\(\int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx\) [542]

Optimal result

Integrand size = 15, antiderivative size = 152 \[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=-\frac {8 a^3 x^2}{39 b^3 \sqrt [4]{a+b x^4}}+\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}+\frac {8 a^{7/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^4}} \] Output:

-8/39*a^3*x^2/b^3/(b*x^4+a)^(1/4)+4/39*a^2*x^2*(b*x^4+a)^(3/4)/b^3-10/117* 
a*x^6*(b*x^4+a)^(3/4)/b^2+1/13*x^10*(b*x^4+a)^(3/4)/b+8/39*a^(7/2)*(1+b*x^ 
4/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/b^(7/2) 
/(b*x^4+a)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 6.41 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.60 \[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^2 \left (12 a^3+2 a^2 b x^4-a b^2 x^8+9 b^3 x^{12}-12 a^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{117 b^3 \sqrt [4]{a+b x^4}} \] Input:

Integrate[x^13/(a + b*x^4)^(1/4),x]
 

Output:

(x^2*(12*a^3 + 2*a^2*b*x^4 - a*b^2*x^8 + 9*b^3*x^12 - 12*a^3*(1 + (b*x^4)/ 
a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(117*b^3*(a + b* 
x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {807, 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.

\(\displaystyle \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{2} \int \frac {x^{12}}{\sqrt [4]{b x^4+a}}dx^2\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \int \frac {x^8}{\sqrt [4]{b x^4+a}}dx^2}{13 b}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx^2}{3 b}\right )}{13 b}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{5 b}\right )}{3 b}\right )}{13 b}\right )\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )}{13 b}\right )\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )}{13 b}\right )\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {1}{2} \left (\frac {2 x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {10 a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )}{13 b}\right )\)

Input:

Int[x^13/(a + b*x^4)^(1/4),x]
 

Output:

((2*x^10*(a + b*x^4)^(3/4))/(13*b) - (10*a*((2*x^6*(a + b*x^4)^(3/4))/(9*b 
) - (2*a*((2*x^2*(a + b*x^4)^(3/4))/(5*b) - (2*a*(1 + (b*x^4)/a)^(1/4)*((2 
*x^2)/(1 + (b*x^4)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sq 
rt[a]]/2, 2])/Sqrt[b]))/(5*b*(a + b*x^4)^(1/4))))/(3*b)))/(13*b))/2
 

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]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

Maple [F]

\[\int \frac {x^{13}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]

Input:

int(x^13/(b*x^4+a)^(1/4),x)
 

Output:

int(x^13/(b*x^4+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="fricas")
 

Output:

integral(x^13/(b*x^4 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.18 \[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^{14} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{14 \sqrt [4]{a}} \] Input:

integrate(x**13/(b*x**4+a)**(1/4),x)
 

Output:

x**14*hyper((1/4, 7/2), (9/2,), b*x**4*exp_polar(I*pi)/a)/(14*a**(1/4))
 

Maxima [F]

\[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate(x^13/(b*x^4 + a)^(1/4), x)
 

Giac [F]

\[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(x^13/(b*x^4 + a)^(1/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^{13}}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:

int(x^13/(a + b*x^4)^(1/4),x)
 

Output:

int(x^13/(a + b*x^4)^(1/4), x)
 

Reduce [F]

\[ \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^{13}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \] Input:

int(x^13/(b*x^4+a)^(1/4),x)
 

Output:

int(x**13/(a + b*x**4)**(1/4),x)