\(\int \frac {A+B x^2}{x^{5/2} (b x^2+c x^4)^{3/2}} \, dx\) [272]

Optimal result

Integrand size = 28, antiderivative size = 405 \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}+\frac {9 b B-11 A c}{9 b^2 x^{3/2} \sqrt {b x^2+c x^4}}-\frac {7 c^{3/2} (9 b B-11 A c) x^{3/2} \left (b+c x^2\right )}{15 b^4 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {7 (9 b B-11 A c) \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}+\frac {7 c (9 b B-11 A c) \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {7 c^{5/4} (9 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{15/4} \sqrt {b x^2+c x^4}}-\frac {7 c^{5/4} (9 b B-11 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{30 b^{15/4} \sqrt {b x^2+c x^4}} \] Output:

-2/9*A/b/x^(7/2)/(c*x^4+b*x^2)^(1/2)+1/9*(-11*A*c+9*B*b)/b^2/x^(3/2)/(c*x^ 
4+b*x^2)^(1/2)-7/15*c^(3/2)*(-11*A*c+9*B*b)*x^(3/2)*(c*x^2+b)/b^4/(b^(1/2) 
+c^(1/2)*x)/(c*x^4+b*x^2)^(1/2)-7/45*(-11*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b 
^3/x^(7/2)+7/15*c*(-11*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b^4/x^(3/2)+7/15*c^( 
5/4)*(-11*A*c+9*B*b)*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^( 
15/4)/(c*x^4+b*x^2)^(1/2)-7/30*c^(5/4)*(-11*A*c+9*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^(15/4)/(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 = 79, normalized size of antiderivative = 0.20 \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-10 A b+2 (-9 b B+11 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},-\frac {c x^2}{b}\right )}{45 b^2 x^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(-10*A*b + 2*(-9*b*B + 11*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[- 
5/4, 3/2, -1/4, -((c*x^2)/b)])/(45*b^2*x^(7/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 387, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1944, 1428, 1430, 1430, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(9 b B-11 A c) \int \frac {1}{\sqrt {x} \left (c x^4+b x^2\right )^{3/2}}dx}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1428

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \int \frac {1}{x^{5/2} \sqrt {c x^4+b x^2}}dx}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \int \frac {1}{\sqrt {x} \sqrt {c x^4+b x^2}}dx}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {c \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{b}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {c x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {2 c x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {(9 b B-11 A c) \left (\frac {7 \left (-\frac {3 c \left (\frac {2 c x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{b x^{3/2}}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^{7/2}}\right )}{2 b}+\frac {1}{b x^{3/2} \sqrt {b x^2+c x^4}}\right )}{9 b}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}\)

Input:

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

Output:

(-2*A)/(9*b*x^(7/2)*Sqrt[b*x^2 + c*x^4]) + ((9*b*B - 11*A*c)*(1/(b*x^(3/2) 
*Sqrt[b*x^2 + c*x^4]) + (7*((-2*Sqrt[b*x^2 + c*x^4])/(5*b*x^(7/2)) - (3*c* 
((-2*Sqrt[b*x^2 + c*x^4])/(b*x^(3/2)) + (2*c*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)*S 
qrt[x])/b^(1/4)], 1/2])/(c^(1/4)*Sqrt[b + c*x^2]))/Sqrt[c]) + (b^(1/4)*(Sq 
rt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*A 
rcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[b + c*x^2])))/(b*S 
qrt[b*x^2 + c*x^4])))/(5*b)))/(2*b)))/(9*b)
 

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 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]
 

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]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {\left (c \,x^{2}+b \right ) \left (462 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-231 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-378 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}+189 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}-462 A \,c^{3} x^{6}+378 B b \,c^{2} x^{6}-308 A b \,c^{2} x^{4}+252 x^{4} B \,b^{2} c +44 A \,b^{2} c \,x^{2}-36 x^{2} B \,b^{3}-20 A \,b^{3}\right )}{90 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} x^{\frac {3}{2}} b^{4}}\)	\(450\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (93 x^{4} A \,c^{2}-72 x^{4} B b c -16 A b c \,x^{2}+9 x^{2} B \,b^{2}+5 b^{2} A \right )}{45 b^{4} x^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c^{2} \left (\frac {\left (31 A c -24 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{c \sqrt {c \,x^{3}+b x}}-15 b \left (A c -B b \right ) \left (\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}-\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(467\)

Input:

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

Output:

1/90/(c*x^4+b*x^2)^(3/2)/x^(3/2)*(c*x^2+b)*(462*A*((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^2*x^4-231*A*((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*c^2*x^4-378*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)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1 
/2))^(1/2),1/2*2^(1/2))*b^2*c*x^4+189*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^4-462*A*c^3*x^6+378*B*b*c^2*x^6-308*A*b*c^2*x^4+252*x^4*B*b^2*c+44*A*b 
^2*c*x^2-36*x^2*B*b^3-20*A*b^3)/b^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.40 \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {21 \, {\left ({\left (9 \, B b c^{2} - 11 \, A c^{3}\right )} x^{8} + {\left (9 \, B b^{2} c - 11 \, A b c^{2}\right )} x^{6}\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (21 \, {\left (9 \, B b c^{2} - 11 \, A c^{3}\right )} x^{6} + 14 \, {\left (9 \, B b^{2} c - 11 \, A b c^{2}\right )} x^{4} - 10 \, A b^{3} - 2 \, {\left (9 \, B b^{3} - 11 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{45 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \] Input:

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

Output:

1/45*(21*((9*B*b*c^2 - 11*A*c^3)*x^8 + (9*B*b^2*c - 11*A*b*c^2)*x^6)*sqrt( 
c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) + (21*(9* 
B*b*c^2 - 11*A*c^3)*x^6 + 14*(9*B*b^2*c - 11*A*b*c^2)*x^4 - 10*A*b^3 - 2*( 
9*B*b^3 - 11*A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^4*c*x^8 + b^5*x 
^6)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, b +11 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{10}+2 b c \,x^{8}+b^{2} x^{6}}d x \right ) a b c \,x^{4}+11 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{10}+2 b c \,x^{8}+b^{2} x^{6}}d x \right ) a \,c^{2} x^{6}-9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{10}+2 b c \,x^{8}+b^{2} x^{6}}d x \right ) b^{3} x^{4}-9 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{10}+2 b c \,x^{8}+b^{2} x^{6}}d x \right ) b^{2} c \,x^{6}}{11 \sqrt {x}\, c \,x^{4} \left (c \,x^{2}+b \right )} \] Input:

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

Output:

( - 2*sqrt(b + c*x**2)*b + 11*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2 
*x**6 + 2*b*c*x**8 + c**2*x**10),x)*a*b*c*x**4 + 11*sqrt(x)*int((sqrt(x)*s 
qrt(b + c*x**2))/(b**2*x**6 + 2*b*c*x**8 + c**2*x**10),x)*a*c**2*x**6 - 9* 
sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**6 + 2*b*c*x**8 + c**2*x**1 
0),x)*b**3*x**4 - 9*sqrt(x)*int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**6 + 2* 
b*c*x**8 + c**2*x**10),x)*b**2*c*x**6)/(11*sqrt(x)*c*x**4*(b + c*x**2))