\(\int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx\) [259]

Optimal result

Integrand size = 21, antiderivative size = 231 \[ \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (b+c x^2\right )}{\sqrt {c} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 \sqrt [4]{b} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {b x^2+c x^4}}+\frac {\sqrt [4]{b} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{c^{3/4} \sqrt {b x^2+c x^4}} \] Output:

2*x^(3/2)*(c*x^2+b)/c^(1/2)/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2)^(1/2)-2*b^(1 
/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Elliptic 
E(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))/c^(3/4)/(c*x^4+b*x^2 
)^(1/2)+b^(1/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1 
/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(3/4) 
/(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 = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.25 \[ \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{5/2} \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )}{3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(2*x^(5/2)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/ 
b)])/(3*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1431, 266, 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.

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

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 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 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]
 

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 = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {\sqrt {x}\, b \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, c}\)	\(131\)

Input:

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

Output:

1/(c*x^4+b*x^2)^(1/2)*x^(1/2)*b/c*((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) 
*(2*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-Ellipti 
cF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.10 \[ \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right )}{\sqrt {c}} \] Input:

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

Output:

-2*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x))/sqrt(c)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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