\(\int \frac {x^6}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [981]

Optimal result

Integrand size = 24, antiderivative size = 150 \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {c x^3}{3 d (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {c (b c-2 a d) x}{d^2 (b c-a d)^2 \sqrt {c+d x^2}}-\frac {a^{5/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b (b c-a d)^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b d^{5/2}} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(401\) vs. \(2(150)=300\).

Time = 2.70 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.67 \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {c x \left (b c \left (3 c+4 d x^2\right )-a d \left (6 c+7 d x^2\right )\right )}{3 d^2 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {a^{3/2} \left (\sqrt {b} \sqrt {c}+\sqrt {b c-a d}\right ) \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{b d (b c-a d)^{5/2}}+\frac {a^{3/2} \left (-\sqrt {b} \sqrt {c}+\sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{b d (b c-a d)^{5/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{b d^{5/2}} \] Input:

Integrate[x^6/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*(c*x*(b*c*(3*c + 4*d*x^2) - a*d*(6*c + 7*d*x^2)))/(d^2*(b*c - a*d)^2* 
(c + d*x^2)^(3/2)) + (a^(3/2)*(Sqrt[b]*Sqrt[c] + Sqrt[b*c - a*d])*Sqrt[2*b 
*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*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*d*(b*c - a*d)^(5/2)) + (a^(3/2)*(-(Sqrt[b]*Sqrt[c]) + Sqrt[b*c - a* 
d])*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*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*d*(b*c - a*d)^(5/2)) + (2*ArcTanh[(Sqrt[d]*x)/(-Sqrt[c] 
 + Sqrt[c + d*x^2])])/(b*d^(5/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.29, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {372, 27, 440, 25, 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.

\(\displaystyle \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 440

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

Int[x^6/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*(c*x^3)/(d*(b*c - a*d)*(c + d*x^2)^(3/2)) + (-((c*(b*c - 2*a*d)*x)/(d 
*(b*c - a*d)*Sqrt[c + d*x^2])) + (-((a^(5/2)*d^2*ArcTan[(Sqrt[b*c - a*d]*x 
)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*Sqrt[b*c - a*d])) + ((b*c - a*d)^2*ArcTan 
h[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(b*Sqrt[d]))/(d*(b*c - a*d)))/(d*(b*c - a* 
d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 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 440

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

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-c \left (\frac {a^{3} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\left (a d -b c \right )^{2} b c \sqrt {\left (a d -b c \right ) a}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )}{b c \,d^{\frac {5}{2}}}-\frac {x^{3}}{3 \left (a d -b c \right ) d \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {\left (2 a d -b c \right ) x}{\left (a d -b c \right )^{2} d^{2} \sqrt {x^{2} d +c}}\right )\)	\(153\)
default	\(\text {Expression too large to display}\)	\(1567\)

Input:

int(x^6/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-c*(1/(a*d-b*c)^2*a^3/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))-1/b/c/d^(5/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))-1/3/ 
(a*d-b*c)/d/(d*x^2+c)^(3/2)*x^3-(2*a*d-b*c)/(a*d-b*c)^2/d^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. 384 vs. \(2 (128) = 256\).

Time = 0.55 (sec) , antiderivative size = 1733, normalized size of antiderivative = 11.55 \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^6/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(6*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 
 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)*sqrt(d)*l 
og(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(a^2*d^5*x^4 + 2*a^2*c* 
d^4*x^2 + a^2*c^2*d^3)*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*((4*b^2*c^2*d^2 - 7*a*b*c*d^ 
3)*x^3 + 3*(b^2*c^3*d - 2*a*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(b^3*c^4*d^3 - 
2*a*b^2*c^3*d^4 + a^2*b*c^2*d^5 + (b^3*c^2*d^5 - 2*a*b^2*c*d^6 + a^2*b*d^7 
)*x^4 + 2*(b^3*c^3*d^4 - 2*a*b^2*c^2*d^5 + a^2*b*c*d^6)*x^2), -1/12*(12*(b 
^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4) 
*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)*sqrt(-d)*arctan(sqrt 
(-d)*x/sqrt(d*x^2 + c)) - 3*(a^2*d^5*x^4 + 2*a^2*c*d^4*x^2 + a^2*c^2*d^3)* 
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*((4*b^2*c^2*d^2 - 7*a*b*c*d^3)*x^3 + 3*(b^2*c^3*d - 
 2*a*b*c^2*d^2)*x)*sqrt(d*x^2 + c))/(b^3*c^4*d^3 - 2*a*b^2*c^3*d^4 + a^2*b 
*c^2*d^5 + (b^3*c^2*d^5 - 2*a*b^2*c*d^6 + a^2*b*d^7)*x^4 + 2*(b^3*c^3*d^4 
- 2*a*b^2*c^2*d^5 + a^2*b*c*d^6)*x^2), 1/6*(3*(a^2*d^5*x^4 + 2*a^2*c*d^...
 

Sympy [F]

\[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**6/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(x**6/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^6/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x^6/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^6/(b*x^2+a)/(d*x^2+c)^(5/2),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. \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^6}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^6/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(x^6/((a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 1367, normalized size of antiderivative = 9.11 \[ \int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^6/(b*x^2+a)/(d*x^2+c)^(5/2),x)
 

Output:

( - 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*c**2 
*d**3 - 6*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**4*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) 
*a**2*d**5*x**4 - 3*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqr 
t(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**3 - 6*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**2*c*d**4*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)*a**2*d**5*x**4 + 3*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqr 
t(a*d - b*c) + 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**2*c 
**2*d**3 + 6*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) 
 + 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**2*c*d**4*x**2 + 
 3*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt( 
d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**2*d**5*x**4 + 12*sqrt(c + 
 d*x**2)*a**2*b*c**2*d**3*x + 14*sqrt(c + d*x**2)*a**2*b*c*d**4*x**3 - 18* 
sqrt(c + d*x**2)*a*b**2*c**3*d**2*x - 22*sqrt(c + d*x**2)*a*b**2*c**2*d...