3.19.59 \(\int \frac {5 x+22 x^2+13 x^3+2 x^4+(5+4 x+5 x^2+4 x^3) \log (e^x (1+2 x^2+x^4))}{1+x^2} \, dx\)

Optimal. Leaf size=19 \[ x (5+2 x) \log \left (e^x \left (1+x^2\right )^2\right ) \]

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Rubi [B]  time = 0.20, antiderivative size = 39, normalized size of antiderivative = 2.05, number of steps used = 13, number of rules used = 7, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6725, 1802, 635, 203, 260, 2551, 1629} \begin {gather*} \frac {1}{8} (4 x+5)^2 \log \left (e^x \left (x^2+1\right )^2\right )-\frac {25}{4} \log \left (x^2+1\right )-\frac {25 x}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x + 22*x^2 + 13*x^3 + 2*x^4 + (5 + 4*x + 5*x^2 + 4*x^3)*Log[E^x*(1 + 2*x^2 + x^4)])/(1 + x^2),x]

[Out]

(-25*x)/8 - (25*Log[1 + x^2])/4 + ((5 + 4*x)^2*Log[E^x*(1 + x^2)^2])/8

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 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 1629

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

Rule 1802

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

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

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 {x \left (5+22 x+13 x^2+2 x^3\right )}{1+x^2}+(5+4 x) \log \left (e^x \left (1+x^2\right )^2\right )\right ) \, dx\\ &=\int \frac {x \left (5+22 x+13 x^2+2 x^3\right )}{1+x^2} \, dx+\int (5+4 x) \log \left (e^x \left (1+x^2\right )^2\right ) \, dx\\ &=\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {1}{8} \int \frac {(5+4 x)^2 \left (1+4 x+x^2\right )}{1+x^2} \, dx+\int \left (20+13 x+2 x^2-\frac {4 (5+2 x)}{1+x^2}\right ) \, dx\\ &=20 x+\frac {13 x^2}{2}+\frac {2 x^3}{3}+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {1}{8} \int \left (185+104 x+16 x^2-\frac {4 (40-9 x)}{1+x^2}\right ) \, dx-4 \int \frac {5+2 x}{1+x^2} \, dx\\ &=-\frac {25 x}{8}+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )+\frac {1}{2} \int \frac {40-9 x}{1+x^2} \, dx-8 \int \frac {x}{1+x^2} \, dx-20 \int \frac {1}{1+x^2} \, dx\\ &=-\frac {25 x}{8}-20 \tan ^{-1}(x)-4 \log \left (1+x^2\right )+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )-\frac {9}{2} \int \frac {x}{1+x^2} \, dx+20 \int \frac {1}{1+x^2} \, dx\\ &=-\frac {25 x}{8}-\frac {25}{4} \log \left (1+x^2\right )+\frac {1}{8} (5+4 x)^2 \log \left (e^x \left (1+x^2\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 19, normalized size = 1.00 \begin {gather*} x (5+2 x) \log \left (e^x \left (1+x^2\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x + 22*x^2 + 13*x^3 + 2*x^4 + (5 + 4*x + 5*x^2 + 4*x^3)*Log[E^x*(1 + 2*x^2 + x^4)])/(1 + x^2),x]

[Out]

x*(5 + 2*x)*Log[E^x*(1 + x^2)^2]

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fricas [A]  time = 0.73, size = 24, normalized size = 1.26 \begin {gather*} {\left (2 \, x^{2} + 5 \, x\right )} \log \left ({\left (x^{4} + 2 \, x^{2} + 1\right )} e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+5*x^2+4*x+5)*log((x^4+2*x^2+1)*exp(x))+2*x^4+13*x^3+22*x^2+5*x)/(x^2+1),x, algorithm="fricas
")

[Out]

(2*x^2 + 5*x)*log((x^4 + 2*x^2 + 1)*e^x)

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giac [A]  time = 0.25, size = 28, normalized size = 1.47 \begin {gather*} 2 \, x^{3} + 5 \, x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+5*x^2+4*x+5)*log((x^4+2*x^2+1)*exp(x))+2*x^4+13*x^3+22*x^2+5*x)/(x^2+1),x, algorithm="giac")

[Out]

2*x^3 + 5*x^2 + 2*(2*x^2 + 5*x)*log(x^2 + 1)

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maple [B]  time = 0.40, size = 38, normalized size = 2.00

method	result	size
default	\(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x\)	\(38\)
norman	\(2 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x^{2}+5 \ln \left (\left (x^{4}+2 x^{2}+1\right ) {\mathrm e}^{x}\right ) x\)	\(38\)
risch	\(\left (2 x^{2}+5 x \right ) \ln \left ({\mathrm e}^{x}\right )+4 x^{2} \ln \left (x^{2}+1\right )+10 x \ln \left (x^{2}+1\right )+2 i \pi \,x^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )\right ) \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )^{2}+\frac {5 i \pi x \,\mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}}{2}-\frac {5 i \pi x \mathrm {csgn}\left (i \left (x^{2}+1\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )}{2}+5 i \pi x \,\mathrm {csgn}\left (i \left (x^{2}+1\right )\right ) \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )^{2}+i \pi \,x^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}-\frac {5 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )}{2}-i \pi \,x^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )-i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{3}+i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}-i \pi \,x^{2} \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )^{3}+\frac {5 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{2}}{2}-\frac {5 i \pi x \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right )^{3}}{2}-\frac {5 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )^{3}}{2}-i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i \left (x^{2}+1\right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x} \left (x^{2}+1\right )^{2}\right )\)	\(417\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+5*x^2+4*x+5)*ln((x^4+2*x^2+1)*exp(x))+2*x^4+13*x^3+22*x^2+5*x)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

2*ln((x^4+2*x^2+1)*exp(x))*x^2+5*ln((x^4+2*x^2+1)*exp(x))*x

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maxima [B]  time = 0.64, size = 37, normalized size = 1.95 \begin {gather*} 2 \, x^{3} + 5 \, x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x + 2\right )} \log \left (x^{2} + 1\right ) - 4 \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+5*x^2+4*x+5)*log((x^4+2*x^2+1)*exp(x))+2*x^4+13*x^3+22*x^2+5*x)/(x^2+1),x, algorithm="maxima
")

[Out]

2*x^3 + 5*x^2 + 2*(2*x^2 + 5*x + 2)*log(x^2 + 1) - 4*log(x^2 + 1)

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mupad [B]  time = 1.32, size = 23, normalized size = 1.21 \begin {gather*} \left (2\,x^2+5\,x\right )\,\left (x+\ln \left (x^4+2\,x^2+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(exp(x)*(2*x^2 + x^4 + 1))*(4*x + 5*x^2 + 4*x^3 + 5) + 22*x^2 + 13*x^3 + 2*x^4)/(x^2 + 1),x)

[Out]

(5*x + 2*x^2)*(x + log(2*x^2 + x^4 + 1))

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sympy [A]  time = 0.38, size = 34, normalized size = 1.79 \begin {gather*} - x + \left (2 x^{2} + 5 x + 1\right ) \log {\left (\left (x^{4} + 2 x^{2} + 1\right ) e^{x} \right )} - 2 \log {\left (x^{2} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+5*x**2+4*x+5)*ln((x**4+2*x**2+1)*exp(x))+2*x**4+13*x**3+22*x**2+5*x)/(x**2+1),x)

[Out]

-x + (2*x**2 + 5*x + 1)*log((x**4 + 2*x**2 + 1)*exp(x)) - 2*log(x**2 + 1)

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