\(\int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx\) [495]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^2}+\frac {2 \sqrt {\frac {a}{x^3}} x^2 \sqrt {1+x^2}}{1+x}-\frac {2 \sqrt {\frac {a}{x^3}} x^{3/2} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} E\left (2 \arctan \left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {1+x^2}}+\frac {\sqrt {\frac {a}{x^3}} x^{3/2} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{\sqrt {1+x^2}} \] Output:

-2*(a/x^3)^(1/2)*x*(x^2+1)^(1/2)+2*(a/x^3)^(1/2)*x^2*(x^2+1)^(1/2)/(1+x)-2 
*(a/x^3)^(1/2)*x^(3/2)*(1+x)*((x^2+1)/(1+x)^2)^(1/2)*EllipticE(sin(2*arcta 
n(x^(1/2))),1/2*2^(1/2))/(x^2+1)^(1/2)+(a/x^3)^(1/2)*x^(3/2)*(1+x)*((x^2+1 
)/(1+x)^2)^(1/2)*InverseJacobiAM(2*arctan(x^(1/2)),1/2*2^(1/2))/(x^2+1)^(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 = 27, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=-2 \sqrt {\frac {a}{x^3}} x \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-x^2\right ) \] Input:

Integrate[Sqrt[a/x^3]/Sqrt[1 + x^2],x]
 

Output:

-2*Sqrt[a/x^3]*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -x^2]
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {34, 264, 266, 834, 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.

\(\displaystyle \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {x^2+1}} \, dx\)

\(\Big \downarrow \) 34

\(\displaystyle x^{3/2} \sqrt {\frac {a}{x^3}} \int \frac {1}{x^{3/2} \sqrt {x^2+1}}dx\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[Sqrt[a/x^3]/Sqrt[1 + x^2],x]
 

Output:

Sqrt[a/x^3]*x^(3/2)*((-2*Sqrt[1 + x^2])/Sqrt[x] + 2*((Sqrt[x]*Sqrt[1 + x^2 
])/(1 + x) - ((1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticE[2*ArcTan[Sqrt[x] 
], 1/2])/Sqrt[1 + x^2] + ((1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*Ar 
cTan[Sqrt[x]], 1/2])/(2*Sqrt[1 + x^2])))
 

Defintions of rubi rules used

rule 34

Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F 
racPart[p]/x^(m*FracPart[p]))   Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x 
] &&  !IntegerQ[p]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, 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 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 [A] (verified)

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.14

method	result	size
meijerg	\(-2 \sqrt {\frac {a}{x^{3}}}\, x \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{4}\right ], -x^{2}\right )\)	\(22\)
default	\(\frac {\sqrt {\frac {a}{x^{3}}}\, x \left (2 \sqrt {2}\, \sqrt {-i \left (x +i\right )}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \operatorname {EllipticE}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {2}\, \sqrt {-i \left (x +i\right )}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-2 x^{2}-2\right )}{\sqrt {x^{2}+1}}\)	\(116\)
risch	\(-2 \sqrt {\frac {a}{x^{3}}}\, x \sqrt {x^{2}+1}+\frac {i \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {i \left (x -i\right )}\, \sqrt {i x}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {a}{x^{3}}}\, x \sqrt {a x \left (x^{2}+1\right )}}{\sqrt {a \,x^{3}+x a}\, \sqrt {x^{2}+1}}\)	\(122\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=-2 \, \sqrt {x^{2} + 1} x \sqrt {\frac {a}{x^{3}}} - 2 \, \sqrt {a} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, x\right )\right ) \] Input:

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

Output:

-2*sqrt(x^2 + 1)*x*sqrt(a/x^3) - 2*sqrt(a)*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, x))
 

Sympy [F]

\[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}}\, dx \] Input:

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

Output:

Integral(sqrt(a/x**3)/sqrt(x**2 + 1), x)
 

Maxima [F]

\[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}} \,d x } \] Input:

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

Output:

integrate(sqrt(a/x^3)/sqrt(x^2 + 1), x)
 

Giac [F]

\[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}} \,d x } \] Input:

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

Output:

integrate(sqrt(a/x^3)/sqrt(x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {x^2+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*int((sqrt(x)*sqrt(x**2 + 1))/(x**4 + x**2),x)