3.23.67 \(\int \frac {32 x+60 x^2-8 x^3+e^4 (16 x+32 x^2)+(34+34 x^2+e^4 (16+16 x^2)) \log (1+x^2)}{64-16 x+65 x^2-16 x^3+x^4+e^8 (16+16 x^2)+e^4 (64-8 x+64 x^2-8 x^3)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\left (4+\frac {2}{x}\right ) x \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x} \]

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Rubi [B]  time = 0.74, antiderivative size = 172, normalized size of antiderivative = 6.14, number of steps used = 14, number of rules used = 8, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6688, 6725, 1629, 635, 203, 260, 2463, 801} \begin {gather*} \frac {2 \left (17+8 e^4\right ) \log \left (x^2+1\right )}{4 \left (2+e^4\right )-x}-\frac {8 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (x^2+1\right )}{65+64 e^4+16 e^8}+\frac {4 \left (3+2 e^4\right ) \log \left (x^2+1\right )}{65+64 e^4+16 e^8}-\frac {16 \left (34+33 e^4+8 e^8\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}+\frac {16 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*x + 60*x^2 - 8*x^3 + E^4*(16*x + 32*x^2) + (34 + 34*x^2 + E^4*(16 + 16*x^2))*Log[1 + x^2])/(64 - 16*x
+ 65*x^2 - 16*x^3 + x^4 + E^8*(16 + 16*x^2) + E^4*(64 - 8*x + 64*x^2 - 8*x^3)),x]

[Out]

(16*(2 + E^4)*(17 + 8*E^4)*Log[4*(2 + E^4) - x])/(65 + 64*E^4 + 16*E^8) - (16*(34 + 33*E^4 + 8*E^8)*Log[4*(2 +
E^4) - x])/(65 + 64*E^4 + 16*E^8) + (4*(3 + 2*E^4)*Log[1 + x^2])/(65 + 64*E^4 + 16*E^8) - (8*(2 + E^4)*(17 +
8*E^4)*Log[1 + x^2])/(65 + 64*E^4 + 16*E^8) + (2*(17 + 8*E^4)*Log[1 + x^2])/(4*(2 + E^4) - x)

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
+ g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -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 {4 x \left (8+15 x-2 x^2+e^4 (4+8 x)\right )+2 \left (17+8 e^4\right ) \left (1+x^2\right ) \log \left (1+x^2\right )}{\left (8+4 e^4-x\right )^2 \left (1+x^2\right )} \, dx\\ &=\int \left (\frac {4 x (1+2 x)}{\left (8+4 e^4-x\right ) \left (1+x^2\right )}+\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{\left (8+4 e^4-x\right )^2}\right ) \, dx\\ &=4 \int \frac {x (1+2 x)}{\left (8+4 e^4-x\right ) \left (1+x^2\right )} \, dx+\left (2 \left (17+8 e^4\right )\right ) \int \frac {\log \left (1+x^2\right )}{\left (8+4 e^4-x\right )^2} \, dx\\ &=\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x}+4 \int \left (\frac {4 \left (34+33 e^4+8 e^8\right )}{\left (65+64 e^4+16 e^8\right ) \left (8+4 e^4-x\right )}+\frac {-17-8 e^4+2 \left (3+2 e^4\right ) x}{\left (65+64 e^4+16 e^8\right ) \left (1+x^2\right )}\right ) \, dx-\left (4 \left (17+8 e^4\right )\right ) \int \frac {x}{\left (8+4 e^4-x\right ) \left (1+x^2\right )} \, dx\\ &=-\frac {16 \left (34+33 e^4+8 e^8\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}+\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x}-\left (4 \left (17+8 e^4\right )\right ) \int \left (\frac {4 \left (2+e^4\right )}{\left (65+64 e^4+16 e^8\right ) \left (8+4 e^4-x\right )}+\frac {-1+4 \left (2+e^4\right ) x}{\left (65+64 e^4+16 e^8\right ) \left (1+x^2\right )}\right ) \, dx+\frac {4 \int \frac {-17-8 e^4+2 \left (3+2 e^4\right ) x}{1+x^2} \, dx}{65+64 e^4+16 e^8}\\ &=\frac {16 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}-\frac {16 \left (34+33 e^4+8 e^8\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}+\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x}+\frac {\left (8 \left (3+2 e^4\right )\right ) \int \frac {x}{1+x^2} \, dx}{65+64 e^4+16 e^8}-\frac {\left (4 \left (17+8 e^4\right )\right ) \int \frac {1}{1+x^2} \, dx}{65+64 e^4+16 e^8}-\frac {\left (4 \left (17+8 e^4\right )\right ) \int \frac {-1+4 \left (2+e^4\right ) x}{1+x^2} \, dx}{65+64 e^4+16 e^8}\\ &=-\frac {4 \left (17+8 e^4\right ) \tan ^{-1}(x)}{65+64 e^4+16 e^8}+\frac {16 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}-\frac {16 \left (34+33 e^4+8 e^8\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}+\frac {4 \left (3+2 e^4\right ) \log \left (1+x^2\right )}{65+64 e^4+16 e^8}+\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x}+\frac {\left (4 \left (17+8 e^4\right )\right ) \int \frac {1}{1+x^2} \, dx}{65+64 e^4+16 e^8}-\frac {\left (16 \left (2+e^4\right ) \left (17+8 e^4\right )\right ) \int \frac {x}{1+x^2} \, dx}{65+64 e^4+16 e^8}\\ &=\frac {16 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}-\frac {16 \left (34+33 e^4+8 e^8\right ) \log \left (4 \left (2+e^4\right )-x\right )}{65+64 e^4+16 e^8}+\frac {4 \left (3+2 e^4\right ) \log \left (1+x^2\right )}{65+64 e^4+16 e^8}-\frac {8 \left (2+e^4\right ) \left (17+8 e^4\right ) \log \left (1+x^2\right )}{65+64 e^4+16 e^8}+\frac {2 \left (17+8 e^4\right ) \log \left (1+x^2\right )}{4 \left (2+e^4\right )-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 25, normalized size = 0.89 \begin {gather*} \frac {2 (1+2 x) \log \left (1+x^2\right )}{8+4 e^4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x + 60*x^2 - 8*x^3 + E^4*(16*x + 32*x^2) + (34 + 34*x^2 + E^4*(16 + 16*x^2))*Log[1 + x^2])/(64 -
16*x + 65*x^2 - 16*x^3 + x^4 + E^8*(16 + 16*x^2) + E^4*(64 - 8*x + 64*x^2 - 8*x^3)),x]

[Out]

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

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fricas [A]  time = 0.79, size = 22, normalized size = 0.79 \begin {gather*} -\frac {2 \, {\left (2 \, x + 1\right )} \log \left (x^{2} + 1\right )}{x - 4 \, e^{4} - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+16)*exp(4)+34*x^2+34)*log(x^2+1)+(32*x^2+16*x)*exp(4)-8*x^3+60*x^2+32*x)/((16*x^2+16)*exp(
4)^2+(-8*x^3+64*x^2-8*x+64)*exp(4)+x^4-16*x^3+65*x^2-16*x+64),x, algorithm="fricas")

[Out]

-2*(2*x + 1)*log(x^2 + 1)/(x - 4*e^4 - 8)

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giac [A]  time = 0.43, size = 27, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (2 \, x \log \left (x^{2} + 1\right ) + \log \left (x^{2} + 1\right )\right )}}{x - 4 \, e^{4} - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+16)*exp(4)+34*x^2+34)*log(x^2+1)+(32*x^2+16*x)*exp(4)-8*x^3+60*x^2+32*x)/((16*x^2+16)*exp(
4)^2+(-8*x^3+64*x^2-8*x+64)*exp(4)+x^4-16*x^3+65*x^2-16*x+64),x, algorithm="giac")

[Out]

-2*(2*x*log(x^2 + 1) + log(x^2 + 1))/(x - 4*e^4 - 8)

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maple [A]  time = 0.25, size = 31, normalized size = 1.11

method	result	size
norman	\(\frac {2 \ln \left (x^{2}+1\right )+4 x \ln \left (x^{2}+1\right )}{8+4 \,{\mathrm e}^{4}-x}\)	\(31\)
risch	\(\frac {2 \left (8 \,{\mathrm e}^{4}+17\right ) \ln \left (x^{2}+1\right )}{8+4 \,{\mathrm e}^{4}-x}-4 \ln \left (x^{2}+1\right )\)	\(35\)
default	\(-\frac {128 \ln \left (-4 \,{\mathrm e}^{4}+x -8\right ) \left ({\mathrm e}^{4}\right )^{2}}{16 \left ({\mathrm e}^{4}\right )^{2}+64 \,{\mathrm e}^{4}+65}-\frac {528 \ln \left (-4 \,{\mathrm e}^{4}+x -8\right ) {\mathrm e}^{4}}{16 \left ({\mathrm e}^{4}\right )^{2}+64 \,{\mathrm e}^{4}+65}-\frac {544 \ln \left (-4 \,{\mathrm e}^{4}+x -8\right )}{16 \left ({\mathrm e}^{4}\right )^{2}+64 \,{\mathrm e}^{4}+65}+\frac {8 \ln \left (x^{2}+1\right ) {\mathrm e}^{4}}{16 \,{\mathrm e}^{8}+64 \,{\mathrm e}^{4}+65}-\frac {32 \arctan \relax (x ) {\mathrm e}^{4}}{16 \,{\mathrm e}^{8}+64 \,{\mathrm e}^{4}+65}+\frac {12 \ln \left (x^{2}+1\right )}{16 \,{\mathrm e}^{8}+64 \,{\mathrm e}^{4}+65}-\frac {68 \arctan \relax (x )}{16 \,{\mathrm e}^{8}+64 \,{\mathrm e}^{4}+65}+\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{2}+\left (-8 \,{\mathrm e}^{4}-16\right ) \textit {\_Z} +64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}+64\right )}{\sum }\left (\frac {\underline {\hspace {1.25 ex}}\alpha }{\left ({\mathrm e}^{4}\right )^{2}-{\mathrm e}^{8}}-\frac {4 \left (2+{\mathrm e}^{4}\right )}{\left ({\mathrm e}^{4}\right )^{2}-{\mathrm e}^{8}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}+1\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha -\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha +\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\dilog \left (\frac {-\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha -\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\dilog \left (\frac {\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha +\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )\right )\right ) {\mathrm e}^{4}}{2}+\frac {17 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{2}+\left (-8 \,{\mathrm e}^{4}-16\right ) \textit {\_Z} +64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}+64\right )}{\sum }\left (\frac {\underline {\hspace {1.25 ex}}\alpha }{\left ({\mathrm e}^{4}\right )^{2}-{\mathrm e}^{8}}-\frac {4 \left (2+{\mathrm e}^{4}\right )}{\left ({\mathrm e}^{4}\right )^{2}-{\mathrm e}^{8}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}+1\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha -\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha +\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\dilog \left (\frac {-\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha -\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )-\dilog \left (\frac {\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}+x}{\underline {\hspace {1.25 ex}}\alpha +\sqrt {\underline {\hspace {1.25 ex}}\alpha ^{2}-8 \underline {\hspace {1.25 ex}}\alpha \,{\mathrm e}^{4}+64 \,{\mathrm e}^{4}+16 \,{\mathrm e}^{8}-16 \underline {\hspace {1.25 ex}}\alpha +63}}\right )\right )\right )}{16}\)	\(792\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^2+16)*exp(4)+34*x^2+34)*ln(x^2+1)+(32*x^2+16*x)*exp(4)-8*x^3+60*x^2+32*x)/((16*x^2+16)*exp(4)^2+(-
8*x^3+64*x^2-8*x+64)*exp(4)+x^4-16*x^3+65*x^2-16*x+64),x,method=_RETURNVERBOSE)

[Out]

(2*ln(x^2+1)+4*x*ln(x^2+1))/(8+4*exp(4)-x)

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maxima [B]  time = 1.95, size = 763, normalized size = 27.25 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2+16)*exp(4)+34*x^2+34)*log(x^2+1)+(32*x^2+16*x)*exp(4)-8*x^3+60*x^2+32*x)/((16*x^2+16)*exp(
4)^2+(-8*x^3+64*x^2-8*x+64)*exp(4)+x^4-16*x^3+65*x^2-16*x+64),x, algorithm="maxima")

[Out]

-32*((16*e^8 + 64*e^4 + 63)*arctan(x)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) + 4*(e^4 + 2)*log(x^
2 + 1)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) - 8*(e^4 + 2)*log(x - 4*e^4 - 8)/(256*e^16 + 2048*e
^12 + 6176*e^8 + 8320*e^4 + 4225) + 16*(e^8 + 4*e^4 + 4)/(x*(16*e^8 + 64*e^4 + 65) - 64*e^12 - 384*e^8 - 772*e
^4 - 520))*e^4 - 8*(16*(e^4 + 2)*arctan(x)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) - (16*e^8 + 64*
e^4 + 63)*log(x^2 + 1)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) + 2*(16*e^8 + 64*e^4 + 63)*log(x -
4*e^4 - 8)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) + 8*(e^4 + 2)/(x*(16*e^8 + 64*e^4 + 65) - 64*e^
12 - 384*e^8 - 772*e^4 - 520))*e^4 - 60*(16*e^8 + 64*e^4 + 63)*arctan(x)/(256*e^16 + 2048*e^12 + 6176*e^8 + 83
20*e^4 + 4225) + 4*(8*e^4 + 17)*arctan(x)/(16*e^8 + 64*e^4 + 65) - 320*(e^4 + 2)*arctan(x)/(256*e^16 + 2048*e^
12 + 6176*e^8 + 8320*e^4 + 4225) + 20*(16*e^8 + 64*e^4 + 63)*log(x^2 + 1)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8
320*e^4 + 4225) - 8*(8*e^8 + 33*e^4 + 34)*log(x^2 + 1)/(16*e^8 + 64*e^4 + 65) - 2*(8*e^4 + 17)*log(x^2 + 1)/(x
- 4*e^4 - 8) - 240*(e^4 + 2)*log(x^2 + 1)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) - 128*(16*e^16
+ 128*e^12 + 387*e^8 + 524*e^4 + 268)*log(x - 4*e^4 - 8)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) -
32*(16*e^8 + 64*e^4 + 63)*log(x - 4*e^4 - 8)/(256*e^16 + 2048*e^12 + 6176*e^8 + 8320*e^4 + 4225) + 16*(8*e^8
+ 33*e^4 + 34)*log(x - 4*e^4 - 8)/(16*e^8 + 64*e^4 + 65) + 480*(e^4 + 2)*log(x - 4*e^4 - 8)/(256*e^16 + 2048*e
^12 + 6176*e^8 + 8320*e^4 + 4225) + 512*(e^12 + 6*e^8 + 12*e^4 + 8)/(x*(16*e^8 + 64*e^4 + 65) - 64*e^12 - 384*
e^8 - 772*e^4 - 520) - 960*(e^8 + 4*e^4 + 4)/(x*(16*e^8 + 64*e^4 + 65) - 64*e^12 - 384*e^8 - 772*e^4 - 520) -
128*(e^4 + 2)/(x*(16*e^8 + 64*e^4 + 65) - 64*e^12 - 384*e^8 - 772*e^4 - 520)

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mupad [B]  time = 1.61, size = 24, normalized size = 0.86 \begin {gather*} \frac {2\,\ln \left (x^2+1\right )\,\left (2\,x+1\right )}{4\,{\mathrm {e}}^4-x+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + exp(4)*(16*x + 32*x^2) + 60*x^2 - 8*x^3 + log(x^2 + 1)*(exp(4)*(16*x^2 + 16) + 34*x^2 + 34))/(exp(
8)*(16*x^2 + 16) - 16*x - exp(4)*(8*x - 64*x^2 + 8*x^3 - 64) + 65*x^2 - 16*x^3 + x^4 + 64),x)

[Out]

(2*log(x^2 + 1)*(2*x + 1))/(4*exp(4) - x + 8)

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sympy [A]  time = 0.24, size = 31, normalized size = 1.11 \begin {gather*} - 4 \log {\left (x^{2} + 1 \right )} + \frac {\left (- 16 e^{4} - 34\right ) \log {\left (x^{2} + 1 \right )}}{x - 4 e^{4} - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**2+16)*exp(4)+34*x**2+34)*ln(x**2+1)+(32*x**2+16*x)*exp(4)-8*x**3+60*x**2+32*x)/((16*x**2+16
)*exp(4)**2+(-8*x**3+64*x**2-8*x+64)*exp(4)+x**4-16*x**3+65*x**2-16*x+64),x)

[Out]

-4*log(x**2 + 1) + (-16*exp(4) - 34)*log(x**2 + 1)/(x - 4*exp(4) - 8)

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