\(\int \frac {a+b x}{\sqrt [4]{-1-x^2} (2+x^2)} \, dx\) [70]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\frac {a \arctan \left (\frac {x}{\sqrt {2} \sqrt [4]{-1-x^2}}\right )}{2 \sqrt {2}}+b \arctan \left (\sqrt [4]{-1-x^2}\right )+\frac {a \text {arctanh}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1-x^2}}\right )}{2 \sqrt {2}}-b \text {arctanh}\left (\sqrt [4]{-1-x^2}\right ) \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 10.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.84 \[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\frac {x \left (b x \sqrt [4]{1+x^2} \operatorname {AppellF1}\left (1,\frac {1}{4},1,2,-x^2,-\frac {x^2}{2}\right )-\frac {24 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-x^2,-\frac {x^2}{2}\right )}{\left (2+x^2\right ) \left (-6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-x^2,-\frac {x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-x^2,-\frac {x^2}{2}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-x^2,-\frac {x^2}{2}\right )\right )\right )}\right )}{4 \sqrt [4]{-1-x^2}} \] Input:

Integrate[(a + b*x)/((-1 - x^2)^(1/4)*(2 + x^2)),x]
 

Output:

(x*(b*x*(1 + x^2)^(1/4)*AppellF1[1, 1/4, 1, 2, -x^2, -1/2*x^2] - (24*a*App 
ellF1[1/2, 1/4, 1, 3/2, -x^2, -1/2*x^2])/((2 + x^2)*(-6*AppellF1[1/2, 1/4, 
 1, 3/2, -x^2, -1/2*x^2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, -x^2, -1/2*x^ 
2] + AppellF1[3/2, 5/4, 1, 5/2, -x^2, -1/2*x^2])))))/(4*(-1 - x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1343, 309, 353, 73, 827, 216, 219}

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.

\(\displaystyle \int \frac {a+b x}{\sqrt [4]{-x^2-1} \left (x^2+2\right )} \, dx\)

\(\Big \downarrow \) 1343

\(\displaystyle a \int \frac {1}{\sqrt [4]{-x^2-1} \left (x^2+2\right )}dx+b \int \frac {x}{\sqrt [4]{-x^2-1} \left (x^2+2\right )}dx\)

\(\Big \downarrow \) 309

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 73

\(\displaystyle a \left (\frac {\arctan \left (\frac {x}{\sqrt {2} \sqrt [4]{-x^2-1}}\right )}{2 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt {2} \sqrt [4]{-x^2-1}}\right )}{2 \sqrt {2}}\right )-2 b \int \frac {x^4}{1-x^8}d\sqrt [4]{-x^2-1}\)

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[(a + b*x)/((-1 - x^2)^(1/4)*(2 + x^2)),x]
 

Output:

a*(ArcTan[x/(Sqrt[2]*(-1 - x^2)^(1/4))]/(2*Sqrt[2]) + ArcTanh[x/(Sqrt[2]*( 
-1 - x^2)^(1/4))]/(2*Sqrt[2])) - 2*b*(-1/2*ArcTan[(-1 - x^2)^(1/4)] + ArcT 
anh[(-1 - x^2)^(1/4)]/2)
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

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

rule 219

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

rule 309

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit 
h[{q = Rt[-b^2/a, 4]}, Simp[(b/(2*Sqrt[2]*a*d*q))*ArcTan[q*(x/(Sqrt[2]*(a + 
 b*x^2)^(1/4)))], x] + Simp[(b/(2*Sqrt[2]*a*d*q))*ArcTanh[q*(x/(Sqrt[2]*(a 
+ b*x^2)^(1/4)))], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && 
NegQ[b^2/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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

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]

\[\int \frac {b x +a}{\left (-x^{2}-1\right )^{\frac {1}{4}} \left (x^{2}+2\right )}d x\]

Input:

int((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x)
 

Output:

int((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\text {Timed out} \] Input:

integrate((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\int \frac {a + b x}{\sqrt [4]{- x^{2} - 1} \left (x^{2} + 2\right )}\, dx \] Input:

integrate((b*x+a)/(-x**2-1)**(1/4)/(x**2+2),x)
 

Output:

Integral((a + b*x)/((-x**2 - 1)**(1/4)*(x**2 + 2)), x)
 

Maxima [F]

\[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 2\right )} {\left (-x^{2} - 1\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x, algorithm="maxima")
 

Output:

integrate((b*x + a)/((x^2 + 2)*(-x^2 - 1)^(1/4)), x)
 

Giac [F]

\[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 2\right )} {\left (-x^{2} - 1\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x, algorithm="giac")
 

Output:

integrate((b*x + a)/((x^2 + 2)*(-x^2 - 1)^(1/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=\int \frac {a+b\,x}{{\left (-x^2-1\right )}^{1/4}\,\left (x^2+2\right )} \,d x \] Input:

int((a + b*x)/((- x^2 - 1)^(1/4)*(x^2 + 2)),x)
 

Output:

int((a + b*x)/((- x^2 - 1)^(1/4)*(x^2 + 2)), x)
 

Reduce [F]

\[ \int \frac {a+b x}{\sqrt [4]{-1-x^2} \left (2+x^2\right )} \, dx=-\left (\left (\int \frac {x}{\left (x^{2}+1\right )^{\frac {1}{4}} x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right ) b +\left (\int \frac {1}{\left (x^{2}+1\right )^{\frac {1}{4}} x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right ) a \right ) \left (-1\right )^{\frac {3}{4}} \] Input:

int((b*x+a)/(-x^2-1)^(1/4)/(x^2+2),x)
 

Output:

(int(x/((x**2 + 1)**(1/4)*x**2 + 2*(x**2 + 1)**(1/4)),x)*b + int(1/((x**2 
+ 1)**(1/4)*x**2 + 2*(x**2 + 1)**(1/4)),x)*a)/( - 1)**(1/4)