Integrand size = 32, antiderivative size = 116 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \sqrt {-1+d x} \sqrt {1+d x}}{3 x^3}+\frac {b \sqrt {-1+d x} \sqrt {1+d x}}{2 x^2}+\frac {\left (3 c+2 a d^2\right ) \sqrt {-1+d x} \sqrt {1+d x}}{3 x}+\frac {1}{2} b d^2 \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \]
1/2*b*d^2*arctan((d*x-1)^(1/2)*(d*x+1)^(1/2))+1/3*a*(d*x-1)^(1/2)*(d*x+1)^ (1/2)/x^3+1/2*b*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x^2+1/3*(2*a*d^2+3*c)*(d*x-1)^ (1/2)*(d*x+1)^(1/2)/x
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {-1+d x} \sqrt {1+d x} \left (3 x (b+2 c x)+a \left (2+4 d^2 x^2\right )\right )}{6 x^3}+b d^2 \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right ) \]
(Sqrt[-1 + d*x]*Sqrt[1 + d*x]*(3*x*(b + 2*c*x) + a*(2 + 4*d^2*x^2)))/(6*x^ 3) + b*d^2*ArcTan[Sqrt[(-1 + d*x)/(1 + d*x)]]
Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2113, 2338, 539, 534, 243, 73, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{x^4 \sqrt {d x-1} \sqrt {d x+1}} \, dx\) |
\(\Big \downarrow \) 2113 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \int \frac {c x^2+b x+a}{x^4 \sqrt {d^2 x^2-1}}dx}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \int \frac {3 b+\left (2 a d^2+3 c\right ) x}{x^3 \sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 539 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {3 b x d^2+2 \left (2 a d^2+3 c\right )}{x^2 \sqrt {d^2 x^2-1}}dx+\frac {3 b \sqrt {d^2 x^2-1}}{2 x^2}\right )+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 b d^2 \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+\frac {2 \sqrt {d^2 x^2-1} \left (2 a d^2+3 c\right )}{x}\right )+\frac {3 b \sqrt {d^2 x^2-1}}{2 x^2}\right )+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {3}{2} b d^2 \int \frac {1}{x^2 \sqrt {d^2 x^2-1}}dx^2+\frac {2 \sqrt {d^2 x^2-1} \left (2 a d^2+3 c\right )}{x}\right )+\frac {3 b \sqrt {d^2 x^2-1}}{2 x^2}\right )+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 b \int \frac {1}{\frac {x^4}{d^2}+\frac {1}{d^2}}d\sqrt {d^2 x^2-1}+\frac {2 \sqrt {d^2 x^2-1} \left (2 a d^2+3 c\right )}{x}\right )+\frac {3 b \sqrt {d^2 x^2-1}}{2 x^2}\right )+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {2 \sqrt {d^2 x^2-1} \left (2 a d^2+3 c\right )}{x}+3 b d^2 \arctan \left (\sqrt {d^2 x^2-1}\right )\right )+\frac {3 b \sqrt {d^2 x^2-1}}{2 x^2}\right )+\frac {a \sqrt {d^2 x^2-1}}{3 x^3}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
(Sqrt[-1 + d^2*x^2]*((a*Sqrt[-1 + d^2*x^2])/(3*x^3) + ((3*b*Sqrt[-1 + d^2* x^2])/(2*x^2) + ((2*(3*c + 2*a*d^2)*Sqrt[-1 + d^2*x^2])/x + 3*b*d^2*ArcTan [Sqrt[-1 + d^2*x^2]])/2)/3))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x])
3.2.59.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_. )*(x_))^(p_.), x_Symbol] :> Simp[(a + b*x)^FracPart[m]*((c + d*x)^FracPart[ m]/(a*c + b*d*x^2)^FracPart[m]) Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a *d, 0] && EqQ[m, n] && !IntegerQ[m]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 5.56 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {\sqrt {d x +1}\, \sqrt {d x -1}\, \left (4 a \,d^{2} x^{2}+6 c \,x^{2}+3 b x +2 a \right )}{6 x^{3}}-\frac {b \,d^{2} \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \sqrt {\left (d x +1\right ) \left (d x -1\right )}}{2 \sqrt {d x -1}\, \sqrt {d x +1}}\) | \(89\) |
default | \(-\frac {\sqrt {d x -1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )^{2} \left (3 \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) b \,d^{2} x^{3}-4 \sqrt {d^{2} x^{2}-1}\, a \,d^{2} x^{2}-6 \sqrt {d^{2} x^{2}-1}\, c \,x^{2}-3 \sqrt {d^{2} x^{2}-1}\, b x -2 \sqrt {d^{2} x^{2}-1}\, a \right )}{6 \sqrt {d^{2} x^{2}-1}\, x^{3}}\) | \(123\) |
1/6*(d*x+1)^(1/2)*(d*x-1)^(1/2)*(4*a*d^2*x^2+6*c*x^2+3*b*x+2*a)/x^3-1/2*b* d^2*arctan(1/(d^2*x^2-1)^(1/2))*((d*x+1)*(d*x-1))^(1/2)/(d*x-1)^(1/2)/(d*x +1)^(1/2)
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {6 \, b d^{2} x^{3} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + 2 \, {\left (2 \, a d^{3} + 3 \, c d\right )} x^{3} + {\left (2 \, {\left (2 \, a d^{2} + 3 \, c\right )} x^{2} + 3 \, b x + 2 \, a\right )} \sqrt {d x + 1} \sqrt {d x - 1}}{6 \, x^{3}} \]
1/6*(6*b*d^2*x^3*arctan(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)) + 2*(2*a*d^3 + 3*c*d)*x^3 + (2*(2*a*d^2 + 3*c)*x^2 + 3*b*x + 2*a)*sqrt(d*x + 1)*sqrt(d*x - 1))/x^3
Timed out. \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {1}{2} \, b d^{2} \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {2 \, \sqrt {d^{2} x^{2} - 1} a d^{2}}{3 \, x} + \frac {\sqrt {d^{2} x^{2} - 1} c}{x} + \frac {\sqrt {d^{2} x^{2} - 1} b}{2 \, x^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} a}{3 \, x^{3}} \]
-1/2*b*d^2*arcsin(1/(d*abs(x))) + 2/3*sqrt(d^2*x^2 - 1)*a*d^2/x + sqrt(d^2 *x^2 - 1)*c/x + 1/2*sqrt(d^2*x^2 - 1)*b/x^2 + 1/3*sqrt(d^2*x^2 - 1)*a/x^3
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (92) = 184\).
Time = 0.33 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.70 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {3 \, b d^{3} \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, b d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{10} - 12 \, c d^{2} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{8} - 96 \, a d^{4} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} - 96 \, c d^{2} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} - 48 \, b d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2} - 128 \, a d^{4} - 192 \, c d^{2}\right )}}{{\left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} + 4\right )}^{3}}}{3 \, d} \]
-1/3*(3*b*d^3*arctan(1/2*(sqrt(d*x + 1) - sqrt(d*x - 1))^2) + 2*(3*b*d^3*( sqrt(d*x + 1) - sqrt(d*x - 1))^10 - 12*c*d^2*(sqrt(d*x + 1) - sqrt(d*x - 1 ))^8 - 96*a*d^4*(sqrt(d*x + 1) - sqrt(d*x - 1))^4 - 96*c*d^2*(sqrt(d*x + 1 ) - sqrt(d*x - 1))^4 - 48*b*d^3*(sqrt(d*x + 1) - sqrt(d*x - 1))^2 - 128*a* d^4 - 192*c*d^2)/((sqrt(d*x + 1) - sqrt(d*x - 1))^4 + 4)^3)/d
Time = 10.68 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.62 \[ \int \frac {a+b x+c x^2}{x^4 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\frac {b\,d^2\,1{}\mathrm {i}}{32}+\frac {b\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {b\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,{\left (\sqrt {d\,x+1}-1\right )}^4}}{\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}}-\frac {b\,d^2\,\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^2\,\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\,1{}\mathrm {i}}{2}+\frac {c\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{x}+\frac {\sqrt {d\,x-1}\,\left (\frac {2\,a\,d^3\,x^3}{3}+\frac {2\,a\,d^2\,x^2}{3}+\frac {a\,d\,x}{3}+\frac {a}{3}\right )}{x^3\,\sqrt {d\,x+1}}+\frac {b\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,{\left (\sqrt {d\,x+1}-1\right )}^2} \]
((b*d^2*1i)/32 + (b*d^2*((d*x - 1)^(1/2) - 1i)^2*1i)/(16*((d*x + 1)^(1/2) - 1)^2) - (b*d^2*((d*x - 1)^(1/2) - 1i)^4*15i)/(32*((d*x + 1)^(1/2) - 1)^4 ))/(((d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + (2*((d*x - 1)^(1/2) - 1i)^4)/((d*x + 1)^(1/2) - 1)^4 + ((d*x - 1)^(1/2) - 1i)^6/((d*x + 1)^(1 /2) - 1)^6) - (b*d^2*log(((d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + 1)*1i)/2 + (b*d^2*log(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1))*1i)/ 2 + (c*(d*x - 1)^(1/2)*(d*x + 1)^(1/2))/x + ((d*x - 1)^(1/2)*(a/3 + (2*a*d ^2*x^2)/3 + (2*a*d^3*x^3)/3 + (a*d*x)/3))/(x^3*(d*x + 1)^(1/2)) + (b*d^2*( (d*x - 1)^(1/2) - 1i)^2*1i)/(32*((d*x + 1)^(1/2) - 1)^2)