3.23.82 \(\int \frac {-15 x^2+6 x^3+3 x^2 \log (3)+(-5+3 x^2+\log (3)) \log (5)+\log ^2(5)}{6 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (x-\frac {\log (5)}{3 x}\right ) (x+x (-6+x+\log (3)+\log (5)))}{2 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 12, 14} \begin {gather*} \frac {x^2}{2}-\frac {1}{2} x (5-\log (15))+\frac {\log (5) (5-\log (15))}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x^2 + 6*x^3 + 3*x^2*Log[3] + (-5 + 3*x^2 + Log[3])*Log[5] + Log[5]^2)/(6*x^2),x]

[Out]

x^2/2 - (x*(5 - Log[15]))/2 + (Log[5]*(5 - Log[15]))/(6*x)

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 {6 x^3+x^2 (-15+3 \log (3))+\left (-5+3 x^2+\log (3)\right ) \log (5)+\log ^2(5)}{6 x^2} \, dx\\ &=\frac {1}{6} \int \frac {6 x^3+x^2 (-15+3 \log (3))+\left (-5+3 x^2+\log (3)\right ) \log (5)+\log ^2(5)}{x^2} \, dx\\ &=\frac {1}{6} \int \left (6 x+3 (-5+\log (15))+\frac {\log (5) (-5+\log (15))}{x^2}\right ) \, dx\\ &=\frac {x^2}{2}-\frac {1}{2} x (5-\log (15))+\frac {\log (5) (5-\log (15))}{6 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.03 \begin {gather*} \frac {x^2}{2}+\frac {1}{2} x (-5+\log (15))-\frac {\log (5) (-5+\log (15))}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x^2 + 6*x^3 + 3*x^2*Log[3] + (-5 + 3*x^2 + Log[3])*Log[5] + Log[5]^2)/(6*x^2),x]

[Out]

x^2/2 + (x*(-5 + Log[15]))/2 - (Log[5]*(-5 + Log[15]))/(6*x)

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fricas [A]  time = 0.73, size = 43, normalized size = 1.48 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} \log \relax (3) - 15 \, x^{2} + {\left (3 \, x^{2} - \log \relax (3) + 5\right )} \log \relax (5) - \log \relax (5)^{2}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(log(5)^2+(log(3)+3*x^2-5)*log(5)+3*x^2*log(3)+6*x^3-15*x^2)/x^2,x, algorithm="fricas")

[Out]

1/6*(3*x^3 + 3*x^2*log(3) - 15*x^2 + (3*x^2 - log(3) + 5)*log(5) - log(5)^2)/x

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giac [A]  time = 1.50, size = 38, normalized size = 1.31 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {1}{2} \, x \log \relax (5) + \frac {1}{2} \, x \log \relax (3) - \frac {5}{2} \, x - \frac {\log \relax (5)^{2} + \log \relax (5) \log \relax (3) - 5 \, \log \relax (5)}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(log(5)^2+(log(3)+3*x^2-5)*log(5)+3*x^2*log(3)+6*x^3-15*x^2)/x^2,x, algorithm="giac")

[Out]

1/2*x^2 + 1/2*x*log(5) + 1/2*x*log(3) - 5/2*x - 1/6*(log(5)^2 + log(5)*log(3) - 5*log(5))/x

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maple [A]  time = 0.05, size = 33, normalized size = 1.14

method	result	size
default	\(\frac {x^{2}}{2}+\frac {x \ln \relax (5)}{2}+\frac {x \ln \relax (3)}{2}-\frac {5 x}{2}-\frac {\ln \relax (5) \left (\ln \relax (5)+\ln \relax (3)-5\right )}{6 x}\)	\(33\)
norman	\(\frac {\left (\frac {\ln \relax (5)}{2}+\frac {\ln \relax (3)}{2}-\frac {5}{2}\right ) x^{2}+\frac {x^{3}}{2}-\frac {\ln \relax (5)^{2}}{6}-\frac {\ln \relax (3) \ln \relax (5)}{6}+\frac {5 \ln \relax (5)}{6}}{x}\)	\(41\)
gosper	\(-\frac {-3 x^{2} \ln \relax (5)-3 x^{2} \ln \relax (3)-3 x^{3}+\ln \relax (5)^{2}+\ln \relax (3) \ln \relax (5)+15 x^{2}-5 \ln \relax (5)}{6 x}\)	\(44\)
risch	\(\frac {x \ln \relax (5)}{2}+\frac {x \ln \relax (3)}{2}+\frac {x^{2}}{2}-\frac {5 x}{2}-\frac {\ln \relax (5)^{2}}{6 x}-\frac {\ln \relax (5) \ln \relax (3)}{6 x}+\frac {5 \ln \relax (5)}{6 x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(ln(5)^2+(ln(3)+3*x^2-5)*ln(5)+3*x^2*ln(3)+6*x^3-15*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x^2+1/2*x*ln(5)+1/2*x*ln(3)-5/2*x-1/6*ln(5)*(ln(5)+ln(3)-5)/x

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maxima [A]  time = 0.43, size = 32, normalized size = 1.10 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {1}{2} \, x {\left (\log \relax (5) + \log \relax (3) - 5\right )} - \frac {{\left (\log \relax (3) - 5\right )} \log \relax (5) + \log \relax (5)^{2}}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(log(5)^2+(log(3)+3*x^2-5)*log(5)+3*x^2*log(3)+6*x^3-15*x^2)/x^2,x, algorithm="maxima")

[Out]

1/2*x^2 + 1/2*x*(log(5) + log(3) - 5) - 1/6*((log(3) - 5)*log(5) + log(5)^2)/x

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mupad [B]  time = 0.07, size = 36, normalized size = 1.24 \begin {gather*} x\,\left (\frac {\ln \left (15\right )}{2}-\frac {5}{2}\right )-\frac {\frac {\ln \relax (3)\,\ln \relax (5)}{6}-\frac {5\,\ln \relax (5)}{6}+\frac {{\ln \relax (5)}^2}{6}}{x}+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*log(3))/2 + log(5)^2/6 + (log(5)*(log(3) + 3*x^2 - 5))/6 - (5*x^2)/2 + x^3)/x^2,x)

[Out]

x*(log(15)/2 - 5/2) - ((log(3)*log(5))/6 - (5*log(5))/6 + log(5)^2/6)/x + x^2/2

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sympy [A]  time = 0.14, size = 39, normalized size = 1.34 \begin {gather*} \frac {x^{2}}{2} + \frac {x \left (-15 + 3 \log {\relax (3 )} + 3 \log {\relax (5 )}\right )}{6} + \frac {- \log {\relax (5 )}^{2} - \log {\relax (3 )} \log {\relax (5 )} + 5 \log {\relax (5 )}}{6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(ln(5)**2+(ln(3)+3*x**2-5)*ln(5)+3*x**2*ln(3)+6*x**3-15*x**2)/x**2,x)

[Out]

x**2/2 + x*(-15 + 3*log(3) + 3*log(5))/6 + (-log(5)**2 - log(3)*log(5) + 5*log(5))/(6*x)

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