\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x} \, dx\) [174]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=-\frac {b (3 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^2}-\frac {(3 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}+\frac {b^3 (3 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{5/2}} \] Output:

-1/128*b*(-8*A*c+3*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^2-1/48*(-8*A*c+3 
*B*b)*(c*x^4+b*x^2)^(3/2)/c+1/8*B*(c*x^4+b*x^2)^(5/2)/c/x^2+1/128*b^3*(-8* 
A*c+3*B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (\frac {\sqrt {c} x \left (-9 b^3 B+6 b^2 c \left (4 A+B x^2\right )+16 c^3 x^4 \left (4 A+3 B x^2\right )+8 b c^2 x^2 \left (14 A+9 B x^2\right )\right )}{b+c x^2}+\frac {6 b^3 (3 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{\left (b+c x^2\right )^{3/2}}\right )}{384 c^{5/2} x^3} \] Input:

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

Output:

((x^2*(b + c*x^2))^(3/2)*((Sqrt[c]*x*(-9*b^3*B + 6*b^2*c*(4*A + B*x^2) + 1 
6*c^3*x^4*(4*A + 3*B*x^2) + 8*b*c^2*x^2*(14*A + 9*B*x^2)))/(b + c*x^2) + ( 
6*b^3*(3*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])])/( 
b + c*x^2)^(3/2)))/(384*c^(5/2)*x^3)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1221, 1131, 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.

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

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((B*(b*x^2 + c*x^4)^(5/2))/(4*c*x^2) - ((3*b*B - 8*A*c)*((b*x^2 + c*x^4)^( 
3/2)/3 + (b*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqr 
t[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/2))/(8*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 1131

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(a + 
 b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b 
*d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

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 = 140, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {\left (48 B \,c^{3} x^{6}+64 A \,c^{3} x^{4}+72 B b \,c^{2} x^{4}+112 A b \,c^{2} x^{2}+6 x^{2} B \,b^{2} c +24 A \,b^{2} c -9 B \,b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{384 c^{2}}-\frac {b^{3} \left (8 A c -3 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\)	\(140\)
pseudoelliptic	\(\frac {\frac {\left (-A \,b^{3} c +\frac {3}{8} B \,b^{4}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{2}+\left (b^{2} \left (\frac {B \,x^{2}}{4}+A \right ) c^{\frac {3}{2}}+\frac {14 x^{2} \left (\frac {9 B \,x^{2}}{14}+A \right ) b \,c^{\frac {5}{2}}}{3}+2 \left (B \,x^{6}+\frac {4}{3} A \,x^{4}\right ) c^{\frac {7}{2}}-\frac {3 B \sqrt {c}\, b^{3}}{8}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {3 B b}{8}\right ) b^{3}}{2}}{16 c^{\frac {5}{2}}}\)	\(142\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (48 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} x^{3}+64 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} x -24 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, b x -16 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b x +6 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{2} x -24 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{2} x +9 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{3} x -24 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c +9 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4}\right )}{384 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}}}\)	\(202\)

Input:

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

Output:

1/384/c^2*(48*B*c^3*x^6+64*A*c^3*x^4+72*B*b*c^2*x^4+112*A*b*c^2*x^2+6*B*b^ 
2*c*x^2+24*A*b^2*c-9*B*b^3)*(x^2*(c*x^2+b))^(1/2)-1/128*b^3*(8*A*c-3*B*b)/ 
c^(5/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1 
/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\left [-\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{6} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{3}}, -\frac {3 \, {\left (3 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (48 \, B c^{4} x^{6} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{3}}\right ] \] Input:

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

Output:

[-1/768*(3*(3*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + 
 b*x^2)*sqrt(c)) - 2*(48*B*c^4*x^6 - 9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c 
^3 + 8*A*c^4)*x^4 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2)) 
/c^3, -1/384*(3*(3*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)* 
sqrt(-c)/(c*x^2 + b)) - (48*B*c^4*x^6 - 9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B* 
b*c^3 + 8*A*c^4)*x^4 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^ 
2))/c^3]
 

Sympy [A] (verification not implemented)

Time = 4.30 (sec) , antiderivative size = 551, normalized size of antiderivative = 3.83 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

A*b*Piecewise((-b**2*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))/(8*c) + (b/(4*c) + x**2/2)*sqrt(b*x**2 
+ c*x**4), Ne(c, 0)), (2*(b*x**2)**(3/2)/(3*b), Ne(b, 0)), (0, True))/2 + 
A*c*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 + B*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)/sqrt(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 + B*c*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
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.50 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {1}{96} \, {\left (12 \, \sqrt {c x^{4} + b x^{2}} b x^{2} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + 16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c}\right )} A + \frac {1}{256} \, {\left (32 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {12 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{2}} + \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c}\right )} B \] Input:

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

Output:

1/96*(12*sqrt(c*x^4 + b*x^2)*b*x^2 - 3*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 
+ b*x^2)*sqrt(c))/c^(3/2) + 16*(c*x^4 + b*x^2)^(3/2) + 6*sqrt(c*x^4 + b*x^ 
2)*b^2/c)*A + 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)*B
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {9 \, B b c^{6} \mathrm {sgn}\left (x\right ) + 8 \, A c^{7} \mathrm {sgn}\left (x\right )}{c^{6}}\right )} x^{2} + \frac {3 \, B b^{2} c^{5} \mathrm {sgn}\left (x\right ) + 56 \, A b c^{6} \mathrm {sgn}\left (x\right )}{c^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, B b^{3} c^{4} \mathrm {sgn}\left (x\right ) - 8 \, A b^{2} c^{5} \mathrm {sgn}\left (x\right )\right )}}{c^{6}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (3 \, B b^{4} \mathrm {sgn}\left (x\right ) - 8 \, A b^{3} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{128 \, c^{\frac {5}{2}}} + \frac {{\left (3 \, B b^{4} \log \left ({\left | b \right |}\right ) - 8 \, A b^{3} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {5}{2}}} \] Input:

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

Output:

1/384*(2*(4*(6*B*c*x^2*sgn(x) + (9*B*b*c^6*sgn(x) + 8*A*c^7*sgn(x))/c^6)*x 
^2 + (3*B*b^2*c^5*sgn(x) + 56*A*b*c^6*sgn(x))/c^6)*x^2 - 3*(3*B*b^3*c^4*sg 
n(x) - 8*A*b^2*c^5*sgn(x))/c^6)*sqrt(c*x^2 + b)*x - 1/128*(3*B*b^4*sgn(x) 
- 8*A*b^3*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(5/2) + 1/256 
*(3*B*b^4*log(abs(b)) - 8*A*b^3*c*log(abs(b)))*sgn(x)/c^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.27 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {24 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x +112 \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{3}+64 \sqrt {c \,x^{2}+b}\, a \,c^{4} x^{5}-9 \sqrt {c \,x^{2}+b}\, b^{4} c x +6 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{3}+72 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{5}+48 \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{7}-24 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{3} c +9 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{5}}{384 c^{3}} \] Input:

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

Output:

(24*sqrt(b + c*x**2)*a*b**2*c**2*x + 112*sqrt(b + c*x**2)*a*b*c**3*x**3 + 
64*sqrt(b + c*x**2)*a*c**4*x**5 - 9*sqrt(b + c*x**2)*b**4*c*x + 6*sqrt(b + 
 c*x**2)*b**3*c**2*x**3 + 72*sqrt(b + c*x**2)*b**2*c**3*x**5 + 48*sqrt(b + 
 c*x**2)*b*c**4*x**7 - 24*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt( 
b))*a*b**3*c + 9*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**5) 
/(384*c**3)