\(\int \frac {1}{(a+\frac {b}{x^{4/3}})^{3/2}} \, dx\) [160]

Optimal result

Integrand size = 13, antiderivative size = 143 \[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=-\frac {3 x}{2 a \sqrt {a+\frac {b}{x^{4/3}}}}+\frac {5 \sqrt {a+\frac {b}{x^{4/3}}} x}{2 a^2}+\frac {5 b^{3/4} \sqrt {\frac {a+\frac {b}{x^{4/3}}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^{2/3}}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^{2/3}}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [3]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4 a^{9/4} \sqrt {a+\frac {b}{x^{4/3}}}} \] Output:

-3/2*x/a/(a+b/x^(4/3))^(1/2)+5/2*(a+b/x^(4/3))^(1/2)*x/a^2+5/4*b^(3/4)*((a 
+b/x^(4/3))/(a^(1/2)+b^(1/2)/x^(2/3))^2)^(1/2)*(a^(1/2)+b^(1/2)/x^(2/3))*I 
nverseJacobiAM(2*arccot(a^(1/4)*x^(1/3)/b^(1/4)),1/2*2^(1/2))/a^(9/4)/(a+b 
/x^(4/3))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=\frac {\sqrt {1+\frac {b}{a x^{4/3}}} x \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {b}{a x^{4/3}}\right )}{a \sqrt {a+\frac {b}{x^{4/3}}}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {774, 858, 819, 847, 761}

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.

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

\(\Big \downarrow \) 774

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

Defintions of rubi rules used

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 774

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, 
Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre 
eQ[{a, b, p}, x] && FractionQ[n]
 

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 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 [C] (verified)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\left (b +a \,x^{\frac {4}{3}}\right ) \left (2 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{\frac {5}{3}}-5 b \sqrt {\frac {-i \sqrt {a}\, x^{\frac {2}{3}}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{\frac {2}{3}}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \,x^{\frac {1}{3}}\right )}{2 \left (\frac {b +a \,x^{\frac {4}{3}}}{x^{\frac {4}{3}}}\right )^{\frac {3}{2}} x^{2} a^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(134\)
default	\(\frac {\left (b +a \,x^{\frac {4}{3}}\right ) \left (2 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{\frac {5}{3}}-5 b \sqrt {-\frac {i \sqrt {a}\, x^{\frac {2}{3}}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{\frac {2}{3}}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \,x^{\frac {1}{3}}\right )}{2 \left (\frac {b +a \,x^{\frac {4}{3}}}{x^{\frac {4}{3}}}\right )^{\frac {3}{2}} x^{2} a^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(137\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=- \frac {3 x \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{\frac {4}{3}}}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Valueintegrate(sageVARx^3*sign(sageVARx)^2/ 
(((sageVARx*
 

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.26 \[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=\frac {x\,{\left (\frac {b}{a\,x^{4/3}}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {3}{2};\ \frac {1}{4};\ -\frac {b}{a\,x^{4/3}}\right )}{{\left (a+\frac {b}{x^{4/3}}\right )}^{3/2}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^{4/3}}\right )^{3/2}} \, dx=\frac {3 x^{\frac {5}{3}} \sqrt {x^{\frac {4}{3}} a +b}\, a +15 x^{\frac {1}{3}} \sqrt {x^{\frac {4}{3}} a +b}\, b -5 x^{\frac {4}{3}} \left (\int \frac {\sqrt {x^{\frac {4}{3}} a +b}}{x^{\frac {2}{3}} b^{2}+x^{\frac {10}{3}} a^{2}+2 a b \,x^{2}}d x \right ) a \,b^{2}-5 \left (\int \frac {\sqrt {x^{\frac {4}{3}} a +b}}{x^{\frac {2}{3}} b^{2}+x^{\frac {10}{3}} a^{2}+2 a b \,x^{2}}d x \right ) b^{3}}{3 a^{2} \left (x^{\frac {4}{3}} a +b \right )} \] Input:

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

Output:

(3*x**(2/3)*sqrt(x**(1/3)*a*x + b)*a*x + 15*x**(1/3)*sqrt(x**(1/3)*a*x + b 
)*b - 5*x**(1/3)*int(sqrt(x**(1/3)*a*x + b)/(x**(2/3)*b**2 + x**(1/3)*a**2 
*x**3 + 2*a*b*x**2),x)*a*b**2*x - 5*int(sqrt(x**(1/3)*a*x + b)/(x**(2/3)*b 
**2 + x**(1/3)*a**2*x**3 + 2*a*b*x**2),x)*b**3)/(3*a**2*(x**(1/3)*a*x + b) 
)