Integrand size = 32, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \sqrt {-1+d x} \sqrt {1+d x}}{x}+\frac {c \text {arccosh}(d x)}{d}+b \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \]
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \sqrt {-1+d x} \sqrt {1+d x}}{x}+2 b \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right )+\frac {2 c \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{d} \]
(a*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/x + 2*b*ArcTan[Sqrt[(-1 + d*x)/(1 + d*x)] ] + (2*c*ArcTanh[Sqrt[(-1 + d*x)/(1 + d*x)]])/d
Time = 0.40 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.62, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2113, 2338, 538, 224, 219, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{x^2 \sqrt {d x-1} \sqrt {d x+1}} \, dx\) |
\(\Big \downarrow \) 2113 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \int \frac {c x^2+b x+a}{x^2 \sqrt {d^2 x^2-1}}dx}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\int \frac {b+c x}{x \sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{x}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (b \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+c \int \frac {1}{\sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{x}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (b \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+c \int \frac {1}{1-\frac {d^2 x^2}{d^2 x^2-1}}d\frac {x}{\sqrt {d^2 x^2-1}}+\frac {a \sqrt {d^2 x^2-1}}{x}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (b \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{x}+\frac {c \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} b \int \frac {1}{x^2 \sqrt {d^2 x^2-1}}dx^2+\frac {a \sqrt {d^2 x^2-1}}{x}+\frac {c \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {b \int \frac {1}{\frac {x^4}{d^2}+\frac {1}{d^2}}d\sqrt {d^2 x^2-1}}{d^2}+\frac {a \sqrt {d^2 x^2-1}}{x}+\frac {c \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {a \sqrt {d^2 x^2-1}}{x}+b \arctan \left (\sqrt {d^2 x^2-1}\right )+\frac {c \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
(Sqrt[-1 + d^2*x^2]*((a*Sqrt[-1 + d^2*x^2])/x + b*ArcTan[Sqrt[-1 + d^2*x^2 ]] + (c*ArcTanh[(d*x)/Sqrt[-1 + d^2*x^2]])/d))/(Sqrt[-1 + d*x]*Sqrt[1 + d* x])
3.1.37.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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[(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[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
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 = 1.65 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73
method | result | size |
risch | \(\frac {a \sqrt {d x -1}\, \sqrt {d x +1}}{x}+\frac {\left (\frac {c \ln \left (\frac {x \,d^{2}}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-1}\right )}{\sqrt {d^{2}}}-b \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right )\right ) \sqrt {\left (d x +1\right ) \left (d x -1\right )}}{\sqrt {d x -1}\, \sqrt {d x +1}}\) | \(95\) |
default | \(\frac {\left (-\arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \operatorname {csgn}\left (d \right ) d b x +\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right ) d a +\ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) c x \right ) \sqrt {d x -1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )}{\sqrt {d^{2} x^{2}-1}\, x d}\) | \(96\) |
a*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x+(c*ln(x*d^2/(d^2)^(1/2)+(d^2*x^2-1)^(1/2)) /(d^2)^(1/2)-b*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 = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a d^{2} x + 2 \, b d x \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + \sqrt {d x + 1} \sqrt {d x - 1} a d - c x \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right )}{d x} \]
(a*d^2*x + 2*b*d*x*arctan(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)) + sqrt(d*x + 1)*sqrt(d*x - 1)*a*d - c*x*log(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)))/(d*x)
Result contains complex when optimal does not.
Time = 28.63 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.93 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=- \frac {a d {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {3}{2}, \frac {3}{2}, 2 \\1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2 & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {i a d {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 1 & \\\frac {3}{4}, \frac {5}{4} & \frac {1}{2}, 1, 1, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {b {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} \]
-a*d*meijerg(((5/4, 7/4, 1), (3/2, 3/2, 2)), ((1, 5/4, 3/2, 7/4, 2), (0,)) , 1/(d**2*x**2))/(4*pi**(3/2)) - I*a*d*meijerg(((1/2, 3/4, 1, 5/4, 3/2, 1) , ()), ((3/4, 5/4), (1/2, 1, 1, 0)), exp_polar(2*I*pi)/(d**2*x**2))/(4*pi* *(3/2)) - b*meijerg(((3/4, 5/4, 1), (1, 1, 3/2)), ((1/2, 3/4, 1, 5/4, 3/2) , (0,)), 1/(d**2*x**2))/(4*pi**(3/2)) + I*b*meijerg(((0, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)), exp_polar(2*I*pi)/(d**2*x**2))/( 4*pi**(3/2)) + c*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/ 4, 1, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d) - I*c*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), exp_polar(2*I*pi)/(d **2*x**2))/(4*pi**(3/2)*d)
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-b \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {c \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} a}{x} \]
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {2 \, b d \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) - \frac {8 \, a d^{2}}{{\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} + 4} + c \log \left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right )}{d} \]
-(2*b*d*arctan(1/2*(sqrt(d*x + 1) - sqrt(d*x - 1))^2) - 8*a*d^2/((sqrt(d*x + 1) - sqrt(d*x - 1))^4 + 4) + c*log((sqrt(d*x + 1) - sqrt(d*x - 1))^2))/ d
Time = 5.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.15 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{x}-\frac {4\,c\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-b\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i} \]