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

Optimal. Leaf size=369 \[ \frac {2 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-13 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {4 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-13 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {4 b^2 x^{3/2} \left (b+c x^2\right ) (7 b B-13 A c)}{195 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b \sqrt {x} \sqrt {b x^2+c x^4} (7 b B-13 A c)}{585 c^2}-\frac {2 x^{5/2} \sqrt {b x^2+c x^4} (7 b B-13 A c)}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c} \]

[Out]

2/13*B*(c*x^4+b*x^2)^(3/2)*x^(1/2)/c+4/195*b^2*(-13*A*c+7*B*b)*x^(3/2)*(c*x^2+b)/c^(5/2)/(b^(1/2)+x*c^(1/2))/(
c*x^4+b*x^2)^(1/2)-2/117*(-13*A*c+7*B*b)*x^(5/2)*(c*x^4+b*x^2)^(1/2)/c-4/585*b*(-13*A*c+7*B*b)*x^(1/2)*(c*x^4+
b*x^2)^(1/2)/c^2-4/195*b^(9/4)*(-13*A*c+7*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^(11/4)/(c*x^4+b*x^2)^(1/2)+2/195*b^(9/4)*(-13*A*c+7*B*b)*x*(cos(2*arc
tan(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^(11/4)/(c*x^4+b*x^
2)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {4 b^2 x^{3/2} \left (b+c x^2\right ) (7 b B-13 A c)}{195 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {2 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-13 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {4 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-13 A c) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {4 b \sqrt {x} \sqrt {b x^2+c x^4} (7 b B-13 A c)}{585 c^2}-\frac {2 x^{5/2} \sqrt {b x^2+c x^4} (7 b B-13 A c)}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*(7*b*B - 13*A*c)*x^(3/2)*(b + c*x^2))/(195*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (4*b*(7
*b*B - 13*A*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(585*c^2) - (2*(7*b*B - 13*A*c)*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(117*
c) + (2*B*Sqrt[x]*(b*x^2 + c*x^4)^(3/2))/(13*c) - (4*b^(9/4)*(7*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^(11/4)*Sqrt[b*x^
2 + c*x^4]) + (2*b^(9/4)*(7*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^(11/4)*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 ) \sqrt {b x^2+c x^4} \, dx &=\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {\left (2 \left (\frac {7 b B}{2}-\frac {13 A c}{2}\right )\right ) \int x^{3/2} \sqrt {b x^2+c x^4} \, dx}{13 c}\\ &=-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(2 b (7 b B-13 A c)) \int \frac {x^{7/2}}{\sqrt {b x^2+c x^4}} \, dx}{117 c}\\ &=-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}+\frac {\left (2 b^2 (7 b B-13 A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{195 c^2}\\ &=-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}+\frac {\left (2 b^2 (7 b B-13 A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{195 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}+\frac {\left (4 b^2 (7 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^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}+\frac {\left (4 b^{5/2} (7 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^{5/2} \sqrt {b x^2+c x^4}}-\frac {\left (4 b^{5/2} (7 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^{5/2} \sqrt {b x^2+c x^4}}\\ &=\frac {4 b^2 (7 b B-13 A c) x^{3/2} \left (b+c x^2\right )}{195 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {4 b^{9/4} (7 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^{11/4} \sqrt {b x^2+c x^4}}+\frac {2 b^{9/4} (7 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^{11/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 111, normalized size = 0.30 \[ \frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (b (7 b B-13 A c) \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{b}\right )-\left (b+c x^2\right ) \sqrt {\frac {c x^2}{b}+1} \left (-13 A c+7 b B-9 B c x^2\right )\right )}{117 c^2 \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)*Sqrt[1 + (c*x^2)/b]*(7*b*B - 13*A*c - 9*B*c*x^2)) + b*(7*b*B -
13*A*c)*Hypergeometric2F1[-1/2, 3/4, 7/4, -((c*x^2)/b)]))/(117*c^2*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{4} + b x^{2}} {\left (B x^{3} + A x\right )} \sqrt {x}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 446, normalized size = 1.21 \[ -\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (-45 B \,c^{4} x^{8}-65 A \,c^{4} x^{6}-55 B b \,c^{3} x^{6}-91 A b \,c^{3} x^{4}+4 B \,b^{2} c^{2} x^{4}-26 A \,b^{2} c^{2} x^{2}+14 B \,b^{3} c \,x^{2}+78 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{3} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-39 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{3} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-42 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+21 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{585 \left (c \,x^{2}+b \right ) c^{3} x^{\frac {3}{2}}} \]

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)^(1/2),x)

[Out]

-2/585*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)/c^3*(-45*B*x^8*c^4-65*A*x^6*c^4-55*B*x^6*b*c^3+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)*
EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-39*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))-42*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
))+21*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))-91*A*x^4*b*c^3+4*B*b^2*c^2*x
^4-26*A*x^2*b^2*c^2+14*B*x^2*b^3*c)

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

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)^(1/2),x)

[Out]

int(x^(3/2)*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2), x)

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

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)**(1/2),x)

[Out]

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

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