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

Optimal. Leaf size=369 \[ -\frac {4 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-13 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{7/4} \sqrt {b x^2+c x^4}}+\frac {8 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-13 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{7/4} \sqrt {b x^2+c x^4}}-\frac {8 b^2 x^{3/2} \left (b+c x^2\right ) (3 b B-13 A c)}{195 c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b \sqrt {x} \sqrt {b x^2+c x^4} (3 b B-13 A c)}{195 c}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (3 b B-13 A c)}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}} \]

[Out]

-2/117*(-13*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/c/x^(3/2)+2/13*B*(c*x^4+b*x^2)^(5/2)/c/x^(7/2)-8/195*b^2*(-13*A*c+3
*B*b)*x^(3/2)*(c*x^2+b)/c^(3/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)-4/195*b*(-13*A*c+3*B*b)*x^(1/2)*(c*x^4
+b*x^2)^(1/2)/c+8/195*b^(9/4)*(-13*A*c+3*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*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)/c^(7/4)/(c*x^4+b*x^2)^(1/2)-4/195*b^(9/4)*(-13*A*c+3*B*b)*x*(cos(2*arcta
n(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)/c^(7/4)/(c*x^4+b*x^2)^
(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.45, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2039, 2021, 2032, 329, 305, 220, 1196} \[ -\frac {8 b^2 x^{3/2} \left (b+c x^2\right ) (3 b B-13 A c)}{195 c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-13 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{7/4} \sqrt {b x^2+c x^4}}+\frac {8 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-13 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{7/4} \sqrt {b x^2+c x^4}}-\frac {2 \left (b x^2+c x^4\right )^{3/2} (3 b B-13 A c)}{117 c x^{3/2}}-\frac {4 b \sqrt {x} \sqrt {b x^2+c x^4} (3 b B-13 A c)}{195 c}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(3*b*B - 13*A*c)*x^(3/2)*(b + c*x^2))/(195*c^(3/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (4*b*(
3*b*B - 13*A*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(195*c) - (2*(3*b*B - 13*A*c)*(b*x^2 + c*x^4)^(3/2))/(117*c*x^(3/
2)) + (2*B*(b*x^2 + c*x^4)^(5/2))/(13*c*x^(7/2)) + (8*b^(9/4)*(3*b*B - 13*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])/(195*c^(7/4)*Sqrt[b*x^
2 + c*x^4]) - (4*b^(9/4)*(3*b*B - 13*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*El
lipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(195*c^(7/4)*Sqrt[b*x^2 + c*x^4])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 329

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*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]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 2039

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

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx &=\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {\left (2 \left (\frac {3 b B}{2}-\frac {13 A c}{2}\right )\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx}{13 c}\\ &=-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {(2 b (3 b B-13 A c)) \int \frac {\sqrt {b x^2+c x^4}}{\sqrt {x}} \, dx}{39 c}\\ &=-\frac {4 b (3 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {\left (4 b^2 (3 b B-13 A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{195 c}\\ &=-\frac {4 b (3 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {\left (4 b^2 (3 b B-13 A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{195 c \sqrt {b x^2+c x^4}}\\ &=-\frac {4 b (3 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {\left (8 b^2 (3 b B-13 A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{195 c \sqrt {b x^2+c x^4}}\\ &=-\frac {4 b (3 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}-\frac {\left (8 b^{5/2} (3 b B-13 A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{195 c^{3/2} \sqrt {b x^2+c x^4}}+\frac {\left (8 b^{5/2} (3 b B-13 A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{195 c^{3/2} \sqrt {b x^2+c x^4}}\\ &=-\frac {8 b^2 (3 b B-13 A c) x^{3/2} \left (b+c x^2\right )}{195 c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b (3 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{195 c}-\frac {2 (3 b B-13 A c) \left (b x^2+c x^4\right )^{3/2}}{117 c x^{3/2}}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{13 c x^{7/2}}+\frac {8 b^{9/4} (3 b B-13 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 )}{195 c^{7/4} \sqrt {b x^2+c x^4}}-\frac {4 b^{9/4} (3 b B-13 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 )}{195 c^{7/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.11, size = 98, normalized size = 0.27 \[ \frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (b (13 A c-3 b B) \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{b}\right )+3 B \sqrt {\frac {c x^2}{b}+1} \left (b+c x^2\right )^2\right )}{39 c \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*(3*B*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] + b*(-3*b*B + 13*A*c)*Hypergeometric2F
1[-3/2, 3/4, 7/4, -((c*x^2)/b)]))/(39*c*Sqrt[1 + (c*x^2)/b])

________________________________________________________________________________________

fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{4} + {\left (B b + A c\right )} x^{2} + A b\right )} \sqrt {c x^{4} + b x^{2}}}{\sqrt {x}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 446, normalized size = 1.21 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (45 B \,c^{4} x^{8}+65 A \,c^{4} x^{6}+120 B b \,c^{3} x^{6}+208 A b \,c^{3} x^{4}+87 B \,b^{2} c^{2} x^{4}+143 A \,b^{2} c^{2} x^{2}+12 B \,b^{3} c \,x^{2}+156 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{3} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-78 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{3} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-36 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+18 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{585 \left (c \,x^{2}+b \right )^{2} c^{2} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/585*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c^2*(45*B*c^4*x^8+65*A*c^4*x^6+120*B*b*c^3*x^6+156*A*b^3*c*((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)*(-1/(-b*c)^(1/2)*c*x)^(1/2
)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-78*A*b^3*c*((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)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2
))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-36*B*b^4*((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)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1
/2))+18*B*b^4*((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)*(-1/(-b
*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+208*A*b*c^3*x^4+87*B*b^2*c
^2*x^4+143*A*b^2*c^2*x^2+12*B*b^3*c*x^2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________