\(\int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx\) [12]

Optimal result

Integrand size = 32, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c \sqrt {-1+d x} \sqrt {1+d x}}{d^2}+\frac {b \text {arccosh}(d x)}{d}+a \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \] Output:

c*(d*x-1)^(1/2)*(d*x+1)^(1/2)/d^2+b*arccosh(d*x)/d+a*arctan((d*x-1)^(1/2)* 
(d*x+1)^(1/2))
 

Mathematica [A] (warning: unable to verify)

Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c \sqrt {-1+d x} \sqrt {1+d x}}{d^2}+2 a \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right )+\frac {2 b \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{d} \] Input:

Integrate[(a + b*x + c*x^2)/(x*Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
 

Output:

(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
 

Rubi [A] (verified)

Time = 0.41 (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

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (a \arctan \left (\sqrt {d^2 x^2-1}\right )+\frac {b \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d}+\frac {c \sqrt {d^2 x^2-1}}{d^2}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

Input:

Int[(a + b*x + c*x^2)/(x*Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
 

Output:

(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])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

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]
 

rule 218

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]
 

rule 219

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])
 

rule 224

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]
 

rule 243

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]
 

rule 538

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]
 

rule 2113

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]
 

rule 2340

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])
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.70 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {\left (-\arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \operatorname {csgn}\left (d \right ) a \,d^{2}+\operatorname {csgn}\left (d \right ) \sqrt {d^{2} x^{2}-1}\, c +\ln \left (\left (\sqrt {\left (x d +1\right ) \left (x d -1\right )}\, \operatorname {csgn}\left (d \right )+x d \right ) \operatorname {csgn}\left (d \right )\right ) b d \right ) \sqrt {x d -1}\, \sqrt {x d +1}\, \operatorname {csgn}\left (d \right )}{d^{2} \sqrt {d^{2} x^{2}-1}}\)	\(95\)

Input:

int((c*x^2+b*x+a)/x/(d*x-1)^(1/2)/(d*x+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

(-arctan(1/(d^2*x^2-1)^(1/2))*csgn(d)*a*d^2+csgn(d)*(d^2*x^2-1)^(1/2)*c+ln 
((((d*x+1)*(d*x-1))^(1/2)*csgn(d)+x*d)*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)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {2 \, a d^{2} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) - b d \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + \sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \] Input:

integrate((c*x^2+b*x+a)/x/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fricas 
")
 

Output:

(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
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 30.25 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.36 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=- \frac {a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} \] Input:

integrate((c*x**2+b*x+a)/x/(d*x-1)**(1/2)/(d*x+1)**(1/2),x)
 

Output:

-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)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-a \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} c}{d^{2}} \] Input:

integrate((c*x^2+b*x+a)/x/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima 
")
 

Output:

-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
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-2 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) - \frac {b \log \left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right )}{d} + \frac {\sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \] Input:

integrate((c*x^2+b*x+a)/x/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")
 

Output:

-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
 

Mupad [B] (verification not implemented)

Time = 4.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.15 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {c\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}-\frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-a\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i} \] Input:

int((a + b*x + c*x^2)/(x*(d*x - 1)^(1/2)*(d*x + 1)^(1/2)),x)
 

Output:

(c*(d*x - 1)^(1/2)*(d*x + 1)^(1/2))/d^2 - (4*b*atan((d*((d*x - 1)^(1/2) - 
1i))/(((d*x + 1)^(1/2) - 1)*(-d^2)^(1/2))))/(-d^2)^(1/2) - a*(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
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53 \[ \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {2 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}-1\right ) a \,d^{2}-2 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}+1\right ) a \,d^{2}+\sqrt {d x +1}\, \sqrt {d x -1}\, c +2 \,\mathrm {log}\left (\frac {\sqrt {d x -1}+\sqrt {d x +1}}{\sqrt {2}}\right ) b d}{d^{2}} \] Input:

int((c*x^2+b*x+a)/x/(d*x-1)^(1/2)/(d*x+1)^(1/2),x)
 

Output:

(2*atan(sqrt(d*x - 1) + sqrt(d*x + 1) - 1)*a*d**2 - 2*atan(sqrt(d*x - 1) + 
 sqrt(d*x + 1) + 1)*a*d**2 + sqrt(d*x + 1)*sqrt(d*x - 1)*c + 2*log((sqrt(d 
*x - 1) + sqrt(d*x + 1))/sqrt(2))*b*d)/d**2