\(\int \frac {(c+d x^2)^{5/2}}{(a+b x^2)^2} \, dx\) [1009]

Optimal result

Integrand size = 21, antiderivative size = 174 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {d (b c-2 a d) x \sqrt {c+d x^2}}{2 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^{3/2}}{2 a b \left (a+b x^2\right )}+\frac {(b c-a d)^{3/2} (b c+4 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} b^3}+\frac {d^{3/2} (5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^3} \] Output:

-1/2*d*(-2*a*d+b*c)*x*(d*x^2+c)^(1/2)/a/b^2+1/2*(-a*d+b*c)*x*(d*x^2+c)^(3/ 
2)/a/b/(b*x^2+a)+1/2*(-a*d+b*c)^(3/2)*(4*a*d+b*c)*arctan((-a*d+b*c)^(1/2)* 
x/a^(1/2)/(d*x^2+c)^(1/2))/a^(3/2)/b^3+1/2*d^(3/2)*(-4*a*d+5*b*c)*arctanh( 
d^(1/2)*x/(d*x^2+c)^(1/2))/b^3
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {b x \sqrt {c+d x^2} \left (b^2 c^2+2 a^2 d^2+a b d \left (-2 c+d x^2\right )\right )}{a \left (a+b x^2\right )}-\frac {\sqrt {b c-a d} \left (b^2 c^2+3 a b c d-4 a^2 d^2\right ) \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{3/2}}+d^{3/2} (-5 b c+4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 b^3} \] Input:

Integrate[(c + d*x^2)^(5/2)/(a + b*x^2)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {315, 403, 27, 398, 224, 219, 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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx\)

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {\int \frac {2 \left (a d^2 (5 b c-4 a d) x^2+c \left (b^2 c^2+2 a b d c-2 a^2 d^2\right )\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{2 b}-\frac {d x \sqrt {c+d x^2} (b c-2 a d)}{b}}{2 a b}+\frac {x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a d^2 (5 b c-4 a d) x^2+c \left (b^2 c^2+2 a b d c-2 a^2 d^2\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}-\frac {d x \sqrt {c+d x^2} (b c-2 a d)}{b}}{2 a b}+\frac {x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )}\)

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {(4 a d+b c) (b c-a d)^2 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}+\frac {a d^{3/2} (5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b}}{b}-\frac {d x \sqrt {c+d x^2} (b c-2 a d)}{b}}{2 a b}+\frac {x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )}\)

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {(b c-a d)^{3/2} (4 a d+b c) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} b}+\frac {a d^{3/2} (5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b}}{b}-\frac {d x \sqrt {c+d x^2} (b c-2 a d)}{b}}{2 a b}+\frac {x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )}\)

Input:

Int[(c + d*x^2)^(5/2)/(a + b*x^2)^2,x]
 

Output:

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

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 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 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 315

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

rule 398

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

rule 403

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(-\frac {\frac {\left (a d -b c \right )^{2} \left (-\frac {\sqrt {x^{2} d +c}\, b x}{b \,x^{2}+a}-\frac {\left (4 a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{a}-d^{2} \left (\sqrt {x^{2} d +c}\, b x -\frac {\left (4 a d -5 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )}{\sqrt {d}}\right )}{2 b^{3}}\)	\(142\)
risch	\(\text {Expression too large to display}\)	\(1028\)
default	\(\text {Expression too large to display}\)	\(5290\)

Input:

int((d*x^2+c)^(5/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2/b^3*((a*d-b*c)^2/a*(-(d*x^2+c)^(1/2)*b*x/(b*x^2+a)-(4*a*d+b*c)/((a*d- 
b*c)*a)^(1/2)*arctanh((d*x^2+c)^(1/2)/x*a/((a*d-b*c)*a)^(1/2)))-d^2*((d*x^ 
2+c)^(1/2)*b*x-(4*a*d-5*b*c)/d^(1/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 1228, normalized size of antiderivative = 7.06 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

[-1/8*(2*(5*a^2*b*c*d - 4*a^3*d^2 + (5*a*b^2*c*d - 4*a^2*b*d^2)*x^2)*sqrt( 
d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + (a*b^2*c^2 + 3*a^2*b* 
c*d - 4*a^3*d^2 + (b^3*c^2 + 3*a*b^2*c*d - 4*a^2*b*d^2)*x^2)*sqrt(-(b*c - 
a*d)/a)*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*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*s 
qrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(a*b^2*d^2*x^3 + (b^ 
3*c^2 - 2*a*b^2*c*d + 2*a^2*b*d^2)*x)*sqrt(d*x^2 + c))/(a*b^4*x^2 + a^2*b^ 
3), -1/8*(4*(5*a^2*b*c*d - 4*a^3*d^2 + (5*a*b^2*c*d - 4*a^2*b*d^2)*x^2)*sq 
rt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (a*b^2*c^2 + 3*a^2*b*c*d - 4*a 
^3*d^2 + (b^3*c^2 + 3*a*b^2*c*d - 4*a^2*b*d^2)*x^2)*sqrt(-(b*c - a*d)/a)*l 
og(((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*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c 
 - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(a*b^2*d^2*x^3 + (b^3*c^2 - 2 
*a*b^2*c*d + 2*a^2*b*d^2)*x)*sqrt(d*x^2 + c))/(a*b^4*x^2 + a^2*b^3), 1/4*( 
(a*b^2*c^2 + 3*a^2*b*c*d - 4*a^3*d^2 + (b^3*c^2 + 3*a*b^2*c*d - 4*a^2*b*d^ 
2)*x^2)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^ 
2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) - (5 
*a^2*b*c*d - 4*a^3*d^2 + (5*a*b^2*c*d - 4*a^2*b*d^2)*x^2)*sqrt(d)*log(-2*d 
*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(a*b^2*d^2*x^3 + (b^3*c^2 - 2* 
a*b^2*c*d + 2*a^2*b*d^2)*x)*sqrt(d*x^2 + c))/(a*b^4*x^2 + a^2*b^3), -1/...
 

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \] Input:

integrate((d*x**2+c)**(5/2)/(b*x**2+a)**2,x)
 

Output:

Integral((c + d*x**2)**(5/2)/(a + b*x**2)**2, x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:

integrate((d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^(5/2)/(b*x^2 + a)^2, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (146) = 292\).

Time = 0.16 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.34 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {d x^{2} + c} d^{2} x}{2 \, b^{2}} - \frac {{\left (5 \, b c d^{\frac {3}{2}} - 4 \, a d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, b^{3}} - \frac {{\left (b^{3} c^{3} \sqrt {d} + 2 \, a b^{2} c^{2} d^{\frac {3}{2}} - 7 \, a^{2} b c d^{\frac {5}{2}} + 4 \, a^{3} d^{\frac {7}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a b^{3}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} d^{\frac {7}{2}} - b^{3} c^{4} \sqrt {d} + 2 \, a b^{2} c^{3} d^{\frac {3}{2}} - a^{2} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a b^{3}} \] Input:

integrate((d*x^2+c)^(5/2)/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

1/2*sqrt(d*x^2 + c)*d^2*x/b^2 - 1/4*(5*b*c*d^(3/2) - 4*a*d^(5/2))*log((sqr 
t(d)*x - sqrt(d*x^2 + c))^2)/b^3 - 1/2*(b^3*c^3*sqrt(d) + 2*a*b^2*c^2*d^(3 
/2) - 7*a^2*b*c*d^(5/2) + 4*a^3*d^(7/2))*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))/(sqrt(a*b*c*d - a^2*d 
^2)*a*b^3) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^3*sqrt(d) - 4*(sqrt(d) 
*x - sqrt(d*x^2 + c))^2*a*b^2*c^2*d^(3/2) + 5*(sqrt(d)*x - sqrt(d*x^2 + c) 
)^2*a^2*b*c*d^(5/2) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^3*d^(7/2) - b^3* 
c^4*sqrt(d) + 2*a*b^2*c^3*d^(3/2) - a^2*b*c^2*d^(5/2))/(((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)*a*b^3)
 

Mupad [F(-1)]

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

int((c + d*x^2)^(5/2)/(a + b*x^2)^2,x)
 

Output:

int((c + d*x^2)^(5/2)/(a + b*x^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 1373, normalized size of antiderivative = 7.89 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)^(5/2)/(b*x^2+a)^2,x)
 

Output:

(4*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**3*d**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)*a**2*b*c*d + 
4*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*b*d**2*x 
**2 - 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*b**2*c* 
*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)*a*b**2*c 
*d*x**2 - 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**3* 
c**2*x**2 + 4*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**3 
*d**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)*a**2*b*c 
*d + 4*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*b*d**2 
*x**2 - 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*b**2*...