\(\int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx\) [549]

Optimal result

Integrand size = 15, antiderivative size = 104 \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=-\frac {5 a x \left (a+b x^4\right )^{3/4}}{32 b^2}+\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}+\frac {5 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}}+\frac {5 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}} \] Output:

-5/32*a*x*(b*x^4+a)^(3/4)/b^2+1/8*x^5*(b*x^4+a)^(3/4)/b+5/64*a^2*arctan(b^ 
(1/4)*x/(b*x^4+a)^(1/4))/b^(9/4)+5/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4 
))/b^(9/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4} \left (-5 a+4 b x^4\right )+5 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+5 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{9/4}} \] Input:

Integrate[x^8/(a + b*x^4)^(1/4),x]
 

Output:

(2*b^(1/4)*x*(a + b*x^4)^(3/4)*(-5*a + 4*b*x^4) + 5*a^2*ArcTan[(b^(1/4)*x) 
/(a + b*x^4)^(1/4)] + 5*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^ 
(9/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {843, 843, 770, 756, 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.

\(\displaystyle \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{8 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{8 b}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{8 b}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{8 b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\)

Input:

Int[x^8/(a + b*x^4)^(1/4),x]
 

Output:

(x^5*(a + b*x^4)^(3/4))/(8*b) - (5*a*((x*(a + b*x^4)^(3/4))/(4*b) - (a*(Ar 
cTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + 
 b*x^4)^(1/4)]/(2*b^(1/4))))/(4*b)))/(8*b)
 

Defintions of rubi rules used

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 756

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

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {16 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{\frac {5}{4}} x^{5}-20 a x \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{\frac {1}{4}}-10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2}+5 \ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2}}{128 b^{\frac {9}{4}}}\)	\(102\)

Input:

int(x^8/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/128*(16*(b*x^4+a)^(3/4)*b^(5/4)*x^5-20*a*x*(b*x^4+a)^(3/4)*b^(1/4)-10*ar 
ctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^2+5*ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b 
^(1/4)*x+(b*x^4+a)^(1/4)))*a^2)/b^(9/4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.16 \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\frac {5 \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {125 \, {\left (b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 5 \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) + 5 i \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (i \, b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 5 i \, b^{2} \left (\frac {a^{8}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {125 \, {\left (-i \, b^{7} x \left (\frac {a^{8}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) + 4 \, {\left (4 \, b x^{5} - 5 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, b^{2}} \] Input:

integrate(x^8/(b*x^4+a)^(1/4),x, algorithm="fricas")
 

Output:

1/128*(5*b^2*(a^8/b^9)^(1/4)*log(125*(b^7*x*(a^8/b^9)^(3/4) + (b*x^4 + a)^ 
(1/4)*a^6)/x) - 5*b^2*(a^8/b^9)^(1/4)*log(-125*(b^7*x*(a^8/b^9)^(3/4) - (b 
*x^4 + a)^(1/4)*a^6)/x) + 5*I*b^2*(a^8/b^9)^(1/4)*log(-125*(I*b^7*x*(a^8/b 
^9)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) - 5*I*b^2*(a^8/b^9)^(1/4)*log(-125*( 
-I*b^7*x*(a^8/b^9)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) + 4*(4*b*x^5 - 5*a*x) 
*(b*x^4 + a)^(3/4))/b^2
                                                                                    
                                                                                    
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.36 \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate(x**8/(b*x**4+a)**(1/4),x)
 

Output:

x**9*gamma(9/4)*hyper((1/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a* 
*(1/4)*gamma(13/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.45 \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=-\frac {5 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{128 \, b^{2}} + \frac {\frac {9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b}{x^{3}} - \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{x^{7}}}{32 \, {\left (b^{4} - \frac {2 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}}\right )}} \] Input:

integrate(x^8/(b*x^4+a)^(1/4),x, algorithm="maxima")
 

Output:

-5/128*a^2*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4 
) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/4))/b^2 + 1 
/32*(9*(b*x^4 + a)^(3/4)*a^2*b/x^3 - 5*(b*x^4 + a)^(7/4)*a^2/x^7)/(b^4 - 2 
*(b*x^4 + a)*b^3/x^4 + (b*x^4 + a)^2*b^2/x^8)
 

Giac [F]

\[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^8/(b*x^4+a)^(1/4),x, algorithm="giac")
 

Output:

integrate(x^8/(b*x^4 + a)^(1/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^8}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:

int(x^8/(a + b*x^4)^(1/4),x)
 

Output:

int(x^8/(a + b*x^4)^(1/4), x)
 

Reduce [F]

\[ \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \] Input:

int(x^8/(b*x^4+a)^(1/4),x)
 

Output:

int(x**8/(a + b*x**4)**(1/4),x)