3.64.61 \(\int \frac {20+35 x+15 x^2+(-20-15 x) \log (2)+(35 x+30 x^2-15 x \log (2)) \log (x)}{x} \, dx\)

Optimal. Leaf size=16 \[ 5 (4+3 x) (1+x-\log (2)) \log (x) \]

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Rubi [A]  time = 0.06, antiderivative size = 29, normalized size of antiderivative = 1.81, number of steps used = 6, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 76, 2313} \begin {gather*} 5 \left (3 x^2+x (7-\log (8))\right ) \log (x)+20 (1-\log (2)) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + 35*x + 15*x^2 + (-20 - 15*x)*Log[2] + (35*x + 30*x^2 - 15*x*Log[2])*Log[x])/x,x]

[Out]

20*(1 - Log[2])*Log[x] + 5*(3*x^2 + x*(7 - Log[8]))*Log[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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {5 (4+3 x) (1+x-\log (2))}{x}+5 (7+6 x-\log (8)) \log (x)\right ) \, dx\\ &=5 \int \frac {(4+3 x) (1+x-\log (2))}{x} \, dx+5 \int (7+6 x-\log (8)) \log (x) \, dx\\ &=5 \left (3 x^2+x (7-\log (8))\right ) \log (x)-5 \int (7+3 x-\log (8)) \, dx+5 \int \left (7+3 x-\frac {4 (-1+\log (2))}{x}-\log (8)\right ) \, dx\\ &=20 (1-\log (2)) \log (x)+5 \left (3 x^2+x (7-\log (8))\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 40, normalized size = 2.50 \begin {gather*} -15 x \log (2)+5 x \log (8)+20 \log (x)+35 x \log (x)+15 x^2 \log (x)-20 \log (2) \log (x)-5 x \log (8) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 35*x + 15*x^2 + (-20 - 15*x)*Log[2] + (35*x + 30*x^2 - 15*x*Log[2])*Log[x])/x,x]

[Out]

-15*x*Log[2] + 5*x*Log[8] + 20*Log[x] + 35*x*Log[x] + 15*x^2*Log[x] - 20*Log[2]*Log[x] - 5*x*Log[8]*Log[x]

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fricas [A]  time = 0.51, size = 23, normalized size = 1.44 \begin {gather*} 5 \, {\left (3 \, x^{2} - {\left (3 \, x + 4\right )} \log \relax (2) + 7 \, x + 4\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x*log(2)+30*x^2+35*x)*log(x)+(-15*x-20)*log(2)+15*x^2+35*x+20)/x,x, algorithm="fricas")

[Out]

5*(3*x^2 - (3*x + 4)*log(2) + 7*x + 4)*log(x)

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giac [A]  time = 0.18, size = 28, normalized size = 1.75 \begin {gather*} 5 \, {\left (3 \, x^{2} - x {\left (3 \, \log \relax (2) - 7\right )}\right )} \log \relax (x) - 20 \, {\left (\log \relax (2) - 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x*log(2)+30*x^2+35*x)*log(x)+(-15*x-20)*log(2)+15*x^2+35*x+20)/x,x, algorithm="giac")

[Out]

5*(3*x^2 - x*(3*log(2) - 7))*log(x) - 20*(log(2) - 1)*log(x)

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maple [A]  time = 0.05, size = 28, normalized size = 1.75

method	result	size
norman	\(\left (-20 \ln \relax (2)+20\right ) \ln \relax (x )+\left (-15 \ln \relax (2)+35\right ) x \ln \relax (x )+15 x^{2} \ln \relax (x )\)	\(28\)
risch	\(\left (-15 x \ln \relax (2)+15 x^{2}+35 x \right ) \ln \relax (x )-20 \ln \relax (2) \ln \relax (x )+20 \ln \relax (x )\)	\(29\)
default	\(-15 \ln \relax (2) \left (x \ln \relax (x )-x \right )+15 x^{2} \ln \relax (x )+35 x \ln \relax (x )-15 x \ln \relax (2)-20 \ln \relax (2) \ln \relax (x )+20 \ln \relax (x )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x*ln(2)+30*x^2+35*x)*ln(x)+(-15*x-20)*ln(2)+15*x^2+35*x+20)/x,x,method=_RETURNVERBOSE)

[Out]

(-20*ln(2)+20)*ln(x)+(-15*ln(2)+35)*x*ln(x)+15*x^2*ln(x)

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maxima [B]  time = 0.37, size = 40, normalized size = 2.50 \begin {gather*} 15 \, x^{2} \log \relax (x) - 15 \, {\left (x \log \relax (x) - x\right )} \log \relax (2) - 15 \, x \log \relax (2) + 35 \, x \log \relax (x) - 20 \, \log \relax (2) \log \relax (x) + 20 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x*log(2)+30*x^2+35*x)*log(x)+(-15*x-20)*log(2)+15*x^2+35*x+20)/x,x, algorithm="maxima")

[Out]

15*x^2*log(x) - 15*(x*log(x) - x)*log(2) - 15*x*log(2) + 35*x*log(x) - 20*log(2)*log(x) + 20*log(x)

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mupad [B]  time = 4.26, size = 16, normalized size = 1.00 \begin {gather*} 5\,\ln \relax (x)\,\left (3\,x+4\right )\,\left (x-\ln \relax (2)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((35*x - log(2)*(15*x + 20) + 15*x^2 + log(x)*(35*x - 15*x*log(2) + 30*x^2) + 20)/x,x)

[Out]

5*log(x)*(3*x + 4)*(x - log(2) + 1)

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sympy [A]  time = 0.13, size = 27, normalized size = 1.69 \begin {gather*} \left (15 x^{2} - 15 x \log {\relax (2 )} + 35 x\right ) \log {\relax (x )} + \left (20 - 20 \log {\relax (2 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x*ln(2)+30*x**2+35*x)*ln(x)+(-15*x-20)*ln(2)+15*x**2+35*x+20)/x,x)

[Out]

(15*x**2 - 15*x*log(2) + 35*x)*log(x) + (20 - 20*log(2))*log(x)

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