\(\int \frac {x}{(1+x^2) (1+2 x^2)^{3/4}} \, dx\) [1467]

Optimal result

Integrand size = 20, antiderivative size = 80 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {-1+\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Input:

Integrate[x/((1 + x^2)*(1 + 2*x^2)^(3/4)),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {353, 73, 755, 27, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {x}{\left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}}dx^2\)

\(\Big \downarrow \) 73

\(\displaystyle \int \frac {1}{\frac {x^8}{2}+\frac {1}{2}}d\sqrt [4]{2 x^2+1}\)

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {1-x^4}{x^8+1}d\sqrt [4]{2 x^2+1}+\int \frac {x^4+1}{x^8+1}d\sqrt [4]{2 x^2+1}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{2} \int \frac {1}{x^4-\sqrt {2} \sqrt [4]{2 x^2+1}+1}d\sqrt [4]{2 x^2+1}+\frac {1}{2} \int \frac {1}{x^4+\sqrt {2} \sqrt [4]{2 x^2+1}+1}d\sqrt [4]{2 x^2+1}+\int \frac {1-x^4}{x^8+1}d\sqrt [4]{2 x^2+1}\)

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[x/((1 + x^2)*(1 + 2*x^2)^(3/4)),x]
 

Output:

-(ArcTan[1 - Sqrt[2]*(1 + 2*x^2)^(1/4)]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*(1 + 
 2*x^2)^(1/4)]/Sqrt[2] - Log[1 + x^4 - Sqrt[2]*(1 + 2*x^2)^(1/4)]/(2*Sqrt[ 
2]) + Log[1 + x^4 + Sqrt[2]*(1 + 2*x^2)^(1/4)]/(2*Sqrt[2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 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 755

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

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 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {2 x^{2}+1}-1}{\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {2 x^{2}+1}-1}\right )+2 \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1\right )+2 \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-1\right )\right )}{4}\)	\(100\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}+x^{2}}{x^{2}+1}\right )}{2}\)	\(153\)

Input:

int(x/(x^2+1)/(2*x^2+1)^(3/4),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4*2^(1/2)*(ln((-(2*x^2+1)^(1/4)*2^(1/2)-(2*x^2+1)^(1/2)-1)/((2*x^2+1)^(1 
/4)*2^(1/2)-(2*x^2+1)^(1/2)-1))+2*arctan((2*x^2+1)^(1/4)*2^(1/2)+1)+2*arct 
an((2*x^2+1)^(1/4)*2^(1/2)-1))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) \] Input:

integrate(x/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="fricas")
 

Output:

1/2*sqrt(2)*arctan(sqrt(2)*(2*x^2 + 1)^(1/4) + 1) + 1/2*sqrt(2)*arctan(sqr 
t(2)*(2*x^2 + 1)^(1/4) - 1) + 1/4*sqrt(2)*log(sqrt(2)*(2*x^2 + 1)^(1/4) + 
sqrt(2*x^2 + 1) + 1) - 1/4*sqrt(2)*log(-sqrt(2)*(2*x^2 + 1)^(1/4) + sqrt(2 
*x^2 + 1) + 1)
 

Sympy [F]

\[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {x}{\left (x^{2} + 1\right ) \left (2 x^{2} + 1\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(x/(x**2+1)/(2*x**2+1)**(3/4),x)
 

Output:

Integral(x/((x**2 + 1)*(2*x**2 + 1)**(3/4)), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) \] Input:

integrate(x/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="maxima")
 

Output:

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^2 + 1)^(1/4))) + 1/2*sqrt 
(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^(1/4))) + 1/4*sqrt(2)*log 
(sqrt(2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1) - 1/4*sqrt(2)*log(-sqrt( 
2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) \] Input:

integrate(x/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="giac")
 

Output:

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^2 + 1)^(1/4))) + 1/2*sqrt 
(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^(1/4))) + 1/4*sqrt(2)*log 
(sqrt(2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1) - 1/4*sqrt(2)*log(-sqrt( 
2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(x/((x^2 + 1)*(2*x^2 + 1)^(3/4)),x)
 

Output:

2^(1/2)*atan(2^(1/2)*(2*x^2 + 1)^(1/4)*(1/2 - 1i/2))*(1/2 + 1i/2) + 2^(1/2 
)*atan(2^(1/2)*(2*x^2 + 1)^(1/4)*(1/2 + 1i/2))*(1/2 - 1i/2)
 

Reduce [F]

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

int(x/(x^2+1)/(2*x^2+1)^(3/4),x)
 

Output:

int(x/((2*x**2 + 1)**(3/4)*x**2 + (2*x**2 + 1)**(3/4)),x)