Integrand size = 20, antiderivative size = 468 \[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\frac {1}{6} \sqrt [3]{\frac {1}{2} \left (-5+3 \sqrt {3}\right )} \arctan \left (\frac {-\frac {2^{2/3}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x-\frac {\sqrt [3]{1+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x^2}}\right )-\frac {1}{6} \sqrt [3]{\frac {1}{2} \left (5+3 \sqrt {3}\right )} \arctan \left (\frac {\frac {2^{2/3}}{\sqrt {3}}+\frac {1}{3} 2^{2/3} x+\frac {\sqrt [3]{1+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x^2}}\right )+\frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (3\ 2^{2/3}+2^{2/3} \sqrt {3} x-6 \sqrt [3]{1+x^2}\right )}{6\ 3^{2/3}}+\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (-3 2^{2/3}+2^{2/3} \sqrt {3} x+6 \sqrt [3]{1+x^2}\right )}{6\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9-5 \sqrt {3}\right )} \log \left (-3 \sqrt [3]{2}+2 \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{1+x^2}+2^{2/3} \sqrt {3} x \sqrt [3]{1+x^2}-6 \left (1+x^2\right )^{2/3}\right )}{12\ 3^{2/3}}-\frac {\sqrt [3]{\frac {1}{2} \left (9+5 \sqrt {3}\right )} \log \left (3 \sqrt [3]{2}+2 \sqrt [3]{2} \sqrt {3} x+\sqrt [3]{2} x^2+3\ 2^{2/3} \sqrt [3]{1+x^2}+2^{2/3} \sqrt {3} x \sqrt [3]{1+x^2}+6 \left (1+x^2\right )^{2/3}\right )}{12\ 3^{2/3}} \]
1/6*(-5/2+3/2*3^(1/2))^(1/3)*arctan((-1/3*2^(2/3)*3^(1/2)+1/3*2^(2/3)*x-1/ 3*(x^2+1)^(1/3)*3^(1/2))/(x^2+1)^(1/3))-1/6*(5/2+3/2*3^(1/2))^(1/3)*arctan ((1/3*2^(2/3)*3^(1/2)+1/3*2^(2/3)*x+1/3*(x^2+1)^(1/3)*3^(1/2))/(x^2+1)^(1/ 3))+1/18*(9/2+5/2*3^(1/2))^(1/3)*ln(3*2^(2/3)+2^(2/3)*x*3^(1/2)-6*(x^2+1)^ (1/3))*3^(1/3)+1/18*(9/2-5/2*3^(1/2))^(1/3)*ln(-3*2^(2/3)+2^(2/3)*x*3^(1/2 )+6*(x^2+1)^(1/3))*3^(1/3)-1/36*(9/2-5/2*3^(1/2))^(1/3)*ln(-3*2^(1/3)+2*2^ (1/3)*x*3^(1/2)-2^(1/3)*x^2-3*2^(2/3)*(x^2+1)^(1/3)+2^(2/3)*3^(1/2)*x*(x^2 +1)^(1/3)-6*(x^2+1)^(2/3))*3^(1/3)-1/36*(9/2+5/2*3^(1/2))^(1/3)*ln(3*2^(1/ 3)+2*2^(1/3)*x*3^(1/2)+2^(1/3)*x^2+3*2^(2/3)*(x^2+1)^(1/3)+2^(2/3)*3^(1/2) *x*(x^2+1)^(1/3)+6*(x^2+1)^(2/3))*3^(1/3)
Time = 5.60 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.87 \[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\frac {6 \sqrt [3]{-5+3 \sqrt {3}} \arctan \left (\frac {2^{2/3} x-\sqrt {3} \left (2^{2/3}+\sqrt [3]{1+x^2}\right )}{3 \sqrt [3]{1+x^2}}\right )-6 \sqrt [3]{5+3 \sqrt {3}} \arctan \left (\frac {2^{2/3} x+\sqrt {3} \left (2^{2/3}+\sqrt [3]{1+x^2}\right )}{3 \sqrt [3]{1+x^2}}\right )+\sqrt [3]{3} \left (2 \sqrt [3]{9+5 \sqrt {3}} \log \left (3\ 2^{2/3}+2^{2/3} \sqrt {3} x-6 \sqrt [3]{1+x^2}\right )+2 \sqrt [3]{9-5 \sqrt {3}} \log \left (-3 2^{2/3}+2^{2/3} \sqrt {3} x+6 \sqrt [3]{1+x^2}\right )-\sqrt [3]{9-5 \sqrt {3}} \log \left (-\sqrt [3]{2} x^2+\sqrt [3]{2} \sqrt {3} x \left (2+\sqrt [3]{2} \sqrt [3]{1+x^2}\right )-3 \left (\sqrt [3]{2}+2^{2/3} \sqrt [3]{1+x^2}+2 \left (1+x^2\right )^{2/3}\right )\right )-\sqrt [3]{9+5 \sqrt {3}} \log \left (\sqrt [3]{2} x^2+\sqrt [3]{2} \sqrt {3} x \left (2+\sqrt [3]{2} \sqrt [3]{1+x^2}\right )+3 \left (\sqrt [3]{2}+2^{2/3} \sqrt [3]{1+x^2}+2 \left (1+x^2\right )^{2/3}\right )\right )\right )}{36 \sqrt [3]{2}} \]
(6*(-5 + 3*Sqrt[3])^(1/3)*ArcTan[(2^(2/3)*x - Sqrt[3]*(2^(2/3) + (1 + x^2) ^(1/3)))/(3*(1 + x^2)^(1/3))] - 6*(5 + 3*Sqrt[3])^(1/3)*ArcTan[(2^(2/3)*x + Sqrt[3]*(2^(2/3) + (1 + x^2)^(1/3)))/(3*(1 + x^2)^(1/3))] + 3^(1/3)*(2*( 9 + 5*Sqrt[3])^(1/3)*Log[3*2^(2/3) + 2^(2/3)*Sqrt[3]*x - 6*(1 + x^2)^(1/3) ] + 2*(9 - 5*Sqrt[3])^(1/3)*Log[-3*2^(2/3) + 2^(2/3)*Sqrt[3]*x + 6*(1 + x^ 2)^(1/3)] - (9 - 5*Sqrt[3])^(1/3)*Log[-(2^(1/3)*x^2) + 2^(1/3)*Sqrt[3]*x*( 2 + 2^(1/3)*(1 + x^2)^(1/3)) - 3*(2^(1/3) + 2^(2/3)*(1 + x^2)^(1/3) + 2*(1 + x^2)^(2/3))] - (9 + 5*Sqrt[3])^(1/3)*Log[2^(1/3)*x^2 + 2^(1/3)*Sqrt[3]* x*(2 + 2^(1/3)*(1 + x^2)^(1/3)) + 3*(2^(1/3) + 2^(2/3)*(1 + x^2)^(1/3) + 2 *(1 + x^2)^(2/3))]))/(36*2^(1/3))
Time = 0.28 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.41, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1343, 25, 304, 353, 67, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x+1}{\left (x^2-3\right ) \sqrt [3]{x^2+1}} \, dx\) |
| \(\Big \downarrow \) 1343 |
| \(\displaystyle \int -\frac {1}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx+\int -\frac {x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx-\int \frac {x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx\) |
| \(\Big \downarrow \) 304 |
| \(\displaystyle -\int \frac {x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx-\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle -\frac {1}{2} \int \frac {1}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx^2-\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {1}{2} \left (-\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{x^2+1}}d\sqrt [3]{x^2+1}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2}}d\sqrt [3]{x^2+1}-\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}\right )-\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2}}d\sqrt [3]{x^2+1}-\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )-\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{x^2+1}+1\right )}{2^{2/3}}-\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )-\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}+\frac {1}{2} \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )+\frac {\arctan (x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\) |
ArcTan[x]/(6*2^(2/3)) - ArcTan[x/(1 + 2^(1/3)*(1 + x^2)^(1/3))]/(2*2^(2/3) ) + ArcTanh[Sqrt[3]/x]/(2*2^(2/3)*Sqrt[3]) + ArcTanh[(Sqrt[3]*(1 - 2^(1/3) *(1 + x^2)^(1/3)))/x]/(2*2^(2/3)*Sqrt[3]) + ((Sqrt[3]*ArcTan[(1 + 2^(1/3)* (1 + x^2)^(1/3))/Sqrt[3]])/2^(2/3) - Log[3 - x^2]/(2*2^(2/3)) + (3*Log[2^( 2/3) - (1 + x^2)^(1/3)])/(2*2^(2/3)))/2
3.31.56.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[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]
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])
Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit h[{q = Rt[b/a, 2]}, Simp[q*(ArcTanh[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/ 3)*d)), x] + (-Simp[q*(ArcTan[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a + b*x^2)^ (1/3))]/(2*2^(2/3)*a^(1/3)*d)), x] + Simp[q*(ArcTan[q*x]/(6*2^(2/3)*a^(1/3) *d)), x] + Simp[q*(ArcTanh[Sqrt[3]*((a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3))/( a^(1/3)*q*x))]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && PosQ[b/a]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q _), x_Symbol] :> Simp[g Int[(a + c*x^2)^p*(d + f*x^2)^q, x], x] + Simp[h Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h, p, q}, x]
\[\int \frac {1+x}{\left (x^{2}-3\right ) \left (x^{2}+1\right )^{\frac {1}{3}}}d x\]
Exception generated. \[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
\[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int \frac {x + 1}{\left (x^{2} - 3\right ) \sqrt [3]{x^{2} + 1}}\, dx \]
\[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int { \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}} \,d x } \]
\[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int { \frac {x + 1}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}} \,d x } \]
Timed out. \[ \int \frac {1+x}{\left (-3+x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int \frac {x+1}{{\left (x^2+1\right )}^{1/3}\,\left (x^2-3\right )} \,d x \]