\(\int \frac {(a-b x^4)^{3/2}}{x^2} \, dx\) [201]

Optimal result

Integrand size = 16, antiderivative size = 151 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^2} \, dx=-\frac {a \sqrt {a-b x^4}}{x}-\frac {1}{5} b x^3 \sqrt {a-b x^4}-\frac {12 a^{7/4} \sqrt [4]{b} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 \sqrt {a-b x^4}}+\frac {12 a^{7/4} \sqrt [4]{b} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{5 \sqrt {a-b x^4}} \] Output:

-a*(-b*x^4+a)^(1/2)/x-1/5*b*x^3*(-b*x^4+a)^(1/2)-12/5*a^(7/4)*b^(1/4)*(1-b 
*x^4/a)^(1/2)*EllipticE(b^(1/4)*x/a^(1/4),I)/(-b*x^4+a)^(1/2)+12/5*a^(7/4) 
*b^(1/4)*(1-b*x^4/a)^(1/2)*EllipticF(b^(1/4)*x/a^(1/4),I)/(-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 = 51, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^2} \, dx=-\frac {a \sqrt {a-b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},\frac {b x^4}{a}\right )}{x \sqrt {1-\frac {b x^4}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {809, 811, 836, 27, 765, 762, 1390, 1389, 327}

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.

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

\(\Big \downarrow \) 809

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 765

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1390

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

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

Output:

-((a - b*x^4)^(3/2)/x) - 6*b*((x^3*Sqrt[a - b*x^4])/5 + (2*a*((a^(3/4)*Sqr 
t[1 - (b*x^4)/a]*EllipticE[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(b^(3/4)*Sqrt 
[a - b*x^4]) - (a^(3/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a 
^(1/4)], -1])/(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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

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 836

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

rule 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && GtQ[a, 0]
 

rule 1390

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[Sqrt 
[1 + c*(x^4/a)]/Sqrt[a + c*x^4]   Int[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], x], 
x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] &&  !GtQ 
[a, 0] &&  !(LtQ[a, 0] && GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\sqrt {-b \,x^{4}+a}\, \left (b \,x^{4}+5 a \right )}{5 x}+\frac {12 a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(113\)
default	\(-\frac {a \sqrt {-b \,x^{4}+a}}{x}-\frac {b \,x^{3} \sqrt {-b \,x^{4}+a}}{5}+\frac {12 a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(121\)
elliptic	\(-\frac {a \sqrt {-b \,x^{4}+a}}{x}-\frac {b \,x^{3} \sqrt {-b \,x^{4}+a}}{5}+\frac {12 a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(121\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((-b*x^4 + a)^(3/2)/x^2, x)
 

Sympy [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^2} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^2} \, dx=\frac {{\left (a-b\,x^4\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ \frac {a}{b\,x^4}\right )}{5\,x\,{\left (1-\frac {a}{b\,x^4}\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^2} \, dx=\frac {7 \sqrt {-b \,x^{4}+a}\, a -\sqrt {-b \,x^{4}+a}\, b \,x^{4}+12 \left (\int \frac {\sqrt {-b \,x^{4}+a}}{-b \,x^{6}+a \,x^{2}}d x \right ) a^{2} x}{5 x} \] Input:

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

Output:

(7*sqrt(a - b*x**4)*a - sqrt(a - b*x**4)*b*x**4 + 12*int(sqrt(a - b*x**4)/ 
(a*x**2 - b*x**6),x)*a**2*x)/(5*x)