\(\int \frac {A+B x^2}{x^{5/2} (b x^2+c x^4)^{3/2}} \, dx\) [268]

Optimal result

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

[Out]

-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)+x*c^(1/2))/(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*(cos(2*arct
an(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)/b^(15/4)/(c*x^4+b*x^2
)^(1/2)-7/30*c^(5/4)*(-11*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)))*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^(15/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2063, 2048, 2050, 2057, 335, 311, 226, 1210} \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {7 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (9 b B-11 A c) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{30 b^{15/4} \sqrt {b x^2+c x^4}}+\frac {7 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (9 b B-11 A c) E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{15/4} \sqrt {b x^2+c x^4}}-\frac {7 c^{3/2} x^{3/2} \left (b+c x^2\right ) (9 b B-11 A c)}{15 b^4 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {7 c \sqrt {b x^2+c x^4} (9 b B-11 A c)}{15 b^4 x^{3/2}}-\frac {7 \sqrt {b x^2+c x^4} (9 b B-11 A c)}{45 b^3 x^{7/2}}+\frac {9 b B-11 A c}{9 b^2 x^{3/2} \sqrt {b x^2+c x^4}}-\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}} \]

[In]

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

[Out]

(-2*A)/(9*b*x^(7/2)*Sqrt[b*x^2 + c*x^4]) + (9*b*B - 11*A*c)/(9*b^2*x^(3/2)*Sqrt[b*x^2 + c*x^4]) - (7*c^(3/2)*(
9*b*B - 11*A*c)*x^(3/2)*(b + c*x^2))/(15*b^4*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (7*(9*b*B - 11*A*c)*
Sqrt[b*x^2 + c*x^4])/(45*b^3*x^(7/2)) + (7*c*(9*b*B - 11*A*c)*Sqrt[b*x^2 + c*x^4])/(15*b^4*x^(3/2)) + (7*c^(5/
4)*(9*b*B - 11*A*c)*x*(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])/(15*b^(15/4)*Sqrt[b*x^2 + c*x^4]) - (7*c^(5/4)*(9*b*B - 11*A*c)*x*(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])/(30*b^(1
5/4)*Sqrt[b*x^2 + c*x^4])

Rule 226

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 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2048

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

Rule 2050

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(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] /; F
reeQ[{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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A}{9 b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {\left (2 \left (-\frac {9 b B}{2}+\frac {11 A c}{2}\right )\right ) \int \frac {1}{\sqrt {x} \left (b x^2+c x^4\right )^{3/2}} \, dx}{9 b} \\ & = -\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 (9 b B-11 A c)) \int \frac {1}{x^{5/2} \sqrt {b x^2+c x^4}} \, dx}{18 b^2} \\ & = -\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 (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)) \int \frac {1}{\sqrt {x} \sqrt {b x^2+c x^4}} \, dx}{30 b^3} \\ & = -\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 (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 {\left (7 c^2 (9 b B-11 A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{30 b^4} \\ & = -\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 (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 {\left (7 c^2 (9 b B-11 A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{30 b^4 \sqrt {b x^2+c x^4}} \\ & = -\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 (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 {\left (7 c^2 (9 b B-11 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^4 \sqrt {b x^2+c x^4}} \\ & = -\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 (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 {\left (7 c^{3/2} (9 b B-11 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^{7/2} \sqrt {b x^2+c x^4}}+\frac {\left (7 c^{3/2} (9 b B-11 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^{7/2} \sqrt {b x^2+c x^4}} \\ & = -\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 \tan ^{-1}\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}} F\left (2 \tan ^{-1}\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}} \\ \end{align*}

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 \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-10 A b+2 (-9 b B+11 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},-\frac {1}{4},-\frac {c x^2}{b}\right )}{45 b^2 x^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

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

[Out]

(-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)])

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.11

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

[In]

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

[Out]

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)*(-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*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)*(-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*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)*(-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^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)*(-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^4-462*A*c^3*x^6+378*x^6*B*b*c^2-308*A*b*c^2*x^4+252*x^4*B*b^2*
c+44*A*b^2*c*x^2-36*b^3*B*x^2-20*b^3*A)/b^4

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.40 \[ \int \frac {A+B x^2}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {21 \, {\left ({\left (9 \, B b c^{2} - 11 \, A c^{3}\right )} x^{8} + {\left (9 \, B b^{2} c - 11 \, A b c^{2}\right )} x^{6}\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (21 \, {\left (9 \, B b c^{2} - 11 \, A c^{3}\right )} x^{6} + 14 \, {\left (9 \, B b^{2} c - 11 \, A b c^{2}\right )} x^{4} - 10 \, A b^{3} - 2 \, {\left (9 \, B b^{3} - 11 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{45 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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