\(\int \frac {\sqrt {x}}{(b x^2+c x^4)^{3/2}} \, dx\) [275]

Optimal result

Integrand size = 21, antiderivative size = 145 \[ \int \frac {\sqrt {x}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {5 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{6 b^{9/4} \sqrt {b x^2+c x^4}} \] Output:

1/b/x^(1/2)/(c*x^4+b*x^2)^(1/2)-5/3*(c*x^4+b*x^2)^(1/2)/b^2/x^(5/2)-5/6*c^ 
(3/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Invers 
eJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/b^(9/4)/(c*x^4+b* 
x^2)^(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 = 60, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {x}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {c x^2}{b}\right )}{3 b \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[Sqrt[x]/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1428, 1430, 1431, 266, 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.

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

\(\Big \downarrow \) 1428

\(\displaystyle \frac {5 \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx}{2 b}+\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[Sqrt[x]/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

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 1428

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(-d)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] + Simp[d^2* 
((m + 4*p + 3)/(2*b*(p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^(p + 1), 
x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[p, -1]
 

rule 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, 
 x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[m + 2*p + 1, 0 
]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) \left (5 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) x +10 c \,x^{2}+4 b \right )}{6 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{2}}\)	\(127\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right )}{3 b^{2} \sqrt {x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {c \left (\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}+3 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(305\)

Input:

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

Output:

-1/6/(c*x^4+b*x^2)^(3/2)*x^(3/2)*(c*x^2+b)*(5*(-b*c)^(1/2)*((c*x+(-b*c)^(1 
/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)* 
(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2 
),1/2*2^(1/2))*x+10*c*x^2+4*b)/b^2
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {x}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {5 \, {\left (c x^{5} + b x^{3}\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} {\left (5 \, c x^{2} + 2 \, b\right )} \sqrt {x}}{3 \, {\left (b^{2} c x^{5} + b^{3} x^{3}\right )}} \] Input:

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

Output:

-1/3*(5*(c*x^5 + b*x^3)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c 
*x^4 + b*x^2)*(5*c*x^2 + 2*b)*sqrt(x))/(b^2*c*x^5 + b^3*x^3)
 

Sympy [F]

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

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

Output:

Integral(sqrt(x)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(x)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

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

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

Output:

integrate(sqrt(x)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**3 + 2*b*c*x**5 + c**2*x**7),x)