3.96.71 \(\int \frac {-1+192 x+6 x^2+15 x^4+(-4-4 x^2) \log (1+x^2)}{2+2 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{2} \left (5 x^3+(24-x) \left (1+4 \log \left (1+x^2\right )\right )\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6725, 1810, 635, 203, 260, 2448, 321} \begin {gather*} \frac {5 x^3}{2}-2 x \log \left (x^2+1\right )+48 \log \left (x^2+1\right )-\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 192*x + 6*x^2 + 15*x^4 + (-4 - 4*x^2)*Log[1 + x^2])/(2 + 2*x^2),x]

[Out]

-1/2*x + (5*x^3)/2 + 48*Log[1 + x^2] - 2*x*Log[1 + x^2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1810

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

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1+192 x+6 x^2+15 x^4}{2 \left (1+x^2\right )}-2 \log \left (1+x^2\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {-1+192 x+6 x^2+15 x^4}{1+x^2} \, dx-2 \int \log \left (1+x^2\right ) \, dx\\ &=-2 x \log \left (1+x^2\right )+\frac {1}{2} \int \left (-9+15 x^2+\frac {8 (1+24 x)}{1+x^2}\right ) \, dx+4 \int \frac {x^2}{1+x^2} \, dx\\ &=-\frac {x}{2}+\frac {5 x^3}{2}-2 x \log \left (1+x^2\right )-4 \int \frac {1}{1+x^2} \, dx+4 \int \frac {1+24 x}{1+x^2} \, dx\\ &=-\frac {x}{2}+\frac {5 x^3}{2}-4 \tan ^{-1}(x)-2 x \log \left (1+x^2\right )+4 \int \frac {1}{1+x^2} \, dx+96 \int \frac {x}{1+x^2} \, dx\\ &=-\frac {x}{2}+\frac {5 x^3}{2}+48 \log \left (1+x^2\right )-2 x \log \left (1+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.03, size = 48, normalized size = 1.85 \begin {gather*} \frac {1}{2} \left (-x+5 x^3-8 \tan ^{-1}(x)+(96-4 i) \log (i-x)+(96+4 i) \log (i+x)-4 x \log \left (1+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 192*x + 6*x^2 + 15*x^4 + (-4 - 4*x^2)*Log[1 + x^2])/(2 + 2*x^2),x]

[Out]

(-x + 5*x^3 - 8*ArcTan[x] + (96 - 4*I)*Log[I - x] + (96 + 4*I)*Log[I + x] - 4*x*Log[1 + x^2])/2

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fricas [A]  time = 0.64, size = 20, normalized size = 0.77 \begin {gather*} \frac {5}{2} \, x^{3} - 2 \, {\left (x - 24\right )} \log \left (x^{2} + 1\right ) - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-4)*log(x^2+1)+15*x^4+6*x^2+192*x-1)/(2*x^2+2),x, algorithm="fricas")

[Out]

5/2*x^3 - 2*(x - 24)*log(x^2 + 1) - 1/2*x

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giac [A]  time = 0.18, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{2} \, x^{3} - 2 \, x \log \left (x^{2} + 1\right ) - \frac {1}{2} \, x + 48 \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-4)*log(x^2+1)+15*x^4+6*x^2+192*x-1)/(2*x^2+2),x, algorithm="giac")

[Out]

5/2*x^3 - 2*x*log(x^2 + 1) - 1/2*x + 48*log(x^2 + 1)

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maple [A]  time = 0.07, size = 27, normalized size = 1.04

method	result	size
default	\(\frac {5 x^{3}}{2}-\frac {x}{2}+48 \ln \left (x^{2}+1\right )-2 x \ln \left (x^{2}+1\right )\)	\(27\)
norman	\(\frac {5 x^{3}}{2}-\frac {x}{2}+48 \ln \left (x^{2}+1\right )-2 x \ln \left (x^{2}+1\right )\)	\(27\)
risch	\(\frac {5 x^{3}}{2}-\frac {x}{2}+48 \ln \left (x^{2}+1\right )-2 x \ln \left (x^{2}+1\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2-4)*ln(x^2+1)+15*x^4+6*x^2+192*x-1)/(2*x^2+2),x,method=_RETURNVERBOSE)

[Out]

5/2*x^3-1/2*x+48*ln(x^2+1)-2*x*ln(x^2+1)

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maxima [A]  time = 0.54, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{2} \, x^{3} - 2 \, x \log \left (x^{2} + 1\right ) - \frac {1}{2} \, x + 48 \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-4)*log(x^2+1)+15*x^4+6*x^2+192*x-1)/(2*x^2+2),x, algorithm="maxima")

[Out]

5/2*x^3 - 2*x*log(x^2 + 1) - 1/2*x + 48*log(x^2 + 1)

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mupad [B]  time = 0.20, size = 27, normalized size = 1.04 \begin {gather*} 48\,\ln \left (x^2+1\right )-x\,\left (2\,\ln \left (x^2+1\right )+\frac {1}{2}\right )+\frac {5\,x^3}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((192*x - log(x^2 + 1)*(4*x^2 + 4) + 6*x^2 + 15*x^4 - 1)/(2*x^2 + 2),x)

[Out]

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

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sympy [A]  time = 0.13, size = 27, normalized size = 1.04 \begin {gather*} \frac {5 x^{3}}{2} - 2 x \log {\left (x^{2} + 1 \right )} - \frac {x}{2} + 48 \log {\left (x^{2} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2-4)*ln(x**2+1)+15*x**4+6*x**2+192*x-1)/(2*x**2+2),x)

[Out]

5*x**3/2 - 2*x*log(x**2 + 1) - x/2 + 48*log(x**2 + 1)

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