Integrand size = 35, antiderivative size = 124 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {a \sqrt {d-e x} \sqrt {d+e x}}{5 d^2 x^5}-\frac {\left (5 b d^2+4 a e^2\right ) \sqrt {d-e x} \sqrt {d+e x}}{15 d^4 x^3}-\frac {\left (15 c d^4+10 b d^2 e^2+8 a e^4\right ) \sqrt {d-e x} \sqrt {d+e x}}{15 d^6 x} \] Output:
-1/5*a*(-e*x+d)^(1/2)*(e*x+d)^(1/2)/d^2/x^5-1/15*(4*a*e^2+5*b*d^2)*(-e*x+d )^(1/2)*(e*x+d)^(1/2)/d^4/x^3-1/15*(8*a*e^4+10*b*d^2*e^2+15*c*d^4)*(-e*x+d )^(1/2)*(e*x+d)^(1/2)/d^6/x
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.70 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {d-e x} \sqrt {d+e x} \left (15 c d^4 x^4+5 b d^2 x^2 \left (d^2+2 e^2 x^2\right )+a \left (3 d^4+4 d^2 e^2 x^2+8 e^4 x^4\right )\right )}{15 d^6 x^5} \] Input:
Integrate[(a + b*x^2 + c*x^4)/(x^6*Sqrt[d - e*x]*Sqrt[d + e*x]),x]
Output:
-1/15*(Sqrt[d - e*x]*Sqrt[d + e*x]*(15*c*d^4*x^4 + 5*b*d^2*x^2*(d^2 + 2*e^ 2*x^2) + a*(3*d^4 + 4*d^2*e^2*x^2 + 8*e^4*x^4)))/(d^6*x^5)
Time = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1905, 1588, 25, 359, 242}
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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1905 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \int \frac {c x^4+b x^2+a}{x^6 \sqrt {d^2-e^2 x^2}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 1588 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (-\frac {\int -\frac {5 c x^2 d^2+5 b d^2+4 a e^2}{x^4 \sqrt {d^2-e^2 x^2}}dx}{5 d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{5 d^2 x^5}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {\int \frac {5 c x^2 d^2+5 b d^2+4 a e^2}{x^4 \sqrt {d^2-e^2 x^2}}dx}{5 d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{5 d^2 x^5}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {\frac {\left (8 a e^4+10 b d^2 e^2+15 c d^4\right ) \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx}{3 d^2}-\frac {\sqrt {d^2-e^2 x^2} \left (\frac {4 a e^2}{d^2}+5 b\right )}{3 x^3}}{5 d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{5 d^2 x^5}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {-\frac {\sqrt {d^2-e^2 x^2} \left (8 a e^4+10 b d^2 e^2+15 c d^4\right )}{3 d^4 x}-\frac {\sqrt {d^2-e^2 x^2} \left (\frac {4 a e^2}{d^2}+5 b\right )}{3 x^3}}{5 d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{5 d^2 x^5}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
Input:
Int[(a + b*x^2 + c*x^4)/(x^6*Sqrt[d - e*x]*Sqrt[d + e*x]),x]
Output:
(Sqrt[d^2 - e^2*x^2]*(-1/5*(a*Sqrt[d^2 - e^2*x^2])/(d^2*x^5) + (-1/3*((5*b + (4*a*e^2)/d^2)*Sqrt[d^2 - e^2*x^2])/x^3 - ((15*c*d^4 + 10*b*d^2*e^2 + 8 *a*e^4)*Sqrt[d^2 - e^2*x^2])/(3*d^4*x))/(5*d^2)))/(Sqrt[d - e*x]*Sqrt[d + e*x])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f ^2*(m + 1)) Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x ) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]
Int[((f_.)*(x_))^(m_.)*((d1_) + (e1_.)*(x_)^(non2_.))^(q_.)*((d2_) + (e2_.) *(x_)^(non2_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x _Symbol] :> Simp[(d1 + e1*x^(n/2))^FracPart[q]*((d2 + e2*x^(n/2))^FracPart[ q]/(d1*d2 + e1*e2*x^n)^FracPart[q]) Int[(f*x)^m*(d1*d2 + e1*e2*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[non2, n/2] && EqQ[d2*e1 + d1*e2, 0]
Time = 0.88 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.66
method | result | size |
gosper | \(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (8 a \,e^{4} x^{4}+10 b \,d^{2} e^{2} x^{4}+15 c \,d^{4} x^{4}+4 a \,d^{2} e^{2} x^{2}+5 b \,d^{4} x^{2}+3 a \,d^{4}\right )}{15 x^{5} d^{6}}\) | \(82\) |
risch | \(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (8 a \,e^{4} x^{4}+10 b \,d^{2} e^{2} x^{4}+15 c \,d^{4} x^{4}+4 a \,d^{2} e^{2} x^{2}+5 b \,d^{4} x^{2}+3 a \,d^{4}\right )}{15 x^{5} d^{6}}\) | \(82\) |
orering | \(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (8 a \,e^{4} x^{4}+10 b \,d^{2} e^{2} x^{4}+15 c \,d^{4} x^{4}+4 a \,d^{2} e^{2} x^{2}+5 b \,d^{4} x^{2}+3 a \,d^{4}\right )}{15 x^{5} d^{6}}\) | \(82\) |
default | \(-\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \operatorname {csgn}\left (e \right )^{2} \left (8 a \,e^{4} x^{4}+10 b \,d^{2} e^{2} x^{4}+15 c \,d^{4} x^{4}+4 a \,d^{2} e^{2} x^{2}+5 b \,d^{4} x^{2}+3 a \,d^{4}\right )}{15 d^{6} x^{5}}\) | \(86\) |
Input:
int((c*x^4+b*x^2+a)/x^6/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x,method=_RETURNVERBO SE)
Output:
-1/15*(e*x+d)^(1/2)*(-e*x+d)^(1/2)*(8*a*e^4*x^4+10*b*d^2*e^2*x^4+15*c*d^4* x^4+4*a*d^2*e^2*x^2+5*b*d^4*x^2+3*a*d^4)/x^5/d^6
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {{\left (3 \, a d^{4} + {\left (15 \, c d^{4} + 10 \, b d^{2} e^{2} + 8 \, a e^{4}\right )} x^{4} + {\left (5 \, b d^{4} + 4 \, a d^{2} e^{2}\right )} x^{2}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{15 \, d^{6} x^{5}} \] Input:
integrate((c*x^4+b*x^2+a)/x^6/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="f ricas")
Output:
-1/15*(3*a*d^4 + (15*c*d^4 + 10*b*d^2*e^2 + 8*a*e^4)*x^4 + (5*b*d^4 + 4*a* d^2*e^2)*x^2)*sqrt(e*x + d)*sqrt(-e*x + d)/(d^6*x^5)
Timed out. \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\text {Timed out} \] Input:
integrate((c*x**4+b*x**2+a)/x**6/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.19 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} c}{d^{2} x} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} b e^{2}}{3 \, d^{4} x} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} a e^{4}}{15 \, d^{6} x} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b}{3 \, d^{2} x^{3}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} a e^{2}}{15 \, d^{4} x^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{5 \, d^{2} x^{5}} \] Input:
integrate((c*x^4+b*x^2+a)/x^6/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="m axima")
Output:
-sqrt(-e^2*x^2 + d^2)*c/(d^2*x) - 2/3*sqrt(-e^2*x^2 + d^2)*b*e^2/(d^4*x) - 8/15*sqrt(-e^2*x^2 + d^2)*a*e^4/(d^6*x) - 1/3*sqrt(-e^2*x^2 + d^2)*b/(d^2 *x^3) - 4/15*sqrt(-e^2*x^2 + d^2)*a*e^2/(d^4*x^3) - 1/5*sqrt(-e^2*x^2 + d^ 2)*a/(d^2*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 1055 vs. \(2 (106) = 212\).
Time = 0.36 (sec) , antiderivative size = 1055, normalized size of antiderivative = 8.51 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx =\text {Too large to display} \] Input:
integrate((c*x^4+b*x^2+a)/x^6/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="g iac")
Output:
-4/15*(15*c*d^4*e^2*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sq rt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^9 + 15*b*d^2*e^4*((sqrt(2) *sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^9 + 15*a*e^6*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e *x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^9 - 240*c*d^4* e^2*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqr t(2)*sqrt(d) - sqrt(-e*x + d)))^7 - 160*b*d^2*e^4*((sqrt(2)*sqrt(d) - sqrt (-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d )))^7 - 80*a*e^6*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt( e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^7 + 1440*c*d^4*e^2*((sqrt(2)* sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^5 + 800*b*d^2*e^4*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sq rt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^5 + 928*a* e^6*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqr t(2)*sqrt(d) - sqrt(-e*x + d)))^5 - 3840*c*d^4*e^2*((sqrt(2)*sqrt(d) - sqr t(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^3 - 2560*b*d^2*e^4*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^3 - 1280*a*e^6*((sqrt( 2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d ) - sqrt(-e*x + d)))^3 + 3840*c*d^4*e^2*((sqrt(2)*sqrt(d) - sqrt(-e*x +...
Time = 4.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.18 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {d-e\,x}\,\left (\frac {a}{5\,d}+\frac {x^4\,\left (15\,c\,d^5+10\,b\,d^3\,e^2+8\,a\,d\,e^4\right )}{15\,d^6}+\frac {x^5\,\left (15\,c\,d^4\,e+10\,b\,d^2\,e^3+8\,a\,e^5\right )}{15\,d^6}+\frac {x^2\,\left (5\,b\,d^5+4\,a\,d^3\,e^2\right )}{15\,d^6}+\frac {x^3\,\left (5\,b\,d^4\,e+4\,a\,d^2\,e^3\right )}{15\,d^6}+\frac {a\,e\,x}{5\,d^2}\right )}{x^5\,\sqrt {d+e\,x}} \] Input:
int((a + b*x^2 + c*x^4)/(x^6*(d + e*x)^(1/2)*(d - e*x)^(1/2)),x)
Output:
-((d - e*x)^(1/2)*(a/(5*d) + (x^4*(15*c*d^5 + 10*b*d^3*e^2 + 8*a*d*e^4))/( 15*d^6) + (x^5*(8*a*e^5 + 10*b*d^2*e^3 + 15*c*d^4*e))/(15*d^6) + (x^2*(5*b *d^5 + 4*a*d^3*e^2))/(15*d^6) + (x^3*(4*a*d^2*e^3 + 5*b*d^4*e))/(15*d^6) + (a*e*x)/(5*d^2)))/(x^5*(d + e*x)^(1/2))
Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.64 \[ \int \frac {a+b x^2+c x^4}{x^6 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (-8 a \,e^{4} x^{4}-10 b \,d^{2} e^{2} x^{4}-15 c \,d^{4} x^{4}-4 a \,d^{2} e^{2} x^{2}-5 b \,d^{4} x^{2}-3 a \,d^{4}\right )}{15 d^{6} x^{5}} \] Input:
int((c*x^4+b*x^2+a)/x^6/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)
Output:
(sqrt(d + e*x)*sqrt(d - e*x)*( - 3*a*d**4 - 4*a*d**2*e**2*x**2 - 8*a*e**4* x**4 - 5*b*d**4*x**2 - 10*b*d**2*e**2*x**4 - 15*c*d**4*x**4))/(15*d**6*x** 5)