\(\int \frac {x^{7/2}}{(b x^2+c x^4)^{3/2}} \, dx\) [272]

Optimal result

Integrand size = 21, antiderivative size = 260 \[ \int \frac {x^{7/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{5/2}}{b \sqrt {b x^2+c x^4}}-\frac {x^{3/2} \left (b+c x^2\right )}{b \sqrt {c} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{b^{3/4} c^{3/4} \sqrt {b x^2+c x^4}}-\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 b^{3/4} c^{3/4} \sqrt {b x^2+c x^4}} \] Output:

x^(5/2)/b/(c*x^4+b*x^2)^(1/2)-x^(3/2)*(c*x^2+b)/b/c^(1/2)/(b^(1/2)+c^(1/2) 
*x)/(c*x^4+b*x^2)^(1/2)+x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)* 
x)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))/ 
b^(3/4)/c^(3/4)/(c*x^4+b*x^2)^(1/2)-1/2*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))/b^(3/4)/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 = 10.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.23 \[ \int \frac {x^{7/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 x^{5/2} \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {c x^2}{b}\right )}{3 b \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[x^(7/2)/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 272, 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 = {1428, 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.

\(\displaystyle \int \frac {x^{7/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1428

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[x^(7/2)/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

x^(5/2)/(b*Sqrt[b*x^2 + c*x^4]) - (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] + Sq 
rt[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])))/(b*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 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 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 = 0.35 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (2 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b -2 c \,x^{2}\right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c b}\)	\(203\)

Input:

int(x^(7/2)/(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)*(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)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2) 
)*b-((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))*b-2*c*x^2)/c/b
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.25 \[ \int \frac {x^{7/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (c x^{2} + b\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + \sqrt {c x^{4} + b x^{2}} c \sqrt {x}}{b c^{2} x^{2} + b^{2} c} \] Input:

integrate(x^(7/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

((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))/(b*c^2*x^2 + b^2*c)
 

Sympy [F]

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

integrate(x**(7/2)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**(7/2)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [F]

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

integrate(x^(7/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^(7/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

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

integrate(x^(7/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(x^(7/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

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

int(x^(7/2)/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int(x^(7/2)/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

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

int(x^(7/2)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

int((sqrt(x)*sqrt(b + c*x**2))/(b**2 + 2*b*c*x**2 + c**2*x**4),x)