3.20.15 \(\int \frac {1+x}{(3+x) (1+2 x) \sqrt [3]{1+x^2}} \, dx\)

Optimal. Leaf size=185 \[ \frac {\log \left (5 \sqrt [3]{x^2+1}+\sqrt [3]{10} x-2 \sqrt [3]{10}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (10^{2/3} x^2+25 \left (x^2+1\right )^{2/3}+\left (10 \sqrt [3]{10}-5 \sqrt [3]{10} x\right ) \sqrt [3]{x^2+1}-4\ 10^{2/3} x+4\ 10^{2/3}\right )}{10 \sqrt [3]{10}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+1}}{\sqrt {3}}-\frac {2 \sqrt [3]{2} x}{\sqrt {3} 5^{2/3}}+\frac {4 \sqrt [3]{2}}{\sqrt {3} 5^{2/3}}}{\sqrt [3]{x^2+1}}\right )}{5 \sqrt [3]{10}} \]

________________________________________________________________________________________

Rubi [C]  time = 0.57, 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 = {6742, 757, 429, 444, 55, 617, 204, 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 {\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}}+\frac {\sqrt {3} \tan ^{-1}\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} \tan ^{-1}\left (\frac {2\ 2^{2/3} \sqrt [3]{x^2+1}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )}{10 \sqrt [3]{10}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 55

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 429

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 444

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 617

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 757

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 6742

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} \operatorname {Subst}\left (\int \frac {1}{(-9+x) \sqrt [3]{1+x}} \, dx,x,x^2\right )+\frac {1}{5} \operatorname {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} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{10^{2/3}+\sqrt [3]{10} x+x^2} \, dx,x,\sqrt [3]{1+x^2}\right )-\frac {3 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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*}

________________________________________________________________________________________

Mathematica [F]  time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x}{(3+x) (1+2 x) \sqrt [3]{1+x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(1 + x)/((3 + x)*(1 + 2*x)*(1 + x^2)^(1/3)),x]

[Out]

Integrate[(1 + x)/((3 + x)*(1 + 2*x)*(1 + x^2)^(1/3)), x]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.27, size = 185, normalized size = 1.00 \begin {gather*} \frac {\log \left (5 \sqrt [3]{x^2+1}+\sqrt [3]{10} x-2 \sqrt [3]{10}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (10^{2/3} x^2+25 \left (x^2+1\right )^{2/3}+\left (10 \sqrt [3]{10}-5 \sqrt [3]{10} x\right ) \sqrt [3]{x^2+1}-4\ 10^{2/3} x+4\ 10^{2/3}\right )}{10 \sqrt [3]{10}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+1}}{\sqrt {3}}-\frac {2 \sqrt [3]{2} x}{\sqrt {3} 5^{2/3}}+\frac {4 \sqrt [3]{2}}{\sqrt {3} 5^{2/3}}}{\sqrt [3]{x^2+1}}\right )}{5 \sqrt [3]{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x)/((3 + x)*(1 + 2*x)*(1 + x^2)^(1/3)),x]

[Out]

-1/5*(Sqrt[3]*ArcTan[((4*2^(1/3))/(Sqrt[3]*5^(2/3)) - (2*2^(1/3)*x)/(Sqrt[3]*5^(2/3)) + (1 + x^2)^(1/3)/Sqrt[3
])/(1 + x^2)^(1/3)])/10^(1/3) + Log[-2*10^(1/3) + 10^(1/3)*x + 5*(1 + x^2)^(1/3)]/(5*10^(1/3)) - Log[4*10^(2/3
) - 4*10^(2/3)*x + 10^(2/3)*x^2 + (10*10^(1/3) - 5*10^(1/3)*x)*(1 + x^2)^(1/3) + 25*(1 + x^2)^(2/3)]/(10*10^(1
/3))

________________________________________________________________________________________

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )} {\left (x + 3\right )}}\,{d x} \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)

________________________________________________________________________________________

maple [C]  time = 16.31, size = 1371, normalized size = 7.41 \begin {gather*} \text {Expression too large to display} \end {gather*}

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)

[Out]

RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*ln(-(2389977875*RootOf(RootOf(_Z^3-100)^2+50*_Z*Ro
otOf(_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+4779955750*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*Roo
tOf(_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^2-197503327200*(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+28679734500*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x-418576710000
*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x+395006654400*(x^2+1)^(2/3)
*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2-7900133088*(x^2+1)^(1/3)*RootO
f(_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+31600532352*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2*x+1039600705800*(x^2+1)^(1/3)*RootOf(_Z^3-100)*Ro
otOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x-3059171680*RootOf(_Z^3-100)*x^3+44648182400*RootOf
(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^3-31600532352*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2-1039600
705800*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)[phone]
*RootOf(_Z^3-100)*x^2-1698026186900*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2-1351064779
50*(x^2+1)^(2/3)*x-36710060160*RootOf(_Z^3-100)*x+535778188800*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100
)+2500*_Z^2)*x+270212955900*(x^2+1)^(2/3)[phone]*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)+1/50*RootOf(_Z^3-100)*ln(-(348813925*RootOf(RootOf(_Z^3-
100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^3-59749446875*RootOf(RootOf(_Z^3-100)^2+50*_Z*Ro
otOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^3+697627850*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+
2500*_Z^2)*RootOf(_Z^3-100)^3*x^2-119498893750*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*R
ootOf(_Z^3-100)^2*x^2+98751663600*(x^2+1)^(2/3)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*Ro
otOf(_Z^3-100)^2*x+4185767100*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x
-716993362500*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x-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+3950066544*(x^2+1
)^(1/3)*RootOf(_Z^3-100)^2*x^2+67553238975*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*Root
Of(_Z^3-100)+2500*_Z^2)*x^2-15800266176*(x^2+1)^(1/3)*RootOf(_Z^3-100)^2*x-270212955900*(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+1144109674*RootOf(_Z^3-100)*x^3-1959781
85750*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^3+15800266176*(x^2+1)^(1/3)*RootOf(_Z^3-10
0)^2+270212955900*(x^2+1)^(1/3)*RootOf(_Z^3-100)*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)-1
5585006169*RootOf(_Z^3-100)*x^2+2669605286375*RootOf(RootOf(_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x^2+
129950088225*(x^2+1)^(2/3)*x+13729316088*RootOf(_Z^3-100)*x-2351738229000*RootOf(RootOf(_Z^3-100)^2+50*_Z*Root
Of(_Z^3-100)+2500*_Z^2)*x-259900176450*(x^2+1)^(2/3)-17873225517*RootOf(_Z^3-100)[phone]*RootOf(RootOf(
_Z^3-100)^2+50*_Z*RootOf(_Z^3-100)+2500*_Z^2))/(1+2*x)/(3+x)^2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )} {\left (x + 3\right )}}\,{d x} \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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

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)

________________________________________________________________________________________