3.91.85 \(\int \frac {43+16 x+15 x^2}{1+16 x+15 x^2} \, dx\)

Optimal. Leaf size=21 \[ x+3 \log \left (\frac {\left (3+\frac {1}{5 x}\right ) x}{1+x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1657, 616, 31} \begin {gather*} x-3 \log (x+1)+3 \log (15 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(43 + 16*x + 15*x^2)/(1 + 16*x + 15*x^2),x]

[Out]

x - 3*Log[1 + x] + 3*Log[1 + 15*x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {42}{1+16 x+15 x^2}\right ) \, dx\\ &=x+42 \int \frac {1}{1+16 x+15 x^2} \, dx\\ &=x+45 \int \frac {1}{1+15 x} \, dx-45 \int \frac {1}{15+15 x} \, dx\\ &=x-3 \log (1+x)+3 \log (1+15 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} x-3 \log (1+x)+3 \log (1+15 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(43 + 16*x + 15*x^2)/(1 + 16*x + 15*x^2),x]

[Out]

x - 3*Log[1 + x] + 3*Log[1 + 15*x]

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fricas [A]  time = 0.52, size = 16, normalized size = 0.76 \begin {gather*} x + 3 \, \log \left (15 \, x + 1\right ) - 3 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2+16*x+43)/(15*x^2+16*x+1),x, algorithm="fricas")

[Out]

x + 3*log(15*x + 1) - 3*log(x + 1)

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giac [A]  time = 0.12, size = 18, normalized size = 0.86 \begin {gather*} x + 3 \, \log \left ({\left | 15 \, x + 1 \right |}\right ) - 3 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2+16*x+43)/(15*x^2+16*x+1),x, algorithm="giac")

[Out]

x + 3*log(abs(15*x + 1)) - 3*log(abs(x + 1))

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maple [A]  time = 0.56, size = 17, normalized size = 0.81

method	result	size
default	\(x -3 \ln \left (x +1\right )+3 \ln \left (15 x +1\right )\)	\(17\)
norman	\(x -3 \ln \left (x +1\right )+3 \ln \left (15 x +1\right )\)	\(17\)
risch	\(x -3 \ln \left (x +1\right )+3 \ln \left (15 x +1\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x^2+16*x+43)/(15*x^2+16*x+1),x,method=_RETURNVERBOSE)

[Out]

x-3*ln(x+1)+3*ln(15*x+1)

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maxima [A]  time = 0.35, size = 16, normalized size = 0.76 \begin {gather*} x + 3 \, \log \left (15 \, x + 1\right ) - 3 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2+16*x+43)/(15*x^2+16*x+1),x, algorithm="maxima")

[Out]

x + 3*log(15*x + 1) - 3*log(x + 1)

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mupad [B]  time = 7.21, size = 10, normalized size = 0.48 \begin {gather*} x-6\,\mathrm {atanh}\left (\frac {15\,x}{7}+\frac {8}{7}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + 15*x^2 + 43)/(16*x + 15*x^2 + 1),x)

[Out]

x - 6*atanh((15*x)/7 + 8/7)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.71 \begin {gather*} x + 3 \log {\left (x + \frac {1}{15} \right )} - 3 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x**2+16*x+43)/(15*x**2+16*x+1),x)

[Out]

x + 3*log(x + 1/15) - 3*log(x + 1)

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