\(\int x^{3/2} (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [225]

Optimal result

Integrand size = 28, antiderivative size = 369 \[ \int x^{3/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {4 b^2 (7 b B-13 A c) x^{3/2} \left (b+c x^2\right )}{195 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 b (7 b B-13 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{585 c^2}-\frac {2 (7 b B-13 A c) x^{5/2} \sqrt {b x^2+c x^4}}{117 c}+\frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {4 b^{9/4} (7 b B-13 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {2 b^{9/4} (7 b B-13 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{195 c^{11/4} \sqrt {b x^2+c x^4}} \] Output:

4/195*b^2*(-13*A*c+7*B*b)*x^(3/2)*(c*x^2+b)/c^(5/2)/(b^(1/2)+c^(1/2)*x)/(c 
*x^4+b*x^2)^(1/2)-4/585*b*(-13*A*c+7*B*b)*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c^2- 
2/117*(-13*A*c+7*B*b)*x^(5/2)*(c*x^4+b*x^2)^(1/2)/c+2/13*B*x^(1/2)*(c*x^4+ 
b*x^2)^(3/2)/c-4/195*b^(9/4)*(-13*A*c+7*B*b)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2 
+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^ 
(1/4))),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*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*Invers 
eJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(11/4)/(c*x^4+b 
*x^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.30 \[ \int x^{3/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (\left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}} \left (7 b B-13 A c-9 B c x^2\right )\right )+b (7 b B-13 A c) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{117 c^2 \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(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])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1945, 1426, 1429, 1431, 266, 834, 27, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1945

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \int x^{3/2} \sqrt {c x^4+b x^2}dx}{13 c}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \int \frac {x^{7/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2 B \sqrt {x} \left (b x^2+c x^4\right )^{3/2}}{13 c}-\frac {(7 b B-13 A c) \left (\frac {2}{9} b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )+\frac {2}{9} x^{5/2} \sqrt {b x^2+c x^4}\right )}{13 c}\)

Input:

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

Output:

(2*B*Sqrt[x]*(b*x^2 + c*x^4)^(3/2))/(13*c) - ((7*b*B - 13*A*c)*((2*x^(5/2) 
*Sqrt[b*x^2 + c*x^4])/9 + (2*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(5*c) - (6 
*b*x*Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x) 
) + (b^(1/4)*(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])/(c^(1/4)*Sqrt[b + 
c*x^2]))/Sqrt[c]) + (b^(1/4)*(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])/(2 
*c^(3/4)*Sqrt[b + c*x^2])))/(5*c*Sqrt[b*x^2 + c*x^4])))/9))/(13*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

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 834

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

rule 1426

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
 

rule 1429

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b 
*d^2*((m + 2*p - 1)/(c*(m + 4*p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ 
p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[m + 2*p - 1, 
 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

rule 1510

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

rule 1945

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[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] - Simp[(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])
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.72

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

Input:

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

Output:

2/585*x^(1/2)*(45*B*c^2*x^4+65*A*c^2*x^2+10*B*b*c*x^2+26*A*b*c-14*B*b^2)/c 
^2*(x^2*(c*x^2+b))^(1/2)-2/195*b^2*(13*A*c-7*B*b)/c^3*(-b*c)^(1/2)*((x+1/c 
*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))*c/(-b*c)^(1/ 
2))^(1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)/(c*x^3+b*x)^(1/2)*(-2/c*(-b*c)^(1/2)*E 
llipticE(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+1/c*(-b* 
c)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2) 
))*(x^2*(c*x^2+b))^(1/2)/x^(3/2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.28 \[ \int x^{3/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {2 \, {\left (6 \, {\left (7 \, B b^{3} - 13 \, A b^{2} c\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (45 \, B c^{3} x^{4} - 14 \, B b^{2} c + 26 \, A b c^{2} + 5 \, {\left (2 \, B b c^{2} + 13 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{585 \, c^{3}} \] Input:

integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

-2/585*(6*(7*B*b^3 - 13*A*b^2*c)*sqrt(c)*weierstrassZeta(-4*b/c, 0, weiers 
trassPInverse(-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
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^(3/2)*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{3/2} \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b c x}{45}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{3}}{9}-\frac {28 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} x}{585}+\frac {4 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{3}}{117}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{5}}{13}-\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{2}+b}d x \right ) a \,b^{2} c}{15}+\frac {14 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{2}+b}d x \right ) b^{4}}{195}}{c^{2}} \] Input:

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

Output:

(2*(26*sqrt(x)*sqrt(b + c*x**2)*a*b*c*x + 65*sqrt(x)*sqrt(b + c*x**2)*a*c* 
*2*x**3 - 14*sqrt(x)*sqrt(b + c*x**2)*b**3*x + 10*sqrt(x)*sqrt(b + c*x**2) 
*b**2*c*x**3 + 45*sqrt(x)*sqrt(b + c*x**2)*b*c**2*x**5 - 39*int((sqrt(x)*s 
qrt(b + c*x**2))/(b + c*x**2),x)*a*b**2*c + 21*int((sqrt(x)*sqrt(b + c*x** 
2))/(b + c*x**2),x)*b**4))/(585*c**2)