3.77.45 \(\int -\frac {1}{10+2 x+(20+4 x) \log (75+30 x+3 x^2)+(10+2 x) \log ^2(75+30 x+3 x^2)} \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{4 \left (1+\log \left (3 (5+x)^2\right )\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6688, 12, 2390, 2302, 30} \begin {gather*} \frac {1}{4 \left (\log \left (3 (x+5)^2\right )+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-(10 + 2*x + (20 + 4*x)*Log[75 + 30*x + 3*x^2] + (10 + 2*x)*Log[75 + 30*x + 3*x^2]^2)^(-1),x]

[Out]

1/(4*(1 + Log[3*(5 + x)^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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{4 \left (1+\log \left (3 (5+x)^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-(10 + 2*x + (20 + 4*x)*Log[75 + 30*x + 3*x^2] + (10 + 2*x)*Log[75 + 30*x + 3*x^2]^2)^(-1),x]

[Out]

1/(4*(1 + Log[3*(5 + x)^2]))

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fricas [A]  time = 0.99, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{4 \, {\left (\log \left (3 \, x^{2} + 30 \, x + 75\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/((2*x+10)*log(3*x^2+30*x+75)^2+(20+4*x)*log(3*x^2+30*x+75)+2*x+10),x, algorithm="fricas")

[Out]

1/4/(log(3*x^2 + 30*x + 75) + 1)

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giac [A]  time = 0.23, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{4 \, {\left (\log \left (3 \, x^{2} + 30 \, x + 75\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/((2*x+10)*log(3*x^2+30*x+75)^2+(20+4*x)*log(3*x^2+30*x+75)+2*x+10),x, algorithm="giac")

[Out]

1/4/(log(3*x^2 + 30*x + 75) + 1)

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maple [A]  time = 0.03, size = 18, normalized size = 1.12

method	result	size
norman	\(\frac {1}{4 \ln \left (3 x^{2}+30 x +75\right )+4}\)	\(18\)
risch	\(\frac {1}{4 \ln \left (3 x^{2}+30 x +75\right )+4}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((2*x+10)*ln(3*x^2+30*x+75)^2+(20+4*x)*ln(3*x^2+30*x+75)+2*x+10),x,method=_RETURNVERBOSE)

[Out]

1/4/(ln(3*x^2+30*x+75)+1)

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maxima [A]  time = 0.45, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{4 \, {\left (\log \relax (3) + 2 \, \log \left (x + 5\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/((2*x+10)*log(3*x^2+30*x+75)^2+(20+4*x)*log(3*x^2+30*x+75)+2*x+10),x, algorithm="maxima")

[Out]

1/4/(log(3) + 2*log(x + 5) + 1)

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mupad [B]  time = 0.31, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{4\,\left (\ln \left (3\,x^2+30\,x+75\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(2*x + log(30*x + 3*x^2 + 75)*(4*x + 20) + log(30*x + 3*x^2 + 75)^2*(2*x + 10) + 10),x)

[Out]

1/(4*(log(30*x + 3*x^2 + 75) + 1))

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sympy [A]  time = 0.12, size = 15, normalized size = 0.94 \begin {gather*} \frac {1}{4 \log {\left (3 x^{2} + 30 x + 75 \right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/((2*x+10)*ln(3*x**2+30*x+75)**2+(20+4*x)*ln(3*x**2+30*x+75)+2*x+10),x)

[Out]

1/(4*log(3*x**2 + 30*x + 75) + 4)

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