3.8.66 \(\int \frac {15 x^2+6 x^3-3 x^2 \log (3)}{25+30 x+9 x^2+(-10-6 x) \log (3)+\log ^2(3)} \, dx\)

Optimal. Leaf size=17 \[ -36+\frac {x^3}{5+3 x-\log (3)} \]

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Rubi [A]  time = 0.09, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {6, 1593, 1983, 27, 74} \begin {gather*} \frac {x^3}{3 x+5-\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*x^2 + 6*x^3 - 3*x^2*Log[3])/(25 + 30*x + 9*x^2 + (-10 - 6*x)*Log[3] + Log[3]^2),x]

[Out]

x^3/(5 + 3*x - Log[3])

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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1983

Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n*ExpandToSum[w,
x]^p, x] /; FreeQ[{m, n, p}, x] && LinearQ[{u, v}, x] && QuadraticQ[w, x] &&  !(LinearMatchQ[{u, v}, x] && Qua
draticMatchQ[w, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 x^3+x^2 (15-3 \log (3))}{25+30 x+9 x^2+(-10-6 x) \log (3)+\log ^2(3)} \, dx\\ &=\int \frac {x^2 (15+6 x-3 \log (3))}{25+30 x+9 x^2+(-10-6 x) \log (3)+\log ^2(3)} \, dx\\ &=\int \frac {x^2 (6 x+3 (5-\log (3)))}{9 x^2+6 x (5-\log (3))+(-5+\log (3))^2} \, dx\\ &=\int \frac {x^2 (6 x+3 (5-\log (3)))}{(5+3 x-\log (3))^2} \, dx\\ &=\frac {x^3}{5+3 x-\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 36, normalized size = 2.12 \begin {gather*} \frac {9 x^3-3 x (-5+\log (3))^2+(-5+\log (3))^3}{9 (5+3 x-\log (3))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*x^2 + 6*x^3 - 3*x^2*Log[3])/(25 + 30*x + 9*x^2 + (-10 - 6*x)*Log[3] + Log[3]^2),x]

[Out]

(9*x^3 - 3*x*(-5 + Log[3])^2 + (-5 + Log[3])^3)/(9*(5 + 3*x - Log[3]))

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fricas [B]  time = 0.69, size = 45, normalized size = 2.65 \begin {gather*} \frac {27 \, x^{3} - 3 \, {\left (x + 5\right )} \log \relax (3)^{2} + \log \relax (3)^{3} + 15 \, {\left (2 \, x + 5\right )} \log \relax (3) - 75 \, x - 125}{27 \, {\left (3 \, x - \log \relax (3) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*log(3)+6*x^3+15*x^2)/(log(3)^2+(-6*x-10)*log(3)+9*x^2+30*x+25),x, algorithm="fricas")

[Out]

1/27*(27*x^3 - 3*(x + 5)*log(3)^2 + log(3)^3 + 15*(2*x + 5)*log(3) - 75*x - 125)/(3*x - log(3) + 5)

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giac [B]  time = 0.31, size = 43, normalized size = 2.53 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{9} \, x \log \relax (3) - \frac {5}{9} \, x + \frac {\log \relax (3)^{3} - 15 \, \log \relax (3)^{2} + 75 \, \log \relax (3) - 125}{27 \, {\left (3 \, x - \log \relax (3) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*log(3)+6*x^3+15*x^2)/(log(3)^2+(-6*x-10)*log(3)+9*x^2+30*x+25),x, algorithm="giac")

[Out]

1/3*x^2 + 1/9*x*log(3) - 5/9*x + 1/27*(log(3)^3 - 15*log(3)^2 + 75*log(3) - 125)/(3*x - log(3) + 5)

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maple [A]  time = 0.21, size = 15, normalized size = 0.88

method	result	size
gosper	\(-\frac {x^{3}}{\ln \relax (3)-3 x -5}\)	\(15\)
norman	\(-\frac {x^{3}}{\ln \relax (3)-3 x -5}\)	\(15\)
default	\(\frac {x^{2}}{3}+\frac {x \ln \relax (3)}{9}-\frac {5 x}{9}-\frac {3 \left (-\frac {\ln \relax (3)^{3}}{81}+\frac {5 \ln \relax (3)^{2}}{27}-\frac {25 \ln \relax (3)}{27}+\frac {125}{81}\right )}{5+3 x -\ln \relax (3)}\)	\(46\)
risch	\(\frac {x \ln \relax (3)}{9}+\frac {x^{2}}{3}-\frac {5 x}{9}-\frac {\ln \relax (3)^{3}}{27 \left (\ln \relax (3)-3 x -5\right )}+\frac {5 \ln \relax (3)^{2}}{9 \left (\ln \relax (3)-3 x -5\right )}-\frac {25 \ln \relax (3)}{9 \left (\ln \relax (3)-3 x -5\right )}+\frac {125}{27 \left (\ln \relax (3)-3 x -5\right )}\)	\(69\)
meijerg	\(\frac {\left (5-\ln \relax (3)\right )^{2} \left (\frac {x \left (\frac {9 x}{5-\ln \relax (3)}+6\right )}{\left (5-\ln \relax (3)\right ) \left (1+\frac {3 x}{5-\ln \relax (3)}\right )}-2 \ln \left (1+\frac {3 x}{5-\ln \relax (3)}\right )\right )}{9}+\frac {2 \left (5-\ln \relax (3)\right )^{3} \left (-\frac {3 x \left (-\frac {18 x^{2}}{\left (5-\ln \relax (3)\right )^{2}}+\frac {18 x}{5-\ln \relax (3)}+12\right )}{4 \left (5-\ln \relax (3)\right ) \left (1+\frac {3 x}{5-\ln \relax (3)}\right )}+3 \ln \left (1+\frac {3 x}{5-\ln \relax (3)}\right )\right )}{81 \left (-\frac {\ln \relax (3)}{3}+\frac {5}{3}\right )}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2*ln(3)+6*x^3+15*x^2)/(ln(3)^2+(-6*x-10)*ln(3)+9*x^2+30*x+25),x,method=_RETURNVERBOSE)

[Out]

-x^3/(ln(3)-3*x-5)

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maxima [B]  time = 0.44, size = 42, normalized size = 2.47 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {1}{9} \, x {\left (\log \relax (3) - 5\right )} + \frac {\log \relax (3)^{3} - 15 \, \log \relax (3)^{2} + 75 \, \log \relax (3) - 125}{27 \, {\left (3 \, x - \log \relax (3) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*log(3)+6*x^3+15*x^2)/(log(3)^2+(-6*x-10)*log(3)+9*x^2+30*x+25),x, algorithm="maxima")

[Out]

1/3*x^2 + 1/9*x*(log(3) - 5) + 1/27*(log(3)^3 - 15*log(3)^2 + 75*log(3) - 125)/(3*x - log(3) + 5)

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mupad [B]  time = 0.79, size = 103, normalized size = 6.06 \begin {gather*} \frac {x^2}{3}-x\,\left (\frac {\ln \left (27\right )}{9}-\frac {2\,\ln \left (729\right )}{27}+\frac {5}{9}\right )+\frac {2\,\mathrm {atanh}\left (\frac {2\,{\left (\ln \relax (3)-5\right )}^3\,\left (18\,x-\ln \left (729\right )+30\right )}{9\,\sqrt {\ln \left (729\right )-6\,\ln \relax (3)}\,\sqrt {6\,\ln \relax (3)+\ln \left (729\right )-60}\,\left (\frac {50\,\ln \relax (3)}{3}-\frac {10\,{\ln \relax (3)}^2}{3}+\frac {2\,{\ln \relax (3)}^3}{9}-\frac {250}{9}\right )}\right )\,{\left (\ln \relax (3)-5\right )}^3}{9\,\sqrt {\ln \left (729\right )-6\,\ln \relax (3)}\,\sqrt {6\,\ln \relax (3)+\ln \left (729\right )-60}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x^2 - 3*x^2*log(3) + 6*x^3)/(30*x - log(3)*(6*x + 10) + log(3)^2 + 9*x^2 + 25),x)

[Out]

x^2/3 - x*(log(27)/9 - (2*log(729))/27 + 5/9) + (2*atanh((2*(log(3) - 5)^3*(18*x - log(729) + 30))/(9*(log(729
) - 6*log(3))^(1/2)*(6*log(3) + log(729) - 60)^(1/2)*((50*log(3))/3 - (10*log(3)^2)/3 + (2*log(3)^3)/9 - 250/9
)))*(log(3) - 5)^3)/(9*(log(729) - 6*log(3))^(1/2)*(6*log(3) + log(729) - 60)^(1/2))

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sympy [B]  time = 0.27, size = 42, normalized size = 2.47 \begin {gather*} \frac {x^{2}}{3} + x \left (- \frac {5}{9} + \frac {\log {\relax (3 )}}{9}\right ) + \frac {-125 - 15 \log {\relax (3 )}^{2} + \log {\relax (3 )}^{3} + 75 \log {\relax (3 )}}{81 x - 27 \log {\relax (3 )} + 135} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2*ln(3)+6*x**3+15*x**2)/(ln(3)**2+(-6*x-10)*ln(3)+9*x**2+30*x+25),x)

[Out]

x**2/3 + x*(-5/9 + log(3)/9) + (-125 - 15*log(3)**2 + log(3)**3 + 75*log(3))/(81*x - 27*log(3) + 135)

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