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

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

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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} $$\begin{array}{c}5\log (5-\log \left({(x^2+5)}^2\right))-e^{2x}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-20x-e^{2x}(50+10x^2)-e^{2x}(-10-2x^2)\log (25+10x^2+x^4)}{(5+x^2)(5-\log \left({(5+x^2)}^2\right))}dx\\ & =\int (-2e^{2x}+\frac{20x}{(5+x^2)(-5+\log \left({(5+x^2)}^2\right))})dx\\ & =-(2\int e^{2x}dx)+20\int \frac{x}{(5+x^2)(-5+\log \left({(5+x^2)}^2\right))}dx\\ & =-e^{2x}+10\mathrm{Subst}(\int \frac{1}{(5+x)(-5+\log ((5+x{)}^2))}dx,x,x^2)\\ & =-e^{2x}+10\mathrm{Subst}(\int \frac{1}{x(-5+\log \left(x^2\right))}dx,x,5+x^2)\\ & =-e^{2x}+5\mathrm{Subst}(\int \frac{1}{x}dx,x,-5+\log \left({(5+x^2)}^2\right))\\ & =-e^{2x}+5\log (5-\log \left({(5+x^2)}^2\right))\end{array}\end{array}$$

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

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 $$\begin{array}{c}-e^{\left(2x\right)}+5\log (\log (x^4+10x^2+25)-5)\end{array}$$

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 $$\begin{array}{c}-e^{\left(2x\right)}+5\log (\log (x^4+10x^2+25)-5)\end{array}$$

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

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 $$\begin{array}{c}-e^{\left(2x\right)}+5\log (\log (x^2+5)-\frac{5}{2})\end{array}$$

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 $$\begin{array}{c}5\ln (\ln (x^4+10x^2+25)-5)-{\mathrm{e}}^{2x}\end{array}$$

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 $$\begin{array}{c}-e^{2x}+5\log (\log (x^4+10x^2+25)-5)\end{array}$$

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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