3.3.21 \(\int x^{3/2} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [221]

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

[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 = 0.31, 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 = {2064, 2046, 2049, 2057, 335, 311, 226, 1210} \begin {gather*} \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 \text {ArcTan}\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 \text {ArcTan}\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} \end {gather*}

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 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 2046

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 2049

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 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 2064

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 ) \text {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 ) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.12, size = 111, normalized size = 0.30 \begin {gather*} \frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}} \left (7 b B-13 A c-9 B c x^2\right )\right )+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 )\right )}{117 c^2 \sqrt {1+\frac {c x^2}{b}}} \end {gather*}

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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Maple [A]
time = 0.51, size = 446, normalized size = 1.21

method	result	size
risch	\(\frac {2 \sqrt {x}\, \left (45 B \,c^{2} x^{4}+65 A \,c^{2} x^{2}+10 b B \,x^{2} c +26 A b c -14 b^{2} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{585 c^{2}}-\frac {2 b^{2} \left (13 A c -7 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{195 c^{3} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(267\)
default	\(-\frac {2 \sqrt {x^{4} c +b \,x^{2}}\, \left (-45 B \,c^{4} x^{8}-65 A \,c^{4} x^{6}-55 B b \,c^{3} x^{6}+78 A \,b^{3} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-39 A \,b^{3} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-42 B \,b^{4} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+21 B \,b^{4} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-91 A b \,c^{3} x^{4}+4 B \,b^{2} c^{2} x^{4}-26 A \,b^{2} c^{2} x^{2}+14 B \,b^{3} c \,x^{2}\right )}{585 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c^{3}}\)	\(446\)

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,method=_RETURNVERBOSE)

[Out]

-2/585*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)/c^3*(-45*B*c^4*x^8-65*A*c^4*x^6-55*B*b*c^3*x^6+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)*(-x*c/(-b*c)^(1/2))^(1/2)*El
lipticE(((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)*(-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))-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)*(-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))+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)*(-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))-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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.32, size = 103, normalized size = 0.28 \begin {gather*} -\frac {2 \, {\left (6 \, {\left (7 \, B b^{3} - 13 \, A b^{2} c\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (45 \, B c^{3} x^{4} - 14 \, B b^{2} c + 26 \, A b c^{2} + 5 \, {\left (2 \, B b c^{2} + 13 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{585 \, c^{3}} \end {gather*}

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]

-2/585*(6*(7*B*b^3 - 13*A*b^2*c)*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) - (45*B
*c^3*x^4 - 14*B*b^2*c + 26*A*b*c^2 + 5*(2*B*b*c^2 + 13*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/c^3

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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