\(\int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} (a d+e x^2+c d x^4)} \, dx\) [9]

Optimal result

Integrand size = 42, antiderivative size = 52 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b d-e} x}{\sqrt {d} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {d} \sqrt {b d-e}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 12.97 (sec) , antiderivative size = 403, normalized size of antiderivative = 7.75 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\frac {i \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticPi}\left (-\frac {\left (b+\sqrt {b^2-4 a c}\right ) d}{-e+\sqrt {-4 a c d^2+e^2}},i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticPi}\left (\frac {\left (b+\sqrt {b^2-4 a c}\right ) d}{e+\sqrt {-4 a c d^2+e^2}},i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} d \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(I*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 
+ (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*(EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/( 
b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] 
)] - EllipticPi[-(((b + Sqrt[b^2 - 4*a*c])*d)/(-e + Sqrt[-4*a*c*d^2 + e^2] 
)), I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 
4*a*c])/(b - Sqrt[b^2 - 4*a*c])] - EllipticPi[((b + Sqrt[b^2 - 4*a*c])*d)/ 
(e + Sqrt[-4*a*c*d^2 + e^2]), I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a 
*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/(Sqrt[2]*Sqrt 
[c/(b + Sqrt[b^2 - 4*a*c])]*d*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2537, 221}

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.

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

\(\Big \downarrow \) 2537

\(\displaystyle a \int \frac {1}{a d-\frac {a (b d-e) x^2}{c x^4+b x^2+a}}d\frac {x}{\sqrt {c x^4+b x^2+a}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\text {arctanh}\left (\frac {x \sqrt {b d-e}}{\sqrt {d} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {d} \sqrt {b d-e}}\)

Input:

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

Output:

ArcTanh[(Sqrt[b*d - e]*x)/(Sqrt[d]*Sqrt[a + b*x^2 + c*x^4])]/(Sqrt[d]*Sqrt 
[b*d - e])
 

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 2537

Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, 
x, 0], b = Coeff[v, x, 2], c = Coeff[v, x, 4], d = Coeff[1/u, x, 0], e = Co 
eff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Simp[A   Subst[Int[1/(d - (b*d - a*e 
)*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; 
 FreeQ[{A, B}, x] && PolyQ[v, x^2, 2] && PolyQ[1/u, x^2, 2]
 

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {d \sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {\left (b d -e \right ) d}}\right )}{\sqrt {\left (b d -e \right ) d}}\)	\(44\)
elliptic	\(\frac {\operatorname {arctanh}\left (\frac {d \sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {\left (b d -e \right ) d}}\right )}{\sqrt {\left (b d -e \right ) d}}\)	\(44\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {d \sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {\left (b d -e \right ) d}}\right )}{\sqrt {\left (b d -e \right ) d}}\)	\(44\)

Input:

int((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x,method=_RETURNV 
ERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 7.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 5.40 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\left [\frac {\log \left (-\frac {c^{2} d^{2} x^{8} + 2 \, {\left (4 \, b c d^{2} - 3 \, c d e\right )} x^{6} + {\left (2 \, {\left (4 \, b^{2} + a c\right )} d^{2} - 8 \, b d e + e^{2}\right )} x^{4} + a^{2} d^{2} + 2 \, {\left (4 \, a b d^{2} - 3 \, a d e\right )} x^{2} + 4 \, {\left (c d x^{5} + {\left (2 \, b d - e\right )} x^{3} + a d x\right )} \sqrt {c x^{4} + b x^{2} + a} \sqrt {b d^{2} - d e}}{c^{2} d^{2} x^{8} + 2 \, c d e x^{6} + 2 \, a d e x^{2} + {\left (2 \, a c d^{2} + e^{2}\right )} x^{4} + a^{2} d^{2}}\right )}{4 \, \sqrt {b d^{2} - d e}}, -\frac {\sqrt {-b d^{2} + d e} \arctan \left (\frac {2 \, \sqrt {c x^{4} + b x^{2} + a} \sqrt {-b d^{2} + d e} x}{c d x^{4} + {\left (2 \, b d - e\right )} x^{2} + a d}\right )}{2 \, {\left (b d^{2} - d e\right )}}\right ] \] Input:

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

Output:

[1/4*log(-(c^2*d^2*x^8 + 2*(4*b*c*d^2 - 3*c*d*e)*x^6 + (2*(4*b^2 + a*c)*d^ 
2 - 8*b*d*e + e^2)*x^4 + a^2*d^2 + 2*(4*a*b*d^2 - 3*a*d*e)*x^2 + 4*(c*d*x^ 
5 + (2*b*d - e)*x^3 + a*d*x)*sqrt(c*x^4 + b*x^2 + a)*sqrt(b*d^2 - d*e))/(c 
^2*d^2*x^8 + 2*c*d*e*x^6 + 2*a*d*e*x^2 + (2*a*c*d^2 + e^2)*x^4 + a^2*d^2)) 
/sqrt(b*d^2 - d*e), -1/2*sqrt(-b*d^2 + d*e)*arctan(2*sqrt(c*x^4 + b*x^2 + 
a)*sqrt(-b*d^2 + d*e)*x/(c*d*x^4 + (2*b*d - e)*x^2 + a*d))/(b*d^2 - d*e)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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