\(\int \frac {1}{(d+e x^2)^{3/2} (a-c x^4)^3} \, dx\) [364]

Optimal result

Integrand size = 22, antiderivative size = 396 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\frac {e^2 \left (11 c^2 d^4-39 a c d^2 e^2-32 a^2 e^4\right ) x}{32 a^2 d \left (c d^2-a e^2\right )^3 \sqrt {d+e x^2}}+\frac {c x \left (d-e x^2\right )}{8 a \left (c d^2-a e^2\right ) \sqrt {d+e x^2} \left (a-c x^4\right )^2}+\frac {c x \left (d \left (7 c d^2-17 a e^2\right )-2 e \left (2 c d^2-7 a e^2\right ) x^2\right )}{32 a^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x^2} \left (a-c x^4\right )}+\frac {3 \sqrt {c} \left (7 c d^2-22 \sqrt {a} \sqrt {c} d e+20 a e^2\right ) \arctan \left (\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{64 a^{11/4} \left (\sqrt {c} d-\sqrt {a} e\right )^{7/2}}+\frac {3 \sqrt {c} \left (7 c d^2+22 \sqrt {a} \sqrt {c} d e+20 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {c} d+\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{64 a^{11/4} \left (\sqrt {c} d+\sqrt {a} e\right )^{7/2}} \] Output:

1/32*e^2*(-32*a^2*e^4-39*a*c*d^2*e^2+11*c^2*d^4)*x/a^2/d/(-a*e^2+c*d^2)^3/ 
(e*x^2+d)^(1/2)+1/8*c*x*(-e*x^2+d)/a/(-a*e^2+c*d^2)/(e*x^2+d)^(1/2)/(-c*x^ 
4+a)^2+1/32*c*x*(d*(-17*a*e^2+7*c*d^2)-2*e*(-7*a*e^2+2*c*d^2)*x^2)/a^2/(-a 
*e^2+c*d^2)^2/(e*x^2+d)^(1/2)/(-c*x^4+a)+3/64*c^(1/2)*(7*c*d^2-22*a^(1/2)* 
c^(1/2)*d*e+20*a*e^2)*arctan((c^(1/2)*d-a^(1/2)*e)^(1/2)*x/a^(1/4)/(e*x^2+ 
d)^(1/2))/a^(11/4)/(c^(1/2)*d-a^(1/2)*e)^(7/2)+3/64*c^(1/2)*(7*c*d^2+22*a^ 
(1/2)*c^(1/2)*d*e+20*a*e^2)*arctanh((c^(1/2)*d+a^(1/2)*e)^(1/2)*x/a^(1/4)/ 
(e*x^2+d)^(1/2))/a^(11/4)/(c^(1/2)*d+a^(1/2)*e)^(7/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 2.53 (sec) , antiderivative size = 2037, normalized size of antiderivative = 5.14 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\text {Result too large to show} \] Input:

Integrate[1/((d + e*x^2)^(3/2)*(a - c*x^4)^3),x]
 

Output:

((-2*x*(32*a^4*e^6 + 2*a^3*c*e^4*(9*d^2 + 9*d*e*x^2 - 32*e^2*x^4) + c^4*d^ 
4*x^4*(7*d^2 - 4*d*e*x^2 - 11*e^2*x^4) + a^2*c^2*e^2*(21*d^4 - 26*d^3*e*x^ 
2 - 61*d^2*e^2*x^4 - 14*d*e^3*x^6 + 32*e^4*x^8) + a*c^3*d^2*(-11*d^4 + 8*d 
^3*e*x^2 - 2*d^2*e^2*x^4 + 18*d*e^3*x^6 + 39*e^4*x^8)))/(a^2*d*(c*d^2 - a* 
e^2)^3*Sqrt[d + e*x^2]*(a - c*x^4)^2) + (32*e^(11/2)*RootSum[c*d^4 - 4*c*d 
^3*#1 + 6*c*d^2*#1^2 - 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (17*c*d^2*L 
og[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 16*a*e^2*Log[d + 2*e* 
x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 6*c*d*Log[d + 2*e*x^2 - 2*Sqrt[e 
]*x*Sqrt[d + e*x^2] - #1]*#1 + c*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e* 
x^2] - #1]*#1^2)/(c*d^3 - 3*c*d^2*#1 + 8*a*e^2*#1 + 3*c*d*#1^2 - c*#1^3) & 
 ])/(c*d^2 - a*e^2)^3 - (16*e^(7/2)*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^2*# 
1^2 - 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (860*c^4*d^8*Log[d + 2*e*x^2 
 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 6091*a*c^3*d^6*e^2*Log[d + 2*e*x^2 
- 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 13680*a^2*c^2*d^4*e^4*Log[d + 2*e*x^ 
2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 12544*a^3*c*d^2*e^6*Log[d + 2*e*x^ 
2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 4096*a^4*e^8*Log[d + 2*e*x^2 - 2*S 
qrt[e]*x*Sqrt[d + e*x^2] - #1] - 706*c^4*d^7*Log[d + 2*e*x^2 - 2*Sqrt[e]*x 
*Sqrt[d + e*x^2] - #1]*#1 + 2364*a*c^3*d^5*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e] 
*x*Sqrt[d + e*x^2] - #1]*#1 - 2688*a^2*c^2*d^3*e^4*Log[d + 2*e*x^2 - 2*Sqr 
t[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 1024*a^3*c*d*e^6*Log[d + 2*e*x^2 - 2*...
 

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.

\(\displaystyle \int \frac {1}{\left (a-c x^4\right )^3 \left (d+e x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1571

\(\displaystyle \int \frac {1}{\left (a-c x^4\right )^3 \left (d+e x^2\right )^{3/2}}dx\)

Input:

Int[1/((d + e*x^2)^(3/2)*(a - c*x^4)^3),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 1571

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U 
nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
 

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.38

method	result	size
pseudoelliptic	\(\frac {\frac {15 d^{2} \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \left (\left (\frac {19}{10} a^{2} e^{4}-\frac {5}{4} a c \,d^{2} e^{2}+\frac {7}{20} c^{2} d^{4}\right ) \sqrt {d^{2} a c}+a^{3} e^{5}+\frac {a^{2} c \,d^{2} e^{3}}{20}-\frac {c^{2} d^{4} e a}{20}\right ) \left (-c \,x^{4}+a \right )^{2} c \sqrt {e \,x^{2}+d}\, \arctan \left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}}\right )}{16}+\frac {15 \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, \left (d^{2} \left (\left (-\frac {19}{10} a^{2} e^{4}+\frac {5}{4} a c \,d^{2} e^{2}-\frac {7}{20} c^{2} d^{4}\right ) \sqrt {d^{2} a c}+a^{3} e^{5}+\frac {a^{2} c \,d^{2} e^{3}}{20}-\frac {c^{2} d^{4} e a}{20}\right ) \left (-c \,x^{4}+a \right )^{2} c \sqrt {e \,x^{2}+d}\, \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}}\right )+\frac {16 x \left (a^{4} e^{6}+\frac {9 \left (-\frac {32}{9} e^{2} x^{4}+d e \,x^{2}+d^{2}\right ) c \,e^{4} a^{3}}{16}+\frac {21 \left (\frac {32}{21} e^{4} x^{8}-\frac {2}{3} d \,e^{3} x^{6}-\frac {61}{21} d^{2} e^{2} x^{4}-\frac {26}{21} d^{3} e \,x^{2}+d^{4}\right ) c^{2} e^{2} a^{2}}{32}-\frac {11 \left (-\frac {39}{11} e^{3} x^{6}+\frac {21}{11} d \,e^{2} x^{4}-\frac {19}{11} d^{2} e \,x^{2}+d^{3}\right ) d^{2} \left (e \,x^{2}+d \right ) c^{3} a}{32}+\frac {7 x^{4} d^{4} \left (e \,x^{2}+d \right ) c^{4} \left (-\frac {11 e \,x^{2}}{7}+d \right )}{32}\right ) \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \sqrt {d^{2} a c}}{15}\right )}{16}}{d \left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {d^{2} a c}\, \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, a^{2} \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \left (-c \,x^{4}+a \right )^{2} \sqrt {e \,x^{2}+d}}\)	\(548\)
default	\(\text {Expression too large to display}\)	\(9876\)

Input:

int(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x,method=_RETURNVERBOSE)
 

Output:

15/16/(e*x^2+d)^(1/2)/(d^2*a*c)^(1/2)*(d^2*((a*e+(d^2*a*c)^(1/2))*a)^(1/2) 
*((19/10*a^2*e^4-5/4*a*c*d^2*e^2+7/20*c^2*d^4)*(d^2*a*c)^(1/2)+a^3*e^5+1/2 
0*a^2*c*d^2*e^3-1/20*c^2*d^4*e*a)*(-c*x^4+a)^2*c*(e*x^2+d)^(1/2)*arctan((e 
*x^2+d)^(1/2)/x*a/((-a*e+(d^2*a*c)^(1/2))*a)^(1/2))+((-a*e+(d^2*a*c)^(1/2) 
)*a)^(1/2)*(d^2*((-19/10*a^2*e^4+5/4*a*c*d^2*e^2-7/20*c^2*d^4)*(d^2*a*c)^( 
1/2)+a^3*e^5+1/20*a^2*c*d^2*e^3-1/20*c^2*d^4*e*a)*(-c*x^4+a)^2*c*(e*x^2+d) 
^(1/2)*arctanh((e*x^2+d)^(1/2)/x*a/((a*e+(d^2*a*c)^(1/2))*a)^(1/2))+16/15* 
x*(a^4*e^6+9/16*(-32/9*e^2*x^4+d*e*x^2+d^2)*c*e^4*a^3+21/32*(32/21*e^4*x^8 
-2/3*d*e^3*x^6-61/21*d^2*e^2*x^4-26/21*d^3*e*x^2+d^4)*c^2*e^2*a^2-11/32*(- 
39/11*e^3*x^6+21/11*d*e^2*x^4-19/11*d^2*e*x^2+d^3)*d^2*(e*x^2+d)*c^3*a+7/3 
2*x^4*d^4*(e*x^2+d)*c^4*(-11/7*e*x^2+d))*((a*e+(d^2*a*c)^(1/2))*a)^(1/2)*( 
d^2*a*c)^(1/2)))/((a*e+(d^2*a*c)^(1/2))*a)^(1/2)/((-a*e+(d^2*a*c)^(1/2))*a 
)^(1/2)/d/(a*e^2-c*d^2)^3/a^2/(-c*x^4+a)^2
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x**2+d)**(3/2)/(-c*x**4+a)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\int { -\frac {1}{{\left (c x^{4} - a\right )}^{3} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x, algorithm="maxima")
 

Output:

-integrate(1/((c*x^4 - a)^3*(e*x^2 + d)^(3/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\int \frac {1}{{\left (a-c\,x^4\right )}^3\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \] Input:

int(1/((a - c*x^4)^3*(d + e*x^2)^(3/2)),x)
 

Output:

int(1/((a - c*x^4)^3*(d + e*x^2)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a-c x^4\right )^3} \, dx=\int \frac {1}{\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (-c \,x^{4}+a \right )^{3}}d x \] Input:

int(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x)
 

Output:

int(1/(e*x^2+d)^(3/2)/(-c*x^4+a)^3,x)