\(\int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx\) [471]

Optimal result

Integrand size = 15, antiderivative size = 152 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}-\frac {b \sqrt [4]{a+b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{231 a^2 x^7}-\frac {4 b^3 \sqrt [4]{a+b x^4}}{231 a^3 x^3}+\frac {8 b^{9/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{231 a^{7/2} \left (a+b x^4\right )^{3/4}} \] Output:

-1/15*(b*x^4+a)^(1/4)/x^15-1/165*b*(b*x^4+a)^(1/4)/a/x^11+2/231*b^2*(b*x^4 
+a)^(1/4)/a^2/x^7-4/231*b^3*(b*x^4+a)^(1/4)/a^3/x^3+8/231*b^(9/2)*(1+a/b/x 
^4)^(3/4)*x^3*InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^( 
7/2)/(b*x^4+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.34 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=-\frac {\sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {1}{4},-\frac {11}{4},-\frac {b x^4}{a}\right )}{15 x^{15} \sqrt [4]{1+\frac {b x^4}{a}}} \] Input:

Integrate[(a + b*x^4)^(1/4)/x^16,x]
 

Output:

-1/15*((a + b*x^4)^(1/4)*Hypergeometric2F1[-15/4, -1/4, -11/4, -((b*x^4)/a 
)])/(x^15*(1 + (b*x^4)/a)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {809, 847, 847, 847, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx\)

\(\Big \downarrow \) 809

\(\displaystyle \frac {1}{15} b \int \frac {1}{x^{12} \left (b x^4+a\right )^{3/4}}dx-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \int \frac {1}{x^8 \left (b x^4+a\right )^{3/4}}dx}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \left (-\frac {2 b \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{3 a}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \left (-\frac {2 b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {2 b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{3 a \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {1}{15} b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {2 b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{3 a x^3}\right )}{7 a}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}\right )}{11 a}-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}\right )-\frac {\sqrt [4]{a+b x^4}}{15 x^{15}}\)

Input:

Int[(a + b*x^4)^(1/4)/x^16,x]
 

Output:

-1/15*(a + b*x^4)^(1/4)/x^15 + (b*(-1/11*(a + b*x^4)^(1/4)/(a*x^11) - (10* 
b*(-1/7*(a + b*x^4)^(1/4)/(a*x^7) - (6*b*(-1/3*(a + b*x^4)^(1/4)/(a*x^3) + 
 (2*b^(3/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^ 
2)]/2, 2])/(3*a^(3/2)*(a + b*x^4)^(3/4))))/(7*a)))/(11*a)))/15
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, 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]
 

rule 809

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

rule 847

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

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{16}}d x\]

Input:

int((b*x^4+a)^(1/4)/x^16,x)
 

Output:

int((b*x^4+a)^(1/4)/x^16,x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{16}} \,d x } \] Input:

integrate((b*x^4+a)^(1/4)/x^16,x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(1/4)/x^16, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.30 \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\frac {\sqrt [4]{a} \Gamma \left (- \frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {15}{4}, - \frac {1}{4} \\ - \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{15} \Gamma \left (- \frac {11}{4}\right )} \] Input:

integrate((b*x**4+a)**(1/4)/x**16,x)
 

Output:

a**(1/4)*gamma(-15/4)*hyper((-15/4, -1/4), (-11/4,), b*x**4*exp_polar(I*pi 
)/a)/(4*x**15*gamma(-11/4))
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{16}} \,d x } \] Input:

integrate((b*x^4+a)^(1/4)/x^16,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(1/4)/x^16, x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{16}} \,d x } \] Input:

integrate((b*x^4+a)^(1/4)/x^16,x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(1/4)/x^16, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{1/4}}{x^{16}} \,d x \] Input:

int((a + b*x^4)^(1/4)/x^16,x)
 

Output:

int((a + b*x^4)^(1/4)/x^16, x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a+b x^4}}{x^{16}} \, dx=\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-\left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b \,x^{20}+a \,x^{16}}d x \right ) a \,x^{15}}{14 x^{15}} \] Input:

int((b*x^4+a)^(1/4)/x^16,x)
 

Output:

( - ((a + b*x**4)**(1/4) + int((a + b*x**4)**(1/4)/(a*x**16 + b*x**20),x)* 
a*x**15))/(14*x**15)