\(\int (a+\frac {b}{x^4})^{3/2} \, dx\) [534]

Optimal result

Integrand size = 11, antiderivative size = 251 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {12 a \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{5 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+a \sqrt {a+\frac {b}{x^4}} x+\frac {12 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}}-\frac {6 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}} \] Output:

-1/5*b*(a+b/x^4)^(1/2)/x^3-12/5*a*b^(1/2)*(a+b/x^4)^(1/2)/(a^(1/2)+b^(1/2) 
/x^2)/x+a*(a+b/x^4)^(1/2)*x+12/5*a^(5/4)*b^(1/4)*((a+b/x^4)/(a^(1/2)+b^(1/ 
2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*EllipticE(sin(2*arccot(a^(1/4)*x/b^ 
(1/4))),1/2*2^(1/2))/(a+b/x^4)^(1/2)-6/5*a^(5/4)*b^(1/4)*((a+b/x^4)/(a^(1/ 
2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*InverseJacobiAM(2*arccot(a^ 
(1/4)*x/b^(1/4)),1/2*2^(1/2))/(a+b/x^4)^(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 = 52, normalized size of antiderivative = 0.21 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {a x^4}{b}\right )}{5 x^3 \sqrt {1+\frac {a x^4}{b}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {773, 809, 811, 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.

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

\(\Big \downarrow \) 773

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

\(\Big \downarrow \) 809

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

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 773

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

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 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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 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 = 0.62 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.56

method	result	size
risch	\(-\frac {\left (7 a \,x^{4}+b \right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{5 x^{3}}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\)	\(140\)
default	\(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x \left (-12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}+8 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a b \,x^{4}+\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2}\right )}{5 \left (a \,x^{4}+b \right )^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(228\)

Input:

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

Output:

-1/5*(7*a*x^4+b)/x^3*((a*x^4+b)/x^4)^(1/2)+12/5*I*a^(3/2)*b^(1/2)/(I*a^(1/ 
2)/b^(1/2))^(1/2)*(1-I*a^(1/2)/b^(1/2)*x^2)^(1/2)*(1+I*a^(1/2)/b^(1/2)*x^2 
)^(1/2)/(a*x^4+b)*(EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)-EllipticE(x*(I 
*a^(1/2)/b^(1/2))^(1/2),I))*((a*x^4+b)/x^4)^(1/2)*x^2
 

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.17 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=- \frac {a^{\frac {3}{2}} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.15 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {x\,{\left (a+\frac {b}{x^4}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ -\frac {a\,x^4}{b}\right )}{5\,{\left (\frac {a\,x^4}{b}+1\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=\frac {\sqrt {a \,x^{4}+b}\, a \,x^{4}-\sqrt {a \,x^{4}+b}\, b -4 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a \,x^{10}+b \,x^{6}}d x \right ) b^{2} x^{5}}{x^{5}} \] Input:

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

Output:

(sqrt(a*x**4 + b)*a*x**4 - sqrt(a*x**4 + b)*b - 4*int(sqrt(a*x**4 + b)/(a* 
x**10 + b*x**6),x)*b**2*x**5)/x**5