\(\int \frac {x^4}{(a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [1023]

Optimal result

Integrand size = 24, antiderivative size = 130 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}} \] Output:

1/2*(a*d+2*b*c)*x/b/(-a*d+b*c)^2/(d*x^2+c)^(1/2)+1/2*a*x/b/(-a*d+b*c)/(b*x 
^2+a)/(d*x^2+c)^(1/2)-3/2*a^(1/2)*c*arctan((-a*d+b*c)^(1/2)*x/a^(1/2)/(d*x 
^2+c)^(1/2))/(-a*d+b*c)^(5/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(672\) vs. \(2(130)=260\).

Time = 10.50 (sec) , antiderivative size = 672, normalized size of antiderivative = 5.17 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {1}{2} \left (-\frac {x \left (3 a c+2 b c x^2+a d x^2\right ) \left (4 c^2+5 c d x^2+d^2 x^4-4 c^{3/2} \sqrt {c+d x^2}-3 \sqrt {c} d x^2 \sqrt {c+d x^2}\right )}{(b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 c^{3/2}+3 \sqrt {c} d x^2-4 c \sqrt {c+d x^2}-d x^2 \sqrt {c+d x^2}\right )}+\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^2 \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} \sqrt {b} c^{3/2} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2} \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^2 \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} \sqrt {b} c^{3/2} \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2} \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}\right ) \] Input:

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

Output:

(-((x*(3*a*c + 2*b*c*x^2 + a*d*x^2)*(4*c^2 + 5*c*d*x^2 + d^2*x^4 - 4*c^(3/ 
2)*Sqrt[c + d*x^2] - 3*Sqrt[c]*d*x^2*Sqrt[c + d*x^2]))/((b*c - a*d)^2*(a + 
 b*x^2)*(c + d*x^2)*(4*c^(3/2) + 3*Sqrt[c]*d*x^2 - 4*c*Sqrt[c + d*x^2] - d 
*x^2*Sqrt[c + d*x^2]))) + (3*Sqrt[a]*c*ArcTan[(Sqrt[2*b*c - a*d - 2*Sqrt[b 
]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/((b* 
c - a*d)^2*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3*Sqr 
t[a]*Sqrt[b]*c^(3/2)*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c 
 - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/((b*c - a*d)^(5/2)*Sqr 
t[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3*Sqrt[a]*c*ArcTan[ 
(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c 
] - Sqrt[c + d*x^2]))])/((b*c - a*d)^2*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c 
]*Sqrt[b*c - a*d]]) + (3*Sqrt[a]*Sqrt[b]*c^(3/2)*ArcTan[(Sqrt[2*b*c - a*d 
- 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqrt[c + d*x^ 
2]))])/((b*c - a*d)^(5/2)*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - 
a*d]]))/2
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {372, 27, 402, 27, 291, 218}

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.

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(a*x)/(2*b*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - (c*(-(((2*b*c + a*d) 
*x)/(c*(b*c - a*d)*Sqrt[c + d*x^2])) + (3*Sqrt[a]*b*ArcTan[(Sqrt[b*c - a*d 
]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(3/2)))/(2*b*(b*c - a*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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 291

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

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 402

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

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(-\frac {c \left (-a \left (\frac {\sqrt {x^{2} d +c}\, x}{c \left (b \,x^{2}+a \right )}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )-\frac {2 x}{\sqrt {x^{2} d +c}}\right )}{2 \left (a d -b c \right )^{2}}\)	\(95\)
default	\(\text {Expression too large to display}\)	\(1943\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (110) = 220\).

Time = 0.25 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.25 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

integrate(x**4/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)
 

Output:

Integral(x**4/((a + b*x**2)**2*(c + d*x**2)**(3/2)), x)
 

Maxima [F]

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

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

Output:

integrate(x^4/((b*x^2 + a)^2*(d*x^2 + c)^(3/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (110) = 220\).

Time = 0.41 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.29 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {3 \, a c \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {c x}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} \] Input:

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

Output:

3/2*a*c*sqrt(d)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a* 
d)/sqrt(a*b*c*d - a^2*d^2))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d 
- a^2*d^2)) + c*x/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(d*x^2 + c)) - ((sq 
rt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c 
))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 
2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2* 
a*d + b*c^2)*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2))
 

Mupad [F(-1)]

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

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x)
 

Output:

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 1881, normalized size of antiderivative = 14.47 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x)
 

Output:

( - 12*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b* 
c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*c** 
2*d**2 - 12*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d 
 - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a** 
2*c*d**3*x**2 + 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sq 
rt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)* 
x)*a*b*c**3*d - 9*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sq 
rt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)* 
x)*a*b*c**2*d**2*x**2 - 12*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*s 
qrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d) 
*sqrt(b)*x)*a*b*c*d**3*x**4 + 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt 
(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sq 
rt(d)*sqrt(b)*x)*b**2*c**3*d*x**2 + 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt( 
2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2 
) + sqrt(d)*sqrt(b)*x)*b**2*c**2*d**2*x**4 - 12*sqrt(a)*sqrt(a*d - b*c)*lo 
g(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + 
 d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*c**2*d**2 - 12*sqrt(a)*sqrt(a*d - b*c)* 
log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c 
 + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*c*d**3*x**2 + 3*sqrt(a)*sqrt(a*d - b* 
c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*...