\(\int \frac {1}{(d+e x^2)^{5/2} (a-c x^4)} \, dx\) [347]

Optimal result

Integrand size = 22, antiderivative size = 222 \[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=-\frac {e^2 x}{3 d \left (c d^2-a e^2\right ) \left (d+e x^2\right )^{3/2}}-\frac {2 e^2 \left (4 c d^2-a e^2\right ) x}{3 d^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x^2}}+\frac {c \arctan \left (\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{2 a^{3/4} \left (\sqrt {c} d-\sqrt {a} e\right )^{5/2}}+\frac {c \text {arctanh}\left (\frac {\sqrt {\sqrt {c} d+\sqrt {a} e} x}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{2 a^{3/4} \left (\sqrt {c} d+\sqrt {a} e\right )^{5/2}} \] Output:

-1/3*e^2*x/d/(-a*e^2+c*d^2)/(e*x^2+d)^(3/2)-2/3*e^2*(-a*e^2+4*c*d^2)*x/d^2 
/(-a*e^2+c*d^2)^2/(e*x^2+d)^(1/2)+1/2*c*arctan((c^(1/2)*d-a^(1/2)*e)^(1/2) 
*x/a^(1/4)/(e*x^2+d)^(1/2))/a^(3/4)/(c^(1/2)*d-a^(1/2)*e)^(5/2)+1/2*c*arct 
anh((c^(1/2)*d+a^(1/2)*e)^(1/2)*x/a^(1/4)/(e*x^2+d)^(1/2))/a^(3/4)/(c^(1/2 
)*d+a^(1/2)*e)^(5/2)
 

Mathematica [C] (verified)

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

Time = 0.39 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=-\frac {e^2 x \left (9 c d^3-3 a d e^2+8 c d^2 e x^2-2 a e^3 x^2\right )}{3 d^2 \left (c d^2-a e^2\right )^2 \left (d+e x^2\right )^{3/2}}+\frac {c e^{3/2} \text {RootSum}\left [c d^4-4 c d^3 \text {$\#$1}+6 c d^2 \text {$\#$1}^2-16 a e^2 \text {$\#$1}^2-4 c d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {c d^3 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right )-4 c d^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a e^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{c d^3-3 c d^2 \text {$\#$1}+8 a e^2 \text {$\#$1}+3 c d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]}{\left (c d^2-a e^2\right )^2} \] Input:

Integrate[1/((d + e*x^2)^(5/2)*(a - c*x^4)),x]
 

Output:

-1/3*(e^2*x*(9*c*d^3 - 3*a*d*e^2 + 8*c*d^2*e*x^2 - 2*a*e^3*x^2))/(d^2*(c*d 
^2 - a*e^2)^2*(d + e*x^2)^(3/2)) + (c*e^(3/2)*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 & , (c*d^3*Log[d + 2*e 
*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 4*c*d^2*Log[d + 2*e*x^2 - 2*Sqr 
t[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 2*a*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sq 
rt[d + e*x^2] - #1]*#1 + c*d*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)^2
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.65, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1487, 209, 208, 2257, 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.

\(\displaystyle \int \frac {1}{\left (a-c x^4\right ) \left (d+e x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1487

\(\displaystyle \frac {c \int \frac {d-e x^2}{\left (e x^2+d\right )^{3/2} \left (a-c x^4\right )}dx}{c d^2-a e^2}-\frac {e^2 \int \frac {1}{\left (e x^2+d\right )^{5/2}}dx}{c d^2-a e^2}\)

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {c \left (\frac {\left (\sqrt {a} e+\sqrt {c} d\right ) \arctan \left (\frac {x \sqrt {\sqrt {c} d-\sqrt {a} e}}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{2 a^{3/4} \left (\sqrt {c} d-\sqrt {a} e\right )^{3/2}}+\frac {\left (\sqrt {c} d-\sqrt {a} e\right ) \text {arctanh}\left (\frac {x \sqrt {\sqrt {a} e+\sqrt {c} d}}{\sqrt [4]{a} \sqrt {d+e x^2}}\right )}{2 a^{3/4} \left (\sqrt {a} e+\sqrt {c} d\right )^{3/2}}-\frac {e x \left (\sqrt {a} e+\sqrt {c} d\right )}{2 \sqrt {a} \sqrt {c} d \sqrt {d+e x^2} \left (\sqrt {c} d-\sqrt {a} e\right )}+\frac {e x \left (\sqrt {c} d-\sqrt {a} e\right )}{2 \sqrt {a} \sqrt {c} d \sqrt {d+e x^2} \left (\sqrt {a} e+\sqrt {c} d\right )}\right )}{c d^2-a e^2}-\frac {e^2 \left (\frac {2 x}{3 d^2 \sqrt {d+e x^2}}+\frac {x}{3 d \left (d+e x^2\right )^{3/2}}\right )}{c d^2-a e^2}\)

Input:

Int[1/((d + e*x^2)^(5/2)*(a - c*x^4)),x]
 

Output:

-((e^2*(x/(3*d*(d + e*x^2)^(3/2)) + (2*x)/(3*d^2*Sqrt[d + e*x^2])))/(c*d^2 
 - a*e^2)) + (c*((e*(Sqrt[c]*d - Sqrt[a]*e)*x)/(2*Sqrt[a]*Sqrt[c]*d*(Sqrt[ 
c]*d + Sqrt[a]*e)*Sqrt[d + e*x^2]) - (e*(Sqrt[c]*d + Sqrt[a]*e)*x)/(2*Sqrt 
[a]*Sqrt[c]*d*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[d + e*x^2]) + ((Sqrt[c]*d + Sqr 
t[a]*e)*ArcTan[(Sqrt[Sqrt[c]*d - Sqrt[a]*e]*x)/(a^(1/4)*Sqrt[d + e*x^2])]) 
/(2*a^(3/4)*(Sqrt[c]*d - Sqrt[a]*e)^(3/2)) + ((Sqrt[c]*d - Sqrt[a]*e)*ArcT 
anh[(Sqrt[Sqrt[c]*d + Sqrt[a]*e]*x)/(a^(1/4)*Sqrt[d + e*x^2])])/(2*a^(3/4) 
*(Sqrt[c]*d + Sqrt[a]*e)^(3/2))))/(c*d^2 - a*e^2)
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 1487

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Simp[e^2 
/(c*d^2 + a*e^2)   Int[(d + e*x^2)^q, x], x] + Simp[c/(c*d^2 + a*e^2)   Int 
[(d + e*x^2)^(q + 1)*((d - e*x^2)/(a + c*x^4)), x], x] /; FreeQ[{a, c, d, e 
}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q] && LtQ[q, -1]
 

rule 2009

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

rule 2257

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.42

method	result	size
pseudoelliptic	\(-\frac {d^{2} \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \left (\frac {\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {d^{2} a c}}{2}+d^{2} e a c \right ) c \left (e \,x^{2}+d \right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}}\right )+\sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, \left (d^{2} \left (\frac {\left (-a \,e^{2}-c \,d^{2}\right ) \sqrt {d^{2} a c}}{2}+d^{2} e a c \right ) c \left (e \,x^{2}+d \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, a}{x \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}}\right )-x \sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \left (e^{2} \left (\frac {2 e \,x^{2}}{3}+d \right ) a -3 d^{2} \left (\frac {8 e \,x^{2}}{9}+d \right ) c \right ) e^{2} \sqrt {d^{2} a c}\right )}{\sqrt {\left (a e +\sqrt {d^{2} a c}\right ) a}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {d^{2} a c}\, \sqrt {\left (-a e +\sqrt {d^{2} a c}\right ) a}\, \left (a \,e^{2}-c \,d^{2}\right )^{2} d^{2}}\)	\(315\)
default	\(\text {Expression too large to display}\)	\(3366\)

Input:

int(1/(e*x^2+d)^(5/2)/(-c*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

-(d^2*((a*e+(d^2*a*c)^(1/2))*a)^(1/2)*(1/2*(a*e^2+c*d^2)*(d^2*a*c)^(1/2)+d 
^2*e*a*c)*c*(e*x^2+d)^(3/2)*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*(1/2*(-a*e^2-c*d^2)*( 
d^2*a*c)^(1/2)+d^2*e*a*c)*c*(e*x^2+d)^(3/2)*arctanh((e*x^2+d)^(1/2)/x*a/(( 
a*e+(d^2*a*c)^(1/2))*a)^(1/2))-x*((a*e+(d^2*a*c)^(1/2))*a)^(1/2)*(e^2*(2/3 
*e*x^2+d)*a-3*d^2*(8/9*e*x^2+d)*c)*e^2*(d^2*a*c)^(1/2)))/((a*e+(d^2*a*c)^( 
1/2))*a)^(1/2)/(e*x^2+d)^(3/2)/(d^2*a*c)^(1/2)/((-a*e+(d^2*a*c)^(1/2))*a)^ 
(1/2)/(a*e^2-c*d^2)^2/d^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6571 vs. \(2 (174) = 348\).

Time = 45.75 (sec) , antiderivative size = 6571, normalized size of antiderivative = 29.60 \[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x^2+d)^(5/2)/(-c*x^4+a),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=- \int \frac {1}{- a d^{2} \sqrt {d + e x^{2}} - 2 a d e x^{2} \sqrt {d + e x^{2}} - a e^{2} x^{4} \sqrt {d + e x^{2}} + c d^{2} x^{4} \sqrt {d + e x^{2}} + 2 c d e x^{6} \sqrt {d + e x^{2}} + c e^{2} x^{8} \sqrt {d + e x^{2}}}\, dx \] Input:

integrate(1/(e*x**2+d)**(5/2)/(-c*x**4+a),x)
 

Output:

-Integral(1/(-a*d**2*sqrt(d + e*x**2) - 2*a*d*e*x**2*sqrt(d + e*x**2) - a* 
e**2*x**4*sqrt(d + e*x**2) + c*d**2*x**4*sqrt(d + e*x**2) + 2*c*d*e*x**6*s 
qrt(d + e*x**2) + c*e**2*x**8*sqrt(d + e*x**2)), x)
 

Maxima [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=\int { -\frac {1}{{\left (c x^{4} - a\right )} {\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(e*x^2+d)^(5/2)/(-c*x^4+a),x, algorithm="maxima")
 

Output:

-integrate(1/((c*x^4 - a)*(e*x^2 + d)^(5/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x^2+d)^(5/2)/(-c*x^4+a),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=\int \frac {1}{\left (a-c\,x^4\right )\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \] Input:

int(1/((a - c*x^4)*(d + e*x^2)^(5/2)),x)
 

Output:

int(1/((a - c*x^4)*(d + e*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e x^2\right )^{5/2} \left (a-c x^4\right )} \, dx=\int \frac {1}{\sqrt {e \,x^{2}+d}\, a \,d^{2}+2 \sqrt {e \,x^{2}+d}\, a d e \,x^{2}+\sqrt {e \,x^{2}+d}\, a \,e^{2} x^{4}-\sqrt {e \,x^{2}+d}\, c \,d^{2} x^{4}-2 \sqrt {e \,x^{2}+d}\, c d e \,x^{6}-\sqrt {e \,x^{2}+d}\, c \,e^{2} x^{8}}d x \] Input:

int(1/(e*x^2+d)^(5/2)/(-c*x^4+a),x)
 

Output:

int(1/(sqrt(d + e*x**2)*a*d**2 + 2*sqrt(d + e*x**2)*a*d*e*x**2 + sqrt(d + 
e*x**2)*a*e**2*x**4 - sqrt(d + e*x**2)*c*d**2*x**4 - 2*sqrt(d + e*x**2)*c* 
d*e*x**6 - sqrt(d + e*x**2)*c*e**2*x**8),x)