\(\int \frac {a+b x}{\sqrt [4]{1+x^2} (2+x^2)} \, dx\) [72]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {a+b x}{\sqrt [4]{1+x^2} \left (2+x^2\right )} \, dx=-\frac {b \arctan \left (\frac {1-\sqrt {1+x^2}}{\sqrt {2} \sqrt [4]{1+x^2}}\right )}{\sqrt {2}}-\frac {1}{2} a \arctan \left (\frac {1+\sqrt {1+x^2}}{x \sqrt [4]{1+x^2}}\right )-\frac {1}{2} a \text {arctanh}\left (\frac {1-\sqrt {1+x^2}}{x \sqrt [4]{1+x^2}}\right )-\frac {b \text {arctanh}\left (\frac {1+\sqrt {1+x^2}}{\sqrt {2} \sqrt [4]{1+x^2}}\right )}{\sqrt {2}} \] Output:

-1/2*a*arctan((1+(x^2+1)^(1/2))/x/(x^2+1)^(1/4))-1/2*a*arctanh((1-(x^2+1)^ 
(1/2))/x/(x^2+1)^(1/4))-1/2*b*arctan(1/2*(1-(x^2+1)^(1/2))/(x^2+1)^(1/4)*2 
^(1/2))*2^(1/2)-1/2*b*arctanh(1/2*(1+(x^2+1)^(1/2))/(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 = 10.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13 \[ \int \frac {a+b x}{\sqrt [4]{1+x^2} \left (2+x^2\right )} \, dx=\frac {1}{4} b x^2 \operatorname {AppellF1}\left (1,\frac {1}{4},1,2,-x^2,-\frac {x^2}{2}\right )-\frac {6 a x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-x^2,-\frac {x^2}{2}\right )}{\sqrt [4]{1+x^2} \left (2+x^2\right ) \left (-6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-x^2,-\frac {x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-x^2,-\frac {x^2}{2}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-x^2,-\frac {x^2}{2}\right )\right )\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1343, 308, 348}

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

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

\(\Big \downarrow \) 308

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

\(\Big \downarrow \) 348

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 308

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*a*d*q))*ArcTan[(b + q^2*Sqrt[a + b*x^2])/ 
(q^3*x*(a + b*x^2)^(1/4))], x] - Simp[(b/(2*a*d*q))*ArcTanh[(b - q^2*Sqrt[a 
 + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))], x]] /; FreeQ[{a, b, c, d}, x] && EqQ 
[b*c - 2*a*d, 0] && PosQ[b^2/a]
 

rule 348

Int[(x_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> 
Simp[(-(Sqrt[2]*Rt[a, 4]*d)^(-1))*ArcTan[(Rt[a, 4]^2 - Sqrt[a + b*x^2])/(Sq 
rt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] - Simp[(1/(Sqrt[2]*Rt[a, 4]*d))*ArcT 
anh[(Rt[a, 4]^2 + Sqrt[a + b*x^2])/(Sqrt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x 
] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[a]
 

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]

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