\(\int x^{5/2} (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [236]

Optimal result

Integrand size = 28, antiderivative size = 321 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {24 b^4 (13 b B-23 A c) \sqrt {b x^2+c x^4}}{33649 c^4 \sqrt {x}}+\frac {72 b^3 (13 b B-23 A c) x^{3/2} \sqrt {b x^2+c x^4}}{168245 c^3}-\frac {8 b^2 (13 b B-23 A c) x^{7/2} \sqrt {b x^2+c x^4}}{24035 c^2}-\frac {4 b (13 b B-23 A c) x^{11/2} \sqrt {b x^2+c x^4}}{2185 c}-\frac {2 (13 b B-23 A c) x^{7/2} \left (b x^2+c x^4\right )^{3/2}}{437 c}+\frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}+\frac {12 b^{19/4} (13 b B-23 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{33649 c^{17/4} \sqrt {b x^2+c x^4}} \] Output:

-24/33649*b^4*(-23*A*c+13*B*b)*(c*x^4+b*x^2)^(1/2)/c^4/x^(1/2)+72/168245*b 
^3*(-23*A*c+13*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^3-8/24035*b^2*(-23*A*c+1 
3*B*b)*x^(7/2)*(c*x^4+b*x^2)^(1/2)/c^2-4/2185*b*(-23*A*c+13*B*b)*x^(11/2)* 
(c*x^4+b*x^2)^(1/2)/c-2/437*(-23*A*c+13*B*b)*x^(7/2)*(c*x^4+b*x^2)^(3/2)/c 
+2/23*B*x^(3/2)*(c*x^4+b*x^2)^(5/2)/c+12/33649*b^(19/4)*(-23*A*c+13*B*b)*x 
*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobi 
AM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(17/4)/(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 = 10.16 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.50 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}} \left (195 b^3 B-55 c^3 x^4 \left (23 A+19 B x^2\right )+11 b c^2 x^2 \left (69 A+65 B x^2\right )-3 b^2 c \left (115 A+143 B x^2\right )\right )+15 b^4 (13 b B-23 A c) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{24035 c^4 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

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

Output:

(2*Sqrt[x^2*(b + c*x^2)]*(-((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b]*(195*b^3*B - 
 55*c^3*x^4*(23*A + 19*B*x^2) + 11*b*c^2*x^2*(69*A + 65*B*x^2) - 3*b^2*c*( 
115*A + 143*B*x^2))) + 15*b^4*(13*b*B - 23*A*c)*Hypergeometric2F1[-3/2, 1/ 
4, 5/4, -((c*x^2)/b)]))/(24035*c^4*Sqrt[x]*Sqrt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1945, 1426, 1426, 1429, 1429, 1429, 1431, 266, 761}

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.

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

\(\Big \downarrow \) 1945

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \int x^{5/2} \left (c x^4+b x^2\right )^{3/2}dx}{23 c}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \int x^{9/2} \sqrt {c x^4+b x^2}dx+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \int \frac {x^{13/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 B x^{3/2} \left (b x^2+c x^4\right )^{5/2}}{23 c}-\frac {(13 b B-23 A c) \left (\frac {6}{19} b \left (\frac {2}{15} b \left (\frac {2 x^{7/2} \sqrt {b x^2+c x^4}}{11 c}-\frac {9 b \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{11 c}\right )+\frac {2}{15} x^{11/2} \sqrt {b x^2+c x^4}\right )+\frac {2}{19} x^{7/2} \left (b x^2+c x^4\right )^{3/2}\right )}{23 c}\)

Input:

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

Output:

(2*B*x^(3/2)*(b*x^2 + c*x^4)^(5/2))/(23*c) - ((13*b*B - 23*A*c)*((2*x^(7/2 
)*(b*x^2 + c*x^4)^(3/2))/19 + (6*b*((2*x^(11/2)*Sqrt[b*x^2 + c*x^4])/15 + 
(2*b*((2*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(11*c) - (9*b*((2*x^(3/2)*Sqrt[b*x^2 
 + c*x^4])/(7*c) - (5*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)* 
x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Elliptic 
F[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4 
])))/(7*c)))/(11*c)))/15))/19))/(23*c)
 

Defintions of rubi rules used

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 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 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 = 1.85 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {2 \left (7315 B \,c^{5} x^{10}+8855 A \,c^{5} x^{8}+9625 B b \,c^{4} x^{8}+12397 A b \,c^{4} x^{6}+308 B \,b^{2} c^{3} x^{6}+644 A \,b^{2} c^{3} x^{4}-364 B \,b^{3} c^{2} x^{4}-828 A \,b^{3} c^{2} x^{2}+468 B \,b^{4} c \,x^{2}+1380 A \,b^{4} c -780 b^{5} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{168245 c^{4} \sqrt {x}}-\frac {12 b^{5} \left (23 A c -13 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{33649 c^{5} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(289\)
default	\(-\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-7315 B \,c^{7} x^{13}-8855 A \,c^{7} x^{11}-16940 B b \,c^{6} x^{11}-21252 A b \,c^{6} x^{9}-9933 B \,b^{2} c^{5} x^{9}-13041 A \,b^{2} c^{5} x^{7}+56 B \,b^{3} c^{4} x^{7}+690 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{5} c +184 A \,b^{3} c^{4} x^{5}-390 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{6}-104 B \,b^{4} c^{3} x^{5}-552 A \,b^{4} c^{3} x^{3}+312 B \,b^{5} c^{2} x^{3}-1380 A \,b^{5} c^{2} x +780 B \,b^{6} c x \right )}{168245 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{5}}\)	\(355\)

Input:

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

Output:

2/168245/c^4*(7315*B*c^5*x^10+8855*A*c^5*x^8+9625*B*b*c^4*x^8+12397*A*b*c^ 
4*x^6+308*B*b^2*c^3*x^6+644*A*b^2*c^3*x^4-364*B*b^3*c^2*x^4-828*A*b^3*c^2* 
x^2+468*B*b^4*c*x^2+1380*A*b^4*c-780*B*b^5)/x^(1/2)*(x^2*(c*x^2+b))^(1/2)- 
12/33649*b^5/c^5*(23*A*c-13*B*b)*(-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)*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 = 0.12 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.53 \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {2 \, {\left (60 \, {\left (13 \, B b^{6} - 23 \, A b^{5} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (7315 \, B c^{6} x^{10} + 385 \, {\left (25 \, B b c^{5} + 23 \, A c^{6}\right )} x^{8} - 780 \, B b^{5} c + 1380 \, A b^{4} c^{2} + 77 \, {\left (4 \, B b^{2} c^{4} + 161 \, A b c^{5}\right )} x^{6} - 28 \, {\left (13 \, B b^{3} c^{3} - 23 \, A b^{2} c^{4}\right )} x^{4} + 36 \, {\left (13 \, B b^{4} c^{2} - 23 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{168245 \, c^{5} x} \] Input:

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

Output:

2/168245*(60*(13*B*b^6 - 23*A*b^5*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 
 0, x) + (7315*B*c^6*x^10 + 385*(25*B*b*c^5 + 23*A*c^6)*x^8 - 780*B*b^5*c 
+ 1380*A*b^4*c^2 + 77*(4*B*b^2*c^4 + 161*A*b*c^5)*x^6 - 28*(13*B*b^3*c^3 - 
 23*A*b^2*c^4)*x^4 + 36*(13*B*b^4*c^2 - 23*A*b^3*c^3)*x^2)*sqrt(c*x^4 + b* 
x^2)*sqrt(x))/(c^5*x)
 

Sympy [F(-1)]

Timed out. \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )} x^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )} x^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{5/2}\,\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\frac {24 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{4} c}{1463}-\frac {72 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{2} x^{2}}{7315}+\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{3} x^{4}}{1045}+\frac {14 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b \,c^{4} x^{6}}{95}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{5} x^{8}}{19}-\frac {312 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{6}}{33649}+\frac {936 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{5} c \,x^{2}}{168245}-\frac {104 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{4} c^{2} x^{4}}{24035}+\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} c^{3} x^{6}}{2185}+\frac {50 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c^{4} x^{8}}{437}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{10}}{23}-\frac {12 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a \,b^{5} c}{1463}+\frac {156 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{7}}{33649}}{c^{4}} \] Input:

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

Output:

(2*(1380*sqrt(x)*sqrt(b + c*x**2)*a*b**4*c - 828*sqrt(x)*sqrt(b + c*x**2)* 
a*b**3*c**2*x**2 + 644*sqrt(x)*sqrt(b + c*x**2)*a*b**2*c**3*x**4 + 12397*s 
qrt(x)*sqrt(b + c*x**2)*a*b*c**4*x**6 + 8855*sqrt(x)*sqrt(b + c*x**2)*a*c* 
*5*x**8 - 780*sqrt(x)*sqrt(b + c*x**2)*b**6 + 468*sqrt(x)*sqrt(b + c*x**2) 
*b**5*c*x**2 - 364*sqrt(x)*sqrt(b + c*x**2)*b**4*c**2*x**4 + 308*sqrt(x)*s 
qrt(b + c*x**2)*b**3*c**3*x**6 + 9625*sqrt(x)*sqrt(b + c*x**2)*b**2*c**4*x 
**8 + 7315*sqrt(x)*sqrt(b + c*x**2)*b*c**5*x**10 - 690*int((sqrt(x)*sqrt(b 
 + c*x**2))/(b*x + c*x**3),x)*a*b**5*c + 390*int((sqrt(x)*sqrt(b + c*x**2) 
)/(b*x + c*x**3),x)*b**7))/(168245*c**4)