\(\int \frac {x^7}{(1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1424]

Optimal result

Integrand size = 22, antiderivative size = 124 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {7}{12} \left (1+2 x^2\right )^{3/4}-\frac {1}{7} \left (1+2 x^2\right )^{7/4}+\frac {1}{44} \left (1+2 x^2\right )^{11/4}+\frac {\arctan \left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Output:

7/12*(2*x^2+1)^(3/4)-1/7*(2*x^2+1)^(7/4)+1/44*(2*x^2+1)^(11/4)+1/2*arctan( 
1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/(2*x^2+1)^(1/4))*2^(1/2)+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 = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{231} \left (1+2 x^2\right )^{3/4} \left (107-45 x^2+21 x^4\right )-\frac {\arctan \left (\frac {-1+\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Input:

Integrate[x^7/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

((1 + 2*x^2)^(3/4)*(107 - 45*x^2 + 21*x^4))/231 - ArcTan[(-1 + Sqrt[1 + 2* 
x^2])/(Sqrt[2]*(1 + 2*x^2)^(1/4))]/Sqrt[2] + ArcTanh[(Sqrt[2]*(1 + 2*x^2)^ 
(1/4))/(1 + Sqrt[1 + 2*x^2])]/Sqrt[2]
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98, 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.

\(\displaystyle \int \frac {x^7}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\)

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {1}{44} \left (2 x^2+1\right )^{11/4}-\frac {1}{7} \left (2 x^2+1\right )^{7/4}+\frac {7}{12} \left (2 x^2+1\right )^{3/4}\)

Input:

Int[x^7/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

(7*(1 + 2*x^2)^(3/4))/12 - (1 + 2*x^2)^(7/4)/7 + (1 + 2*x^2)^(11/4)/44 + A 
rcTan[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*(1 + 2*x^2)^(1/4))]/Sqrt[2] + 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 = 1.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {\left (-\ln \left (\frac {\sqrt {2 x^{2}+1}-\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}{\sqrt {2 x^{2}+1}+\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}\right )-2 \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-1\right )-2 \arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1\right )\right ) \sqrt {2}}{4}+\frac {\left (21 x^{4}-45 x^{2}+107\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}{231}\)	\(122\)
trager	\(\left (\frac {1}{11} x^{4}-\frac {15}{77} x^{2}+\frac {107}{231}\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}+x^{2}}{x^{2}+1}\right )}{2}\)	\(175\)
risch	\(\frac {\left (21 x^{4}-45 x^{2}+107\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}{231}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}+x^{2}}{x^{2}+1}\right )}{2}\)	\(176\)

Input:

int(x^7/(x^2+1)/(2*x^2+1)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/4*(-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))-2*arctan((2*x^2+1)^(1/4)*2^(1/2)-1)-2*arctan((2*x^ 
2+1)^(1/4)*2^(1/2)+1))*2^(1/2)+1/231*(21*x^4-45*x^2+107)*(2*x^2+1)^(3/4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{231} \, {\left (21 \, x^{4} - 45 \, x^{2} + 107\right )} {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) \] Input:

integrate(x^7/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
 

Output:

1/231*(21*x^4 - 45*x^2 + 107)*(2*x^2 + 1)^(3/4) - 1/2*sqrt(2)*arctan(sqrt( 
2)*(2*x^2 + 1)^(1/4) + 1) - 1/2*sqrt(2)*arctan(sqrt(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]

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

integrate(x**7/(x**2+1)/(2*x**2+1)**(1/4),x)
 

Output:

Integral(x**7/((x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{44} \, {\left (2 \, x^{2} + 1\right )}^{\frac {11}{4}} - \frac {1}{7} \, {\left (2 \, x^{2} + 1\right )}^{\frac {7}{4}} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {7}{12} \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} \] Input:

integrate(x^7/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="maxima")
 

Output:

1/44*(2*x^2 + 1)^(11/4) - 1/7*(2*x^2 + 1)^(7/4) - 1/2*sqrt(2)*arctan(1/2*s 
qrt(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) + s 
qrt(2*x^2 + 1) + 1) + 7/12*(2*x^2 + 1)^(3/4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{44} \, {\left (2 \, x^{2} + 1\right )}^{\frac {11}{4}} - \frac {1}{7} \, {\left (2 \, x^{2} + 1\right )}^{\frac {7}{4}} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {7}{12} \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} \] Input:

integrate(x^7/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
 

Output:

1/44*(2*x^2 + 1)^(11/4) - 1/7*(2*x^2 + 1)^(7/4) - 1/2*sqrt(2)*arctan(1/2*s 
qrt(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) + s 
qrt(2*x^2 + 1) + 1) + 7/12*(2*x^2 + 1)^(3/4)
 

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.66 \[ \int \frac {x^7}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {7\,{\left (2\,x^2+1\right )}^{3/4}}{12}-\frac {{\left (2\,x^2+1\right )}^{7/4}}{7}+\frac {{\left (2\,x^2+1\right )}^{11/4}}{44}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \] Input:

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

Output:

(7*(2*x^2 + 1)^(3/4))/12 - 2^(1/2)*atan(2^(1/2)*(2*x^2 + 1)^(1/4)*(1/2 + 1 
i/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) - (2*x^2 + 1)^(7/4)/7 + (2*x^2 + 1)^(11/4)/44
 

Reduce [F]

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

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

Output:

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