\(\int \frac {1}{x^{12} (a+b x^4)^{3/4}} \, dx\) [592]

Optimal result

Integrand size = 15, antiderivative size = 131 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{11 a x^{11}}+\frac {10 b \sqrt [4]{a+b x^4}}{77 a^2 x^7}-\frac {20 b^2 \sqrt [4]{a+b x^4}}{77 a^3 x^3}+\frac {40 b^{7/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 a^{7/2} \left (a+b x^4\right )^{3/4}} \] Output:

-1/11*(b*x^4+a)^(1/4)/a/x^11+10/77*b*(b*x^4+a)^(1/4)/a^2/x^7-20/77*b^2*(b* 
x^4+a)^(1/4)/a^3/x^3+40/77*b^(7/2)*(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 = 0.39 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},\frac {3}{4},-\frac {7}{4},-\frac {b x^4}{a}\right )}{11 x^{11} \left (a+b x^4\right )^{3/4}} \] Input:

Integrate[1/(x^12*(a + b*x^4)^(3/4)),x]
 

Output:

-1/11*((1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[-11/4, 3/4, -7/4, -((b*x^4) 
/a)])/(x^11*(a + b*x^4)^(3/4))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {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.

\(\displaystyle \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[1/(x^12*(a + b*x^4)^(3/4)),x]
 

Output:

-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)
 

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 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]

\[\int \frac {1}{x^{12} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x\]

Input:

int(1/x^12/(b*x^4+a)^(3/4),x)
 

Output:

int(1/x^12/(b*x^4+a)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{12}} \,d x } \] Input:

integrate(1/x^12/(b*x^4+a)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(1/4)/(b*x^16 + a*x^12), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\frac {\Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, \frac {3}{4} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} x^{11} \Gamma \left (- \frac {7}{4}\right )} \] Input:

integrate(1/x**12/(b*x**4+a)**(3/4),x)
 

Output:

gamma(-11/4)*hyper((-11/4, 3/4), (-7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a** 
(3/4)*x**11*gamma(-7/4))
 

Maxima [F]

\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{12}} \,d x } \] Input:

integrate(1/x^12/(b*x^4+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^4 + a)^(3/4)*x^12), x)
 

Giac [F]

\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{12}} \,d x } \] Input:

integrate(1/x^12/(b*x^4+a)^(3/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(1/((b*x^4 + a)^(3/4)*x^12), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\int \frac {1}{x^{12}\,{\left (b\,x^4+a\right )}^{3/4}} \,d x \] Input:

int(1/(x^12*(a + b*x^4)^(3/4)),x)
 

Output:

int(1/(x^12*(a + b*x^4)^(3/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{12} \left (a+b x^4\right )^{3/4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{12}}d x \] Input:

int(1/x^12/(b*x^4+a)^(3/4),x)
 

Output:

int(1/((a + b*x**4)**(3/4)*x**12),x)