\(\int \frac {1}{x^{14} (a+b x^4)^{5/4}} \, dx\) [623]

Optimal result

Integrand size = 15, antiderivative size = 153 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {112 b^{7/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 a^{9/2} \sqrt [4]{a+b x^4}} \] Output:

-1/13/a/x^13/(b*x^4+a)^(1/4)+14/117*b/a^2/x^9/(b*x^4+a)^(1/4)-28/117*b^2/a 
^3/x^5/(b*x^4+a)^(1/4)+56/39*b^3/a^4/x/(b*x^4+a)^(1/4)-112/39*b^(7/2)*(1+a 
/b/x^4)^(1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^ 
(9/2)/(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 = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {13}{4},\frac {5}{4},-\frac {9}{4},-\frac {b x^4}{a}\right )}{13 a x^{13} \sqrt [4]{a+b x^4}} \] Input:

Integrate[1/(x^14*(a + b*x^4)^(5/4)),x]
 

Output:

-1/13*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-13/4, 5/4, -9/4, -((b*x^4) 
/a)])/(a*x^13*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {816, 816, 816, 816, 813, 858, 807, 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.

\(\displaystyle \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 816

\(\displaystyle -\frac {14 b \int \frac {1}{x^{10} \left (b x^4+a\right )^{5/4}}dx}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 816

\(\displaystyle -\frac {14 b \left (-\frac {10 b \int \frac {1}{x^6 \left (b x^4+a\right )^{5/4}}dx}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 816

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \int \frac {1}{x^2 \left (b x^4+a\right )^{5/4}}dx}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 816

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \left (-\frac {2 b \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{a}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 813

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \left (-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {14 b \left (-\frac {10 b \left (-\frac {6 b \left (\frac {2 \sqrt {b} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\right )}{13 a}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}\)

Input:

Int[1/(x^14*(a + b*x^4)^(5/4)),x]
 

Output:

-1/13*1/(a*x^13*(a + b*x^4)^(1/4)) - (14*b*(-1/9*1/(a*x^9*(a + b*x^4)^(1/4 
)) - (10*b*(-1/5*1/(a*x^5*(a + b*x^4)^(1/4)) - (6*b*(-(1/(a*x*(a + b*x^4)^ 
(1/4))) + (2*Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqr 
t[b]*x^2)]/2, 2])/(a^(3/2)*(a + b*x^4)^(1/4))))/(5*a)))/(9*a)))/(13*a)
 

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 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 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 816

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

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]

\[\int \frac {1}{x^{14} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

Input:

int(1/x^14/(b*x^4+a)^(5/4),x)
 

Output:

int(1/x^14/(b*x^4+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{14}} \,d x } \] Input:

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)/(b^2*x^22 + 2*a*b*x^18 + a^2*x^14), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\frac {\Gamma \left (- \frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {13}{4}, \frac {5}{4} \\ - \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x^{13} \Gamma \left (- \frac {9}{4}\right )} \] Input:

integrate(1/x**14/(b*x**4+a)**(5/4),x)
 

Output:

gamma(-13/4)*hyper((-13/4, 5/4), (-9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a** 
(5/4)*x**13*gamma(-9/4))
 

Maxima [F]

\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{14}} \,d x } \] Input:

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^4 + a)^(5/4)*x^14), x)
 

Giac [F]

\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{14}} \,d x } \] Input:

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(1/((b*x^4 + a)^(5/4)*x^14), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{x^{14}\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:

int(1/(x^14*(a + b*x^4)^(5/4)),x)
 

Output:

int(1/(x^14*(a + b*x^4)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{14}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{18}}d x \] Input:

int(1/x^14/(b*x^4+a)^(5/4),x)
 

Output:

int(1/((a + b*x**4)**(1/4)*a*x**14 + (a + b*x**4)**(1/4)*b*x**18),x)