3.3.55 \(\int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx\) [255]

3.3.55.1 Optimal result

Integrand size = 28, antiderivative size = 369 \[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 c^{3/2} (9 b B-7 A c) x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{11/2}}-\frac {2 (9 b B-7 A c) \sqrt {b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac {2 c (9 b B-7 A c) \sqrt {b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac {2 c^{5/4} (9 b B-7 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {b x^2+c x^4}}-\frac {c^{5/4} (9 b B-7 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 b^{11/4} \sqrt {b x^2+c x^4}} \]

output

-2/15*c^(3/2)*(-7*A*c+9*B*b)*x^(3/2)*(c*x^2+b)/b^3/(b^(1/2)+x*c^(1/2))/(c* 
x^4+b*x^2)^(1/2)-2/9*A*(c*x^4+b*x^2)^(1/2)/b/x^(11/2)-2/45*(-7*A*c+9*B*b)* 
(c*x^4+b*x^2)^(1/2)/b^2/x^(7/2)+2/15*c*(-7*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/ 
b^3/x^(3/2)+2/15*c^(5/4)*(-7*A*c+9*B*b)*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*ar 
ctan(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)/b^(11/4)/(c*x^4+b*x^2)^(1/2)-1/15*c^(5/4)*(- 
7*A*c+9*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arct 
an(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)/b^(11/4)/(c*x^4+b*x^2)^(1/2)
 

3.3.55.2 Mathematica [C] (verified)

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

Time = 10.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.23 \[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=-\frac {2 \left (5 A \left (b+c x^2\right )+(9 b B-7 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},-\frac {c x^2}{b}\right )\right )}{45 b x^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

input

Integrate[(A + B*x^2)/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]
 

output

(-2*(5*A*(b + c*x^2) + (9*b*B - 7*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeomet 
ric2F1[-5/4, 1/2, -1/4, -((c*x^2)/b)]))/(45*b*x^(7/2)*Sqrt[x^2*(b + c*x^2) 
])
 

3.3.55.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1944, 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.

\(\displaystyle \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

input

Int[(A + B*x^2)/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]
 

output

(-2*A*Sqrt[b*x^2 + c*x^4])/(9*b*x^(11/2)) + ((9*b*B - 7*A*c)*((-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)*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)/(Sqr 
t[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])))/(5*b)))/(9*b)
 

3.3.55.3.1 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 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]
 

3.3.55.4 Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (21 A \,c^{2} x^{4}-27 x^{4} B b c -7 A b c \,x^{2}+9 b^{2} B \,x^{2}+5 b^{2} A \right )}{45 b^{3} x^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c \left (7 A c -9 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 b^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(266\)
default	\(\frac {42 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-21 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-54 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}+27 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}-42 A \,c^{3} x^{6}+54 x^{6} B b \,c^{2}-28 A b \,c^{2} x^{4}+36 x^{4} B \,b^{2} c +4 A \,b^{2} c \,x^{2}-18 b^{3} B \,x^{2}-10 b^{3} A}{45 \sqrt {x^{4} c +b \,x^{2}}\, x^{\frac {7}{2}} b^{3}}\)	\(443\)

input

int((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

output

-2/45*(c*x^2+b)*(21*A*c^2*x^4-27*B*b*c*x^4-7*A*b*c*x^2+9*B*b^2*x^2+5*A*b^2 
)/b^3/x^(7/2)/(x^2*(c*x^2+b))^(1/2)+1/15*c*(7*A*c-9*B*b)/b^3*(-b*c)^(1/2)* 
((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))*c/(-b 
*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)/(c*x^3+b*x)^(1/2)*(-2/c*(-b*c)^ 
(1/2)*EllipticE(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+1 
/c*(-b*c)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2* 
2^(1/2)))*x^(1/2)/(x^2*(c*x^2+b))^(1/2)*(x*(c*x^2+b))^(1/2)
 

3.3.55.5 Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.28 \[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left (3 \, {\left (9 \, B b c - 7 \, A c^{2}\right )} \sqrt {c} x^{6} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (3 \, {\left (9 \, B b c - 7 \, A c^{2}\right )} x^{4} - 5 \, A b^{2} - {\left (9 \, B b^{2} - 7 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{45 \, b^{3} x^{6}} \]

input

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

output

2/45*(3*(9*B*b*c - 7*A*c^2)*sqrt(c)*x^6*weierstrassZeta(-4*b/c, 0, weierst 
rassPInverse(-4*b/c, 0, x)) + (3*(9*B*b*c - 7*A*c^2)*x^4 - 5*A*b^2 - (9*B* 
b^2 - 7*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^3*x^6)
 

3.3.55.6 Sympy [F]

\[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x^{\frac {9}{2}} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

input

integrate((B*x**2+A)/x**(9/2)/(c*x**4+b*x**2)**(1/2),x)
 

output

Integral((A + B*x**2)/(x**(9/2)*sqrt(x**2*(b + c*x**2))), x)
 

3.3.55.7 Maxima [F]

\[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2}} x^{\frac {9}{2}}} \,d x } \]

input

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

output

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)
 

3.3.55.8 Giac [F]

\[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2}} x^{\frac {9}{2}}} \,d x } \]

input

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

output

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)
 

3.3.55.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx=\int \frac {B\,x^2+A}{x^{9/2}\,\sqrt {c\,x^4+b\,x^2}} \,d x \]

input

int((A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)^(1/2)),x)
 

output

int((A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)^(1/2)), x)