\(\int \frac {A+B x^2+C x^4}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\) [158]

Optimal result

Integrand size = 38, antiderivative size = 449 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\frac {\left (C d^2-e (B d-A e)\right ) x}{d \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {\left (B c d-b C d-A c e+a C e+\frac {b^2 C d+2 c (A c d-a C d+a B e)-b (B c d+A c e+a C e)}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \left (c d^2-e (b d-a e)\right )}+\frac {\left (B c d-b C d-A c e+a C e-\frac {b^2 C d+2 c (A c d-a C d+a B e)-b (B c d+A c e+a C e)}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \left (c d^2-e (b d-a e)\right )} \] Output:

(C*d^2-e*(-A*e+B*d))*x/d/(a*e^2-b*d*e+c*d^2)/(e*x^2+d)^(1/2)+(B*c*d-C*b*d- 
A*c*e+C*a*e+(b^2*C*d+2*c*(A*c*d+B*a*e-C*a*d)-b*(A*c*e+B*c*d+C*a*e))/(-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))/(b-(-4*a*c+b^2)^(1/2))^(1/2)/(2*c*d-(b-( 
-4*a*c+b^2)^(1/2))*e)^(1/2)/(c*d^2-e*(-a*e+b*d))+(B*c*d-C*b*d-A*c*e+C*a*e- 
(b^2*C*d+2*c*(A*c*d+B*a*e-C*a*d)-b*(A*c*e+B*c*d+C*a*e))/(-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))/(b+(-4*a*c+b^2)^(1/2))^(1/2)/(2*c*d-(b+(-4*a*c+b^2)^ 
(1/2))*e)^(1/2)/(c*d^2-e*(-a*e+b*d))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(16421\) vs. \(2(449)=898\).

Time = 21.51 (sec) , antiderivative size = 16421, normalized size of antiderivative = 36.57 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Result too large to show} \] Input:

Integrate[(A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2256, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx\)

\(\Big \downarrow \) 2256

\(\displaystyle \int \left (\frac {-a C+A c+x^2 (B c-b C)}{c \left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )}+\frac {C}{c \left (d+e x^2\right )^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2}}+\frac {2 \left (\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right ) \arctan \left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2}}-\frac {e x \left (-\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right )}{c d \sqrt {d+e x^2} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {e x \left (\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right )}{c d \sqrt {d+e x^2} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {C x}{c d \sqrt {d+e x^2}}\)

Input:

Int[(A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]
 

Output:

(C*x)/(c*d*Sqrt[d + e*x^2]) - ((B*c - b*C - (b*B*c - b^2*C - 2*c*(A*c - a* 
C))/Sqrt[b^2 - 4*a*c])*e*x)/(c*d*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*Sqrt[ 
d + e*x^2]) - ((B*c - b*C + (b*B*c - b^2*C - 2*c*(A*c - a*C))/Sqrt[b^2 - 4 
*a*c])*e*x)/(c*d*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*Sqrt[d + e*x^2]) + (2 
*(B*c - b*C - (b*B*c - b^2*C - 2*c*(A*c - a*C))/Sqrt[b^2 - 4*a*c])*ArcTan[ 
(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*S 
qrt[d + e*x^2])])/(Sqrt[b - Sqrt[b^2 - 4*a*c]]*(2*c*d - (b - Sqrt[b^2 - 4* 
a*c])*e)^(3/2)) + (2*(B*c - b*C + (b*B*c - b^2*C - 2*c*(A*c - a*C))/Sqrt[b 
^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + 
Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*(2*c*d 
- (b + Sqrt[b^2 - 4*a*c])*e)^(3/2))
 

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 = 0.68 (sec) , antiderivative size = 522, normalized size of antiderivative = 1.16

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {e \,x^{2}+d}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (\left (\left (-B e +C d \right ) a +A \left (e b -c d \right )\right ) \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}+2 \left (-a^{2} C e +\left (\left (-B c +\frac {b C}{2}\right ) d +e \left (A c +\frac {B b}{2}\right )\right ) a -\frac {A b \left (e b -c d \right )}{2}\right ) d \right ) d \sqrt {2}\, \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}}\right )}{2}+\sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (\frac {\sqrt {e \,x^{2}+d}\, \left (\left (\left (-B e +C d \right ) a +A \left (e b -c d \right )\right ) \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}-2 \left (-a^{2} C e +\left (\left (-B c +\frac {b C}{2}\right ) d +e \left (A c +\frac {B b}{2}\right )\right ) a -\frac {A b \left (e b -c d \right )}{2}\right ) d \right ) d \sqrt {2}\, \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}}\right )}{2}+\left (A \,e^{2}-B d e +C \,d^{2}\right ) \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, x \right )}{\sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \sqrt {e \,x^{2}+d}\, \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (a \,e^{2}-b d e +c \,d^{2}\right ) d}\)	\(522\)
default	\(\frac {C x}{c d \sqrt {e \,x^{2}+d}}+\frac {-\frac {\sqrt {e \,x^{2}+d}\, c \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (\left (\left (-B e +C d \right ) a +A \left (e b -c d \right )\right ) \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}+2 \left (-a^{2} C e +\left (\left (-B c +\frac {b C}{2}\right ) d +e \left (A c +\frac {B b}{2}\right )\right ) a -\frac {A b \left (e b -c d \right )}{2}\right ) d \right ) d \sqrt {2}\, \operatorname {arctanh}\left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}}\right )}{2}+\sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (\frac {\sqrt {e \,x^{2}+d}\, c \left (\left (\left (-B e +C d \right ) a +A \left (e b -c d \right )\right ) \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}-2 \left (-a^{2} C e +\left (\left (-B c +\frac {b C}{2}\right ) d +e \left (A c +\frac {B b}{2}\right )\right ) a -\frac {A b \left (e b -c d \right )}{2}\right ) d \right ) d \sqrt {2}\, \arctan \left (\frac {a \sqrt {e \,x^{2}+d}\, \sqrt {2}}{x \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}}\right )}{2}+e \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (-C a e +\left (-B c +b C \right ) d +A c e \right ) x \right )}{c \sqrt {\left (2 a e -b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \sqrt {e \,x^{2}+d}\, \sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\, \sqrt {\left (-2 a e +b d +\sqrt {-4 \left (a c -\frac {b^{2}}{4}\right ) d^{2}}\right ) a}\, \left (a \,e^{2}-b d e +c \,d^{2}\right ) d}\)	\(551\)

Input:

int((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOS 
E)
 

Output:

1/((2*a*e-b*d+(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2)/(e*x^2+d)^(1/2)*(-1/2 
*(e*x^2+d)^(1/2)*((-2*a*e+b*d+(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2)*(((-B 
*e+C*d)*a+A*(b*e-c*d))*(-4*(a*c-1/4*b^2)*d^2)^(1/2)+2*(-a^2*C*e+((-B*c+1/2 
*b*C)*d+e*(A*c+1/2*B*b))*a-1/2*A*b*(b*e-c*d))*d)*d*2^(1/2)*arctanh(a*(e*x^ 
2+d)^(1/2)/x*2^(1/2)/((2*a*e-b*d+(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2))+( 
(2*a*e-b*d+(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2)*(1/2*(e*x^2+d)^(1/2)*((( 
-B*e+C*d)*a+A*(b*e-c*d))*(-4*(a*c-1/4*b^2)*d^2)^(1/2)-2*(-a^2*C*e+((-B*c+1 
/2*b*C)*d+e*(A*c+1/2*B*b))*a-1/2*A*b*(b*e-c*d))*d)*d*2^(1/2)*arctan(a*(e*x 
^2+d)^(1/2)/x*2^(1/2)/((-2*a*e+b*d+(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2)) 
+(A*e^2-B*d*e+C*d^2)*(-4*(a*c-1/4*b^2)*d^2)^(1/2)*((-2*a*e+b*d+(-4*(a*c-1/ 
4*b^2)*d^2)^(1/2))*a)^(1/2)*x))/(-4*(a*c-1/4*b^2)*d^2)^(1/2)/((-2*a*e+b*d+ 
(-4*(a*c-1/4*b^2)*d^2)^(1/2))*a)^(1/2)/(a*e^2-b*d*e+c*d^2)/d
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fr 
icas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate((C*x**4+B*x**2+A)/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {C x^{4} + B x^{2} + A}{{\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="ma 
xima")
 

Output:

integrate((C*x^4 + B*x^2 + A)/((c*x^4 + b*x^2 + a)*(e*x^2 + d)^(3/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="gi 
ac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {C\,x^4+B\,x^2+A}{{\left (e\,x^2+d\right )}^{3/2}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \] Input:

int((A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x)
 

Output:

int((A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.09 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx=\frac {\sqrt {e \,x^{2}+d}\, e x +\sqrt {e}\, d +\sqrt {e}\, e \,x^{2}}{d e \left (e \,x^{2}+d \right )} \] Input:

int((C*x^4+B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)
 

Output:

(sqrt(d + e*x**2)*e*x + sqrt(e)*d + sqrt(e)*e*x**2)/(d*e*(d + e*x**2))