2))+2*x^2+8*
x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 38, normalized size = 1.52

method	result	size
risch	\(\frac {\left (2 x^{2}-x 2^{5 x +5}-4 \,2^{5 x +5}\right ) \ln \relax (x )}{2 x}+4 \ln \relax (x )\)	\(38\)
default	\(\frac {-4 \ln \relax (x ) {\mathrm e}^{\left (5 x +5\right ) \ln \relax (2)}-x \,{\mathrm e}^{\left (5 x +5\right ) \ln \relax (2)} \ln \relax (x )}{2 x}+4 \ln \relax (x )+x \ln \relax (x )\)	\(43\)
norman	\(\frac {x^{2} \ln \relax (x )+4 x \ln \relax (x )-2 \ln \relax (x ) {\mathrm e}^{\left (5 x +5\right ) \ln \relax (2)}-\frac {x \,{\mathrm e}^{\left (5 x +5\right ) \ln \relax (2)} \ln \relax (x )}{2}}{x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((((-5*x^2-20*x)*ln(2)+4)*exp((5*x+5)*ln(2))+2*x^2)*ln(x)+(-x-4)*exp((5*x+5)*ln(2))+2*x^2+8*x)/x^2,x,m
ethod=_RETURNVERBOSE)

[Out]

1/2*(2*x^2-x*2^(5*x+5)-4*2^(5*x+5))/x*ln(x)+4*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -320 \, \Gamma \left (-1, -5 \, x \log \relax (2)\right ) \log \relax (2) - \frac {1}{2} \cdot 2^{5 \, x + 5} \log \relax (x) + x \log \relax (x) - \frac {64 \cdot 2^{5 \, x} \log \relax (x)}{x} + 64 \, \int \frac {2^{5 \, x}}{x^{2}}\,{d x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x)+(-x-4)*exp((5*x+5)*log(2))+2*x^2+8*
x)/x^2,x, algorithm="maxima")

[Out]

-320*gamma(-1, -5*x*log(2))*log(2) - 1/2*2^(5*x + 5)*log(x) + x*log(x) - 64*2^(5*x)*log(x)/x + 64*integrate(2^
(5*x)/x^2, x) + 4*log(x)

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mupad [B]  time = 2.28, size = 18, normalized size = 0.72 \begin {gather*} \frac {\ln \relax (x)\,\left (x-16\,2^{5\,x}\right )\,\left (x+4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - (log(x)*(exp(log(2)*(5*x + 5))*(log(2)*(20*x + 5*x^2) - 4) - 2*x^2))/2 - (exp(log(2)*(5*x + 5))*(x
+ 4))/2 + x^2)/x^2,x)

[Out]

(log(x)*(x - 16*2^(5*x))*(x + 4))/x

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sympy [A]  time = 0.32, size = 34, normalized size = 1.36 \begin {gather*} x \log {\relax (x )} + 4 \log {\relax (x )} + \frac {\left (- x \log {\relax (x )} - 4 \log {\relax (x )}\right ) e^{\left (5 x + 5\right ) \log {\relax (2 )}}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((-5*x**2-20*x)*ln(2)+4)*exp((5*x+5)*ln(2))+2*x**2)*ln(x)+(-x-4)*exp((5*x+5)*ln(2))+2*x**2+8*x
)/x**2,x)

[Out]

x*log(x) + 4*log(x) + (-x*log(x) - 4*log(x))*exp((5*x + 5)*log(2))/(2*x)

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