\(\int \frac {1}{x^5 (1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1430]

Optimal result

Integrand size = 22, antiderivative size = 147 \[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {\left (1+2 x^2\right )^{3/4}}{4 x^4}+\frac {9 \left (1+2 x^2\right )^{3/4}}{8 x^2}+\frac {17}{8} \arctan \left (\sqrt [4]{1+2 x^2}\right )+\frac {\arctan \left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {17}{8} \text {arctanh}\left (\sqrt [4]{1+2 x^2}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Output:

-1/4*(2*x^2+1)^(3/4)/x^4+9/8*(2*x^2+1)^(3/4)/x^2+17/8*arctan((2*x^2+1)^(1/ 
4))+1/2*arctan(1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/(2*x^2+1)^(1/4))*2^(1/2)-17 
/8*arctanh((2*x^2+1)^(1/4))+1/2*arctanh(2^(1/2)*(2*x^2+1)^(1/4)/(1+(2*x^2+ 
1)^(1/2)))*2^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{8} \left (\frac {3 \sqrt [4]{1+2 x^2}}{x^2}-\frac {2 \left (1+2 x^2\right )^{3/4}}{x^4}+\frac {6 \left (1+2 x^2\right )^{3/4}}{x^2}+\frac {6 \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}+17 \arctan \left (\sqrt [4]{1+2 x^2}\right )-4 \sqrt {2} \arctan \left (\frac {-1+\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )-17 \text {arctanh}\left (\sqrt [4]{1+2 x^2}\right )+4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )\right ) \] Input:

Integrate[1/(x^5*(1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

((3*(1 + 2*x^2)^(1/4))/x^2 - (2*(1 + 2*x^2)^(3/4))/x^4 + (6*(1 + 2*x^2)^(3 
/4))/x^2 + (6*(1 + 2*x^2)^(1/4))/(1 + Sqrt[1 + 2*x^2]) + 17*ArcTan[(1 + 2* 
x^2)^(1/4)] - 4*Sqrt[2]*ArcTan[(-1 + Sqrt[1 + 2*x^2])/(Sqrt[2]*(1 + 2*x^2) 
^(1/4))] - 17*ArcTanh[(1 + 2*x^2)^(1/4)] + 4*Sqrt[2]*ArcTanh[(Sqrt[2]*(1 + 
 2*x^2)^(1/4))/(1 + Sqrt[1 + 2*x^2])])/8
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {349, 2009}

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.

\(\displaystyle \int \frac {1}{x^5 \left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\)

\(\Big \downarrow \) 349

\(\displaystyle \int \left (-\frac {x}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}}+\frac {1}{\sqrt [4]{2 x^2+1} x}+\frac {1}{\sqrt [4]{2 x^2+1} x^5}-\frac {1}{\sqrt [4]{2 x^2+1} x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {17}{8} \arctan \left (\sqrt [4]{2 x^2+1}\right )+\frac {\arctan \left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {17}{8} \text {arctanh}\left (\sqrt [4]{2 x^2+1}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {9 \left (2 x^2+1\right )^{3/4}}{8 x^2}-\frac {\left (2 x^2+1\right )^{3/4}}{4 x^4}\)

Input:

Int[1/(x^5*(1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

-1/4*(1 + 2*x^2)^(3/4)/x^4 + (9*(1 + 2*x^2)^(3/4))/(8*x^2) + (17*ArcTan[(1 
 + 2*x^2)^(1/4)])/8 + ArcTan[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*(1 + 2*x^2)^(1 
/4))]/Sqrt[2] - (17*ArcTanh[(1 + 2*x^2)^(1/4)])/8 + ArcTanh[(1 + Sqrt[1 + 
2*x^2])/(Sqrt[2]*(1 + 2*x^2)^(1/4))]/Sqrt[2]
 

Defintions of rubi rules used

rule 349

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(1/4)*(c + d*x^2)), x], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In 
tegerQ[m/2])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 6.93 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.54

method	result	size
pseudoelliptic	\(\frac {-4 \sqrt {2}\, \ln \left (\frac {\sqrt {2 x^{2}+1}-\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}{\sqrt {2 x^{2}+1}+\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}\right ) x^{4}-8 \sqrt {2}\, \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1\right ) x^{4}-8 \sqrt {2}\, \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-1\right ) x^{4}+17 \ln \left (-1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right ) x^{4}+34 \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}}\right ) x^{4}-17 \ln \left (1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right ) x^{4}+18 x^{2} \left (2 x^{2}+1\right )^{\frac {3}{4}}-4 \left (2 x^{2}+1\right )^{\frac {3}{4}}}{4 {\left (-1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right )}^{2} \left (1+\sqrt {2 x^{2}+1}\right )^{2} {\left (1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right )}^{2}}\)	\(226\)
trager	\(\frac {\left (9 x^{2}-2\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}{8 x^{4}}+\frac {17 \ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}-\sqrt {2 x^{2}+1}-x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}-1}{x^{2}}\right )}{16}+\frac {17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {2 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (2 x^{2}+1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{2}}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {2 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) x^{2}-\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \sqrt {2 x^{2}+1}-\left (2 x^{2}+1\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}\)	\(342\)
risch	\(\frac {18 x^{4}+5 x^{2}-2}{8 x^{4} \left (2 x^{2}+1\right )^{\frac {1}{4}}}+\frac {17 \ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}-\sqrt {2 x^{2}+1}-x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}-1}{x^{2}}\right )}{16}+\frac {17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {2 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (2 x^{2}+1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{2}}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {2 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) x^{2}-\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \sqrt {2 x^{2}+1}+\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}\)	\(345\)

Input:

int(1/x^5/(x^2+1)/(2*x^2+1)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/4*(-4*2^(1/2)*ln(((2*x^2+1)^(1/2)-(2*x^2+1)^(1/4)*2^(1/2)+1)/((2*x^2+1)^ 
(1/2)+(2*x^2+1)^(1/4)*2^(1/2)+1))*x^4-8*2^(1/2)*arctan((2*x^2+1)^(1/4)*2^( 
1/2)+1)*x^4-8*2^(1/2)*arctan((2*x^2+1)^(1/4)*2^(1/2)-1)*x^4+17*ln(-1+(2*x^ 
2+1)^(1/4))*x^4+34*arctan((2*x^2+1)^(1/4))*x^4-17*ln(1+(2*x^2+1)^(1/4))*x^ 
4+18*x^2*(2*x^2+1)^(3/4)-4*(2*x^2+1)^(3/4))/(-1+(2*x^2+1)^(1/4))^2/(1+(2*x 
^2+1)^(1/2))^2/(1+(2*x^2+1)^(1/4))^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {8 \, \sqrt {2} x^{4} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + 8 \, \sqrt {2} x^{4} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) - 4 \, \sqrt {2} x^{4} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + 4 \, \sqrt {2} x^{4} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - 34 \, x^{4} \arctan \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right ) + 17 \, x^{4} \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - 17 \, x^{4} \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) - 2 \, {\left (9 \, x^{2} - 2\right )} {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}}}{16 \, x^{4}} \] Input:

integrate(1/x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
 

Output:

-1/16*(8*sqrt(2)*x^4*arctan(sqrt(2)*(2*x^2 + 1)^(1/4) + 1) + 8*sqrt(2)*x^4 
*arctan(sqrt(2)*(2*x^2 + 1)^(1/4) - 1) - 4*sqrt(2)*x^4*log(sqrt(2)*(2*x^2 
+ 1)^(1/4) + sqrt(2*x^2 + 1) + 1) + 4*sqrt(2)*x^4*log(-sqrt(2)*(2*x^2 + 1) 
^(1/4) + sqrt(2*x^2 + 1) + 1) - 34*x^4*arctan((2*x^2 + 1)^(1/4)) + 17*x^4* 
log((2*x^2 + 1)^(1/4) + 1) - 17*x^4*log((2*x^2 + 1)^(1/4) - 1) - 2*(9*x^2 
- 2)*(2*x^2 + 1)^(3/4))/x^4
 

Sympy [F]

\[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{x^{5} \left (x^{2} + 1\right ) \sqrt [4]{2 x^{2} + 1}}\, dx \] Input:

integrate(1/x**5/(x**2+1)/(2*x**2+1)**(1/4),x)
 

Output:

Integral(1/(x**5*(x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )} x^{5}} \,d x } \] Input:

integrate(1/x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((2*x^2 + 1)^(1/4)*(x^2 + 1)*x^5), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {9 \, {\left (2 \, x^{2} + 1\right )}^{\frac {7}{4}} - 13 \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}}}{16 \, x^{4}} + \frac {17}{8} \, \arctan \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {17}{16} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {17}{16} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \] Input:

integrate(1/x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^2 + 1)^(1/4))) - 1/2*sqr 
t(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^(1/4))) + 1/4*sqrt(2)*lo 
g(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) + 1/16*(9*(2*x^2 + 1)^(7/4) - 
 13*(2*x^2 + 1)^(3/4))/x^4 + 17/8*arctan((2*x^2 + 1)^(1/4)) - 17/16*log((2 
*x^2 + 1)^(1/4) + 1) + 17/16*log((2*x^2 + 1)^(1/4) - 1)
 

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {17\,\mathrm {atan}\left ({\left (2\,x^2+1\right )}^{1/4}\right )}{8}+\frac {\mathrm {atan}\left ({\left (2\,x^2+1\right )}^{1/4}\,1{}\mathrm {i}\right )\,17{}\mathrm {i}}{8}+\frac {\frac {13\,{\left (2\,x^2+1\right )}^{3/4}}{4}-\frac {9\,{\left (2\,x^2+1\right )}^{7/4}}{4}}{4\,x^2-{\left (2\,x^2+1\right )}^2+1}-{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (2\,x^2+1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-{\left (-1\right )}^{3/4}\,\mathrm {atan}\left ({\left (-1\right )}^{3/4}\,{\left (2\,x^2+1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i} \] Input:

int(1/(x^5*(x^2 + 1)*(2*x^2 + 1)^(1/4)),x)
 

Output:

(17*atan((2*x^2 + 1)^(1/4)))/8 + (atan((2*x^2 + 1)^(1/4)*1i)*17i)/8 + ((13 
*(2*x^2 + 1)^(3/4))/4 - (9*(2*x^2 + 1)^(7/4))/4)/(4*x^2 - (2*x^2 + 1)^2 + 
1) - (-1)^(1/4)*atan((-1)^(1/4)*(2*x^2 + 1)^(1/4)*1i)*1i - (-1)^(3/4)*atan 
((-1)^(3/4)*(2*x^2 + 1)^(1/4)*1i)*1i
 

Reduce [F]

\[ \int \frac {1}{x^5 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{7}+\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{5}}d x \] Input:

int(1/x^5/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Output:

int(1/((2*x**2 + 1)**(1/4)*x**7 + (2*x**2 + 1)**(1/4)*x**5),x)