\(\int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx\) [871]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 146 \[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=-\frac {16 a^3 x}{39 b^3 \sqrt [4]{a+b x^2}}+\frac {8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac {20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}+\frac {16 a^{7/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^2}} \] Output:

-16/39*a^3*x/b^3/(b*x^2+a)^(1/4)+8/39*a^2*x*(b*x^2+a)^(3/4)/b^3-20/117*a*x 
^3*(b*x^2+a)^(3/4)/b^2+2/13*x^5*(b*x^2+a)^(3/4)/b+16/39*a^(7/2)*(1+b*x^2/a 
)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(7/2)/(b*x 
^2+a)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.94 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.62 \[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\frac {2 \left (12 a^3 x+2 a^2 b x^3-a b^2 x^5+9 b^3 x^7-12 a^3 x \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{117 b^3 \sqrt [4]{a+b x^2}} \] Input:

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

Output:

(2*(12*a^3*x + 2*a^2*b*x^3 - a*b^2*x^5 + 9*b^3*x^7 - 12*a^3*x*(1 + (b*x^2) 
/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(117*b^3*(a + b 
*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {262, 262, 262, 227, 225, 212}

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.

\(\displaystyle \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \int \frac {x^4}{\sqrt [4]{b x^2+a}}dx}{13 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^2}{\sqrt [4]{b x^2+a}}dx}{3 b}\right )}{13 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}\right )}{3 b}\right )}{13 b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}\)

Input:

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

Output:

(2*x^5*(a + b*x^2)^(3/4))/(13*b) - (10*a*((2*x^3*(a + b*x^2)^(3/4))/(9*b) 
- (2*a*((2*x*(a + b*x^2)^(3/4))/(5*b) - (2*a*(1 + (b*x^2)/a)^(1/4)*((2*x)/ 
(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2 
, 2])/Sqrt[b]))/(5*b*(a + b*x^2)^(1/4))))/(3*b)))/(13*b)
 

Defintions of rubi rules used

rule 212
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 262
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 
Maple [F]

\[\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]

Input:

int(x^6/(b*x^2+a)^(1/4),x)
 

Output:

int(x^6/(b*x^2+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^6/(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

integral(x^6/(b*x^2 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.18 \[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\frac {x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 \sqrt [4]{a}} \] Input:

integrate(x**6/(b*x**2+a)**(1/4),x)
 

Output:

x**7*hyper((1/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(1/4))
 

Maxima [F]

\[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^6/(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate(x^6/(b*x^2 + a)^(1/4), x)
 

Giac [F]

\[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^6/(b*x^2+a)^(1/4),x, algorithm="giac")
 

Output:

integrate(x^6/(b*x^2 + a)^(1/4), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\int \frac {x^6}{{\left (b\,x^2+a\right )}^{1/4}} \,d x \] Input:

int(x^6/(a + b*x^2)^(1/4),x)
 

Output:

int(x^6/(a + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx=\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \] Input:

int(x^6/(b*x^2+a)^(1/4),x)
 

Output:

int(x**6/(a + b*x**2)**(1/4),x)