\(\int x (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [173]

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^4 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \] Output:

3/256*b^2*(-2*A*c+B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3-1/32*(-2*A*c+B* 
b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(3/2)/c^2+1/10*B*(c*x^4+b*x^2)^(5/2)/c-3/256* 
b^4*(-2*A*c+B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.34 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (15 b^4 B-30 A b^3 c-10 b^3 B c x^2+20 A b^2 c^2 x^2+8 b^2 B c^2 x^4+240 A b c^3 x^4+176 b B c^3 x^6+160 A c^4 x^6+128 B c^4 x^8\right )}{1280 c^3 x^2 \left (b+c x^2\right )}-\frac {3 b^4 (b B-2 A c) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{128 c^{7/2} x^3 \left (b+c x^2\right )^{3/2}} \] Input:

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

Output:

((x^2*(b + c*x^2))^(3/2)*(15*b^4*B - 30*A*b^3*c - 10*b^3*B*c*x^2 + 20*A*b^ 
2*c^2*x^2 + 8*b^2*B*c^2*x^4 + 240*A*b*c^3*x^4 + 176*b*B*c^3*x^6 + 160*A*c^ 
4*x^6 + 128*B*c^4*x^8))/(1280*c^3*x^2*(b + c*x^2)) - (3*b^4*(b*B - 2*A*c)* 
(x^2*(b + c*x^2))^(3/2)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]) 
/(128*c^(7/2)*x^3*(b + c*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1940, 1160, 1087, 1087, 1091, 219}

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.

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

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}dx^2\)

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^4+b x^2}dx^2}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {(b B-2 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{2 c}\right )\)

Input:

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

Output:

((B*(b*x^2 + c*x^4)^(5/2))/(5*c) - ((b*B - 2*A*c)*(((b + 2*c*x^2)*(b*x^2 + 
 c*x^4)^(3/2))/(8*c) - (3*b^2*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - 
 (b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/(16*c)))/( 
2*c))/2
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

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

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (-\frac {1}{2} A \,b^{4} c +\frac {1}{4} b^{5} B \right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (b^{3} \left (\frac {B \,x^{2}}{3}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} b^{2} \left (\frac {2 B \,x^{2}}{5}+A \right ) c^{\frac {5}{2}}}{3}-8 \left (\frac {11 B \,x^{2}}{15}+A \right ) x^{4} b \,c^{\frac {7}{2}}-\frac {16 \left (\frac {4 B \,x^{2}}{5}+A \right ) x^{6} c^{\frac {9}{2}}}{3}-\frac {B \sqrt {c}\, b^{4}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {B b}{2}\right ) b^{4}}{2}\right )}{128 c^{\frac {7}{2}}}\)	\(159\)
risch	\(-\frac {\left (-128 B \,x^{8} c^{4}-160 A \,x^{6} c^{4}-176 B \,x^{6} b \,c^{3}-240 A \,x^{4} b \,c^{3}-8 B \,x^{4} b^{2} c^{2}-20 A \,b^{2} c^{2} x^{2}+10 B \,b^{3} c \,x^{2}+30 A \,b^{3} c -15 B \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 c^{3}}+\frac {3 b^{4} \left (2 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\)	\(164\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (128 B \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} x^{5}+160 A \,c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} x^{3}-80 B \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{3}-80 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} b x +20 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x +40 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} x +30 A \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{3} x -10 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x -15 B \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{4} x +30 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4} c -15 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{1280 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\)	\(244\)

Input:

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

Output:

-3/128*((-1/2*A*b^4*c+1/4*b^5*B)*ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/ 
2)+b)/c^(1/2))+(b^3*(1/3*B*x^2+A)*c^(3/2)-2/3*x^2*b^2*(2/5*B*x^2+A)*c^(5/2 
)-8*(11/15*B*x^2+A)*x^4*b*c^(7/2)-16/3*(4/5*B*x^2+A)*x^6*c^(9/2)-1/2*B*c^( 
1/2)*b^4)*(x^2*(c*x^2+b))^(1/2)+1/2*ln(2)*(A*c-1/2*B*b)*b^4)/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.14 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \] Input:

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

Output:

[-1/2560*(15*(B*b^5 - 2*A*b^4*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + 
 b*x^2)*sqrt(c)) - 2*(128*B*c^5*x^8 + 16*(11*B*b*c^4 + 10*A*c^5)*x^6 + 15* 
B*b^4*c - 30*A*b^3*c^2 + 8*(B*b^2*c^3 + 30*A*b*c^4)*x^4 - 10*(B*b^3*c^2 - 
2*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4, 1/1280*(15*(B*b^5 - 2*A*b^4*c) 
*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (128*B*c^5*x^ 
8 + 16*(11*B*b*c^4 + 10*A*c^5)*x^6 + 15*B*b^4*c - 30*A*b^3*c^2 + 8*(B*b^2* 
c^3 + 30*A*b*c^4)*x^4 - 10*(B*b^3*c^2 - 2*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x 
^2))/c^4]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 614, normalized size of antiderivative = 4.15 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

A*b*Piecewise((b**3*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2 
*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sq 
rt(c*(b/(2*c) + x**2)**2), True))/(16*c**2) + sqrt(b*x**2 + c*x**4)*(-b**2 
/(8*c**2) + b*x**2/(12*c) + x**4/3), Ne(c, 0)), (2*(b*x**2)**(5/2)/(5*b**2 
), Ne(b, 0)), (0, True))/2 + A*c*Piecewise((-5*b**4*Piecewise((log(b + 2*s 
qrt(c)*sqrt(b*x**2 + c*x**4) + 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c 
) + x**2)*log(b/(2*c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True))/(128*c** 
3) + sqrt(b*x**2 + c*x**4)*(5*b**3/(64*c**3) - 5*b**2*x**2/(96*c**2) + b*x 
**4/(24*c) + x**6/4), Ne(c, 0)), (2*(b*x**2)**(7/2)/(7*b**3), Ne(b, 0)), ( 
0, True))/2 + B*b*Piecewise((-5*b**4*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x 
**2 + c*x**4) + 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b 
/(2*c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True))/(128*c**3) + sqrt(b*x** 
2 + c*x**4)*(5*b**3/(64*c**3) - 5*b**2*x**2/(96*c**2) + b*x**4/(24*c) + x* 
*6/4), Ne(c, 0)), (2*(b*x**2)**(7/2)/(7*b**3), Ne(b, 0)), (0, True))/2 + B 
*c*Piecewise((7*b**5*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 
2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/s 
qrt(c*(b/(2*c) + x**2)**2), True))/(256*c**4) + sqrt(b*x**2 + c*x**4)*(-7* 
b**4/(128*c**4) + 7*b**3*x**2/(192*c**3) - 7*b**2*x**4/(240*c**2) + b*x**6 
/(40*c) + x**8/5), Ne(c, 0)), (2*(b*x**2)**(9/2)/(9*b**4), Ne(b, 0)), (0, 
True))/2
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (128) = 256\).

Time = 0.04 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{256} \, {\left (32 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {12 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{2}} + \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c}\right )} A + \frac {1}{2560} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{2}} - \frac {160 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c} - \frac {15 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{2}} + \frac {256 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{c}\right )} B \] Input:

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

Output:

1/256*(32*(c*x^4 + b*x^2)^(3/2)*x^2 - 12*sqrt(c*x^4 + b*x^2)*b^2*x^2/c + 3 
*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 6*sqrt(c*x 
^4 + b*x^2)*b^3/c^2 + 16*(c*x^4 + b*x^2)^(3/2)*b/c)*A + 1/2560*(60*sqrt(c* 
x^4 + b*x^2)*b^3*x^2/c^2 - 160*(c*x^4 + b*x^2)^(3/2)*b*x^2/c - 15*b^5*log( 
2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x 
^2)*b^4/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 256*(c*x^4 + b*x^2)^(5/2) 
/c)*B
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.40 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {11 \, B b c^{8} \mathrm {sgn}\left (x\right ) + 10 \, A c^{9} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} + \frac {B b^{2} c^{7} \mathrm {sgn}\left (x\right ) + 30 \, A b c^{8} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} - \frac {5 \, {\left (B b^{3} c^{6} \mathrm {sgn}\left (x\right ) - 2 \, A b^{2} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} x^{2} + \frac {15 \, {\left (B b^{4} c^{5} \mathrm {sgn}\left (x\right ) - 2 \, A b^{3} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x + \frac {3 \, {\left (B b^{5} \mathrm {sgn}\left (x\right ) - 2 \, A b^{4} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, {\left (B b^{5} \log \left ({\left | b \right |}\right ) - 2 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/1280*(2*(4*(2*(8*B*c*x^2*sgn(x) + (11*B*b*c^8*sgn(x) + 10*A*c^9*sgn(x))/ 
c^8)*x^2 + (B*b^2*c^7*sgn(x) + 30*A*b*c^8*sgn(x))/c^8)*x^2 - 5*(B*b^3*c^6* 
sgn(x) - 2*A*b^2*c^7*sgn(x))/c^8)*x^2 + 15*(B*b^4*c^5*sgn(x) - 2*A*b^3*c^6 
*sgn(x))/c^8)*sqrt(c*x^2 + b)*x + 3/256*(B*b^5*sgn(x) - 2*A*b^4*c*sgn(x))* 
log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(7/2) - 3/512*(B*b^5*log(abs(b)) 
- 2*A*b^4*c*log(abs(b)))*sgn(x)/c^(7/2)
 

Mupad [B] (verification not implemented)

Time = 9.85 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.59 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {B\,{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}+\frac {A\,{\left (c\,x^4+b\,x^2\right )}^{3/2}\,\left (c\,x^2+\frac {b}{2}\right )}{8\,c}-\frac {3\,A\,b^2\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{32\,c}-\frac {B\,b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \] Input:

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

Output:

(B*(b*x^2 + c*x^4)^(5/2))/(10*c) + (A*(b*x^2 + c*x^4)^(3/2)*(b/2 + c*x^2)) 
/(8*c) - (3*A*b^2*((b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2) - (b^2*log((b/2 
 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(32*c) - (B*b*(( 
x^2*(b*x^2 + c*x^4)^(3/2))/4 - (3*b^2*(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(1/2 
))/(4*c) - (b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^( 
3/2))))/(16*c) + (b*(b*x^2 + c*x^4)^(3/2))/(8*c)))/(4*c)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.50 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {-30 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{2} x +20 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{3} x^{3}+240 \sqrt {c \,x^{2}+b}\, a b \,c^{4} x^{5}+160 \sqrt {c \,x^{2}+b}\, a \,c^{5} x^{7}+15 \sqrt {c \,x^{2}+b}\, b^{5} c x -10 \sqrt {c \,x^{2}+b}\, b^{4} c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, b^{3} c^{3} x^{5}+176 \sqrt {c \,x^{2}+b}\, b^{2} c^{4} x^{7}+128 \sqrt {c \,x^{2}+b}\, b \,c^{5} x^{9}+30 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{4} c -15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{6}}{1280 c^{4}} \] Input:

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

Output:

( - 30*sqrt(b + c*x**2)*a*b**3*c**2*x + 20*sqrt(b + c*x**2)*a*b**2*c**3*x* 
*3 + 240*sqrt(b + c*x**2)*a*b*c**4*x**5 + 160*sqrt(b + c*x**2)*a*c**5*x**7 
 + 15*sqrt(b + c*x**2)*b**5*c*x - 10*sqrt(b + c*x**2)*b**4*c**2*x**3 + 8*s 
qrt(b + c*x**2)*b**3*c**3*x**5 + 176*sqrt(b + c*x**2)*b**2*c**4*x**7 + 128 
*sqrt(b + c*x**2)*b*c**5*x**9 + 30*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c) 
*x)/sqrt(b))*a*b**4*c - 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt 
(b))*b**6)/(1280*c**4)