3.34.22 \(\int \frac {-30+80 x-20 x^2+(15-10 x^2) \log (4 x^2)}{2 x^2} \, dx\)

Optimal. Leaf size=22 \[ 5 \left (2-\left (-4+\frac {3}{2 x}+x\right ) \log \left (4 x^2\right )\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2334} \begin {gather*} 40 \log (x)-\frac {5}{2} \left (2 x+\frac {3}{x}\right ) \log \left (4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + 80*x - 20*x^2 + (15 - 10*x^2)*Log[4*x^2])/(2*x^2),x]

[Out]

40*Log[x] - (5*(3/x + 2*x)*Log[4*x^2])/2

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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-30+80 x-20 x^2+\left (15-10 x^2\right ) \log \left (4 x^2\right )}{x^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {10 \left (3-8 x+2 x^2\right )}{x^2}-\frac {5 \left (-3+2 x^2\right ) \log \left (4 x^2\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {\left (-3+2 x^2\right ) \log \left (4 x^2\right )}{x^2} \, dx\right )-5 \int \frac {3-8 x+2 x^2}{x^2} \, dx\\ &=-\frac {5}{2} \left (\frac {3}{x}+2 x\right ) \log \left (4 x^2\right )+5 \int \left (2+\frac {3}{x^2}\right ) \, dx-5 \int \left (2+\frac {3}{x^2}-\frac {8}{x}\right ) \, dx\\ &=40 \log (x)-\frac {5}{2} \left (\frac {3}{x}+2 x\right ) \log \left (4 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.23 \begin {gather*} 40 \log (x)-\frac {15 \log \left (4 x^2\right )}{2 x}-5 x \log \left (4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 80*x - 20*x^2 + (15 - 10*x^2)*Log[4*x^2])/(2*x^2),x]

[Out]

40*Log[x] - (15*Log[4*x^2])/(2*x) - 5*x*Log[4*x^2]

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fricas [A]  time = 0.57, size = 21, normalized size = 0.95 \begin {gather*} -\frac {5 \, {\left (2 \, x^{2} - 8 \, x + 3\right )} \log \left (4 \, x^{2}\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*x^2+15)*log(4*x^2)-20*x^2+80*x-30)/x^2,x, algorithm="fricas")

[Out]

-5/2*(2*x^2 - 8*x + 3)*log(4*x^2)/x

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giac [A]  time = 0.21, size = 22, normalized size = 1.00 \begin {gather*} -\frac {5}{2} \, {\left (2 \, x + \frac {3}{x}\right )} \log \left (4 \, x^{2}\right ) + 40 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*x^2+15)*log(4*x^2)-20*x^2+80*x-30)/x^2,x, algorithm="giac")

[Out]

-5/2*(2*x + 3/x)*log(4*x^2) + 40*log(x)

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maple [A]  time = 0.03, size = 24, normalized size = 1.09

method	result	size
risch	\(-\frac {5 \left (2 x^{2}+3\right ) \ln \left (4 x^{2}\right )}{2 x}+40 \ln \relax (x )\)	\(24\)
norman	\(\frac {-5 x^{2} \ln \left (4 x^{2}\right )-\frac {15 \ln \left (4 x^{2}\right )}{2}}{x}+40 \ln \relax (x )\)	\(30\)
default	\(40 \ln \relax (x )-10 x \ln \relax (2)-\frac {15 \ln \relax (2)}{x}-5 x \ln \left (x^{2}\right )-\frac {15 \ln \left (x^{2}\right )}{2 x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-10*x^2+15)*ln(4*x^2)-20*x^2+80*x-30)/x^2,x,method=_RETURNVERBOSE)

[Out]

-5/2*(2*x^2+3)/x*ln(4*x^2)+40*ln(x)

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maxima [A]  time = 0.41, size = 25, normalized size = 1.14 \begin {gather*} -5 \, x \log \left (4 \, x^{2}\right ) - \frac {15 \, \log \left (4 \, x^{2}\right )}{2 \, x} + 40 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*x^2+15)*log(4*x^2)-20*x^2+80*x-30)/x^2,x, algorithm="maxima")

[Out]

-5*x*log(4*x^2) - 15/2*log(4*x^2)/x + 40*log(x)

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mupad [B]  time = 1.93, size = 21, normalized size = 0.95 \begin {gather*} -\frac {5\,\ln \left (4\,x^2\right )\,\left (2\,x^2-8\,x+3\right )}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((log(4*x^2)*(10*x^2 - 15))/2 - 40*x + 10*x^2 + 15)/x^2,x)

[Out]

-(5*log(4*x^2)*(2*x^2 - 8*x + 3))/(2*x)

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sympy [A]  time = 0.13, size = 22, normalized size = 1.00 \begin {gather*} 40 \log {\relax (x )} + \frac {\left (- 10 x^{2} - 15\right ) \log {\left (4 x^{2} \right )}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-10*x**2+15)*ln(4*x**2)-20*x**2+80*x-30)/x**2,x)

[Out]

40*log(x) + (-10*x**2 - 15)*log(4*x**2)/(2*x)

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