3.24.37 \(\int \frac {1+x}{(3+x) (1+2 x) \sqrt [3]{1+x^2}} \, dx\) [2337]
Optimal. Leaf size=185 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\frac {4 \sqrt [3]{2}}{\sqrt {3} 5^{2/3}}-\frac {2 \sqrt [3]{2} x}{\sqrt {3} 5^{2/3}}+\frac {\sqrt [3]{1+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x^2}}\right )}{5 \sqrt [3]{10}}+\frac {\log \left (-2 \sqrt [3]{10}+\sqrt [3]{10} x+5 \sqrt [3]{1+x^2}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (4\ 10^{2/3}-4\ 10^{2/3} x+10^{2/3} x^2+\left (10 \sqrt [3]{10}-5 \sqrt [3]{10} x\right ) \sqrt [3]{1+x^2}+25 \left (1+x^2\right )^{2/3}\right )}{10 \sqrt [3]{10}} \]
[Out]
-1/50*3^(1/2)*arctan((4/15*2^(1/3)*3^(1/2)*5^(1/3)-2/15*2^(1/3)*x*3^(1/2)*5^(1/3)+1/3*(x^2+1)^(1/3)*3^(1/2))/(
x^2+1)^(1/3))*10^(2/3)+1/50*ln(-2*10^(1/3)+10^(1/3)*x+5*(x^2+1)^(1/3))*10^(2/3)-1/100*ln(4*10^(2/3)-4*10^(2/3)
*x+10^(2/3)*x^2+(10*10^(1/3)-5*10^(1/3)*x)*(x^2+1)^(1/3)+25*(x^2+1)^(2/3))*10^(2/3)
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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in
optimal.
time = 0.42, antiderivative size = 238, normalized size of antiderivative = 1.29, number of steps
used = 18, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6874, 771,
440, 455, 57, 631, 210, 31} \begin {gather*} \frac {2}{15} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};\frac {x^2}{9},-x^2\right )+\frac {1}{5} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,-x^2\right )+\frac {\sqrt {3} \text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{x^2+1}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )}{5 \sqrt [3]{10}}+\frac {\sqrt {3} \text {ArcTan}\left (\frac {2\ 2^{2/3} \sqrt [3]{x^2+1}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )}{10 \sqrt [3]{10}}-\frac {\log \left (1-4 x^2\right )}{20 \sqrt [3]{10}}-\frac {\log \left (9-x^2\right )}{10 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-2 \sqrt [3]{x^2+1}\right )}{20 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-\sqrt [3]{x^2+1}\right )}{10 \sqrt [3]{10}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(1 + x)/((3 + x)*(1 + 2*x)*(1 + x^2)^(1/3)),x]
[Out]
(2*x*AppellF1[1/2, 1, 1/3, 3/2, x^2/9, -x^2])/15 + (x*AppellF1[1/2, 1, 1/3, 3/2, 4*x^2, -x^2])/5 + (Sqrt[3]*Ar
cTan[(5^(1/3) + 2^(2/3)*(1 + x^2)^(1/3))/(Sqrt[3]*5^(1/3))])/(5*10^(1/3)) + (Sqrt[3]*ArcTan[(5^(1/3) + 2*2^(2/
3)*(1 + x^2)^(1/3))/(Sqrt[3]*5^(1/3))])/(10*10^(1/3)) - Log[1 - 4*x^2]/(20*10^(1/3)) - Log[9 - x^2]/(10*10^(1/
3)) + (3*Log[10^(1/3) - 2*(1 + x^2)^(1/3)])/(20*10^(1/3)) + (3*Log[10^(1/3) - (1 + x^2)^(1/3)])/(10*10^(1/3))
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 57
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]
Rule 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 440
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 455
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]
Rule 631
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 771
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - e*(x/(d^2 - e^2*x^2)))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]
Rule 6874
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]
Rubi steps
\begin {align*} \int \frac {1+x}{(3+x) (1+2 x) \sqrt [3]{1+x^2}} \, dx &=\int \left (\frac {2}{5 (3+x) \sqrt [3]{1+x^2}}+\frac {1}{5 (1+2 x) \sqrt [3]{1+x^2}}\right ) \, dx\\ &=\frac {1}{5} \int \frac {1}{(1+2 x) \sqrt [3]{1+x^2}} \, dx+\frac {2}{5} \int \frac {1}{(3+x) \sqrt [3]{1+x^2}} \, dx\\ &=\frac {1}{5} \int \left (\frac {1}{\left (1-4 x^2\right ) \sqrt [3]{1+x^2}}+\frac {2 x}{\sqrt [3]{1+x^2} \left (-1+4 x^2\right )}\right ) \, dx+\frac {2}{5} \int \left (-\frac {3}{\left (-9+x^2\right ) \sqrt [3]{1+x^2}}+\frac {x}{\left (-9+x^2\right ) \sqrt [3]{1+x^2}}\right ) \, dx\\ &=\frac {1}{5} \int \frac {1}{\left (1-4 x^2\right ) \sqrt [3]{1+x^2}} \, dx+\frac {2}{5} \int \frac {x}{\left (-9+x^2\right ) \sqrt [3]{1+x^2}} \, dx+\frac {2}{5} \int \frac {x}{\sqrt [3]{1+x^2} \left (-1+4 x^2\right )} \, dx-\frac {6}{5} \int \frac {1}{\left (-9+x^2\right ) \sqrt [3]{1+x^2}} \, dx\\ &=\frac {2}{15} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};\frac {x^2}{9},-x^2\right )+\frac {1}{5} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,-x^2\right )+\frac {1}{5} \text {Subst}\left (\int \frac {1}{(-9+x) \sqrt [3]{1+x}} \, dx,x,x^2\right )+\frac {1}{5} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1+x} (-1+4 x)} \, dx,x,x^2\right )\\ &=\frac {2}{15} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};\frac {x^2}{9},-x^2\right )+\frac {1}{5} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,-x^2\right )-\frac {\log \left (1-4 x^2\right )}{20 \sqrt [3]{10}}-\frac {\log \left (9-x^2\right )}{10 \sqrt [3]{10}}+\frac {3}{40} \text {Subst}\left (\int \frac {1}{\frac {5^{2/3}}{2 \sqrt [3]{2}}+\frac {\sqrt [3]{5} x}{2^{2/3}}+x^2} \, dx,x,\sqrt [3]{1+x^2}\right )+\frac {3}{10} \text {Subst}\left (\int \frac {1}{10^{2/3}+\sqrt [3]{10} x+x^2} \, dx,x,\sqrt [3]{1+x^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{5}}{2^{2/3}}-x} \, dx,x,\sqrt [3]{1+x^2}\right )}{20 \sqrt [3]{10}}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt [3]{10}-x} \, dx,x,\sqrt [3]{1+x^2}\right )}{10 \sqrt [3]{10}}\\ &=\frac {2}{15} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};\frac {x^2}{9},-x^2\right )+\frac {1}{5} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,-x^2\right )-\frac {\log \left (1-4 x^2\right )}{20 \sqrt [3]{10}}-\frac {\log \left (9-x^2\right )}{10 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-2 \sqrt [3]{1+x^2}\right )}{20 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-\sqrt [3]{1+x^2}\right )}{10 \sqrt [3]{10}}-\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2\ 2^{2/3} \sqrt [3]{1+x^2}}{\sqrt [3]{5}}\right )}{10 \sqrt [3]{10}}-\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{1+x^2}}{\sqrt [3]{5}}\right )}{5 \sqrt [3]{10}}\\ &=\frac {2}{15} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};\frac {x^2}{9},-x^2\right )+\frac {1}{5} x F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,-x^2\right )+\frac {\sqrt {3} \tan ^{-1}\left (\frac {5+10^{2/3} \sqrt [3]{1+x^2}}{5 \sqrt {3}}\right )}{5 \sqrt [3]{10}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {5+2\ 10^{2/3} \sqrt [3]{1+x^2}}{5 \sqrt {3}}\right )}{10 \sqrt [3]{10}}-\frac {\log \left (1-4 x^2\right )}{20 \sqrt [3]{10}}-\frac {\log \left (9-x^2\right )}{10 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-2 \sqrt [3]{1+x^2}\right )}{20 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-\sqrt [3]{1+x^2}\right )}{10 \sqrt [3]{10}}\\ \end {align*}
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Mathematica [A]
time = 0.23, size = 148, normalized size = 0.80 \begin {gather*} -\frac {2 \sqrt {3} \text {ArcTan}\left (\frac {4 \sqrt [3]{10}-2 \sqrt [3]{10} x+5 \sqrt [3]{1+x^2}}{5 \sqrt {3} \sqrt [3]{1+x^2}}\right )-2 \log \left (-2 \sqrt [3]{10}+\sqrt [3]{10} x+5 \sqrt [3]{1+x^2}\right )+\log \left (4\ 10^{2/3}-4\ 10^{2/3} x+10^{2/3} x^2-5 \sqrt [3]{10} (-2+x) \sqrt [3]{1+x^2}+25 \left (1+x^2\right )^{2/3}\right )}{10 \sqrt [3]{10}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(1 + x)/((3 + x)*(1 + 2*x)*(1 + x^2)^(1/3)),x]
[Out]
-1/10*(2*Sqrt[3]*ArcTan[(4*10^(1/3) - 2*10^(1/3)*x + 5*(1 + x^2)^(1/3))/(5*Sqrt[3]*(1 + x^2)^(1/3))] - 2*Log[-
2*10^(1/3) + 10^(1/3)*x + 5*(1 + x^2)^(1/3)] + Log[4*10^(2/3) - 4*10^(2/3)*x + 10^(2/3)*x^2 - 5*10^(1/3)*(-2 +
x)*(1 + x^2)^(1/3) + 25*(1 + x^2)^(2/3)])/10^(1/3)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 16.22, size = 2047, normalized size = 11.06
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| method |
result |
size |
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| trager |
\(\text {Expression too large to display}\) |
\(2047\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1+x)/(3+x)/(1+2*x)/(x^2+1)^(1/3),x,method=_RETURNVERBOSE)
[Out]
-1/50*ln(-(-3087605725*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^3-3488
1392500*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^3+197503327200*(x^2
+1)^(2/3)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2*x-6175211450*RootOf(R
ootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^2-69762785000*RootOf(RootOf(_Z^3-100)
^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^2-395006654400*(x^2+1)^(2/3)*RootOf(RootOf(_Z^3-10
0)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2-2702129559*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2*x^2+2599
00176450*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2-370512
68700*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x-418576710000*RootOf(Roo
tOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x+10808518236*(x^2+1)^(1/3)*RootOf(_Z^3
-100)^2*x-1039600705800*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_
Z^2)*x-10127346778*RootOf(_Z^3-100)*x^3-114410967400*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^
2)*x^3+259900176450*(x^2+1)^(2/3)*x-10808518236*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2+1039600705800*(x^2+1)^(1/3)*R
ootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)+137954223793*RootOf(_Z^3-100)*x^2+
1558500616900*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2-519800352900*(x^2+1)^(2/3)-12152
8161336*RootOf(_Z^3-100)*x-1372931608800*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x+1582089
17349*RootOf(_Z^3-100)+1787322551700*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2))/(1+2*x)/(3+x
)^2)*RootOf(_Z^3-100)-ln(-(-3087605725*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3
-100)^3*x^3-34881392500*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^3+1
97503327200*(x^2+1)^(2/3)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2*x-617
5211450*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^2-69762785000*RootOf(
RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^2-395006654400*(x^2+1)^(2/3)*RootO
f(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2-2702129559*(x^2+1)^(1/3)*RootOf(_Z^3
-100)^2*x^2+259900176450*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*
_Z^2)*x^2-37051268700*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x-4185767
10000*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x+10808518236*(x^2+1)^(
1/3)*RootOf(_Z^3-100)^2*x-1039600705800*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(
_Z^3-100)+2500*_Z^2)*x-10127346778*RootOf(_Z^3-100)*x^3-114410967400*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z
^3-100)+2500*_Z^2)*x^3+259900176450*(x^2+1)^(2/3)*x-10808518236*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2+1039600705800
*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)+137954223793*RootO
f(_Z^3-100)*x^2+1558500616900*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2-519800352900*(x^
2+1)^(2/3)-121528161336*RootOf(_Z^3-100)*x-1372931608800*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500
*_Z^2)*x+158208917349*RootOf(_Z^3-100)+1787322551700*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^
2))/(1+2*x)/(3+x)^2)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)+1/50*RootOf(_Z^3-100)*ln((308
7605725*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^3+119498893750*RootOf
(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^3+197503327200*(x^2+1)^(2/3)*Root
Of(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2*x+6175211450*RootOf(RootOf(_Z^3-100
)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^2+238997787500*RootOf(RootOf(_Z^3-100)^2+50*_Z*Root
Of(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^2-395006654400*(x^2+1)^(2/3)*RootOf(RootOf(_Z^3-100)^2+50*_Z*Ro
otOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2-5198003529*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2*x^2+135106477950*(x^2
+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2+37051268700*RootOf(
RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x+1433986725000*RootOf(RootOf(_Z^3-100
)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x+20792014116*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2*x-54
0425911800*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x-395213
5328*RootOf(_Z^3-100)*x^3-152958584000*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^3+1351064
77950*(x^2+1)^(2/3)*x-20792014116*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2+540425911800*(x^2+1)^(1/3)*RootOf(_Z^3-100)
*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-10...
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(3+x)/(1+2*x)/(x^2+1)^(1/3),x, algorithm="maxima")
[Out]
integrate((x + 1)/((x^2 + 1)^(1/3)*(2*x + 1)*(x + 3)), x)
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(3+x)/(1+2*x)/(x^2+1)^(1/3),x, algorithm="fricas")
[Out]
Exception raised: TypeError >> Error detected within library code: integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\left (x + 3\right ) \left (2 x + 1\right ) \sqrt [3]{x^{2} + 1}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(3+x)/(1+2*x)/(x**2+1)**(1/3),x)
[Out]
Integral((x + 1)/((x + 3)*(2*x + 1)*(x**2 + 1)**(1/3)), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(3+x)/(1+2*x)/(x^2+1)^(1/3),x, algorithm="giac")
[Out]
integrate((x + 1)/((x^2 + 1)^(1/3)*(2*x + 1)*(x + 3)), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x+1}{\left (2\,x+1\right )\,{\left (x^2+1\right )}^{1/3}\,\left (x+3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x + 1)/((2*x + 1)*(x^2 + 1)^(1/3)*(x + 3)),x)
[Out]
int((x + 1)/((2*x + 1)*(x^2 + 1)^(1/3)*(x + 3)), x)
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