\(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [267]

Optimal result

Integrand size = 28, antiderivative size = 137 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^{3/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(b B+A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 b^{5/4} c^{5/4} \sqrt {b x^2+c x^4}} \] Output:

-(-A*c+B*b)*x^(3/2)/b/c/(c*x^4+b*x^2)^(1/2)+1/2*(A*c+B*b)*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^(5/4)/c^(5/4)/(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 = 10.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.55 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (-b B+A c+(b B+A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{b c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1943, 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.

\(\displaystyle \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

-(((b*B - A*c)*x^(3/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + ((b*B + A*c)*x*(Sqrt[ 
b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT 
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*b^(5/4)*c^(5/4)*Sqrt[b*x^2 + c*x^4 
])
 

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 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 1943

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j 
+ 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( 
m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1)))   Int[(e*x)^(m 
 - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, 
n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 
] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c +B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b +2 A \,c^{2} x -2 B b c x \right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b \,c^{2}}\)	\(222\)

Input:

int(x^(5/2)*(B*x^2+A)/(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)*(A*(-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))*c+B*(-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)*El 
lipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b+2*A*c^2*x-2 
*B*b*c*x)/b/c^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.66 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left ({\left (B b c + A c^{2}\right )} x^{3} + {\left (B b^{2} + A b c\right )} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} {\left (B b c - A c^{2}\right )} \sqrt {x}}{b c^{3} x^{3} + b^{2} c^{2} x} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**(5/2)*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**(5/2)*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [F]

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

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

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

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

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

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

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