Integrand size = 37, antiderivative size = 33 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2+c x^4}}\right )}{\sqrt {b} d} \] Output:
arctanh(b^(1/2)*x/(c*x^4+b*x^2+a)^(1/2))/b^(1/2)/d
Time = 1.86 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2+c x^4}}\right )}{\sqrt {b} d} \] Input:
Integrate[(a - c*x^4)/(Sqrt[a + b*x^2 + c*x^4]*(a*d + c*d*x^4)),x]
Output:
ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2 + c*x^4]]/(Sqrt[b]*d)
Time = 0.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, 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.
\(\displaystyle \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx\) |
\(\Big \downarrow \) 2537 |
\(\displaystyle a \int \frac {1}{a d-\frac {a b d x^2}{c x^4+b x^2+a}}d\frac {x}{\sqrt {c x^4+b x^2+a}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2+c x^4}}\right )}{\sqrt {b} d}\) |
Input:
Int[(a - c*x^4)/(Sqrt[a + b*x^2 + c*x^4]*(a*d + c*d*x^4)),x]
Output:
ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2 + c*x^4]]/(Sqrt[b]*d)
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]
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]
Time = 1.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91
method | result | size |
default | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {b}}\right )}{d \sqrt {b}}\) | \(30\) |
elliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {b}}\right )}{d \sqrt {b}}\) | \(30\) |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {b}}\right )}{d \sqrt {b}}\) | \(30\) |
Input:
int((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+a*d),x,method=_RETURNVERBOSE )
Output:
1/d/b^(1/2)*arctanh((c*x^4+b*x^2+a)^(1/2)/x/b^(1/2))
Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 4.73 \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\left [\frac {\log \left (-\frac {c^{2} x^{8} + 8 \, b c x^{6} + 2 \, {\left (4 \, b^{2} + a c\right )} x^{4} + 8 \, a b x^{2} + a^{2} + 4 \, {\left (c x^{5} + 2 \, b x^{3} + a x\right )} \sqrt {c x^{4} + b x^{2} + a} \sqrt {b}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}\right )}{4 \, \sqrt {b} d}, -\frac {\sqrt {-b} \arctan \left (\frac {2 \, \sqrt {c x^{4} + b x^{2} + a} \sqrt {-b} x}{c x^{4} + 2 \, b x^{2} + a}\right )}{2 \, b d}\right ] \] Input:
integrate((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+a*d),x, algorithm="fri cas")
Output:
[1/4*log(-(c^2*x^8 + 8*b*c*x^6 + 2*(4*b^2 + a*c)*x^4 + 8*a*b*x^2 + a^2 + 4 *(c*x^5 + 2*b*x^3 + a*x)*sqrt(c*x^4 + b*x^2 + a)*sqrt(b))/(c^2*x^8 + 2*a*c *x^4 + a^2))/(sqrt(b)*d), -1/2*sqrt(-b)*arctan(2*sqrt(c*x^4 + b*x^2 + a)*s qrt(-b)*x/(c*x^4 + 2*b*x^2 + a))/(b*d)]
\[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=- \frac {\int \left (- \frac {a}{a \sqrt {a + b x^{2} + c x^{4}} + c x^{4} \sqrt {a + b x^{2} + c x^{4}}}\right )\, dx + \int \frac {c x^{4}}{a \sqrt {a + b x^{2} + c x^{4}} + c x^{4} \sqrt {a + b x^{2} + c x^{4}}}\, dx}{d} \] Input:
integrate((-c*x**4+a)/(c*x**4+b*x**2+a)**(1/2)/(c*d*x**4+a*d),x)
Output:
-(Integral(-a/(a*sqrt(a + b*x**2 + c*x**4) + c*x**4*sqrt(a + b*x**2 + c*x* *4)), x) + Integral(c*x**4/(a*sqrt(a + b*x**2 + c*x**4) + c*x**4*sqrt(a + b*x**2 + c*x**4)), x))/d
\[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\int { -\frac {c x^{4} - a}{{\left (c d x^{4} + a d\right )} \sqrt {c x^{4} + b x^{2} + a}} \,d x } \] Input:
integrate((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+a*d),x, algorithm="max ima")
Output:
-integrate((c*x^4 - a)/((c*d*x^4 + a*d)*sqrt(c*x^4 + b*x^2 + a)), x)
\[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\int { -\frac {c x^{4} - a}{{\left (c d x^{4} + a d\right )} \sqrt {c x^{4} + b x^{2} + a}} \,d x } \] Input:
integrate((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+a*d),x, algorithm="gia c")
Output:
integrate(-(c*x^4 - a)/((c*d*x^4 + a*d)*sqrt(c*x^4 + b*x^2 + a)), x)
Timed out. \[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\int \frac {a-c\,x^4}{\left (c\,d\,x^4+a\,d\right )\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:
int((a - c*x^4)/((a*d + c*d*x^4)*(a + b*x^2 + c*x^4)^(1/2)),x)
Output:
int((a - c*x^4)/((a*d + c*d*x^4)*(a + b*x^2 + c*x^4)^(1/2)), x)
\[ \int \frac {a-c x^4}{\sqrt {a+b x^2+c x^4} \left (a d+c d x^4\right )} \, dx=\frac {\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c^{2} x^{8}+b c \,x^{6}+2 a c \,x^{4}+a b \,x^{2}+a^{2}}d x \right ) a -\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{4}}{c^{2} x^{8}+b c \,x^{6}+2 a c \,x^{4}+a b \,x^{2}+a^{2}}d x \right ) c}{d} \] Input:
int((-c*x^4+a)/(c*x^4+b*x^2+a)^(1/2)/(c*d*x^4+a*d),x)
Output:
(int(sqrt(a + b*x**2 + c*x**4)/(a**2 + a*b*x**2 + 2*a*c*x**4 + b*c*x**6 + c**2*x**8),x)*a - int((sqrt(a + b*x**2 + c*x**4)*x**4)/(a**2 + a*b*x**2 + 2*a*c*x**4 + b*c*x**6 + c**2*x**8),x)*c)/d