3.102.73 \(\int \frac {34 x^2+20 x^3+14 x^4+10 x^5+e^x (-12+12 x+11 x^2+4 x^3+5 x^4)+(12+4 x^2) \log (\frac {1}{4} (3+x^2))}{3 x^2+x^4} \, dx\)

Optimal. Leaf size=33 \[ (4+5 x) \left (2+\frac {e^x}{x}+x-\frac {\log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x}\right ) \]

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Rubi [A]  time = 0.90, antiderivative size = 47, normalized size of antiderivative = 1.42, number of steps used = 18, number of rules used = 12, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1593, 6725, 2199, 2194, 2177, 2178, 6688, 1810, 635, 203, 260, 2455} \begin {gather*} 5 x^2-5 \log \left (x^2+3\right )-\frac {4 \log \left (\frac {x^2}{4}+\frac {3}{4}\right )}{x}+14 x+5 e^x+\frac {4 e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(34*x^2 + 20*x^3 + 14*x^4 + 10*x^5 + E^x*(-12 + 12*x + 11*x^2 + 4*x^3 + 5*x^4) + (12 + 4*x^2)*Log[(3 + x^2
)/4])/(3*x^2 + x^4),x]

[Out]

5*E^x + (4*E^x)/x + 14*x + 5*x^2 - (4*Log[3/4 + x^2/4])/x - 5*Log[3 + 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 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 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 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2455

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {34 x^2+20 x^3+14 x^4+10 x^5+e^x \left (-12+12 x+11 x^2+4 x^3+5 x^4\right )+\left (12+4 x^2\right ) \log \left (\frac {1}{4} \left (3+x^2\right )\right )}{x^2 \left (3+x^2\right )} \, dx\\ &=\int \left (\frac {e^x \left (-4+4 x+5 x^2\right )}{x^2}+\frac {2 \left (17 x^2+10 x^3+7 x^4+5 x^5+6 \log \left (\frac {1}{4} \left (3+x^2\right )\right )+2 x^2 \log \left (\frac {1}{4} \left (3+x^2\right )\right )\right )}{x^2 \left (3+x^2\right )}\right ) \, dx\\ &=2 \int \frac {17 x^2+10 x^3+7 x^4+5 x^5+6 \log \left (\frac {1}{4} \left (3+x^2\right )\right )+2 x^2 \log \left (\frac {1}{4} \left (3+x^2\right )\right )}{x^2 \left (3+x^2\right )} \, dx+\int \frac {e^x \left (-4+4 x+5 x^2\right )}{x^2} \, dx\\ &=2 \int \left (\frac {17+10 x+7 x^2+5 x^3}{3+x^2}+\frac {2 \log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x^2}\right ) \, dx+\int \left (5 e^x-\frac {4 e^x}{x^2}+\frac {4 e^x}{x}\right ) \, dx\\ &=2 \int \frac {17+10 x+7 x^2+5 x^3}{3+x^2} \, dx-4 \int \frac {e^x}{x^2} \, dx+4 \int \frac {e^x}{x} \, dx+4 \int \frac {\log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x^2} \, dx+5 \int e^x \, dx\\ &=5 e^x+\frac {4 e^x}{x}+4 \text {Ei}(x)-\frac {4 \log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x}+2 \int \frac {1}{\frac {3}{4}+\frac {x^2}{4}} \, dx+2 \int \left (7+5 x-\frac {4+5 x}{3+x^2}\right ) \, dx-4 \int \frac {e^x}{x} \, dx\\ &=5 e^x+\frac {4 e^x}{x}+14 x+5 x^2+\frac {8 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {4 \log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x}-2 \int \frac {4+5 x}{3+x^2} \, dx\\ &=5 e^x+\frac {4 e^x}{x}+14 x+5 x^2+\frac {8 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {4 \log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x}-8 \int \frac {1}{3+x^2} \, dx-10 \int \frac {x}{3+x^2} \, dx\\ &=5 e^x+\frac {4 e^x}{x}+14 x+5 x^2-\frac {4 \log \left (\frac {3}{4}+\frac {x^2}{4}\right )}{x}-5 \log \left (3+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.27, size = 107, normalized size = 3.24 \begin {gather*} 14 x+5 x^2+\frac {e^x (4+5 x)}{x}+\frac {8 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} i \left (15 i+4 \sqrt {3}\right ) \log \left (\sqrt {3}+i x\right )-\frac {1}{3} \left (15+4 i \sqrt {3}\right ) \log \left (i \sqrt {3}+x\right )-\frac {4 \log \left (\frac {1}{4} \left (3+x^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(34*x^2 + 20*x^3 + 14*x^4 + 10*x^5 + E^x*(-12 + 12*x + 11*x^2 + 4*x^3 + 5*x^4) + (12 + 4*x^2)*Log[(3
+ x^2)/4])/(3*x^2 + x^4),x]

[Out]

14*x + 5*x^2 + (E^x*(4 + 5*x))/x + (8*ArcTan[x/Sqrt[3]])/Sqrt[3] + (I/3)*(15*I + 4*Sqrt[3])*Log[Sqrt[3] + I*x]
- ((15 + (4*I)*Sqrt[3])*Log[I*Sqrt[3] + x])/3 - (4*Log[(3 + x^2)/4])/x

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fricas [A]  time = 0.53, size = 38, normalized size = 1.15 \begin {gather*} \frac {5 \, x^{3} + 14 \, x^{2} + {\left (5 \, x + 4\right )} e^{x} - {\left (5 \, x + 4\right )} \log \left (\frac {1}{4} \, x^{2} + \frac {3}{4}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+12)*log(3/4+1/4*x^2)+(5*x^4+4*x^3+11*x^2+12*x-12)*exp(x)+10*x^5+14*x^4+20*x^3+34*x^2)/(x^4+3
*x^2),x, algorithm="fricas")

[Out]

(5*x^3 + 14*x^2 + (5*x + 4)*e^x - (5*x + 4)*log(1/4*x^2 + 3/4))/x

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giac [A]  time = 0.16, size = 43, normalized size = 1.30 \begin {gather*} \frac {5 \, x^{3} + 14 \, x^{2} + 5 \, x e^{x} - 5 \, x \log \left (x^{2} + 3\right ) + 4 \, e^{x} - 4 \, \log \left (\frac {1}{4} \, x^{2} + \frac {3}{4}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+12)*log(3/4+1/4*x^2)+(5*x^4+4*x^3+11*x^2+12*x-12)*exp(x)+10*x^5+14*x^4+20*x^3+34*x^2)/(x^4+3
*x^2),x, algorithm="giac")

[Out]

(5*x^3 + 14*x^2 + 5*x*e^x - 5*x*log(x^2 + 3) + 4*e^x - 4*log(1/4*x^2 + 3/4))/x

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maple [A]  time = 0.21, size = 47, normalized size = 1.42

method	result	size
default	\(\frac {4 \,{\mathrm e}^{x}}{x}+5 \,{\mathrm e}^{x}-\frac {4 \ln \left (x^{2}+3\right )}{x}+\frac {8 \ln \relax (2)}{x}+5 x^{2}+14 x -5 \ln \left (x^{2}+3\right )\)	\(47\)
risch	\(-\frac {4 \ln \left (\frac {3}{4}+\frac {x^{2}}{4}\right )}{x}-\frac {-5 x^{3}+5 \ln \left (x^{2}+3\right ) x -14 x^{2}-5 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}}{x}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+12)*ln(3/4+1/4*x^2)+(5*x^4+4*x^3+11*x^2+12*x-12)*exp(x)+10*x^5+14*x^4+20*x^3+34*x^2)/(x^4+3*x^2),x
,method=_RETURNVERBOSE)

[Out]

4*exp(x)/x+5*exp(x)-4/x*ln(x^2+3)+8*ln(2)/x+5*x^2+14*x-5*ln(x^2+3)

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maxima [A]  time = 0.48, size = 42, normalized size = 1.27 \begin {gather*} 5 \, x^{2} + 14 \, x + \frac {{\left (5 \, x + 4\right )} e^{x} + 8 \, \log \relax (2) - 4 \, \log \left (x^{2} + 3\right )}{x} - 5 \, \log \left (x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+12)*log(3/4+1/4*x^2)+(5*x^4+4*x^3+11*x^2+12*x-12)*exp(x)+10*x^5+14*x^4+20*x^3+34*x^2)/(x^4+3
*x^2),x, algorithm="maxima")

[Out]

5*x^2 + 14*x + ((5*x + 4)*e^x + 8*log(2) - 4*log(x^2 + 3))/x - 5*log(x^2 + 3)

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mupad [B]  time = 6.98, size = 41, normalized size = 1.24 \begin {gather*} 14\,x-5\,\ln \left (x^2+3\right )-\frac {4\,\ln \left (\frac {x^2}{4}+\frac {3}{4}\right )}{x}+5\,x^2+\frac {{\mathrm {e}}^x\,\left (5\,x+4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(12*x + 11*x^2 + 4*x^3 + 5*x^4 - 12) + 34*x^2 + 20*x^3 + 14*x^4 + 10*x^5 + log(x^2/4 + 3/4)*(4*x^2
+ 12))/(3*x^2 + x^4),x)

[Out]

14*x - 5*log(x^2 + 3) - (4*log(x^2/4 + 3/4))/x + 5*x^2 + (exp(x)*(5*x + 4))/x

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sympy [A]  time = 0.37, size = 39, normalized size = 1.18 \begin {gather*} 5 x^{2} + 14 x - 5 \log {\left (x^{2} + 3 \right )} + \frac {\left (5 x + 4\right ) e^{x}}{x} - \frac {4 \log {\left (\frac {x^{2}}{4} + \frac {3}{4} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+12)*ln(3/4+1/4*x**2)+(5*x**4+4*x**3+11*x**2+12*x-12)*exp(x)+10*x**5+14*x**4+20*x**3+34*x**2
)/(x**4+3*x**2),x)

[Out]

5*x**2 + 14*x - 5*log(x**2 + 3) + (5*x + 4)*exp(x)/x - 4*log(x**2/4 + 3/4)/x

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