\(\int \frac {(a+b x^2)^{3/2}}{c+d x} \, dx\) [1078]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\frac {\left (2 \left (b c^2+a d^2\right )-b c d x\right ) \sqrt {a+b x^2}}{2 d^3}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}-\frac {\sqrt {b} c \left (2 b c^2+3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 d^4}-\frac {\left (b c^2+a d^2\right )^{3/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^4} \] Output:

1/2*(-b*c*d*x+2*a*d^2+2*b*c^2)*(b*x^2+a)^(1/2)/d^3+1/3*(b*x^2+a)^(3/2)/d-1 
/2*b^(1/2)*c*(3*a*d^2+2*b*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/d^4-(a*d 
^2+b*c^2)^(3/2)*arctanh((-b*c*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/ 
d^4
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\frac {d \sqrt {a+b x^2} \left (6 b c^2+8 a d^2-3 b c d x+2 b d^2 x^2\right )-12 \left (-b c^2-a d^2\right )^{3/2} \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )+3 \sqrt {b} c \left (2 b c^2+3 a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 d^4} \] Input:

Integrate[(a + b*x^2)^(3/2)/(c + d*x),x]
 

Output:

(d*Sqrt[a + b*x^2]*(6*b*c^2 + 8*a*d^2 - 3*b*c*d*x + 2*b*d^2*x^2) - 12*(-(b 
*c^2) - a*d^2)^(3/2)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[- 
(b*c^2) - a*d^2]] + 3*Sqrt[b]*c*(2*b*c^2 + 3*a*d^2)*Log[-(Sqrt[b]*x) + Sqr 
t[a + b*x^2]])/(6*d^4)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {493, 682, 27, 719, 224, 219, 488, 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.

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

\(\Big \downarrow \) 493

\(\displaystyle \frac {\int \frac {(a d-b c x) \sqrt {b x^2+a}}{c+d x}dx}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 682

\(\displaystyle \frac {\frac {\int \frac {b \left (a d \left (b c^2+2 a d^2\right )-b c \left (2 b c^2+3 a d^2\right ) x\right )}{(c+d x) \sqrt {b x^2+a}}dx}{2 b d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a d \left (b c^2+2 a d^2\right )-b c \left (2 b c^2+3 a d^2\right ) x}{(c+d x) \sqrt {b x^2+a}}dx}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {\frac {2 \left (a d^2+b c^2\right )^2 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {b c \left (3 a d^2+2 b c^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {2 \left (a d^2+b c^2\right )^2 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {b c \left (3 a d^2+2 b c^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 \left (a d^2+b c^2\right )^2 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a d^2+2 b c^2\right )}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\frac {-\frac {2 \left (a d^2+b c^2\right )^2 \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{d}-\frac {\sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a d^2+2 b c^2\right )}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {2 \left (a d^2+b c^2\right )^{3/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{d}-\frac {\sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (3 a d^2+2 b c^2\right )}{d}}{2 d^2}+\frac {\sqrt {a+b x^2} \left (2 \left (a d^2+b c^2\right )-b c d x\right )}{2 d^2}}{d}+\frac {\left (a+b x^2\right )^{3/2}}{3 d}\)

Input:

Int[(a + b*x^2)^(3/2)/(c + d*x),x]
 

Output:

(a + b*x^2)^(3/2)/(3*d) + (((2*(b*c^2 + a*d^2) - b*c*d*x)*Sqrt[a + b*x^2]) 
/(2*d^2) + (-((Sqrt[b]*c*(2*b*c^2 + 3*a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + 
b*x^2]])/d) - (2*(b*c^2 + a*d^2)^(3/2)*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + 
 a*d^2]*Sqrt[a + b*x^2])])/d)/(2*d^2))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 488

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

rule 493

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] + Simp[2*(p/(d*(n + 
 2*p + 1)))   Int[(c + d*x)^n*(a + b*x^2)^(p - 1)*(a*d - b*c*x), x], x] /; 
FreeQ[{a, b, c, d, n}, x] && GtQ[p, 0] && NeQ[n + 2*p + 1, 0] && ( !Rationa 
lQ[n] || LtQ[n, 1]) &&  !ILtQ[n + 2*p, 0] && IntQuadraticQ[a, 0, b, c, d, n 
, p, x]
 

rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p 
+ 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p 
+ 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)))   Int[(d + e*x) 
^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* 
d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x 
], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  ! 
RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 719

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.50

method	result	size
risch	\(\frac {\left (2 b \,x^{2} d^{2}-3 b c d x +8 a \,d^{2}+6 b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{6 d^{3}}-\frac {\frac {2 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {\sqrt {b}\, c \left (3 a \,d^{2}+2 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d}}{2 d^{3}}\)	\(238\)
default	\(\frac {\frac {\left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}{3}-\frac {b c \left (\frac {\left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{4 b}+\frac {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\right )}{8 b^{\frac {3}{2}}}\right )}{d}+\frac {\left (a \,d^{2}+b \,c^{2}\right ) \left (\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}-\frac {\sqrt {b}\, c \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\right )}{d}-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{d^{2}}}{d}\)	\(496\)

Input:

int((b*x^2+a)^(3/2)/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

1/6*(2*b*d^2*x^2-3*b*c*d*x+8*a*d^2+6*b*c^2)*(b*x^2+a)^(1/2)/d^3-1/2/d^3*(2 
*(a^2*d^4+2*a*b*c^2*d^2+b^2*c^4)/d^2/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^ 
2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b* 
c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d))+b^(1/2)*c*(3*a*d^2+2*b*c^2) 
/d*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 774, normalized size of antiderivative = 4.87 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx =\text {Too large to display} \] Input:

integrate((b*x^2+a)^(3/2)/(d*x+c),x, algorithm="fricas")
 

Output:

[1/12*(3*(2*b*c^3 + 3*a*c*d^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sq 
rt(b)*x - a) + 6*(b*c^2 + a*d^2)^(3/2)*log((2*a*b*c*d*x - a*b*c^2 - 2*a^2* 
d^2 - (2*b^2*c^2 + a*b*d^2)*x^2 - 2*sqrt(b*c^2 + a*d^2)*(b*c*x - a*d)*sqrt 
(b*x^2 + a))/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(2*b*d^3*x^2 - 3*b*c*d^2*x + 6 
*b*c^2*d + 8*a*d^3)*sqrt(b*x^2 + a))/d^4, 1/6*(3*(2*b*c^3 + 3*a*c*d^2)*sqr 
t(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + 3*(b*c^2 + a*d^2)^(3/2)*log((2* 
a*b*c*d*x - a*b*c^2 - 2*a^2*d^2 - (2*b^2*c^2 + a*b*d^2)*x^2 - 2*sqrt(b*c^2 
 + a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a))/(d^2*x^2 + 2*c*d*x + c^2)) + (2*b 
*d^3*x^2 - 3*b*c*d^2*x + 6*b*c^2*d + 8*a*d^3)*sqrt(b*x^2 + a))/d^4, -1/12* 
(12*(b*c^2 + a*d^2)*sqrt(-b*c^2 - a*d^2)*arctan(sqrt(-b*c^2 - a*d^2)*(b*c* 
x - a*d)*sqrt(b*x^2 + a)/(a*b*c^2 + a^2*d^2 + (b^2*c^2 + a*b*d^2)*x^2)) - 
3*(2*b*c^3 + 3*a*c*d^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x 
 - a) - 2*(2*b*d^3*x^2 - 3*b*c*d^2*x + 6*b*c^2*d + 8*a*d^3)*sqrt(b*x^2 + a 
))/d^4, -1/6*(6*(b*c^2 + a*d^2)*sqrt(-b*c^2 - a*d^2)*arctan(sqrt(-b*c^2 - 
a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a)/(a*b*c^2 + a^2*d^2 + (b^2*c^2 + a*b*d 
^2)*x^2)) - 3*(2*b*c^3 + 3*a*c*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 
+ a)) - (2*b*d^3*x^2 - 3*b*c*d^2*x + 6*b*c^2*d + 8*a*d^3)*sqrt(b*x^2 + a)) 
/d^4]
 

Sympy [F]

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

integrate((b*x**2+a)**(3/2)/(d*x+c),x)
 

Output:

Integral((a + b*x**2)**(3/2)/(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx=-\frac {\sqrt {b x^{2} + a} b c x}{2 \, d^{2}} - \frac {b^{\frac {3}{2}} c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{d^{4}} - \frac {3 \, a \sqrt {b} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, d^{2}} + \frac {{\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{d} + \frac {\sqrt {b x^{2} + a} b c^{2}}{d^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{3 \, d} + \frac {\sqrt {b x^{2} + a} a}{d} \] Input:

integrate((b*x^2+a)^(3/2)/(d*x+c),x, algorithm="maxima")
 

Output:

-1/2*sqrt(b*x^2 + a)*b*c*x/d^2 - b^(3/2)*c^3*arcsinh(b*x/sqrt(a*b))/d^4 - 
3/2*a*sqrt(b)*c*arcsinh(b*x/sqrt(a*b))/d^2 + (a + b*c^2/d^2)^(3/2)*arcsinh 
(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/d + sqrt(b 
*x^2 + a)*b*c^2/d^3 + 1/3*(b*x^2 + a)^(3/2)/d + sqrt(b*x^2 + a)*a/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x^2+a)^(3/2)/(d*x+c),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

int((a + b*x^2)^(3/2)/(c + d*x),x)
 

Output:

int((a + b*x^2)^(3/2)/(c + d*x), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1948, normalized size of antiderivative = 12.25 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{c+d x} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^(3/2)/(d*x+c),x)
 

Output:

( - 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)* 
sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt( 
b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a*c*d**2 - 6*sqrt(b)*sqrt 
(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c* 
*2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b 
*c**2)*c - a*d**2 - 2*b*c**2))*b*c**3 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c 
**2)*c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2 
*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a**2*d**4 - 12*sqrt 
(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x 
**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2* 
b*c**2))*a*b*c**2*d**2 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 
 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a 
*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*b**2*c**4 - 3*sqrt(b)*sqrt(2*sqrt( 
b)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2)*sqrt(a*d**2 + b*c**2)*log( 
 - sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2) + sqrt(a + 
b*x**2)*d + sqrt(b)*d*x)*a*c*d**2 - 3*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + 
 b*c**2)*c + a*d**2 + 2*b*c**2)*sqrt(a*d**2 + b*c**2)*log( - sqrt(2*sqrt(b 
)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2) + sqrt(a + b*x**2)*d + sqrt 
(b)*d*x)*b*c**3 + 3*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c + a*d** 
2 + 2*b*c**2)*sqrt(a*d**2 + b*c**2)*log(sqrt(2*sqrt(b)*sqrt(a*d**2 + b*...