Integrand size = 37, antiderivative size = 35 \[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d} \sqrt {a+c x^4}}\right )}{\sqrt {d} \sqrt {e}} \] Output:
arctan(e^(1/2)*x/d^(1/2)/(c*x^4+a)^(1/2))/d^(1/2)/e^(1/2)
Time = 1.90 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d} \sqrt {a+c x^4}}\right )}{\sqrt {d} \sqrt {e}} \] Input:
Integrate[(a - c*x^4)/(Sqrt[a + c*x^4]*(a*d + e*x^2 + c*d*x^4)),x]
Output:
ArcTan[(Sqrt[e]*x)/(Sqrt[d]*Sqrt[a + c*x^4])]/(Sqrt[d]*Sqrt[e])
Time = 0.49 (sec) , antiderivative size = 35, 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, 218}
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.
\(\displaystyle \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+c d x^4+e x^2\right )} \, dx\) |
\(\Big \downarrow \) 2537 |
\(\displaystyle a \int \frac {1}{\frac {a e x^2}{c x^4+a}+a d}d\frac {x}{\sqrt {c x^4+a}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d} \sqrt {a+c x^4}}\right )}{\sqrt {d} \sqrt {e}}\) |
Input:
Int[(a - c*x^4)/(Sqrt[a + c*x^4]*(a*d + e*x^2 + c*d*x^4)),x]
Output:
ArcTan[(Sqrt[e]*x)/(Sqrt[d]*Sqrt[a + c*x^4])]/(Sqrt[d]*Sqrt[e])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 = 0.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {\arctan \left (\frac {d \sqrt {c \,x^{4}+a}}{x \sqrt {d e}}\right )}{\sqrt {d e}}\) | \(28\) |
elliptic | \(-\frac {\arctan \left (\frac {d \sqrt {c \,x^{4}+a}}{x \sqrt {d e}}\right )}{\sqrt {d e}}\) | \(28\) |
pseudoelliptic | \(-\frac {\arctan \left (\frac {d \sqrt {c \,x^{4}+a}}{x \sqrt {d e}}\right )}{\sqrt {d e}}\) | \(28\) |
Input:
int((-c*x^4+a)/(c*x^4+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x,method=_RETURNVERBOSE )
Output:
-1/(d*e)^(1/2)*arctan(d*(c*x^4+a)^(1/2)/x/(d*e)^(1/2))
Time = 1.52 (sec) , antiderivative size = 201, normalized size of antiderivative = 5.74 \[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\left [-\frac {\sqrt {-d e} \log \left (-\frac {c^{2} d^{2} x^{8} - 6 \, c d e x^{6} - 6 \, a d e x^{2} + {\left (2 \, a c d^{2} + e^{2}\right )} x^{4} + a^{2} d^{2} + 4 \, {\left (c d x^{5} - e x^{3} + a d x\right )} \sqrt {c x^{4} + a} \sqrt {-d e}}{c^{2} d^{2} x^{8} + 2 \, c d e x^{6} + 2 \, a d e x^{2} + {\left (2 \, a c d^{2} + e^{2}\right )} x^{4} + a^{2} d^{2}}\right )}{4 \, d e}, \frac {\sqrt {d e} \arctan \left (\frac {2 \, \sqrt {c x^{4} + a} \sqrt {d e} x}{c d x^{4} - e x^{2} + a d}\right )}{2 \, d e}\right ] \] Input:
integrate((-c*x^4+a)/(c*x^4+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x, algorithm="fri cas")
Output:
[-1/4*sqrt(-d*e)*log(-(c^2*d^2*x^8 - 6*c*d*e*x^6 - 6*a*d*e*x^2 + (2*a*c*d^ 2 + e^2)*x^4 + a^2*d^2 + 4*(c*d*x^5 - e*x^3 + a*d*x)*sqrt(c*x^4 + a)*sqrt( -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))/(d*e), 1/2*sqrt(d*e)*arctan(2*sqrt(c*x^4 + a)*sqrt(d*e)*x/(c*d*x ^4 - e*x^2 + a*d))/(d*e)]
\[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=- \int \left (- \frac {a}{a d \sqrt {a + c x^{4}} + c d x^{4} \sqrt {a + c x^{4}} + e x^{2} \sqrt {a + c x^{4}}}\right )\, dx - \int \frac {c x^{4}}{a d \sqrt {a + c x^{4}} + c d x^{4} \sqrt {a + c x^{4}} + e x^{2} \sqrt {a + c x^{4}}}\, dx \] Input:
integrate((-c*x**4+a)/(c*x**4+a)**(1/2)/(c*d*x**4+e*x**2+a*d),x)
Output:
-Integral(-a/(a*d*sqrt(a + c*x**4) + c*d*x**4*sqrt(a + c*x**4) + e*x**2*sq rt(a + c*x**4)), x) - Integral(c*x**4/(a*d*sqrt(a + c*x**4) + c*d*x**4*sqr t(a + c*x**4) + e*x**2*sqrt(a + c*x**4)), x)
\[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\int { -\frac {c x^{4} - a}{{\left (c d x^{4} + e x^{2} + a d\right )} \sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((-c*x^4+a)/(c*x^4+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x, algorithm="max ima")
Output:
-integrate((c*x^4 - a)/((c*d*x^4 + e*x^2 + a*d)*sqrt(c*x^4 + a)), x)
\[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\int { -\frac {c x^{4} - a}{{\left (c d x^{4} + e x^{2} + a d\right )} \sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((-c*x^4+a)/(c*x^4+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x, algorithm="gia c")
Output:
integrate(-(c*x^4 - a)/((c*d*x^4 + e*x^2 + a*d)*sqrt(c*x^4 + a)), x)
Timed out. \[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\int \frac {a-c\,x^4}{\sqrt {c\,x^4+a}\,\left (c\,d\,x^4+e\,x^2+a\,d\right )} \,d x \] Input:
int((a - c*x^4)/((a + c*x^4)^(1/2)*(a*d + e*x^2 + c*d*x^4)),x)
Output:
int((a - c*x^4)/((a + c*x^4)^(1/2)*(a*d + e*x^2 + c*d*x^4)), x)
\[ \int \frac {a-c x^4}{\sqrt {a+c x^4} \left (a d+e x^2+c d x^4\right )} \, dx=\left (\int \frac {\sqrt {c \,x^{4}+a}}{c^{2} d \,x^{8}+c e \,x^{6}+2 a c d \,x^{4}+a e \,x^{2}+a^{2} d}d x \right ) a -\left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{4}}{c^{2} d \,x^{8}+c e \,x^{6}+2 a c d \,x^{4}+a e \,x^{2}+a^{2} d}d x \right ) c \] Input:
int((-c*x^4+a)/(c*x^4+a)^(1/2)/(c*d*x^4+e*x^2+a*d),x)
Output:
int(sqrt(a + c*x**4)/(a**2*d + 2*a*c*d*x**4 + a*e*x**2 + c**2*d*x**8 + c*e *x**6),x)*a - int((sqrt(a + c*x**4)*x**4)/(a**2*d + 2*a*c*d*x**4 + a*e*x** 2 + c**2*d*x**8 + c*e*x**6),x)*c