3.41.41 \(\int \frac {20 x+e^{2 x} (50+10 x^2)+e^{2 x} (-10-2 x^2) \log (25+10 x^2+x^4)}{-25-5 x^2+(5+x^2) \log (25+10 x^2+x^4)} \, dx\)

Optimal. Leaf size=23 \[ -e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \]

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Rubi [A]  time = 0.56, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6741, 6725, 2194, 2475, 2390, 2302, 29} \begin {gather*} 5 \log \left (5-\log \left (\left (x^2+5\right )^2\right )\right )-e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*x + E^(2*x)*(50 + 10*x^2) + E^(2*x)*(-10 - 2*x^2)*Log[25 + 10*x^2 + x^4])/(-25 - 5*x^2 + (5 + x^2)*Log
[25 + 10*x^2 + x^4]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x-e^{2 x} \left (50+10 x^2\right )-e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{\left (5+x^2\right ) \left (5-\log \left (\left (5+x^2\right )^2\right )\right )} \, dx\\ &=\int \left (-2 e^{2 x}+\frac {20 x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )}\right ) \, dx\\ &=-\left (2 \int e^{2 x} \, dx\right )+20 \int \frac {x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )} \, dx\\ &=-e^{2 x}+10 \operatorname {Subst}\left (\int \frac {1}{(5+x) \left (-5+\log \left ((5+x)^2\right )\right )} \, dx,x,x^2\right )\\ &=-e^{2 x}+10 \operatorname {Subst}\left (\int \frac {1}{x \left (-5+\log \left (x^2\right )\right )} \, dx,x,5+x^2\right )\\ &=-e^{2 x}+5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log \left (\left (5+x^2\right )^2\right )\right )\\ &=-e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 23, normalized size = 1.00 \begin {gather*} -e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x + E^(2*x)*(50 + 10*x^2) + E^(2*x)*(-10 - 2*x^2)*Log[25 + 10*x^2 + x^4])/(-25 - 5*x^2 + (5 + x^
2)*Log[25 + 10*x^2 + x^4]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 23, normalized size = 1.00 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-10)*exp(2*x)*log(x^4+10*x^2+25)+(10*x^2+50)*exp(2*x)+20*x)/((x^2+5)*log(x^4+10*x^2+25)-5*x^
2-25),x, algorithm="fricas")

[Out]

-e^(2*x) + 5*log(log(x^4 + 10*x^2 + 25) - 5)

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giac [A]  time = 0.17, size = 23, normalized size = 1.00 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-10)*exp(2*x)*log(x^4+10*x^2+25)+(10*x^2+50)*exp(2*x)+20*x)/((x^2+5)*log(x^4+10*x^2+25)-5*x^
2-25),x, algorithm="giac")

[Out]

-e^(2*x) + 5*log(log(x^4 + 10*x^2 + 25) - 5)

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maple [A]  time = 0.17, size = 24, normalized size = 1.04

method	result	size
default	\(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\)	\(24\)
norman	\(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\)	\(24\)
risch	\(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{2}+5\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \left (x^{2}+5\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (x^{2}+5\right )\right ) \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (x^{2}+5\right )^{2}\right )^{3}-10 i\right )}{4}\right )\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-10)*exp(2*x)*ln(x^4+10*x^2+25)+(10*x^2+50)*exp(2*x)+20*x)/((x^2+5)*ln(x^4+10*x^2+25)-5*x^2-25),x,
method=_RETURNVERBOSE)

[Out]

-exp(2*x)+5*ln(ln(x^4+10*x^2+25)-5)

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maxima [A]  time = 0.42, size = 18, normalized size = 0.78 \begin {gather*} -e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{2} + 5\right ) - \frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-10)*exp(2*x)*log(x^4+10*x^2+25)+(10*x^2+50)*exp(2*x)+20*x)/((x^2+5)*log(x^4+10*x^2+25)-5*x^
2-25),x, algorithm="maxima")

[Out]

-e^(2*x) + 5*log(log(x^2 + 5) - 5/2)

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mupad [B]  time = 3.26, size = 23, normalized size = 1.00 \begin {gather*} 5\,\ln \left (\ln \left (x^4+10\,x^2+25\right )-5\right )-{\mathrm {e}}^{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + exp(2*x)*(10*x^2 + 50) - exp(2*x)*log(10*x^2 + x^4 + 25)*(2*x^2 + 10))/(5*x^2 - log(10*x^2 + x^4
+ 25)*(x^2 + 5) + 25),x)

[Out]

5*log(log(10*x^2 + x^4 + 25) - 5) - exp(2*x)

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sympy [A]  time = 0.39, size = 20, normalized size = 0.87 \begin {gather*} - e^{2 x} + 5 \log {\left (\log {\left (x^{4} + 10 x^{2} + 25 \right )} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-10)*exp(2*x)*ln(x**4+10*x**2+25)+(10*x**2+50)*exp(2*x)+20*x)/((x**2+5)*ln(x**4+10*x**2+25)
-5*x**2-25),x)

[Out]

-exp(2*x) + 5*log(log(x**4 + 10*x**2 + 25) - 5)

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