Integrand size = 24, antiderivative size = 198 \[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=-\frac {a \arctan (x)}{6\ 2^{2/3}}+\frac {a \arctan \left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1+x^2}}\right )}{2\ 2^{2/3}}-\frac {\sqrt {3} b \arctan \left (\frac {1+\sqrt [3]{2} \sqrt [3]{1+x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {a \text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {b \log \left (3-x^2\right )}{4\ 2^{2/3}}-\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1+x^2}\right )}{4\ 2^{2/3}} \]
-1/12*a*arctan(x)*2^(1/3)+1/4*a*arctan(x/(1+2^(1/3)*(x^2+1)^(1/3)))*2^(1/3 )+1/8*b*ln(-x^2+3)*2^(1/3)-3/8*b*ln(2^(2/3)-(x^2+1)^(1/3))*2^(1/3)-1/12*a* arctanh(3^(1/2)/x)*2^(1/3)*3^(1/2)-1/12*a*arctanh((1-2^(1/3)*(x^2+1)^(1/3) )*3^(1/2)/x)*2^(1/3)*3^(1/2)-1/4*b*arctan(1/3*(1+2^(1/3)*(x^2+1)^(1/3))*3^ (1/2))*3^(1/2)*2^(1/3)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 10.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.77 \[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=\frac {1}{6} b x^2 \operatorname {AppellF1}\left (1,\frac {1}{3},1,2,-x^2,\frac {x^2}{3}\right )-\frac {9 a x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-x^2,\frac {x^2}{3}\right )}{\left (-3+x^2\right ) \sqrt [3]{1+x^2} \left (9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},-x^2,\frac {x^2}{3}\right )+2 x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},-x^2,\frac {x^2}{3}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},-x^2,\frac {x^2}{3}\right )\right )\right )} \]
(b*x^2*AppellF1[1, 1/3, 1, 2, -x^2, x^2/3])/6 - (9*a*x*AppellF1[1/2, 1/3, 1, 3/2, -x^2, x^2/3])/((-3 + x^2)*(1 + x^2)^(1/3)*(9*AppellF1[1/2, 1/3, 1, 3/2, -x^2, x^2/3] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, -x^2, x^2/3] - Appe llF1[3/2, 4/3, 1, 5/2, -x^2, x^2/3])))
Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1343, 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.
| \(\displaystyle \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}} \, dx\) |
| \(\Big \downarrow \) 1343 |
| \(\displaystyle a \int \frac {1}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx+b \int \frac {x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx\) |
| \(\Big \downarrow \) 304 |
| \(\displaystyle b \int \frac {x}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx+a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {1}{2} b \int \frac {1}{\left (3-x^2\right ) \sqrt [3]{x^2+1}}dx^2+a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {1}{2} b \left (\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{x^2+1}}d\sqrt [3]{x^2+1}}{2\ 2^{2/3}}-\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2}}d\sqrt [3]{x^2+1}+\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} b \left (-\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{x^2+1}+2 \sqrt [3]{2}}d\sqrt [3]{x^2+1}+\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} b \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{x^2+1}+1\right )}{2^{2/3}}+\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle a \left (\frac {\arctan \left (\frac {x}{\sqrt [3]{2} \sqrt [3]{x^2+1}+1}\right )}{2\ 2^{2/3}}-\frac {\arctan (x)}{6\ 2^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{x^2+1}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}\right )+\frac {1}{2} b \left (-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )}{2^{2/3}}+\frac {\log \left (3-x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}-\sqrt [3]{x^2+1}\right )}{2\ 2^{2/3}}\right )\) |
a*(-1/6*ArcTan[x]/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])) + (b*(-((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
3.1.54.3.1 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]
\[\int \frac {b x +a}{\left (-x^{2}+3\right ) \left (x^{2}+1\right )^{\frac {1}{3}}}d x\]
Exception generated. \[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
\[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=- \int \frac {a}{x^{2} \sqrt [3]{x^{2} + 1} - 3 \sqrt [3]{x^{2} + 1}}\, dx - \int \frac {b x}{x^{2} \sqrt [3]{x^{2} + 1} - 3 \sqrt [3]{x^{2} + 1}}\, dx \]
-Integral(a/(x**2*(x**2 + 1)**(1/3) - 3*(x**2 + 1)**(1/3)), x) - Integral( b*x/(x**2*(x**2 + 1)**(1/3) - 3*(x**2 + 1)**(1/3)), x)
\[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int { -\frac {b x + a}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}} \,d x } \]
\[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int { -\frac {b x + a}{{\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 3\right )}} \,d x } \]
Timed out. \[ \int \frac {a+b x}{\left (3-x^2\right ) \sqrt [3]{1+x^2}} \, dx=\int -\frac {a+b\,x}{{\left (x^2+1\right )}^{1/3}\,\left (x^2-3\right )} \,d x \]