Integrand size = 27, antiderivative size = 618 \[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=-\frac {e}{a d^2 \sqrt {a+c x^2}}+\frac {1}{a d x \sqrt {a+c x^2}}+\frac {a e \left (a f^2+c \left (e^2-2 d f\right )\right )+c d \left (a f^2+c \left (e^2-d f\right )\right ) x}{a d^2 \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {2 \sqrt {a+c x^2}}{a^2 d x}+\frac {f \left (e \left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {f \left (e \left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}+\frac {e \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2} \] Output:
-e/a/d^2/(c*x^2+a)^(1/2)+1/a/d/x/(c*x^2+a)^(1/2)+(a*e*(a*f^2+c*(-2*d*f+e^2 ))+c*d*(a*f^2+c*(-d*f+e^2))*x)/a/d^2/(a*c*e^2+(-a*f+c*d)^2)/(c*x^2+a)^(1/2 )-2*(c*x^2+a)^(1/2)/a^2/d/x+1/2*f*(e*(e-(-4*d*f+e^2)^(1/2))*(a*f^2+c*(-2*d *f+e^2))-2*a*f^2*(-d*f+e^2)-2*c*(d^2*f^2-3*d*e^2*f+e^4))*arctanh(1/2*(2*a* f-c*(e-(-4*d*f+e^2)^(1/2))*x)*2^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2) ^(1/2)))^(1/2)/(c*x^2+a)^(1/2))*2^(1/2)/d^2/(-4*d*f+e^2)^(1/2)/(a*c*e^2+(- a*f+c*d)^2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)-1/2*f*(e*(e +(-4*d*f+e^2)^(1/2))*(a*f^2+c*(-2*d*f+e^2))-2*a*f^2*(-d*f+e^2)-2*c*(d^2*f^ 2-3*d*e^2*f+e^4))*arctanh(1/2*(2*a*f-c*(e+(-4*d*f+e^2)^(1/2))*x)*2^(1/2)/( 2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2)))^(1/2)/(c*x^2+a)^(1/2))*2^(1/2) /d^2/(-4*d*f+e^2)^(1/2)/(a*c*e^2+(-a*f+c*d)^2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4 *d*f+e^2)^(1/2)))^(1/2)+e*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/d^2
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.76 (sec) , antiderivative size = 684, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=-\frac {\frac {d \left (a^3 f^2+2 c^3 d^2 x^2+a c^2 \left (d^2+e^2 x^2+d x (e-3 f x)\right )+a^2 c \left (e^2+f \left (-2 d+f x^2\right )\right )\right )}{a^2 \left (c^2 d^2+a^2 f^2+a c \left (e^2-2 d f\right )\right ) x \sqrt {a+c x^2}}+\frac {2 e \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c e^3 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-2 a c d e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^2 e f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{3/2} e^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-6 c^{3/2} d e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 c^{3/2} d^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} e^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a \sqrt {c} d f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c e^3 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 c d e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a e f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{c^2 d^2+a^2 f^2+a c \left (e^2-2 d f\right )}}{d^2} \] Input:
Integrate[1/(x^2*(a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
Output:
-(((d*(a^3*f^2 + 2*c^3*d^2*x^2 + a*c^2*(d^2 + e^2*x^2 + d*x*(e - 3*f*x)) + a^2*c*(e^2 + f*(-2*d + f*x^2))))/(a^2*(c^2*d^2 + a^2*f^2 + a*c*(e^2 - 2*d *f))*x*Sqrt[a + c*x^2]) + (2*e*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[ a]])/a^(3/2) + RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (a*c*e^3*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x ^2] - #1] - 2*a*c*d*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + a^2*e *f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(3/2)*e^4*Log[-(Sqrt[c ]*x) + Sqrt[a + c*x^2] - #1]*#1 - 6*c^(3/2)*d*e^2*f*Log[-(Sqrt[c]*x) + Sqr t[a + c*x^2] - #1]*#1 + 2*c^(3/2)*d^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^ 2] - #1]*#1 + 2*a*Sqrt[c]*e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] *#1 - 2*a*Sqrt[c]*d*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c*e^ 3*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 2*c*d*e*f^2*Log[-(Sqrt [c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - a*e*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ]/(c^2*d^2 + a^2*f^2 + a*c*(e^2 - 2*d*f)))/d^2)
Time = 2.01 (sec) , antiderivative size = 618, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7279, 2009}
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.
\(\displaystyle \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-d f+e^2+e f x}{d^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}-\frac {e}{d^2 x \left (a+c x^2\right )^{3/2}}+\frac {1}{d x^2 \left (a+c x^2\right )^{3/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^2}-\frac {2 c x}{a^2 d \sqrt {a+c x^2}}+\frac {f \left (e \left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {f \left (e \left (\sqrt {e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {c d x \left (a f^2+c \left (e^2-d f\right )\right )+a e \left (a f^2+c \left (e^2-2 d f\right )\right )}{a d^2 \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )}-\frac {e}{a d^2 \sqrt {a+c x^2}}-\frac {1}{a d x \sqrt {a+c x^2}}\) |
Input:
Int[1/(x^2*(a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
Output:
-(e/(a*d^2*Sqrt[a + c*x^2])) - 1/(a*d*x*Sqrt[a + c*x^2]) - (2*c*x)/(a^2*d* Sqrt[a + c*x^2]) + (a*e*(a*f^2 + c*(e^2 - 2*d*f)) + c*d*(a*f^2 + c*(e^2 - d*f))*x)/(a*d^2*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2]) + (f*(e*(e - Sq rt[e^2 - 4*d*f])*(a*f^2 + c*(e^2 - 2*d*f)) - 2*(a*f^2*(e^2 - d*f) + c*(e^4 - 3*d*e^2*f + d^2*f^2)))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(S qrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^ 2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^ 2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) - (f*(e*(e + Sqrt[e^2 - 4*d*f] )*(a*f^2 + c*(e^2 - 2*d*f)) - 2*(a*f^2*(e^2 - d*f) + c*(e^4 - 3*d*e^2*f + d^2*f^2)))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a *f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]* d^2*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2 - 2* d*f + e*Sqrt[e^2 - 4*d*f])]) + (e*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(a^(3/ 2)*d^2)
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1638\) vs. \(2(565)=1130\).
Time = 2.24 (sec) , antiderivative size = 1639, normalized size of antiderivative = 2.65
method | result | size |
default | \(\text {Expression too large to display}\) | \(1639\) |
risch | \(\text {Expression too large to display}\) | \(1788\) |
Input:
int(1/x^2/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
-4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))*(-1/a/x/(c*x^2+a)^(1/2 )-2*c/a^2*x/(c*x^2+a)^(1/2))+4*f^2/(-e+(-4*d*f+e^2)^(1/2))^2/(-4*d*f+e^2)^ (1/2)*(2/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)*f^2/(c*(x-1/2/f*( -e+(-4*d*f+e^2)^(1/2)))^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+ e^2)^(1/2)))+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2 )+2*c*(e-(-4*d*f+e^2)^(1/2))*f/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c* e^2)*(2*c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))-c*(e-(-4*d*f+e^2)^(1/2))/f)/(2 *c*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2-c^2*(e-(-4*d*f+e^2) ^(1/2))^2/f^2)/(c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2-c*(e-(-4*d*f+e^2)^(1 /2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a* f^2-2*d*f*c+c*e^2)/f^2)^(1/2)-2/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c *e^2)*f^2*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1 /2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2-c*(e-(-4*d*f+e ^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((-(-4*d*f+e^2) ^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x-1/2/f*(-e+(-4*d*f+e^2 )^(1/2)))^2-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2))) +2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)/f^2)^(1/2))/(x-1/2/f*(- e+(-4*d*f+e^2)^(1/2)))))-4*f^2/(e+(-4*d*f+e^2)^(1/2))^2/(-4*d*f+e^2)^(1/2) *(2/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*d*f*c+c*e^2)*f^2/(c*(x+1/2*(e+(-4*d* f+e^2)^(1/2))/f)^2-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1...
Timed out. \[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^{2} \left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:
integrate(1/x**2/(c*x**2+a)**(3/2)/(f*x**2+e*x+d),x)
Output:
Integral(1/(x**2*(a + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)
\[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int { \frac {1}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (f x^{2} + e x + d\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="maxima")
Output:
integrate(1/((c*x^2 + a)^(3/2)*(f*x^2 + e*x + d)*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^2\,{\left (c\,x^2+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:
int(1/(x^2*(a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x)
Output:
int(1/(x^2*(a + c*x^2)^(3/2)*(d + e*x + f*x^2)), x)
\[ \int \frac {1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {too large to display} \] Input:
int(1/x^2/(c*x^2+a)^(3/2)/(f*x^2+e*x+d),x)
Output:
( - 4*sqrt(a + c*x**2)*a*d*e**2*f - 2*sqrt(a + c*x**2)*a*d*e*f**2*x + 2*sq rt(a + c*x**2)*a*e**4 - 8*sqrt(a + c*x**2)*c*d*e**2*f*x**2 + 4*sqrt(a + c* x**2)*c*e**4*x**2 - sqrt(a)*log(sqrt(a + c*x**2) - sqrt(a))*a*d*e*f**2*x - sqrt(a)*log(sqrt(a + c*x**2) - sqrt(a))*c*d*e*f**2*x**3 + sqrt(a)*log(sqr t(a + c*x**2) + sqrt(a))*a*d*e*f**2*x + sqrt(a)*log(sqrt(a + c*x**2) + sqr t(a))*c*d*e*f**2*x**3 + 2*int(sqrt(a + c*x**2)/(a**2*d*x**2 + a**2*e*x**3 + a**2*f*x**4 + 2*a*c*d*x**4 + 2*a*c*e*x**5 + 2*a*c*f*x**6 + c**2*d*x**6 + c**2*e*x**7 + c**2*f*x**8),x)*a**3*d**3*f**2*x - 4*int(sqrt(a + c*x**2)/( a**2*d*x**2 + a**2*e*x**3 + a**2*f*x**4 + 2*a*c*d*x**4 + 2*a*c*e*x**5 + 2* a*c*f*x**6 + c**2*d*x**6 + c**2*e*x**7 + c**2*f*x**8),x)*a**3*d**2*e**2*f* x + 2*int(sqrt(a + c*x**2)/(a**2*d*x**2 + a**2*e*x**3 + a**2*f*x**4 + 2*a* c*d*x**4 + 2*a*c*e*x**5 + 2*a*c*f*x**6 + c**2*d*x**6 + c**2*e*x**7 + c**2* f*x**8),x)*a**3*d*e**4*x + 2*int(sqrt(a + c*x**2)/(a**2*d*x**2 + a**2*e*x* *3 + a**2*f*x**4 + 2*a*c*d*x**4 + 2*a*c*e*x**5 + 2*a*c*f*x**6 + c**2*d*x** 6 + c**2*e*x**7 + c**2*f*x**8),x)*a**2*c*d**3*f**2*x**3 - 4*int(sqrt(a + c *x**2)/(a**2*d*x**2 + a**2*e*x**3 + a**2*f*x**4 + 2*a*c*d*x**4 + 2*a*c*e*x **5 + 2*a*c*f*x**6 + c**2*d*x**6 + c**2*e*x**7 + c**2*f*x**8),x)*a**2*c*d* *2*e**2*f*x**3 + 2*int(sqrt(a + c*x**2)/(a**2*d*x**2 + a**2*e*x**3 + a**2* f*x**4 + 2*a*c*d*x**4 + 2*a*c*e*x**5 + 2*a*c*f*x**6 + c**2*d*x**6 + c**2*e *x**7 + c**2*f*x**8),x)*a**2*c*d*e**4*x**3 + 2*int(sqrt(a + c*x**2)/(a*...