\(\int \frac {a+b x}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\) [53]

Optimal result

Integrand size = 24, antiderivative size = 198 \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\sqrt {3} b \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \] Output:

-1/12*a*arctanh(x)*2^(1/3)+1/4*a*arctanh(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*arctan(3^(1/2)/x)*2^(1/3)*3^(1/2)+1/12*a*arctan((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*x^2+2)^(1/3))*3^(1/ 
2))*3^(1/2)*2^(1/3)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.73 \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{6} b x^2 \operatorname {AppellF1}\left (1,\frac {1}{3},1,2,x^2,-\frac {x^2}{3}\right )-\frac {9 a x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{1-x^2} \left (3+x^2\right ) \left (-9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )+2 x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},x^2,-\frac {x^2}{3}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},x^2,-\frac {x^2}{3}\right )\right )\right )} \] Input:

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

Output:

(b*x^2*AppellF1[1, 1/3, 1, 2, x^2, -1/3*x^2])/6 - (9*a*x*AppellF1[1/2, 1/3 
, 1, 3/2, x^2, -1/3*x^2])/((1 - x^2)^(1/3)*(3 + x^2)*(-9*AppellF1[1/2, 1/3 
, 1, 3/2, x^2, -1/3*x^2] + 2*x^2*(AppellF1[3/2, 1/3, 2, 5/2, x^2, -1/3*x^2 
] - AppellF1[3/2, 4/3, 1, 5/2, x^2, -1/3*x^2])))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1343, 305, 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.

\(\displaystyle \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (x^2+3\right )} \, dx\)

\(\Big \downarrow \) 1343

\(\displaystyle a \int \frac {1}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx+b \int \frac {x}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx\)

\(\Big \downarrow \) 305

\(\displaystyle b \int \frac {x}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx+a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} b \int \frac {1}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2+a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{2} b \left (-\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{1-x^2}}d\sqrt [3]{1-x^2}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} b \left (\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{2} b \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{1-x^2}+1\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )+a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle a \left (\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}\right )+\frac {1}{2} b \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )\)

Input:

Int[(a + b*x)/((1 - x^2)^(1/3)*(3 + x^2)),x]
 

Output:

a*(ArcTan[Sqrt[3]/x]/(2*2^(2/3)*Sqrt[3]) + ArcTan[(Sqrt[3]*(1 - 2^(1/3)*(1 
 - x^2)^(1/3)))/x]/(2*2^(2/3)*Sqrt[3]) - ArcTanh[x]/(6*2^(2/3)) + ArcTanh[ 
x/(1 + 2^(1/3)*(1 - x^2)^(1/3))]/(2*2^(2/3))) + (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
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 67

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]
 

rule 217

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 305

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit 
h[{q = Rt[-b/a, 2]}, Simp[q*(ArcTan[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/ 
3)*d)), x] + (Simp[q*(ArcTanh[(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*(ArcTanh[q*x]/(6*2^(2/3)*a^(1/3 
)*d)), x] + Simp[q*(ArcTan[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] && NegQ[b/a]
 

rule 353

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]
 

rule 1082

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]
 

rule 1343

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]
 

Maple [F]

\[\int \frac {b x +a}{\left (-x^{2}+1\right )^{\frac {1}{3}} \left (x^{2}+3\right )}d x\]

Input:

int((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x)
 

Output:

int((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (trace 0)
 

Sympy [F]

\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int \frac {a + b x}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \] Input:

integrate((b*x+a)/(-x**2+1)**(1/3)/(x**2+3),x)
 

Output:

Integral((a + b*x)/((-(x - 1)*(x + 1))**(1/3)*(x**2 + 3)), x)
 

Maxima [F]

\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="maxima")
 

Output:

integrate((b*x + a)/((x^2 + 3)*(-x^2 + 1)^(1/3)), x)
 

Giac [F]

\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="giac")
 

Output:

integrate((b*x + a)/((x^2 + 3)*(-x^2 + 1)^(1/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int \frac {a+b\,x}{{\left (1-x^2\right )}^{1/3}\,\left (x^2+3\right )} \,d x \] Input:

int((a + b*x)/((1 - x^2)^(1/3)*(x^2 + 3)),x)
 

Output:

int((a + b*x)/((1 - x^2)^(1/3)*(x^2 + 3)), x)
 

Reduce [F]

\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\left (\int \frac {x}{\left (-x^{2}+1\right )^{\frac {1}{3}} x^{2}+3 \left (-x^{2}+1\right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {1}{\left (-x^{2}+1\right )^{\frac {1}{3}} x^{2}+3 \left (-x^{2}+1\right )^{\frac {1}{3}}}d x \right ) a \] Input:

int((b*x+a)/(-x^2+1)^(1/3)/(x^2+3),x)
 

Output:

int(x/(( - x**2 + 1)**(1/3)*x**2 + 3*( - x**2 + 1)**(1/3)),x)*b + int(1/(( 
 - x**2 + 1)**(1/3)*x**2 + 3*( - x**2 + 1)**(1/3)),x)*a