\(\int \frac {(d+e x^2)^{5/2} (A+B x^2+C x^4)}{(a+b x^2+c x^4)^2} \, dx\) [160]

Optimal result

Integrand size = 38, antiderivative size = 1335 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Output:

-1/2*e*(A*c^2*(-2*a*e+b*d)-a*(B*c*(-b*e+2*c*d)-C*(6*a*c*e-2*b^2*e+b*c*d))) 
*x*(e*x^2+d)^(1/2)/a/c^2/(-4*a*c+b^2)-1/2*(A*b*c-2*B*a*c+C*a*b)*e*x*(e*x^2 
+d)^(3/2)/a/c/(-4*a*c+b^2)+1/2*x*(A*(-2*a*c+b^2)-a*(B*b-2*C*a)+(A*b*c-2*B* 
a*c+C*a*b)*x^2)*(e*x^2+d)^(5/2)/a/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-1/2*(A*c^3* 
(4*a*e*(a*e^2+2*c*d^2)-b*(5*a*d*e^2+c*d^3))+a*(B*c*(2*c^3*d^3+2*b^3*e^3-2* 
c^2*d*e*(-7*a*e+2*b*d)-b*c*e^2*(10*a*e+b*d))+C*(7*b^3*c*d*e^2-4*b^4*e^3-2* 
b^2*c*e*(-10*a*e^2+c*d^2)+4*a*c^2*e*(-3*a*e^2+4*c*d^2)-b*c^2*d*(33*a*e^2+c 
*d^2)))+(A*c^3*(4*a*c*d*(a*e^2+3*c*d^2)-4*a*b*e*(a*e^2+3*c*d^2)-b^2*(-5*a* 
d*e^2+c*d^3))+a*(4*b^5*C*e^3-b^4*c*e^2*(2*B*e+7*C*d)-b^3*c*e*(28*C*a*e^2-c 
*d*(B*e+2*C*d))+4*a*c^3*(c*d^2*(2*B*e+C*d)-a*e^2*(4*B*e+13*C*d))+b^2*c^2*( 
c*d^2*(4*B*e+C*d)+a*e^2*(14*B*e+47*C*d))-4*b*c^2*(a*C*e*(-11*a*e^2+5*c*d^2 
)+B*c*d*(4*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/c^3/( 
-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)-1/2*(A*c^3*(4*a*e*(a*e^2+2*c*d^2)-b*(5*a*d*e^2+c*d^3))+a*(B*c*(2*c^3 
*d^3+2*b^3*e^3-2*c^2*d*e*(-7*a*e+2*b*d)-b*c*e^2*(10*a*e+b*d))+C*(7*b^3*c*d 
*e^2-4*b^4*e^3-2*b^2*c*e*(-10*a*e^2+c*d^2)+4*a*c^2*e*(-3*a*e^2+4*c*d^2)-b* 
c^2*d*(33*a*e^2+c*d^2)))-(A*c^3*(4*a*c*d*(a*e^2+3*c*d^2)-4*a*b*e*(a*e^2+3* 
c*d^2)-b^2*(-5*a*d*e^2+c*d^3))+a*(4*b^5*C*e^3-b^4*c*e^2*(2*B*e+7*C*d)-b^3* 
c*e*(28*C*a*e^2-c*d*(B*e+2*C*d))+4*a*c^3*(c*d^2*(2*B*e+C*d)-a*e^2*(4*B*...
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(63689\) vs. \(2(1335)=2670\).

Time = 18.80 (sec) , antiderivative size = 63689, normalized size of antiderivative = 47.71 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Result too large to show} \] Input:

Integrate[((d + e*x^2)^(5/2)*(A + B*x^2 + C*x^4))/(a + b*x^2 + c*x^4)^2,x]
 

Output:

Result too large to show
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 2256

\(\displaystyle \int \left (\frac {\left (d+e x^2\right )^{5/2} \left (-a C+A c+x^2 (B c-b C)\right )}{c \left (a+b x^2+c x^4\right )^2}+\frac {C \left (d+e x^2\right )^{5/2}}{c \left (a+b x^2+c x^4\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {C e \left (7 c d-2 \left (b-\sqrt {b^2-4 a c}\right ) e\right ) \sqrt {e x^2+d} x}{8 c^2 \sqrt {b^2-4 a c}}-\frac {C e \left (7 c d-2 \left (b+\sqrt {b^2-4 a c}\right ) e\right ) \sqrt {e x^2+d} x}{8 c^2 \sqrt {b^2-4 a c}}+\frac {C \left (2 c^3 d^3-3 c^2 e \left (b d-\sqrt {b^2-4 a c} d+2 a e\right ) d-b^2 \left (b-\sqrt {b^2-4 a c}\right ) e^3+c e^2 \left (3 d b^2-3 \sqrt {b^2-4 a c} d b+3 a e b-a \sqrt {b^2-4 a c} e\right )\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {e x^2+d}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}}-\frac {C \left (2 c^3 d^3-3 c^2 e \left (b d+\sqrt {b^2-4 a c} d+2 a e\right ) d-b^2 \left (b+\sqrt {b^2-4 a c}\right ) e^3+c e^2 \left (3 d b^2+3 \left (\sqrt {b^2-4 a c} d+a e\right ) b+a \sqrt {b^2-4 a c} e\right )\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {e x^2+d}}\right )}{c^3 \sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}+\frac {C \sqrt {e} \left (15 c^2 d^2+4 b \left (b-\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (5 b d-5 \sqrt {b^2-4 a c} d+4 a e\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^2+d}}\right )}{8 c^3 \sqrt {b^2-4 a c}}-\frac {C \sqrt {e} \left (15 c^2 d^2+4 b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (5 b d+5 \sqrt {b^2-4 a c} d+4 a e\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^2+d}}\right )}{8 c^3 \sqrt {b^2-4 a c}}+\frac {(A c-a C) \int \frac {\left (e x^2+d\right )^{5/2}}{\left (c x^4+b x^2+a\right )^2}dx}{c}+\frac {(B c-b C) \int \frac {x^2 \left (e x^2+d\right )^{5/2}}{\left (c x^4+b x^2+a\right )^2}dx}{c}\)

Input:

Int[((d + e*x^2)^(5/2)*(A + B*x^2 + C*x^4))/(a + b*x^2 + c*x^4)^2,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2256

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]
 

Maple [A] (verified)

Time = 69.18 (sec) , antiderivative size = 1214, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1214\)
default	\(\text {Expression too large to display}\)	\(1714\)
risch	\(\text {Expression too large to display}\)	\(1849\)

Input:

int((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+b*x^2+a)^2,x,method=_RETURNVERB 
OSE)
 

Output:

1/2*(1/2*(1/2*(4*a*c-b^2)*e^2*(c*x^4+b*x^2+a)*(2*B*c*e-4*C*b*e+5*C*c*d)*(( 
2*a*e-b*d+(-d^2*(4*a*c-b^2))^(1/2))*a)^(1/2)*a*((-2*a*e+b*d+(-d^2*(4*a*c-b 
^2))^(1/2))*a)^(1/2)*arctanh((e*x^2+d)^(1/2)/x/e^(1/2))-e^(1/2)*(c*((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*C*a^3*c*e^2+((b^2*C-1/2*c*(7*C*x^2+B)*b+c^2*(-2*C*x^4+B* 
x^2+A))*e^2+2*c*e*(-1/2*b*C+c*(C*x^2+B))*d+C*c^2*d^2)*a^2+(1/2*b*(2*b^2*C- 
c*(-C*x^2+B)*b+A*c^2)*x^2*e^2-c*e*(C*b^2*x^2+c*(-B*x^2+A)*b+2*A*c^2*x^2)*d 
-c^2*(1/2*(-C*x^2+B)*b+c*(B*x^2+A))*d^2)*a+1/2*A*b*c^2*d^2*(c*x^2+b))*x*(e 
*x^2+d)^(1/2)-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))*((4*(-B*c^2+2 
*C*b*c)*e^3-13*C*c^2*d*e^2)*a^3+((B*b^2*c-2*C*b^3)*e^3+c*(A*c^2-1/2*b*B*c+ 
7/2*b^2*C)*d*e^2+e*(2*B*c^3-C*b*c^2)*d^2+C*c^3*d^3)*a^2-(A*b*e-3*d*(A*c-1/ 
6*B*b))*c^3*d^2*a-1/2*A*b^2*c^3*d^3)*(c*x^4+b*x^2+a)*2^(1/2)))*(-d^2*(4*a* 
c-b^2))^(1/2)+e^(1/2)*(-3*C*a^4*c^2*e^3+c*((-3/2*b*B*c+3*b^2*C+A*c^2)*e^2+ 
(-5*C*b*c+7/2*B*c^2)*d*e+4*C*c^2*d^2)*e*a^3+(1/4*b^3*(B*c-2*C*b)*e^3-3/2*b 
*c*(A*c^2+1/12*b*B*c-7/12*b^2*C)*d*e^2+2*c^2*(A*c^2-3/4*b*B*c-1/8*b^2*C)*d 
^2*e+1/2*c^3*d^3*(B*c-C*b))*a^2+1/4*b*c^3*(A*b*e-4*(A*c-1/8*B*b)*d)*d^2*a+ 
1/8*A*b^3*c^3*d^3)*(c*x^4+b*x^2+a)*d*2^(1/2)*(arctan(a*(e*x^2+d)^(1/2)/...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**(5/2)*(C*x**4+B*x**2+A)/(c*x**4+b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(e*x^2 + d)^(5/2)/(c*x^4 + b*x^2 + a)^2, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6900 vs. \(2 (1271) = 2542\).

Time = 0.96 (sec) , antiderivative size = 6900, normalized size of antiderivative = 5.17 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm=" 
giac")
 

Output:

1/2*sqrt(e*x^2 + d)*C*e^2*x/c^2 - ((sqrt(e)*x - sqrt(e*x^2 + d))^6*C*a*b*c 
^3*d^3*sqrt(e) - 2*(sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a*c^4*d^3*sqrt(e) + ( 
sqrt(e)*x - sqrt(e*x^2 + d))^6*A*b*c^4*d^3*sqrt(e) - 4*(sqrt(e)*x - sqrt(e 
*x^2 + d))^6*C*a*b^2*c^2*d^2*e^(3/2) + 8*(sqrt(e)*x - sqrt(e*x^2 + d))^6*C 
*a^2*c^3*d^2*e^(3/2) + 4*(sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a*b*c^3*d^2*e^( 
3/2) - 8*(sqrt(e)*x - sqrt(e*x^2 + d))^6*A*a*c^4*d^2*e^(3/2) + 5*(sqrt(e)* 
x - sqrt(e*x^2 + d))^6*C*a*b^3*c*d*e^(5/2) - 15*(sqrt(e)*x - sqrt(e*x^2 + 
d))^6*C*a^2*b*c^2*d*e^(5/2) - 5*(sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a*b^2*c^ 
2*d*e^(5/2) + 10*(sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a^2*c^3*d*e^(5/2) + 5*( 
sqrt(e)*x - sqrt(e*x^2 + d))^6*A*a*b*c^3*d*e^(5/2) - 2*(sqrt(e)*x - sqrt(e 
*x^2 + d))^6*C*a*b^4*e^(7/2) + 8*(sqrt(e)*x - sqrt(e*x^2 + d))^6*C*a^2*b^2 
*c*e^(7/2) + 2*(sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a*b^3*c*e^(7/2) - 4*(sqrt 
(e)*x - sqrt(e*x^2 + d))^6*C*a^3*c^2*e^(7/2) - 6*(sqrt(e)*x - sqrt(e*x^2 + 
 d))^6*B*a^2*b*c^2*e^(7/2) - 2*(sqrt(e)*x - sqrt(e*x^2 + d))^6*A*a*b^2*c^2 
*e^(7/2) + 4*(sqrt(e)*x - sqrt(e*x^2 + d))^6*A*a^2*c^3*e^(7/2) - 3*(sqrt(e 
)*x - sqrt(e*x^2 + d))^4*C*a*b*c^3*d^4*sqrt(e) + 6*(sqrt(e)*x - sqrt(e*x^2 
 + d))^4*B*a*c^4*d^4*sqrt(e) - 3*(sqrt(e)*x - sqrt(e*x^2 + d))^4*A*b*c^4*d 
^4*sqrt(e) + 10*(sqrt(e)*x - sqrt(e*x^2 + d))^4*C*a*b^2*c^2*d^3*e^(3/2) - 
12*(sqrt(e)*x - sqrt(e*x^2 + d))^4*C*a^2*c^3*d^3*e^(3/2) - 14*(sqrt(e)*x - 
 sqrt(e*x^2 + d))^4*B*a*b*c^3*d^3*e^(3/2) + 4*(sqrt(e)*x - sqrt(e*x^2 +...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{5/2}\,\left (C\,x^4+B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \] Input:

int(((d + e*x^2)^(5/2)*(A + B*x^2 + C*x^4))/(a + b*x^2 + c*x^4)^2,x)
 

Output:

int(((d + e*x^2)^(5/2)*(A + B*x^2 + C*x^4))/(a + b*x^2 + c*x^4)^2, x)
 

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}}{c \,x^{4}+b \,x^{2}+a}d x \right ) d^{2}+\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c \,x^{4}+b \,x^{2}+a}d x \right ) e^{2}+2 \left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) d e \] Input:

int((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+b*x^2+a)^2,x)
 

Output:

int(sqrt(d + e*x**2)/(a + b*x**2 + c*x**4),x)*d**2 + int((sqrt(d + e*x**2) 
*x**4)/(a + b*x**2 + c*x**4),x)*e**2 + 2*int((sqrt(d + e*x**2)*x**2)/(a + 
b*x**2 + c*x**4),x)*d*e