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

Optimal. Leaf size=447 \[ -\frac {4 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-21 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {8 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-21 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}-\frac {8 b^4 x^{3/2} \left (b+c x^2\right ) (11 b B-21 A c)}{3315 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {8 b^3 \sqrt {x} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{9945 c^3}-\frac {8 b^2 x^{5/2} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{13923 c^2}-\frac {4 b x^{9/2} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{1547 c}-\frac {2 x^{5/2} \left (b x^2+c x^4\right )^{3/2} (11 b B-21 A c)}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c} \]

[Out]

-2/357*(-21*A*c+11*B*b)*x^(5/2)*(c*x^4+b*x^2)^(3/2)/c+2/21*B*(c*x^4+b*x^2)^(5/2)*x^(1/2)/c-8/3315*b^4*(-21*A*c
+11*B*b)*x^(3/2)*(c*x^2+b)/c^(7/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)-8/13923*b^2*(-21*A*c+11*B*b)*x^(5/2
)*(c*x^4+b*x^2)^(1/2)/c^2-4/1547*b*(-21*A*c+11*B*b)*x^(9/2)*(c*x^4+b*x^2)^(1/2)/c+8/9945*b^3*(-21*A*c+11*B*b)*
x^(1/2)*(c*x^4+b*x^2)^(1/2)/c^3+8/3315*b^(17/4)*(-21*A*c+11*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^(15/4)/(c*x^4+b*x^2)^(1/2)-4/3315*b^(17/4)*(-21*A*c
+11*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(s
in(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^(15/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.59, antiderivative size = 447, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2039, 2021, 2024, 2032, 329, 305, 220, 1196} \[ -\frac {8 b^2 x^{5/2} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{13923 c^2}-\frac {8 b^4 x^{3/2} \left (b+c x^2\right ) (11 b B-21 A c)}{3315 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {8 b^3 \sqrt {x} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{9945 c^3}-\frac {4 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-21 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {8 b^{17/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-21 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}-\frac {4 b x^{9/2} \sqrt {b x^2+c x^4} (11 b B-21 A c)}{1547 c}-\frac {2 x^{5/2} \left (b x^2+c x^4\right )^{3/2} (11 b B-21 A c)}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^4*(11*b*B - 21*A*c)*x^(3/2)*(b + c*x^2))/(3315*c^(7/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (8*b
^3*(11*b*B - 21*A*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(9945*c^3) - (8*b^2*(11*b*B - 21*A*c)*x^(5/2)*Sqrt[b*x^2 + c
*x^4])/(13923*c^2) - (4*b*(11*b*B - 21*A*c)*x^(9/2)*Sqrt[b*x^2 + c*x^4])/(1547*c) - (2*(11*b*B - 21*A*c)*x^(5/
2)*(b*x^2 + c*x^4)^(3/2))/(357*c) + (2*B*Sqrt[x]*(b*x^2 + c*x^4)^(5/2))/(21*c) + (8*b^(17/4)*(11*b*B - 21*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])/(3315*c^(15/4)*Sqrt[b*x^2 + c*x^4]) - (4*b^(17/4)*(11*b*B - 21*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])/(3315*c^(15/4)*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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 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 x^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (2 \left (\frac {11 b B}{2}-\frac {21 A c}{2}\right )\right ) \int x^{3/2} \left (b x^2+c x^4\right )^{3/2} \, dx}{21 c}\\ &=-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {(2 b (11 b B-21 A c)) \int x^{7/2} \sqrt {b x^2+c x^4} \, dx}{119 c}\\ &=-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (4 b^2 (11 b B-21 A c)\right ) \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx}{1547 c}\\ &=-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}+\frac {\left (4 b^3 (11 b B-21 A c)\right ) \int \frac {x^{7/2}}{\sqrt {b x^2+c x^4}} \, dx}{1989 c^2}\\ &=\frac {8 b^3 (11 b B-21 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{9945 c^3}-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (4 b^4 (11 b B-21 A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{3315 c^3}\\ &=\frac {8 b^3 (11 b B-21 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{9945 c^3}-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (4 b^4 (11 b B-21 A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{3315 c^3 \sqrt {b x^2+c x^4}}\\ &=\frac {8 b^3 (11 b B-21 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{9945 c^3}-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (8 b^4 (11 b B-21 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 )}{3315 c^3 \sqrt {b x^2+c x^4}}\\ &=\frac {8 b^3 (11 b B-21 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{9945 c^3}-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}-\frac {\left (8 b^{9/2} (11 b B-21 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 )}{3315 c^{7/2} \sqrt {b x^2+c x^4}}+\frac {\left (8 b^{9/2} (11 b B-21 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 )}{3315 c^{7/2} \sqrt {b x^2+c x^4}}\\ &=-\frac {8 b^4 (11 b B-21 A c) x^{3/2} \left (b+c x^2\right )}{3315 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {8 b^3 (11 b B-21 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{9945 c^3}-\frac {8 b^2 (11 b B-21 A c) x^{5/2} \sqrt {b x^2+c x^4}}{13923 c^2}-\frac {4 b (11 b B-21 A c) x^{9/2} \sqrt {b x^2+c x^4}}{1547 c}-\frac {2 (11 b B-21 A c) x^{5/2} \left (b x^2+c x^4\right )^{3/2}}{357 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{5/2}}{21 c}+\frac {8 b^{17/4} (11 b B-21 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 )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}-\frac {4 b^{17/4} (11 b B-21 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 )}{3315 c^{15/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 138, normalized size = 0.31 \[ \frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (7 b^3 (21 A c-11 b B) \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{b}\right )+\left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1} \left (-b c \left (147 A+143 B x^2\right )+13 c^2 x^2 \left (21 A+17 B x^2\right )+77 b^2 B\right )\right )}{4641 c^3 \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]*(77*b^2*B + 13*c^2*x^2*(21*A + 17*B*x^2) -
b*c*(147*A + 143*B*x^2)) + 7*b^3*(-11*b*B + 21*A*c)*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x^2)/b)]))/(4641*c
^3*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B c x^{7} + {\left (B b + A c\right )} x^{5} + A b x^{3}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*c*x^7 + (B*b + A*c)*x^5 + A*b*x^3)*sqrt(c*x^4 + b*x^2)*sqrt(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 494, normalized size = 1.11 \[ \frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3315 B \,c^{6} x^{12}+4095 A \,c^{6} x^{10}+7800 B b \,c^{5} x^{10}+10080 A b \,c^{5} x^{8}+4665 B \,b^{2} c^{4} x^{8}+6405 A \,b^{2} c^{4} x^{6}-40 B \,b^{3} c^{3} x^{6}-168 A \,b^{3} c^{3} x^{4}+88 B \,b^{4} c^{2} x^{4}-588 A \,b^{4} c^{2} x^{2}+308 B \,b^{5} c \,x^{2}+1764 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{5} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-882 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{5} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-924 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{6} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+462 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{6} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{69615 \left (c \,x^{2}+b \right )^{2} c^{4} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/69615*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c^4*(3315*B*x^12*c^6+4095*A*x^10*c^6+7800*B*x^10*b*c^5+10080*A
*x^8*b*c^5+4665*B*x^8*b^2*c^4+6405*A*x^6*b^2*c^4-40*B*x^6*b^3*c^3+1764*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)*(-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))*b^5*c-882*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)*(-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))*b^5*c-924*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)*(-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))*b^6+462*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)*(-1/(-b*c)^(1/2)*c*x)^(1/2)*Ellipt
icF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^6-168*A*x^4*b^3*c^3+88*B*x^4*b^4*c^2-588*A*x^2*b^4*
c^2+308*B*x^2*b^5*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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