Integrand size = 32, antiderivative size = 83 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \sqrt {-1+d x} \sqrt {1+d x}}{2 x^2}+\frac {b \sqrt {-1+d x} \sqrt {1+d x}}{x}+\frac {1}{2} \left (2 c+a d^2\right ) \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \]
1/2*(a*d^2+2*c)*arctan((d*x-1)^(1/2)*(d*x+1)^(1/2))+1/2*a*(d*x-1)^(1/2)*(d *x+1)^(1/2)/x^2+b*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {(a+2 b x) \sqrt {-1+d x} \sqrt {1+d x}}{2 x^2}+\left (2 c+a d^2\right ) \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right ) \]
((a + 2*b*x)*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/(2*x^2) + (2*c + a*d^2)*ArcTan[ Sqrt[(-1 + d*x)/(1 + d*x)]]
Time = 0.39 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2113, 2338, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{x^3 \sqrt {d x-1} \sqrt {d x+1}} \, dx\) |
\(\Big \downarrow \) 2113 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \int \frac {c x^2+b x+a}{x^3 \sqrt {d^2 x^2-1}}dx}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} \int \frac {2 b+\left (a d^2+2 c\right ) x}{x^2 \sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{2 x^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} \left (\left (a d^2+2 c\right ) \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+\frac {2 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {a \sqrt {d^2 x^2-1}}{2 x^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} \left (\frac {1}{2} \left (a d^2+2 c\right ) \int \frac {1}{x^2 \sqrt {d^2 x^2-1}}dx^2+\frac {2 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {a \sqrt {d^2 x^2-1}}{2 x^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} \left (\frac {\left (a d^2+2 c\right ) \int \frac {1}{\frac {x^4}{d^2}+\frac {1}{d^2}}d\sqrt {d^2 x^2-1}}{d^2}+\frac {2 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {a \sqrt {d^2 x^2-1}}{2 x^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} \left (\left (a d^2+2 c\right ) \arctan \left (\sqrt {d^2 x^2-1}\right )+\frac {2 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {a \sqrt {d^2 x^2-1}}{2 x^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
(Sqrt[-1 + d^2*x^2]*((a*Sqrt[-1 + d^2*x^2])/(2*x^2) + ((2*b*Sqrt[-1 + d^2* x^2])/x + (2*c + a*d^2)*ArcTan[Sqrt[-1 + d^2*x^2]])/2))/(Sqrt[-1 + d*x]*Sq rt[1 + d*x])
3.2.58.3.1 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[(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 = 1.63 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {\sqrt {d x +1}\, \sqrt {d x -1}\, \left (2 b x +a \right )}{2 x^{2}}-\frac {\left (c +\frac {a \,d^{2}}{2}\right ) \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \sqrt {\left (d x +1\right ) \left (d x -1\right )}}{\sqrt {d x -1}\, \sqrt {d x +1}}\) | \(76\) |
default | \(-\frac {\sqrt {d x -1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )^{2} \left (\arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) a \,d^{2} x^{2}+2 \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) c \,x^{2}-2 \sqrt {d^{2} x^{2}-1}\, b x -\sqrt {d^{2} x^{2}-1}\, a \right )}{2 \sqrt {d^{2} x^{2}-1}\, x^{2}}\) | \(103\) |
1/2*(d*x+1)^(1/2)*(d*x-1)^(1/2)*(2*b*x+a)/x^2-(c+1/2*a*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 = 69, normalized size of antiderivative = 0.83 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {2 \, b d x^{2} + 2 \, {\left (a d^{2} + 2 \, c\right )} x^{2} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + {\left (2 \, b x + a\right )} \sqrt {d x + 1} \sqrt {d x - 1}}{2 \, x^{2}} \]
1/2*(2*b*d*x^2 + 2*(a*d^2 + 2*c)*x^2*arctan(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)) + (2*b*x + a)*sqrt(d*x + 1)*sqrt(d*x - 1))/x^2
Timed out. \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {1}{2} \, a d^{2} \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) - c \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {\sqrt {d^{2} x^{2} - 1} b}{x} + \frac {\sqrt {d^{2} x^{2} - 1} a}{2 \, x^{2}} \]
-1/2*a*d^2*arcsin(1/(d*abs(x))) - c*arcsin(1/(d*abs(x))) + sqrt(d^2*x^2 - 1)*b/x + 1/2*sqrt(d^2*x^2 - 1)*a/x^2
Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (67) = 134\).
Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.75 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {{\left (a d^{3} + 2 \, c d\right )} \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) + \frac {2 \, {\left (a d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{6} - 4 \, b d^{2} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} - 4 \, a d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2} - 16 \, b d^{2}\right )}}{{\left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} + 4\right )}^{2}}}{d} \]
-((a*d^3 + 2*c*d)*arctan(1/2*(sqrt(d*x + 1) - sqrt(d*x - 1))^2) + 2*(a*d^3 *(sqrt(d*x + 1) - sqrt(d*x - 1))^6 - 4*b*d^2*(sqrt(d*x + 1) - sqrt(d*x - 1 ))^4 - 4*a*d^3*(sqrt(d*x + 1) - sqrt(d*x - 1))^2 - 16*b*d^2)/((sqrt(d*x + 1) - sqrt(d*x - 1))^4 + 4)^2)/d
Time = 10.45 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.81 \[ \int \frac {a+b x+c x^2}{x^3 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\frac {a\,d^2\,1{}\mathrm {i}}{32}+\frac {a\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,{\left (\sqrt {d\,x+1}-1\right )}^2}-\frac {a\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,{\left (\sqrt {d\,x+1}-1\right )}^4}}{\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}}-c\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i}-\frac {a\,d^2\,\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{2}+\frac {a\,d^2\,\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\,1{}\mathrm {i}}{2}+\frac {b\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{x}+\frac {a\,d^2\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,{\left (\sqrt {d\,x+1}-1\right )}^2} \]
((a*d^2*1i)/32 + (a*d^2*((d*x - 1)^(1/2) - 1i)^2*1i)/(16*((d*x + 1)^(1/2) - 1)^2) - (a*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) - c*(log(((d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + 1) - log(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1)))*1i - (a*d^2*log(((d* x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + 1)*1i)/2 + (a*d^2*log(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1))*1i)/2 + (b*(d*x - 1)^(1/2)*(d*x + 1)^(1/2))/x + (a*d^2*((d*x - 1)^(1/2) - 1i)^2*1i)/(32*((d*x + 1)^(1/2) - 1)^2)