Integrand size = 32, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c \sqrt {-1+d x} \sqrt {1+d x}}{d^2}+\frac {b \text {arccosh}(d x)}{d}+a \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \]
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c \sqrt {-1+d x} \sqrt {1+d x}}{d^2}+2 a \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right )+\frac {2 b \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{d} \]
(c*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/d^2 + 2*a*ArcTan[Sqrt[(-1 + d*x)/(1 + d*x )]] + (2*b*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 = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2113, 2340, 27, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{x \sqrt {d x-1} \sqrt {d x+1}} \, dx\) |
\(\Big \downarrow \) 2113 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \int \frac {c x^2+b x+a}{x \sqrt {d^2 x^2-1}}dx}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {\int \frac {d^2 (a+b x)}{x \sqrt {d^2 x^2-1}}dx}{d^2}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\int \frac {a+b x}{x \sqrt {d^2 x^2-1}}dx+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (a \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+b \int \frac {1}{\sqrt {d^2 x^2-1}}dx+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (a \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+b \int \frac {1}{1-\frac {d^2 x^2}{d^2 x^2-1}}d\frac {x}{\sqrt {d^2 x^2-1}}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (a \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+\frac {b \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{2} a \int \frac {1}{x^2 \sqrt {d^2 x^2-1}}dx^2+\frac {b \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {a \int \frac {1}{\frac {x^4}{d^2}+\frac {1}{d^2}}d\sqrt {d^2 x^2-1}}{d^2}+\frac {b \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (a \arctan \left (\sqrt {d^2 x^2-1}\right )+\frac {b \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\) |
(Sqrt[-1 + d^2*x^2]*((c*Sqrt[-1 + d^2*x^2])/d^2 + a*ArcTan[Sqrt[-1 + d^2*x ^2]] + (b*ArcTanh[(d*x)/Sqrt[-1 + d^2*x^2]])/d))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x])
3.1.36.3.1 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_))^(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 = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.65 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73
method | result | size |
default | \(\frac {\left (-\operatorname {csgn}\left (d \right ) \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) a \,d^{2}+\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right ) c +\ln \left (\left (\sqrt {\left (d x +1\right ) \left (d x -1\right )}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b d \right ) \sqrt {d x -1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )}{d^{2} \sqrt {d^{2} x^{2}-1}}\) | \(95\) |
(-csgn(d)*arctan(1/(d^2*x^2-1)^(1/2))*a*d^2+(d^2*x^2-1)^(1/2)*csgn(d)*c+ln ((((d*x+1)*(d*x-1))^(1/2)*csgn(d)+d*x)*csgn(d))*b*d)*(d*x-1)^(1/2)*(d*x+1) ^(1/2)/d^2*csgn(d)/(d^2*x^2-1)^(1/2)
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {2 \, a d^{2} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) - b d \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + \sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \]
(2*a*d^2*arctan(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)) - b*d*log(-d*x + sqrt( d*x + 1)*sqrt(d*x - 1)) + sqrt(d*x + 1)*sqrt(d*x - 1)*c)/d^2
Result contains complex when optimal does not.
Time = 29.13 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.36 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=- \frac {a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} \]
-a*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*a*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)) + b*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*b*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) + c*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**2) + I*c*meijerg(((- 1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), exp_ polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**2)
Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-a \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} c}{d^{2}} \]
-a*arcsin(1/(d*abs(x))) + b*log(2*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d + sqrt( d^2*x^2 - 1)*c/d^2
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-2 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) - \frac {b \log \left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right )}{d} + \frac {\sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \]
-2*a*arctan(1/2*(sqrt(d*x + 1) - sqrt(d*x - 1))^2) - b*log((sqrt(d*x + 1) - sqrt(d*x - 1))^2)/d + sqrt(d*x + 1)*sqrt(d*x - 1)*c/d^2
Time = 5.82 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.15 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}-\frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-a\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i} \]