\(\int \frac {(a+b x^4)^{5/4}}{x^{11}} \, dx\) [515]

Optimal result

Integrand size = 15, antiderivative size = 122 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=-\frac {b \sqrt [4]{a+b x^4}}{12 x^6}-\frac {b^2 \sqrt [4]{a+b x^4}}{24 a x^2}-\frac {\left (a+b x^4\right )^{5/4}}{10 x^{10}}-\frac {b^{5/2} \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{24 \sqrt {a} \left (a+b x^4\right )^{3/4}} \] Output:

-1/12*b*(b*x^4+a)^(1/4)/x^6-1/24*b^2*(b*x^4+a)^(1/4)/a/x^2-1/10*(b*x^4+a)^ 
(5/4)/x^10-1/24*b^(5/2)*(1+b*x^4/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/ 
2)*x^2/a^(1/2)),2^(1/2))/a^(1/2)/(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 = 52, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=-\frac {a \sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{4},-\frac {3}{2},-\frac {b x^4}{a}\right )}{10 x^{10} \sqrt [4]{1+\frac {b x^4}{a}}} \] Input:

Integrate[(a + b*x^4)^(5/4)/x^11,x]
 

Output:

-1/10*(a*(a + b*x^4)^(1/4)*Hypergeometric2F1[-5/2, -5/4, -3/2, -((b*x^4)/a 
)])/(x^10*(1 + (b*x^4)/a)^(1/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {807, 247, 247, 264, 231, 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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^4+a\right )^{5/4}}{x^{12}}dx^2\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \int \frac {\sqrt [4]{b x^4+a}}{x^8}dx^2-\frac {\left (a+b x^4\right )^{5/4}}{5 x^{10}}\right )\)

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 231

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

\(\Big \downarrow \) 229

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (\frac {1}{6} b \left (-\frac {\sqrt {b} \left (\frac {b x^4}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a+b x^4\right )^{3/4}}-\frac {\sqrt [4]{a+b x^4}}{a x^2}\right )-\frac {\sqrt [4]{a+b x^4}}{3 x^6}\right )-\frac {\left (a+b x^4\right )^{5/4}}{5 x^{10}}\right )\)

Input:

Int[(a + b*x^4)^(5/4)/x^11,x]
 

Output:

(-1/5*(a + b*x^4)^(5/4)/x^10 + (b*(-1/3*(a + b*x^4)^(1/4)/x^6 + (b*(-((a + 
 b*x^4)^(1/4)/(a*x^2)) - (Sqrt[b]*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[( 
Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(Sqrt[a]*(a + b*x^4)^(3/4))))/6))/2)/2
 

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 231

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

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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]

\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{11}}d x\]

Input:

int((b*x^4+a)^(5/4)/x^11,x)
 

Output:

int((b*x^4+a)^(5/4)/x^11,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{11}} \,d x } \] Input:

integrate((b*x^4+a)^(5/4)/x^11,x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(5/4)/x^11, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.94 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.28 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=- \frac {a^{\frac {5}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {5}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 x^{10}} \] Input:

integrate((b*x**4+a)**(5/4)/x**11,x)
 

Output:

-a**(5/4)*hyper((-5/2, -5/4), (-3/2,), b*x**4*exp_polar(I*pi)/a)/(10*x**10 
)
 

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{11}} \,d x } \] Input:

integrate((b*x^4+a)^(5/4)/x^11,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(5/4)/x^11, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{11}} \,d x } \] Input:

integrate((b*x^4+a)^(5/4)/x^11,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^4 + a)^(5/4)/x^11, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{5/4}}{x^{11}} \,d x \] Input:

int((a + b*x^4)^(5/4)/x^11,x)
 

Output:

int((a + b*x^4)^(5/4)/x^11, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^{11}} \, dx=\frac {-4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a -9 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}+5 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b \,x^{15}+a \,x^{11}}d x \right ) a^{2} x^{10}}{45 x^{10}} \] Input:

int((b*x^4+a)^(5/4)/x^11,x)
 

Output:

( - 4*(a + b*x**4)**(1/4)*a - 9*(a + b*x**4)**(1/4)*b*x**4 + 5*int((a + b* 
x**4)**(1/4)/(a*x**11 + b*x**15),x)*a**2*x**10)/(45*x**10)