3.52.94 \(\int \frac {4+(5 x+5 x^3+e^x (10 x^2+5 x^3+10 x^4+5 x^5)) \log ^2(\frac {x^2}{1+x^2})}{(5 x+5 x^3) \log ^2(\frac {x^2}{1+x^2})} \, dx\)

Optimal. Leaf size=25 \[ x+e^x x^2-\frac {2}{5 \log \left (\frac {x}{\frac {1}{x}+x}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.68, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 6, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {1593, 6688, 2196, 2176, 2194, 6686} \begin {gather*} e^x x^2-\frac {2}{5 \log \left (\frac {x^2}{x^2+1}\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + (5*x + 5*x^3 + E^x*(10*x^2 + 5*x^3 + 10*x^4 + 5*x^5))*Log[x^2/(1 + x^2)]^2)/((5*x + 5*x^3)*Log[x^2/(1
+ x^2)]^2),x]

[Out]

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

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$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 2196

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 {4+\left (5 x+5 x^3+e^x \left (10 x^2+5 x^3+10 x^4+5 x^5\right )\right ) \log ^2\left (\frac {x^2}{1+x^2}\right )}{x \left (5+5 x^2\right ) \log ^2\left (\frac {x^2}{1+x^2}\right )} \, dx\\ &=\int \left (1+e^x x (2+x)+\frac {4}{5 x \left (1+x^2\right ) \log ^2\left (\frac {x^2}{1+x^2}\right )}\right ) \, dx\\ &=x+\frac {4}{5} \int \frac {1}{x \left (1+x^2\right ) \log ^2\left (\frac {x^2}{1+x^2}\right )} \, dx+\int e^x x (2+x) \, dx\\ &=x-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )}+\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=x-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )}+2 \int e^x x \, dx+\int e^x x^2 \, dx\\ &=x+2 e^x x+e^x x^2-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )}-2 \int e^x \, dx-2 \int e^x x \, dx\\ &=-2 e^x+x+e^x x^2-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )}+2 \int e^x \, dx\\ &=x+e^x x^2-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 27, normalized size = 1.08 \begin {gather*} x+e^x x^2-\frac {2}{5 \log \left (\frac {x^2}{1+x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + (5*x + 5*x^3 + E^x*(10*x^2 + 5*x^3 + 10*x^4 + 5*x^5))*Log[x^2/(1 + x^2)]^2)/((5*x + 5*x^3)*Log[
x^2/(1 + x^2)]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 40, normalized size = 1.60 \begin {gather*} \frac {5 \, {\left (x^{2} e^{x} + x\right )} \log \left (\frac {x^{2}}{x^{2} + 1}\right ) - 2}{5 \, \log \left (\frac {x^{2}}{x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^5+10*x^4+5*x^3+10*x^2)*exp(x)+5*x^3+5*x)*log(x^2/(x^2+1))^2+4)/(5*x^3+5*x)/log(x^2/(x^2+1))^2
,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.35, size = 52, normalized size = 2.08 \begin {gather*} \frac {5 \, x^{2} e^{x} \log \left (\frac {x^{2}}{x^{2} + 1}\right ) + 5 \, x \log \left (\frac {x^{2}}{x^{2} + 1}\right ) - 2}{5 \, \log \left (\frac {x^{2}}{x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^5+10*x^4+5*x^3+10*x^2)*exp(x)+5*x^3+5*x)*log(x^2/(x^2+1))^2+4)/(5*x^3+5*x)/log(x^2/(x^2+1))^2
,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.23, size = 25, normalized size = 1.00

method	result	size
default	\(x -\frac {2}{5 \ln \left (\frac {x^{2}}{x^{2}+1}\right )}+{\mathrm e}^{x} x^{2}\)	\(25\)
risch	\({\mathrm e}^{x} x^{2}+x -\frac {4 i}{5 \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}+1}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{x^{2}+1}\right )-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{x^{2}+1}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}+1}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{x^{2}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x^{2}}{x^{2}+1}\right )^{3}+4 i \ln \relax (x )-2 i \ln \left (x^{2}+1\right )\right )}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^5+10*x^4+5*x^3+10*x^2)*exp(x)+5*x^3+5*x)*ln(x^2/(x^2+1))^2+4)/(5*x^3+5*x)/ln(x^2/(x^2+1))^2,x,metho
d=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 23, normalized size = 0.92 \begin {gather*} x^{2} e^{x} + x + \frac {2}{5 \, {\left (\log \left (x^{2} + 1\right ) - 2 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^5+10*x^4+5*x^3+10*x^2)*exp(x)+5*x^3+5*x)*log(x^2/(x^2+1))^2+4)/(5*x^3+5*x)/log(x^2/(x^2+1))^2
,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.34, size = 27, normalized size = 1.08 \begin {gather*} x-\frac {2}{5\,\left (\ln \left (x^2\right )-\ln \left (x^2+1\right )\right )}+x^2\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2/(x^2 + 1))^2*(5*x + exp(x)*(10*x^2 + 5*x^3 + 10*x^4 + 5*x^5) + 5*x^3) + 4)/(log(x^2/(x^2 + 1))^2*
(5*x + 5*x^3)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.37, size = 20, normalized size = 0.80 \begin {gather*} x^{2} e^{x} + x - \frac {2}{5 \log {\left (\frac {x^{2}}{x^{2} + 1} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x**5+10*x**4+5*x**3+10*x**2)*exp(x)+5*x**3+5*x)*ln(x**2/(x**2+1))**2+4)/(5*x**3+5*x)/ln(x**2/(x
**2+1))**2,x)

[Out]

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

________________________________________________________________________________________