Integrand size = 38, antiderivative size = 1348 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Output:
1/2*e*(A*(b^3*d*e^2+b*c*d*(-3*a*e^2+c*d^2)+4*a*c*e*(-2*a*e^2+c*d^2)-b^2*(- 2*a*e^3+2*c*d^2*e))+a*d*(C*(a*b*e^2-12*a*c*d*e+2*b^2*d*e+b*c*d^2)-B*(2*c^2 *d^2+3*b^2*e^2-2*c*e*(5*a*e+b*d))))*x/a/(-4*a*c+b^2)/d/(a*e^2-b*d*e+c*d^2) ^2/(e*x^2+d)^(1/2)+1/2*x*(A*(3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*d)-a*(2*B*a*c *e-B*b^2*e+B*b*c*d+C*a*b*e-2*C*a*c*d)+c*(A*(2*a*c*e-b^2*e+b*c*d)-a*(-B*b*e +2*B*c*d+2*C*a*e-C*b*d))*x^2)/a/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(e*x^2+d) ^(1/2)/(c*x^4+b*x^2+a)+1/2*c*(A*(b^3*d*e^2+12*a^2*c*e^3+b*c*d*(a*e^2+c*d^2 )-2*b^2*(2*a*e^3+c*d^2*e))+a*(C*(-4*a^2*e^3+5*a*b*d*e^2+8*a*c*d^2*e-4*b^2* d^2*e+b*c*d^3)-B*(2*c^2*d^3-2*c*d*e*(-7*a*e+2*b*d)-b*e^2*(2*a*e+b*d)))+(A* (b^4*d*e^2+b^2*c*d*(-a*e^2+c*d^2)+20*a*b*c*e*(a*e^2+c*d^2)-12*a*c^2*d*(3*a *e^2+c*d^2)-2*b^3*(2*a*e^3+c*d^2*e))+a*(b^3*d*e*(B*e+4*C*d)+4*a*c*(a*e^2*( -4*B*e+5*C*d)-c*d^2*(-2*B*e+C*d))-b^2*(a*e^2*(-2*B*e+11*C*d)+c*d^2*(8*B*e+ C*d))-4*b*(a*C*e*(-a*e^2+c*d^2)-B*c*d*(2*a*e^2+c*d^2))))/(-4*a*c+b^2)^(1/2 ))*arctan((2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)^(1/2)*x/(b-(-4*a*c+b^2)^(1/2))^ (1/2)/(e*x^2+d)^(1/2))/a/(-4*a*c+b^2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)/(2*c*d- (b-(-4*a*c+b^2)^(1/2))*e)^(1/2)/(a*e^2-b*d*e+c*d^2)^2+1/2*c*(A*(b^3*d*e^2+ 12*a^2*c*e^3+b*c*d*(a*e^2+c*d^2)-2*b^2*(2*a*e^3+c*d^2*e))+a*(C*(-4*a^2*e^3 +5*a*b*d*e^2+8*a*c*d^2*e-4*b^2*d^2*e+b*c*d^3)-B*(2*c^2*d^3-2*c*d*e*(-7*a*e +2*b*d)-b*e^2*(2*a*e+b*d)))-(A*(b^4*d*e^2+b^2*c*d*(-a*e^2+c*d^2)+20*a*b*c* e*(a*e^2+c*d^2)-12*a*c^2*d*(3*a*e^2+c*d^2)-2*b^3*(2*a*e^3+c*d^2*e))+a*(...
Leaf count is larger than twice the leaf count of optimal. \(59143\) vs. \(2(1348)=2696\).
Time = 23.07 (sec) , antiderivative size = 59143, normalized size of antiderivative = 43.87 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Result too large to show} \] Input:
Integrate[(A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^2),x]
Output:
Result too large to show
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.
\(\displaystyle \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 2256 |
\(\displaystyle \int \left (\frac {-a C+A c+x^2 (B c-b C)}{c \left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2}+\frac {C}{c \left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(A c-a C) \int \frac {1}{\left (e x^2+d\right )^{3/2} \left (c x^4+b x^2+a\right )^2}dx}{c}+\frac {(B c-b C) \int \frac {x^2}{\left (e x^2+d\right )^{3/2} \left (c x^4+b x^2+a\right )^2}dx}{c}-\frac {C \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac {C \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}+\frac {C e^2 x}{c d \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}\) |
Input:
Int[(A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^2),x]
Output:
$Aborted
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4 )^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Time = 166.52 (sec) , antiderivative size = 1358, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(\text {Expression too large to display}\) | \(1358\) |
default | \(\text {Expression too large to display}\) | \(1914\) |
Input:
int((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^2,x,method=_RETURNVERB OSE)
Output:
-1/(e*x^2+d)^(1/2)*(((arctanh(a*(e*x^2+d)^(1/2)/x*2^(1/2)/((2*a*e-b*d+(-d^ 2*(4*a*c-b^2))^(1/2))*a)^(1/2))*((-2*a*e+b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^ (1/2)-arctan(a*(e*x^2+d)^(1/2)/x*2^(1/2)/((-2*a*e+b*d+(-d^2*(4*a*c-b^2))^( 1/2))*a)^(1/2))*((2*a*e-b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2))*(-3/8*c^2* (1/3*C*a^2+(A*c-1/6*B*b)*a-1/6*A*b^2)*d^3+5/8*(1/5*(2*B*c+C*b)*a^2+b*(A*c- 1/5*B*b)*a-1/5*A*b^3)*c*e*d^2-9/8*e^2*(-5/9*c*a^3*C+(A*c^2+1/6*b*(B*c+C*b) )*a^2-1/18*a*B*b^3-1/18*A*b^4)*d+(a*c-1/4*b^2)*e^3*a*(-1/2*B*a+A*b))*(c*x^ 4+b*x^2+a)*d*2^(1/2)*(e*x^2+d)^(1/2)-((2*a*e-b*d+(-d^2*(4*a*c-b^2))^(1/2)) *a)^(1/2)*(1/4*c^2*(-C*a^2+((B*c-1/2*b*C)*x^2+A*c+1/2*B*b)*a-1/2*A*b*(c*x^ 2+b))*d^4-3/4*(1/3*(-C*c*x^2-2*B*c-C*b)*a^2+(1/3*(-B*c^2+1/2*C*b*c)*x^4+1/ 3*(A*c^2+1/2*b*B*c)*x^2+b*(A*c+1/3*B*b))*a-1/3*b*A*(-1/2*c*x^2+b)*(c*x^2+b ))*c*e*d^3-1/4*(-5*c*a^3*C+(-6*C*c^2*x^4+(-B*c^2-9/2*C*b*c)*x^2+A*c^2+3/2* b*(B*c+C*b))*a^2+((2*A*c^3+B*b*c^2+C*b^2*c)*x^4+(3/2*c^2*b*A+1/2*B*b^2*c+C *b^3)*x^2-2*A*b^2*c-1/2*B*b^3)*a+1/2*A*b^2*(c*x^2+b)*(-2*c*x^2+b))*e^2*d^2 -1/4*e^3*(4*c*(-1/4*C*x^2+B)*a^3+((5*B*c^2+1/2*C*b*c)*x^4+(A*c^2+11/2*b*B* c+1/2*b^2*C)*x^2-B*b^2)*a^2-2*(3/4*c*(A*c+B*b)*x^2+b*(A*c+3/4*B*b))*b*x^2* a+1/2*A*b^3*x^2*(c*x^2+b))*d+(a*c-1/4*b^2)*A*e^4*(c*x^4+b*x^2+a)*a)*((-2*a *e+b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2)*x)*(-d^2*(4*a*c-b^2))^(1/2)+3/2* (1/3*c^2*(1/2*(-B*c+C*b)*a^2+b*(A*c-1/8*B*b)*a-1/8*A*b^3)*d^3-7/12*c*e*(-8 /7*c*a^3*C+1/7*(-2*B*b*c+5*C*b^2)*a^2+b^2*(A*c-1/7*B*b)*a-1/7*A*b^4)*d^...
Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" fricas")
Output:
Timed out
Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((C*x**4+B*x**2+A)/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a)**2,x)
Output:
Timed out
\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {C x^{4} + B x^{2} + A}{{\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" maxima")
Output:
integrate((C*x^4 + B*x^2 + A)/((c*x^4 + b*x^2 + a)^2*(e*x^2 + d)^(3/2)), x )
Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm=" giac")
Output:
Timed out
Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {C\,x^4+B\,x^2+A}{{\left (e\,x^2+d\right )}^{3/2}\,{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \] Input:
int((A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^2),x)
Output:
int((A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^2), x)
\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {1}{\sqrt {e \,x^{2}+d}\, a d +\sqrt {e \,x^{2}+d}\, a e \,x^{2}+\sqrt {e \,x^{2}+d}\, b d \,x^{2}+\sqrt {e \,x^{2}+d}\, b e \,x^{4}+\sqrt {e \,x^{2}+d}\, c d \,x^{4}+\sqrt {e \,x^{2}+d}\, c e \,x^{6}}d x \] Input:
int((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^2,x)
Output:
int(1/(sqrt(d + e*x**2)*a*d + sqrt(d + e*x**2)*a*e*x**2 + sqrt(d + e*x**2) *b*d*x**2 + sqrt(d + e*x**2)*b*e*x**4 + sqrt(d + e*x**2)*c*d*x**4 + sqrt(d + e*x**2)*c*e*x**6),x)