Integrand size = 35, antiderivative size = 102 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {a \sqrt {d-e x} \sqrt {d+e x}}{d^2 x}+\frac {c x (-d+e x) \sqrt {d+e x}}{2 e^2 \sqrt {d-e x}}-\frac {\left (c d^2+2 b e^2\right ) \arctan \left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right )}{e^3} \]
-(2*b*e^2+c*d^2)*arctan((-e*x+d)^(1/2)/(e*x+d)^(1/2))/e^3+1/2*c*x*(e*x-d)* (e*x+d)^(1/2)/e^2/(-e*x+d)^(1/2)-a*(-e*x+d)^(1/2)*(e*x+d)^(1/2)/d^2/x
Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.84 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {-\frac {e \sqrt {d-e x} \sqrt {d+e x} \left (2 a e^2+c d^2 x^2\right )}{d^2 x}+2 \left (c d^2+2 b e^2\right ) \arctan \left (\frac {\sqrt {d+e x}}{\sqrt {d-e x}}\right )}{2 e^3} \]
(-((e*Sqrt[d - e*x]*Sqrt[d + e*x]*(2*a*e^2 + c*d^2*x^2))/(d^2*x)) + 2*(c*d ^2 + 2*b*e^2)*ArcTan[Sqrt[d + e*x]/Sqrt[d - e*x]])/(2*e^3)
Time = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1905, 1588, 25, 27, 299, 224, 216}
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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1905 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \int \frac {c x^4+b x^2+a}{x^2 \sqrt {d^2-e^2 x^2}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 1588 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (-\frac {\int -\frac {d^2 \left (c x^2+b\right )}{\sqrt {d^2-e^2 x^2}}dx}{d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {\int \frac {d^2 \left (c x^2+b\right )}{\sqrt {d^2-e^2 x^2}}dx}{d^2}-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\int \frac {c x^2+b}{\sqrt {d^2-e^2 x^2}}dx-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {1}{2} \left (2 b+\frac {c d^2}{e^2}\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}-\frac {c x \sqrt {d^2-e^2 x^2}}{2 e^2}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {1}{2} \left (2 b+\frac {c d^2}{e^2}\right ) \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}-\frac {c x \sqrt {d^2-e^2 x^2}}{2 e^2}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (-\frac {a \sqrt {d^2-e^2 x^2}}{d^2 x}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (2 b+\frac {c d^2}{e^2}\right )}{2 e}-\frac {c x \sqrt {d^2-e^2 x^2}}{2 e^2}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
(Sqrt[d^2 - e^2*x^2]*(-((a*Sqrt[d^2 - e^2*x^2])/(d^2*x)) - (c*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + ((2*b + (c*d^2)/e^2)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2] ])/(2*e)))/(Sqrt[d - e*x]*Sqrt[d + e*x])
3.2.41.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
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 = 0.44 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.14
method | result | size |
risch | \(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (c \,d^{2} x^{2}+2 a \,e^{2}\right )}{2 e^{2} d^{2} x}+\frac {\left (2 b \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) \sqrt {\left (e x +d \right ) \left (-e x +d \right )}}{2 e^{2} \sqrt {e^{2}}\, \sqrt {e x +d}\, \sqrt {-e x +d}}\) | \(116\) |
default | \(-\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \left (\operatorname {csgn}\left (e \right ) c \,d^{2} e \,x^{2} \sqrt {-e^{2} x^{2}+d^{2}}-2 \arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) b \,d^{2} e^{2} x -\arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) c \,d^{4} x +2 \,\operatorname {csgn}\left (e \right ) e^{3} \sqrt {-e^{2} x^{2}+d^{2}}\, a \right ) \operatorname {csgn}\left (e \right )}{2 d^{2} e^{3} \sqrt {-e^{2} x^{2}+d^{2}}\, x}\) | \(148\) |
-1/2*(e*x+d)^(1/2)*(-e*x+d)^(1/2)*(c*d^2*x^2+2*a*e^2)/e^2/d^2/x+1/2*(2*b*e ^2+c*d^2)/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))*((e*x +d)*(-e*x+d))^(1/2)/(e*x+d)^(1/2)/(-e*x+d)^(1/2)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {2 \, {\left (c d^{4} + 2 \, b d^{2} e^{2}\right )} x \arctan \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{e x}\right ) + {\left (c d^{2} e x^{2} + 2 \, a e^{3}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{2 \, d^{2} e^{3} x} \]
-1/2*(2*(c*d^4 + 2*b*d^2*e^2)*x*arctan((sqrt(e*x + d)*sqrt(-e*x + d) - d)/ (e*x)) + (c*d^2*e*x^2 + 2*a*e^3)*sqrt(e*x + d)*sqrt(-e*x + d))/(d^2*e^3*x)
Timed out. \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {b \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} + \frac {c d^{2} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}} e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} c x}{2 \, e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{d^{2} x} \]
b*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + 1/2*c*d^2*arcsin(e^2*x/(d*sqrt(e ^2)))/(sqrt(e^2)*e^2) - 1/2*sqrt(-e^2*x^2 + d^2)*c*x/e^2 - sqrt(-e^2*x^2 + d^2)*a/(d^2*x)
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (88) = 176\).
Time = 0.37 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.34 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\frac {8 \, a e^{4} {\left (\frac {\sqrt {2} \sqrt {d} - \sqrt {-e x + d}}{\sqrt {e x + d}} - \frac {\sqrt {e x + d}}{\sqrt {2} \sqrt {d} - \sqrt {-e x + d}}\right )}}{{\left ({\left (\frac {\sqrt {2} \sqrt {d} - \sqrt {-e x + d}}{\sqrt {e x + d}} - \frac {\sqrt {e x + d}}{\sqrt {2} \sqrt {d} - \sqrt {-e x + d}}\right )}^{2} - 4\right )} d^{2}} - {\left (\pi + 2 \, \arctan \left (\frac {\sqrt {e x + d} {\left (\frac {{\left (\sqrt {2} \sqrt {d} - \sqrt {-e x + d}\right )}^{2}}{e x + d} - 1\right )}}{2 \, {\left (\sqrt {2} \sqrt {d} - \sqrt {-e x + d}\right )}}\right )\right )} {\left (c d^{2} + 2 \, b e^{2}\right )} + {\left ({\left (e x + d\right )} c - c d\right )} \sqrt {e x + d} \sqrt {-e x + d}}{2 \, e^{3}} \]
-1/2*(8*a*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)))/((((sqrt(2)*sqrt(d) - sqrt(-e*x + d))/sqrt(e*x + d) - sqrt(e*x + d)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d)))^2 - 4)*d^2) - (pi + 2*arctan(1/2*sqrt(e*x + d)*((sqrt(2)*sqrt(d) - sqrt(-e*x + d))^2/(e*x + d) - 1)/(sqrt(2)*sqrt(d) - sqrt(-e*x + d))))*(c*d^2 + 2*b* e^2) + ((e*x + d)*c - c*d)*sqrt(e*x + d)*sqrt(-e*x + d))/e^3
Time = 11.99 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.00 \[ \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {\frac {14\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^3}-\frac {14\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^5}+\frac {2\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^7}-\frac {2\,c\,d^2\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {d-e\,x}-\sqrt {d}}}{e^3\,{\left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}+1\right )}^4}-\frac {4\,b\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {d-e\,x}-\sqrt {d}\right )}{\sqrt {e^2}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {e^2}}+\frac {2\,c\,d^2\,\mathrm {atan}\left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{e^3}-\frac {\left (\frac {a}{d}+\frac {a\,e\,x}{d^2}\right )\,\sqrt {d-e\,x}}{x\,\sqrt {d+e\,x}} \]
((14*c*d^2*((d + e*x)^(1/2) - d^(1/2))^3)/((d - e*x)^(1/2) - d^(1/2))^3 - (14*c*d^2*((d + e*x)^(1/2) - d^(1/2))^5)/((d - e*x)^(1/2) - d^(1/2))^5 + ( 2*c*d^2*((d + e*x)^(1/2) - d^(1/2))^7)/((d - e*x)^(1/2) - d^(1/2))^7 - (2* c*d^2*((d + e*x)^(1/2) - d^(1/2)))/((d - e*x)^(1/2) - d^(1/2)))/(e^3*(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 + 1)^4) - (4*b*ata n((e*((d - e*x)^(1/2) - d^(1/2)))/((e^2)^(1/2)*((d + e*x)^(1/2) - d^(1/2)) )))/(e^2)^(1/2) + (2*c*d^2*atan(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)^(1/ 2) - d^(1/2))))/e^3 - ((a/d + (a*e*x)/d^2)*(d - e*x)^(1/2))/(x*(d + e*x)^( 1/2))