\(\int \frac {x^5 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [196]

Optimal result

Integrand size = 26, antiderivative size = 139 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {b (5 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^3}-\frac {(5 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {b^2 (5 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}} \] Output:

1/16*b*(-6*A*c+5*B*b)*(c*x^4+b*x^2)^(1/2)/c^3-1/24*(-6*A*c+5*B*b)*x^2*(c*x 
^4+b*x^2)^(1/2)/c^2+1/6*B*x^4*(c*x^4+b*x^2)^(1/2)/c-1/16*b^2*(-6*A*c+5*B*b 
)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (15 b^2 B+4 c^2 x^2 \left (3 A+2 B x^2\right )-2 b c \left (9 A+5 B x^2\right )\right )+6 b^2 (5 b B-6 A c) \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*(Sqrt[c]*x*(b + c*x^2)*(15*b^2*B + 4*c^2*x^2*(3*A + 2*B*x^2) - 2*b*c*(9 
*A + 5*B*x^2)) + 6*b^2*(5*b*B - 6*A*c)*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x) 
/(Sqrt[b] - Sqrt[b + c*x^2])]))/(48*c^(7/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1221, 1134, 1160, 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.

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

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1134

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {B x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\right )\)

Input:

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

Output:

((B*x^4*Sqrt[b*x^2 + c*x^4])/(3*c) - ((5*b*B - 6*A*c)*((x^2*Sqrt[b*x^2 + c 
*x^4])/(2*c) - (3*b*(Sqrt[b*x^2 + c*x^4]/c - (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt 
[b*x^2 + c*x^4]])/c^(3/2)))/(4*c)))/(6*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 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 1134

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

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 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 = 0.46 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (-\frac {1}{2} A \,b^{2} c +\frac {5}{12} B \,b^{3}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (b \left (\frac {5 B \,x^{2}}{9}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} \left (\frac {2 B \,x^{2}}{3}+A \right ) c^{\frac {5}{2}}}{3}-\frac {5 B \sqrt {c}\, b^{2}}{6}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {5 B b}{6}\right ) b^{2}}{2}\right )}{8 c^{\frac {7}{2}}}\)	\(121\)
risch	\(-\frac {x^{2} \left (-8 B \,c^{2} x^{4}-12 A \,c^{2} x^{2}+10 x^{2} B b c +18 A b c -15 B \,b^{2}\right ) \left (c \,x^{2}+b \right )}{48 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (6 A c -5 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{16 c^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(124\)
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (8 B \,c^{\frac {7}{2}} \sqrt {c \,x^{2}+b}\, x^{5}+12 A \,c^{\frac {7}{2}} \sqrt {c \,x^{2}+b}\, x^{3}-10 B \,c^{\frac {5}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{3}-18 A \,c^{\frac {5}{2}} \sqrt {c \,x^{2}+b}\, b x +15 B \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{2} x +18 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c^{2}-15 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c \right )}{48 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {9}{2}}}\)	\(169\)

Input:

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-3/8/c^(7/2)*((-1/2*A*b^2*c+5/12*B*b^3)*ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2 
)*c^(1/2)+b)/c^(1/2))+(b*(5/9*B*x^2+A)*c^(3/2)-2/3*x^2*(2/3*B*x^2+A)*c^(5/ 
2)-5/6*B*c^(1/2)*b^2)*(x^2*(c*x^2+b))^(1/2)+1/2*ln(2)*(A*c-5/6*B*b)*b^2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.63 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{4}}\right ] \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + 
b*x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 
- 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4, 1/48*(3*(5*B*b^3 - 6*A*b^2*c)*sq 
rt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^3*x^4 + 1 
5*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2)) 
/c^4]
 

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.45 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\begin {cases} \frac {3 b^{2} \left (A - \frac {5 B b}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x^{2} + c x^{4}} \left (\frac {B x^{4}}{3 c} - \frac {3 b \left (A - \frac {5 B b}{6 c}\right )}{4 c^{2}} + \frac {x^{2} \left (A - \frac {5 B b}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {\frac {2 A \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {2 B \left (b x^{2}\right )^{\frac {7}{2}}}{7 b^{3}}}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{6}}{3} + \frac {B x^{8}}{4}\right ) & \text {otherwise} \end {cases}}{2} \] Input:

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
 

Output:

Piecewise((3*b**2*(A - 5*B*b/(6*c))*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))/(8*c**2) + sqrt(b*x**2 + 
 c*x**4)*(B*x**4/(3*c) - 3*b*(A - 5*B*b/(6*c))/(4*c**2) + x**2*(A - 5*B*b/ 
(6*c))/(2*c)), Ne(c, 0)), ((2*A*(b*x**2)**(5/2)/(5*b**2) + 2*B*(b*x**2)**( 
7/2)/(7*b**3))/b, Ne(b, 0)), (zoo*(A*x**6/3 + B*x**8/4), True))/2
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.32 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{16} \, {\left (\frac {4 \, \sqrt {c x^{4} + b x^{2}} x^{2}}{c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{c^{2}}\right )} A + \frac {1}{96} \, {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} x^{4}}{c} - \frac {20 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c^{2}} - \frac {15 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{3}}\right )} B \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

1/16*(4*sqrt(c*x^4 + b*x^2)*x^2/c + 3*b^2*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/c^2)*A + 1/96*(16*sqrt( 
c*x^4 + b*x^2)*x^4/c - 20*sqrt(c*x^4 + b*x^2)*b*x^2/c^2 - 15*b^3*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 
^2/c^3)*B
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.08 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{48} \, \sqrt {c x^{2} + b} {\left (2 \, x^{2} {\left (\frac {4 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {5 \, B b c^{3} \mathrm {sgn}\left (x\right ) - 6 \, A c^{4} \mathrm {sgn}\left (x\right )}{c^{5}}\right )} + \frac {3 \, {\left (5 \, B b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 6 \, A b c^{3} \mathrm {sgn}\left (x\right )\right )}}{c^{5}}\right )} x - \frac {{\left (5 \, B b^{3} \log \left ({\left | b \right |}\right ) - 6 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {7}{2}}} + \frac {{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

1/48*sqrt(c*x^2 + b)*(2*x^2*(4*B*x^2/(c*sgn(x)) - (5*B*b*c^3*sgn(x) - 6*A* 
c^4*sgn(x))/c^5) + 3*(5*B*b^2*c^2*sgn(x) - 6*A*b*c^3*sgn(x))/c^5)*x - 1/32 
*(5*B*b^3*log(abs(b)) - 6*A*b^2*c*log(abs(b)))*sgn(x)/c^(7/2) + 1/16*(5*B* 
b^3 - 6*A*b^2*c)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(7/2)*sgn(x))
 

Mupad [F(-1)]

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

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
 

Output:

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {-18 \sqrt {c \,x^{2}+b}\, a b \,c^{2} x +12 \sqrt {c \,x^{2}+b}\, a \,c^{3} x^{3}+15 \sqrt {c \,x^{2}+b}\, b^{3} c x -10 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{5}+18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{2} c -15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{4}}{48 c^{4}} \] Input:

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
 

Output:

( - 18*sqrt(b + c*x**2)*a*b*c**2*x + 12*sqrt(b + c*x**2)*a*c**3*x**3 + 15* 
sqrt(b + c*x**2)*b**3*c*x - 10*sqrt(b + c*x**2)*b**2*c**2*x**3 + 8*sqrt(b 
+ c*x**2)*b*c**3*x**5 + 18*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt 
(b))*a*b**2*c - 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b** 
4)/(48*c**4)