Integrand size = 29, antiderivative size = 77 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {(c+2 b d) \sqrt {-1+d x} \sqrt {1+d x}}{2 d^3}+\frac {c (-1+d x)^{3/2} \sqrt {1+d x}}{2 d^3}+\frac {\left (c+2 a d^2\right ) \text {arccosh}(d x)}{2 d^3} \] Output:
1/2*(2*b*d+c)*(d*x-1)^(1/2)*(d*x+1)^(1/2)/d^3+1/2*c*(d*x-1)^(3/2)*(d*x+1)^ (1/2)/d^3+1/2*(2*a*d^2+c)*arccosh(d*x)/d^3
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {d (2 b+c x) \sqrt {-1+d x} \sqrt {1+d x}+2 \left (c+2 a d^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{2 d^3} \] Input:
Integrate[(a + b*x + c*x^2)/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
Output:
(d*(2*b + c*x)*Sqrt[-1 + d*x]*Sqrt[1 + d*x] + 2*(c + 2*a*d^2)*ArcTanh[Sqrt [(-1 + d*x)/(1 + d*x)]])/(2*d^3)
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1189, 83, 646, 43}
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.
\(\displaystyle \int \frac {a+b x+c x^2}{\sqrt {d x-1} \sqrt {d x+1}} \, dx\) |
\(\Big \downarrow \) 1189 |
\(\displaystyle \int \frac {c x^2+a}{\sqrt {d x-1} \sqrt {d x+1}}dx+b \int \frac {x}{\sqrt {d x-1} \sqrt {d x+1}}dx\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \int \frac {c x^2+a}{\sqrt {d x-1} \sqrt {d x+1}}dx+\frac {b \sqrt {d x-1} \sqrt {d x+1}}{d^2}\) |
\(\Big \downarrow \) 646 |
\(\displaystyle \frac {\left (2 a d^2+c\right ) \int \frac {1}{\sqrt {d x-1} \sqrt {d x+1}}dx}{2 d^2}+\frac {b \sqrt {d x-1} \sqrt {d x+1}}{d^2}+\frac {c x \sqrt {d x-1} \sqrt {d x+1}}{2 d^2}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {\left (2 a d^2+c\right ) \text {arccosh}(d x)}{2 d^3}+\frac {b \sqrt {d x-1} \sqrt {d x+1}}{d^2}+\frac {c x \sqrt {d x-1} \sqrt {d x+1}}{2 d^2}\) |
Input:
Int[(a + b*x + c*x^2)/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
Output:
(b*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/d^2 + (c*x*Sqrt[-1 + d*x]*Sqrt[1 + d*x])/ (2*d^2) + ((c + 2*a*d^2)*ArcCosh[d*x])/(2*d^3)
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) ^2), x_Symbol] :> Simp[b*x*(c + d*x)^(m + 1)*((e + f*x)^(n + 1)/(d*f*(2*m + 3))), x] - Simp[(b*c*e - a*d*f*(2*m + 3))/(d*f*(2*m + 3)) Int[(c + d*x)^ m*(e + f*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && !LtQ[m, -1]
Int[((d_) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[b Int[x*(d + e*x)^m*(f + g*x)^n, x], x] + Int[(d + e*x)^m*(f + g*x)^n*(a + c*x^2), x] /; FreeQ[{a, b, c, d, e, f , g, m, n}, x] && EqQ[m, n] && EqQ[e*f + d*g, 0]
Time = 0.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25
method | result | size |
risch | \(\frac {\left (c x +2 b \right ) \sqrt {x d +1}\, \sqrt {x d -1}}{2 d^{2}}+\frac {\left (2 a \,d^{2}+c \right ) \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-1}\right ) \sqrt {\left (x d +1\right ) \left (x d -1\right )}}{2 d^{2} \sqrt {d^{2}}\, \sqrt {x d -1}\, \sqrt {x d +1}}\) | \(96\) |
default | \(\frac {\sqrt {x d -1}\, \sqrt {x d +1}\, \left (\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-1}\, c x +2 \ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+x d \right ) \operatorname {csgn}\left (d \right )\right ) a \,d^{2}+2 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-1}\, b +\ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+x d \right ) \operatorname {csgn}\left (d \right )\right ) c \right ) \operatorname {csgn}\left (d \right )}{2 d^{3} \sqrt {d^{2} x^{2}-1}}\) | \(120\) |
Input:
int((c*x^2+b*x+a)/(d*x-1)^(1/2)/(d*x+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(c*x+2*b)*(d*x+1)^(1/2)*(d*x-1)^(1/2)/d^2+1/2*(2*a*d^2+c)/d^2*ln(d^2*x /(d^2)^(1/2)+(d^2*x^2-1)^(1/2))/(d^2)^(1/2)*((d*x+1)*(d*x-1))^(1/2)/(d*x-1 )^(1/2)/(d*x+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {{\left (c d x + 2 \, b d\right )} \sqrt {d x + 1} \sqrt {d x - 1} - {\left (2 \, a d^{2} + c\right )} \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right )}{2 \, d^{3}} \] Input:
integrate((c*x^2+b*x+a)/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fricas")
Output:
1/2*((c*d*x + 2*b*d)*sqrt(d*x + 1)*sqrt(d*x - 1) - (2*a*d^2 + c)*log(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)))/d^3
Timed out. \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\text {Timed out} \] Input:
integrate((c*x**2+b*x+a)/(d*x-1)**(1/2)/(d*x+1)**(1/2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} c x}{2 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} b}{d^{2}} + \frac {c \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{2 \, d^{3}} \] Input:
integrate((c*x^2+b*x+a)/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima")
Output:
a*log(2*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d + 1/2*sqrt(d^2*x^2 - 1)*c*x/d^2 + sqrt(d^2*x^2 - 1)*b/d^2 + 1/2*c*log(2*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d^3
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {d x + 1} \sqrt {d x - 1} {\left (\frac {{\left (d x + 1\right )} c}{d^{2}} + \frac {2 \, b d^{5} - c d^{4}}{d^{6}}\right )} - \frac {2 \, {\left (2 \, a d^{2} + c\right )} \log \left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}{d^{2}}}{2 \, d} \] Input:
integrate((c*x^2+b*x+a)/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")
Output:
1/2*(sqrt(d*x + 1)*sqrt(d*x - 1)*((d*x + 1)*c/d^2 + (2*b*d^5 - c*d^4)/d^6) - 2*(2*a*d^2 + c)*log(sqrt(d*x + 1) - sqrt(d*x - 1))/d^2)/d
Time = 11.32 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.05 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {b\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}+\frac {2\,c\,\mathrm {atanh}\left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-\frac {\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,c\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {2\,c\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\sqrt {d\,x+1}-1}}{d^3-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}} \] Input:
int((a + b*x + c*x^2)/((d*x - 1)^(1/2)*(d*x + 1)^(1/2)),x)
Output:
(2*c*atanh(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1)))/d^3 - ((14*c*((d *x - 1)^(1/2) - 1i)^3)/((d*x + 1)^(1/2) - 1)^3 + (14*c*((d*x - 1)^(1/2) - 1i)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*c*((d*x - 1)^(1/2) - 1i)^7)/((d*x + 1) ^(1/2) - 1)^7 + (2*c*((d*x - 1)^(1/2) - 1i))/((d*x + 1)^(1/2) - 1))/(d^3 - (4*d^3*((d*x - 1)^(1/2) - 1i)^2)/((d*x + 1)^(1/2) - 1)^2 + (6*d^3*((d*x - 1)^(1/2) - 1i)^4)/((d*x + 1)^(1/2) - 1)^4 - (4*d^3*((d*x - 1)^(1/2) - 1i) ^6)/((d*x + 1)^(1/2) - 1)^6 + (d^3*((d*x - 1)^(1/2) - 1i)^8)/((d*x + 1)^(1 /2) - 1)^8) - (4*a*atan((d*((d*x - 1)^(1/2) - 1i))/(((d*x + 1)^(1/2) - 1)* (-d^2)^(1/2))))/(-d^2)^(1/2) + (b*(d*x - 1)^(1/2)*(d*x + 1)^(1/2))/d^2
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {a+b x+c x^2}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {2 \sqrt {d x +1}\, \sqrt {d x -1}\, b d +\sqrt {d x +1}\, \sqrt {d x -1}\, c d x +4 \,\mathrm {log}\left (\frac {\sqrt {d x -1}+\sqrt {d x +1}}{\sqrt {2}}\right ) a \,d^{2}+2 \,\mathrm {log}\left (\frac {\sqrt {d x -1}+\sqrt {d x +1}}{\sqrt {2}}\right ) c}{2 d^{3}} \] Input:
int((c*x^2+b*x+a)/(d*x-1)^(1/2)/(d*x+1)^(1/2),x)
Output:
(2*sqrt(d*x + 1)*sqrt(d*x - 1)*b*d + sqrt(d*x + 1)*sqrt(d*x - 1)*c*d*x + 4 *log((sqrt(d*x - 1) + sqrt(d*x + 1))/sqrt(2))*a*d**2 + 2*log((sqrt(d*x - 1 ) + sqrt(d*x + 1))/sqrt(2))*c)/(2*d**3)