Integrand size = 21, antiderivative size = 208 \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}+\frac {(4 b c-5 a d) x \sqrt {a+b x^2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac {(2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{48 c^3 (b c-a d)^2 \left (c+d x^2\right )}+\frac {a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{5/2}} \] Output:
1/6*x*(b*x^2+a)^(1/2)/c/(d*x^2+c)^3+1/24*(-5*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/ c^2/(-a*d+b*c)/(d*x^2+c)^2+1/48*(-5*a*d+2*b*c)*(-3*a*d+4*b*c)*x*(b*x^2+a)^ (1/2)/c^3/(-a*d+b*c)^2/(d*x^2+c)+1/16*a*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*a rctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a)^(1/2))/c^(7/2)/(-a*d+b*c)^(5/2 )
Time = 10.62 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\frac {x \sqrt {a+b x^2} \left ((b c-a d) \left (8 b^2 c^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )-2 a b c d \left (30 c^2+35 c d x^2+13 d^2 x^4\right )+a^2 d^2 \left (33 c^2+40 c d x^2+15 d^2 x^4\right )\right )+\frac {3 a \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}} \left (c+d x^2\right )^3 \text {arctanh}\left (\sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}}\right )}{x^2}\right )}{48 c^3 (b c-a d)^3 \left (c+d x^2\right )^3} \] Input:
Integrate[Sqrt[a + b*x^2]/(c + d*x^2)^4,x]
Output:
(x*Sqrt[a + b*x^2]*((b*c - a*d)*(8*b^2*c^2*(3*c^2 + 3*c*d*x^2 + d^2*x^4) - 2*a*b*c*d*(30*c^2 + 35*c*d*x^2 + 13*d^2*x^4) + a^2*d^2*(33*c^2 + 40*c*d*x ^2 + 15*d^2*x^4)) + (3*a*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]*(c + d*x^2)^3*ArcTanh[Sqrt[((b*c - a*d)*x^2)/( c*(a + b*x^2))]])/x^2))/(48*c^3*(b*c - a*d)^3*(c + d*x^2)^3)
Time = 0.34 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {314, 25, 402, 402, 27, 291, 221}
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.
\(\displaystyle \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 314 |
\(\displaystyle \frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}-\frac {\int -\frac {4 b x^2+5 a}{\sqrt {b x^2+a} \left (d x^2+c\right )^3}dx}{6 c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 b x^2+5 a}{\sqrt {b x^2+a} \left (d x^2+c\right )^3}dx}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\int \frac {2 b (4 b c-5 a d) x^2+a (16 b c-15 a d)}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\frac {\int \frac {3 a \left (8 b^2 c^2-12 a b d c+5 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {3 a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\frac {3 a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {3 a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{3/2}}+\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (4 b c-5 a d)}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2}}{6 c \left (c+d x^2\right )^3}\) |
Input:
Int[Sqrt[a + b*x^2]/(c + d*x^2)^4,x]
Output:
(x*Sqrt[a + b*x^2])/(6*c*(c + d*x^2)^3) + (((4*b*c - 5*a*d)*x*Sqrt[a + b*x ^2])/(4*c*(b*c - a*d)*(c + d*x^2)^2) + (((2*b*c - 5*a*d)*(4*b*c - 3*a*d)*x *Sqrt[a + b*x^2])/(2*c*(b*c - a*d)*(c + d*x^2)) + (3*a*(8*b^2*c^2 - 12*a*b *c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/ (2*c^(3/2)*(b*c - a*d)^(3/2)))/(4*c*(b*c - a*d)))/(6*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] + Simp[1/(2*a* (p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*p + 3) + d *(2*(p + q + 1) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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]
Time = 0.79 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(-\frac {-11 \left (\frac {8 b^{2} c^{4}}{11}-\frac {20 d \left (-\frac {2 b \,x^{2}}{5}+a \right ) b \,c^{3}}{11}+d^{2} \left (-\frac {4 b \,x^{2}}{33}+a \right ) \left (-2 b \,x^{2}+a \right ) c^{2}+\frac {40 a \,d^{3} \left (-\frac {13 b \,x^{2}}{20}+a \right ) x^{2} c}{33}+\frac {5 a^{2} d^{4} x^{4}}{11}\right ) \sqrt {\left (a d -b c \right ) c}\, \sqrt {b \,x^{2}+a}\, x +5 a \left (a^{2} d^{2}-\frac {12}{5} a b c d +\frac {8}{5} b^{2} c^{2}\right ) \left (x^{2} d +c \right )^{3} \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{16 \sqrt {\left (a d -b c \right ) c}\, \left (x^{2} d +c \right )^{3} \left (a d -b c \right )^{2} c^{3}}\) | \(199\) |
default | \(\text {Expression too large to display}\) | \(6475\) |
Input:
int((b*x^2+a)^(1/2)/(d*x^2+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/16/((a*d-b*c)*c)^(1/2)*(-11*(8/11*b^2*c^4-20/11*d*(-2/5*b*x^2+a)*b*c^3+ d^2*(-4/33*b*x^2+a)*(-2*b*x^2+a)*c^2+40/33*a*d^3*(-13/20*b*x^2+a)*x^2*c+5/ 11*a^2*d^4*x^4)*((a*d-b*c)*c)^(1/2)*(b*x^2+a)^(1/2)*x+5*a*(a^2*d^2-12/5*a* b*c*d+8/5*b^2*c^2)*(d*x^2+c)^3*arctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1 /2)))/(d*x^2+c)^3/(a*d-b*c)^2/c^3
Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (184) = 368\).
Time = 0.45 (sec) , antiderivative size = 1220, normalized size of antiderivative = 5.87 \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\text {Too large to display} \] Input:
integrate((b*x^2+a)^(1/2)/(d*x^2+c)^4,x, algorithm="fricas")
Output:
[1/192*(3*(8*a*b^2*c^5 - 12*a^2*b*c^4*d + 5*a^3*c^3*d^2 + (8*a*b^2*c^2*d^3 - 12*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 3*(8*a*b^2*c^3*d^2 - 12*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + 3*(8*a*b^2*c^4*d - 12*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3) *x^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2 *c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x^3 + a*c*x)*sqrt( b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((8*b^3*c ^4*d^2 - 34*a*b^2*c^3*d^3 + 41*a^2*b*c^2*d^4 - 15*a^3*c*d^5)*x^5 + 2*(12*b ^3*c^5*d - 47*a*b^2*c^4*d^2 + 55*a^2*b*c^3*d^3 - 20*a^3*c^2*d^4)*x^3 + 3*( 8*b^3*c^6 - 28*a*b^2*c^5*d + 31*a^2*b*c^4*d^2 - 11*a^3*c^3*d^3)*x)*sqrt(b* x^2 + a))/(b^3*c^10 - 3*a*b^2*c^9*d + 3*a^2*b*c^8*d^2 - a^3*c^7*d^3 + (b^3 *c^7*d^3 - 3*a*b^2*c^6*d^4 + 3*a^2*b*c^5*d^5 - a^3*c^4*d^6)*x^6 + 3*(b^3*c ^8*d^2 - 3*a*b^2*c^7*d^3 + 3*a^2*b*c^6*d^4 - a^3*c^5*d^5)*x^4 + 3*(b^3*c^9 *d - 3*a*b^2*c^8*d^2 + 3*a^2*b*c^7*d^3 - a^3*c^6*d^4)*x^2), -1/96*(3*(8*a* b^2*c^5 - 12*a^2*b*c^4*d + 5*a^3*c^3*d^2 + (8*a*b^2*c^2*d^3 - 12*a^2*b*c*d ^4 + 5*a^3*d^5)*x^6 + 3*(8*a*b^2*c^3*d^2 - 12*a^2*b*c^2*d^3 + 5*a^3*c*d^4) *x^4 + 3*(8*a*b^2*c^4*d - 12*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(-b*c ^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt (b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*((8*b^3 *c^4*d^2 - 34*a*b^2*c^3*d^3 + 41*a^2*b*c^2*d^4 - 15*a^3*c*d^5)*x^5 + 2*(12 *b^3*c^5*d - 47*a*b^2*c^4*d^2 + 55*a^2*b*c^3*d^3 - 20*a^3*c^2*d^4)*x^3 ...
\[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\int \frac {\sqrt {a + b x^{2}}}{\left (c + d x^{2}\right )^{4}}\, dx \] Input:
integrate((b*x**2+a)**(1/2)/(d*x**2+c)**4,x)
Output:
Integral(sqrt(a + b*x**2)/(c + d*x**2)**4, x)
\[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{{\left (d x^{2} + c\right )}^{4}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(d*x^2+c)^4,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + a)/(d*x^2 + c)^4, x)
Leaf count of result is larger than twice the leaf count of optimal. 958 vs. \(2 (184) = 368\).
Time = 0.69 (sec) , antiderivative size = 958, normalized size of antiderivative = 4.61 \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx =\text {Too large to display} \] Input:
integrate((b*x^2+a)^(1/2)/(d*x^2+c)^4,x, algorithm="giac")
Output:
-1/16*(8*a*b^(5/2)*c^2 - 12*a^2*b^(3/2)*c*d + 5*a^3*sqrt(b)*d^2)*arctan(1/ 2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c* d))/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(-b^2*c^2 + a*b*c*d)) - 1/2 4*(24*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a*b^(5/2)*c^2*d^3 - 36*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^2*b^(3/2)*c*d^4 + 15*(sqrt(b)*x - sqrt(b*x^2 + a))^ 10*a^3*sqrt(b)*d^5 + 240*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(7/2)*c^3*d^2 - 480*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^2*b^(5/2)*c^2*d^3 + 330*(sqrt(b)* x - sqrt(b*x^2 + a))^8*a^3*b^(3/2)*c*d^4 - 75*(sqrt(b)*x - sqrt(b*x^2 + a) )^8*a^4*sqrt(b)*d^5 - 256*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(11/2)*c^5 + 1 216*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(9/2)*c^4*d - 2016*(sqrt(b)*x - sq rt(b*x^2 + a))^6*a^2*b^(7/2)*c^3*d^2 + 1736*(sqrt(b)*x - sqrt(b*x^2 + a))^ 6*a^3*b^(5/2)*c^2*d^3 - 800*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^4*b^(3/2)*c* d^4 + 150*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^5*sqrt(b)*d^5 - 384*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(9/2)*c^4*d + 1392*(sqrt(b)*x - sqrt(b*x^2 + a ))^4*a^3*b^(7/2)*c^3*d^2 - 1608*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^4*b^(5/2 )*c^2*d^3 + 780*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^5*b^(3/2)*c*d^4 - 150*(s qrt(b)*x - sqrt(b*x^2 + a))^4*a^6*sqrt(b)*d^5 - 96*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*b^(7/2)*c^3*d^2 + 336*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^5*b^( 5/2)*c^2*d^3 - 300*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^6*b^(3/2)*c*d^4 + 75* (sqrt(b)*x - sqrt(b*x^2 + a))^2*a^7*sqrt(b)*d^5 - 8*a^6*b^(5/2)*c^2*d^3...
Timed out. \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (d\,x^2+c\right )}^4} \,d x \] Input:
int((a + b*x^2)^(1/2)/(c + d*x^2)^4,x)
Output:
int((a + b*x^2)^(1/2)/(c + d*x^2)^4, x)
Time = 85.44 (sec) , antiderivative size = 2942, normalized size of antiderivative = 14.14 \[ \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^4} \, dx =\text {Too large to display} \] Input:
int((b*x^2+a)^(1/2)/(d*x^2+c)^4,x)
Output:
( - 15*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b* x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*c**3*d**4 - 45*sqrt(c)* sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d) *sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*c**2*d**5*x**2 - 45*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x )/(sqrt(c)*sqrt(b)))*a**4*c*d**6*x**4 - 15*sqrt(c)*sqrt(a*d - b*c)*atan((s qrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sq rt(b)))*a**4*d**7*x**6 + 66*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*b* c**4*d**3 + 198*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sq rt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*b*c**3*d**4*x* *2 + 198*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*b*c**2*d**5*x**4 + 66 *sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*b*c*d**6*x**6 - 96*sqrt(c)*sq rt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*s qrt(b)*x)/(sqrt(c)*sqrt(b)))*a**2*b**2*c**5*d**2 - 288*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x) /(sqrt(c)*sqrt(b)))*a**2*b**2*c**4*d**3*x**2 - 288*sqrt(c)*sqrt(a*d - b*c) *atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/...