\(\int \sqrt {x} (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [238]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 282 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {8 b^3 (9 b B-19 A c) \sqrt {b x^2+c x^4}}{4389 c^3 \sqrt {x}}-\frac {8 b^2 (9 b B-19 A c) x^{3/2} \sqrt {b x^2+c x^4}}{7315 c^2}-\frac {4 b (9 b B-19 A c) x^{7/2} \sqrt {b x^2+c x^4}}{1045 c}-\frac {2 (9 b B-19 A c) x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{285 c}+\frac {2 B \left (b x^2+c x^4\right )^{5/2}}{19 c \sqrt {x}}-\frac {4 b^{15/4} (9 b B-19 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4389 c^{13/4} \sqrt {b x^2+c x^4}} \] Output:

8/4389*b^3*(-19*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)-8/7315*b^2*(-19 
*A*c+9*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-4/1045*b*(-19*A*c+9*B*b)*x^(7/ 
2)*(c*x^4+b*x^2)^(1/2)/c-2/285*(-19*A*c+9*B*b)*x^(3/2)*(c*x^4+b*x^2)^(3/2) 
/c+2/19*B*(c*x^4+b*x^2)^(5/2)/c/x^(1/2)-4/4389*b^(15/4)*(-19*A*c+9*B*b)*x* 
(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiA 
M(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/2*2^(1/2))/c^(13/4)/(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 = 10.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.49 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right )^2 \sqrt {1+\frac {c x^2}{b}} \left (45 b^2 B+11 c^2 x^2 \left (19 A+15 B x^2\right )-b c \left (95 A+99 B x^2\right )\right )+5 b^3 (-9 b B+19 A c) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3135 c^3 \sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:

Integrate[Sqrt[x]*(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]*(45*b^2*B + 11 
*c^2*x^2*(19*A + 15*B*x^2) - b*c*(95*A + 99*B*x^2)) + 5*b^3*(-9*b*B + 19*A 
*c)*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/b)]))/(3135*c^3*Sqrt[x]*Sq 
rt[1 + (c*x^2)/b])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1945, 1426, 1426, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1429

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

(2*B*(b*x^2 + c*x^4)^(5/2))/(19*c*Sqrt[x]) - ((9*b*B - 19*A*c)*((2*x^(3/2) 
*(b*x^2 + c*x^4)^(3/2))/15 + (2*b*((2*x^(7/2)*Sqrt[b*x^2 + c*x^4])/11 + (2 
*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]*EllipticF[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))/5))/(19*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 = 1.30 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.94

method result size
risch \(-\frac {2 \left (-1155 B \,x^{8} c^{4}-1463 A \,x^{6} c^{4}-1617 B \,x^{6} b \,c^{3}-2261 A \,x^{4} b \,c^{3}-84 B \,x^{4} b^{2} c^{2}-228 A \,b^{2} c^{2} x^{2}+108 B \,b^{3} c \,x^{2}+380 A \,b^{3} c -180 B \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21945 c^{3} \sqrt {x}}+\frac {4 b^{4} \left (19 A c -9 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{4389 c^{4} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) \(265\)
default \(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (1155 B \,c^{6} x^{11}+1463 A \,c^{6} x^{9}+2772 B b \,c^{5} x^{9}+3724 A b \,c^{5} x^{7}+1701 B \,b^{2} c^{4} x^{7}+190 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-b c}\, b^{4} c +2489 A \,b^{2} c^{4} x^{5}-90 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-b c}\, b^{5}-24 B \,b^{3} c^{3} x^{5}-152 A \,b^{3} c^{3} x^{3}+72 B \,b^{4} c^{2} x^{3}-380 A \,b^{4} c^{2} x +180 B \,b^{5} c x \right )}{21945 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c^{4}}\) \(331\)

Input:

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

Output:

-2/21945/c^3*(-1155*B*c^4*x^8-1463*A*c^4*x^6-1617*B*b*c^3*x^6-2261*A*b*c^3 
*x^4-84*B*b^2*c^2*x^4-228*A*b^2*c^2*x^2+108*B*b^3*c*x^2+380*A*b^3*c-180*B* 
b^4)/x^(1/2)*(x^2*(c*x^2+b))^(1/2)+4/4389*b^4/c^4*(19*A*c-9*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 = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.52 \[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {2 \, {\left (20 \, {\left (9 \, B b^{5} - 19 \, A b^{4} c\right )} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - {\left (1155 \, B c^{5} x^{8} + 77 \, {\left (21 \, B b c^{4} + 19 \, A c^{5}\right )} x^{6} + 180 \, B b^{4} c - 380 \, A b^{3} c^{2} + 7 \, {\left (12 \, B b^{2} c^{3} + 323 \, A b c^{4}\right )} x^{4} - 12 \, {\left (9 \, B b^{3} c^{2} - 19 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{21945 \, c^{4} x} \] Input:

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

Output:

-2/21945*(20*(9*B*b^5 - 19*A*b^4*c)*sqrt(c)*x*weierstrassPInverse(-4*b/c, 
0, x) - (1155*B*c^5*x^8 + 77*(21*B*b*c^4 + 19*A*c^5)*x^6 + 180*B*b^4*c - 3 
80*A*b^3*c^2 + 7*(12*B*b^2*c^3 + 323*A*b*c^4)*x^4 - 12*(9*B*b^3*c^2 - 19*A 
*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^4*x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(x^(1/2)*(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)*sqrt(x), x)
 

Giac [F]

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

integrate(x^(1/2)*(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)*sqrt(x), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {-\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{3} c}{231}+\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x^{2}}{385}+\frac {34 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{4}}{165}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, a \,c^{4} x^{6}}{15}+\frac {24 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{5}}{1463}-\frac {72 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{2}}{7315}+\frac {8 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{4}}{1045}+\frac {14 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{6}}{95}+\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{8}}{19}+\frac {4 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) a \,b^{4} c}{231}-\frac {12 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{6}}{1463}}{c^{3}} \] Input:

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

Output:

(2*( - 380*sqrt(x)*sqrt(b + c*x**2)*a*b**3*c + 228*sqrt(x)*sqrt(b + c*x**2 
)*a*b**2*c**2*x**2 + 2261*sqrt(x)*sqrt(b + c*x**2)*a*b*c**3*x**4 + 1463*sq 
rt(x)*sqrt(b + c*x**2)*a*c**4*x**6 + 180*sqrt(x)*sqrt(b + c*x**2)*b**5 - 1 
08*sqrt(x)*sqrt(b + c*x**2)*b**4*c*x**2 + 84*sqrt(x)*sqrt(b + c*x**2)*b**3 
*c**2*x**4 + 1617*sqrt(x)*sqrt(b + c*x**2)*b**2*c**3*x**6 + 1155*sqrt(x)*s 
qrt(b + c*x**2)*b*c**4*x**8 + 190*int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c* 
x**3),x)*a*b**4*c - 90*int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b* 
*6))/(21945*c**3)