3.4.94 \(\int \frac {x^{11/2}}{(b x^2+c x^4)^{3/2}} \, dx\) [394]

3.4.94.1 Optimal result

Integrand size = 21, antiderivative size = 259 \[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^{5/2}}{c \sqrt {b x^2+c x^4}}+\frac {3 x^{3/2} \left (b+c x^2\right )}{c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {3 \sqrt [4]{b} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt {b x^2+c x^4}}+\frac {3 \sqrt [4]{b} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{7/4} \sqrt {b x^2+c x^4}} \]

output

-x^(5/2)/c/(c*x^4+b*x^2)^(1/2)+3*x^(3/2)*(c*x^2+b)/c^(3/2)/(b^(1/2)+x*c^(1 
/2))/(c*x^4+b*x^2)^(1/2)-3*b^(1/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4) 
))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan( 
c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^( 
1/2)+x*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+b*x^2)^(1/2)+3/2*b^(1/4)*x*(cos(2* 
arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^( 
1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1 
/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+b*x^ 
2)^(1/2)
 

3.4.94.2 Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {2 x^{5/2} \left (-1+\sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{c \sqrt {x^2 \left (b+c x^2\right )}} \]

input

Integrate[x^(11/2)/(b*x^2 + c*x^4)^(3/2),x]
 

output

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

3.4.94.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1427, 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.

\(\displaystyle \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1427

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

input

Int[x^(11/2)/(b*x^2 + c*x^4)^(3/2),x]
 

output

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

3.4.94.3.1 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 1427

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

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]
 

3.4.94.4 Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (6 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b -3 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b -2 c \,x^{2}\right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{2}}\)	\(200\)

input

int(x^(11/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

output

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

3.4.94.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.25 \[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (c x^{2} + b\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + \sqrt {c x^{4} + b x^{2}} c \sqrt {x}}{c^{3} x^{2} + b c^{2}} \]

input

integrate(x^(11/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

output

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

3.4.94.6 Sympy [F]

\[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{\frac {11}{2}}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

input

integrate(x**(11/2)/(c*x**4+b*x**2)**(3/2),x)
 

output

Integral(x**(11/2)/(x**2*(b + c*x**2))**(3/2), x)
 

3.4.94.7 Maxima [F]

\[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {11}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate(x^(11/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

output

integrate(x^(11/2)/(c*x^4 + b*x^2)^(3/2), x)
 

3.4.94.8 Giac [F]

\[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {11}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate(x^(11/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

output

integrate(x^(11/2)/(c*x^4 + b*x^2)^(3/2), x)
 

3.4.94.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^{11/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{11/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]

input

int(x^(11/2)/(b*x^2 + c*x^4)^(3/2),x)
 

output

int(x^(11/2)/(b*x^2 + c*x^4)^(3/2), x)