\(\int \frac {1}{x^6 (a+b x^4)^{3/2}} \, dx\) [395]

Optimal result

Integrand size = 15, antiderivative size = 282 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\frac {1}{2 a x^5 \sqrt {a+b x^4}}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {21 b^{3/2} x \sqrt {a+b x^4}}{10 a^3 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{11/4} \sqrt {a+b x^4}}-\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{20 a^{11/4} \sqrt {a+b x^4}} \] Output:

1/2/a/x^5/(b*x^4+a)^(1/2)-7/10*(b*x^4+a)^(1/2)/a^2/x^5+21/10*b*(b*x^4+a)^( 
1/2)/a^3/x-21/10*b^(3/2)*x*(b*x^4+a)^(1/2)/a^3/(a^(1/2)+b^(1/2)*x^2)+21/10 
*b^(5/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*E 
llipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/a^(11/4)/(b*x^4+a)^ 
(1/2)-21/20*b^(5/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2) 
^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(11/4 
)/(b*x^4+a)^(1/2)
 

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 = 0.19 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},-\frac {b x^4}{a}\right )}{5 a x^5 \sqrt {a+b x^4}} \] Input:

Integrate[1/(x^6*(a + b*x^4)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {819, 847, 847, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {7 \int \frac {1}{x^6 \sqrt {b x^4+a}}dx}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {7 \left (-\frac {3 b \int \frac {1}{x^2 \sqrt {b x^4+a}}dx}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {7 \left (-\frac {3 b \left (\frac {b \int \frac {x^2}{\sqrt {b x^4+a}}dx}{a}-\frac {\sqrt {a+b x^4}}{a x}\right )}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {7 \left (-\frac {3 b \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{a}-\frac {\sqrt {a+b x^4}}{a x}\right )}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7 \left (-\frac {3 b \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{a}-\frac {\sqrt {a+b x^4}}{a x}\right )}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {7 \left (-\frac {3 b \left (\frac {b \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{a}-\frac {\sqrt {a+b x^4}}{a x}\right )}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {7 \left (-\frac {3 b \left (\frac {b \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{a}-\frac {\sqrt {a+b x^4}}{a x}\right )}{5 a}-\frac {\sqrt {a+b x^4}}{5 a x^5}\right )}{2 a}+\frac {1}{2 a x^5 \sqrt {a+b x^4}}\)

Input:

Int[1/(x^6*(a + b*x^4)^(3/2)),x]
 

Output:

1/(2*a*x^5*Sqrt[a + b*x^4]) + (7*(-1/5*Sqrt[a + b*x^4]/(a*x^5) - (3*b*(-(S 
qrt[a + b*x^4]/(a*x)) + (b*(-((-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^ 
2)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b] 
*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + 
b*x^4]))/Sqrt[b]) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr 
t[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b 
^(3/4)*Sqrt[a + b*x^4])))/a))/(5*a)))/(2*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761

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

rule 819

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

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.85 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.56

method	result	size
default	\(-\frac {\sqrt {b \,x^{4}+a}}{5 a^{2} x^{5}}+\frac {8 b \sqrt {b \,x^{4}+a}}{5 a^{3} x}+\frac {b^{2} x^{3}}{2 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {21 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{10 a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(157\)
elliptic	\(-\frac {\sqrt {b \,x^{4}+a}}{5 a^{2} x^{5}}+\frac {8 b \sqrt {b \,x^{4}+a}}{5 a^{3} x}+\frac {b^{2} x^{3}}{2 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {21 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{10 a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(157\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}\, \left (-8 b \,x^{4}+a \right )}{5 a^{3} x^{5}}-\frac {b^{2} \left (3 a \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+8 b \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{5 a^{3}}\)	\(278\)

Input:

int(1/x^6/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/5*(b*x^4+a)^(1/2)/a^2/x^5+8/5*b*(b*x^4+a)^(1/2)/a^3/x+1/2*b^2/a^3*x^3/( 
(x^4+a/b)*b)^(1/2)-21/10*I*b^(3/2)/a^(5/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I* 
b^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*b^(1/2)*x^2/a^(1/2))^(1/2)/(b*x^4+a)^(1/2) 
*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2)) 
^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\frac {21 \, {\left (b^{2} x^{9} + a b x^{5}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 21 \, {\left (b^{2} x^{9} + a b x^{5}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (21 \, b^{2} x^{8} + 14 \, a b x^{4} - 2 \, a^{2}\right )} \sqrt {b x^{4} + a}}{10 \, {\left (a^{3} b x^{9} + a^{4} x^{5}\right )}} \] Input:

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="fricas")
 

Output:

1/10*(21*(b^2*x^9 + a*b*x^5)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/ 
a)^(1/4)), -1) - 21*(b^2*x^9 + a*b*x^5)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(ar 
csin(x*(-b/a)^(1/4)), -1) + (21*b^2*x^8 + 14*a*b*x^4 - 2*a^2)*sqrt(b*x^4 + 
 a))/(a^3*b*x^9 + a^4*x^5)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{5} \Gamma \left (- \frac {1}{4}\right )} \] Input:

integrate(1/x**6/(b*x**4+a)**(3/2),x)
 

Output:

gamma(-5/4)*hyper((-5/4, 3/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3 
/2)*x**5*gamma(-1/4))
 

Maxima [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate(1/((b*x^4 + a)^(3/2)*x^6), x)
 

Giac [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="giac")
 

Output:

integrate(1/((b*x^4 + a)^(3/2)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:

int(1/(x^6*(a + b*x^4)^(3/2)),x)
 

Output:

int(1/(x^6*(a + b*x^4)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{14}+2 a b \,x^{10}+a^{2} x^{6}}d x \] Input:

int(1/x^6/(b*x^4+a)^(3/2),x)
 

Output:

int(sqrt(a + b*x**4)/(a**2*x**6 + 2*a*b*x**10 + b**2*x**14),x)