\(\int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx\) [157]

Optimal result

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

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(55)=110\).

Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.45, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1624, 1821, 858, 223, 212, 272, 65, 211} \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {a \left (1-d^2 x^2\right )}{x \sqrt {d x-1} \sqrt {d x+1}}+\frac {b \sqrt {d^2 x^2-1} \arctan \left (\sqrt {d^2 x^2-1}\right )}{\sqrt {d x-1} \sqrt {d x+1}}+\frac {c \sqrt {d^2 x^2-1} \text {arctanh}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d \sqrt {d x-1} \sqrt {d x+1}} \]

[In]

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

[Out]

-((a*(1 - d^2*x^2))/(x*Sqrt[-1 + d*x]*Sqrt[1 + d*x])) + (b*Sqrt[-1 + d^2*x^2]*ArcTan[Sqrt[-1 + d^2*x^2]])/(Sqr
t[-1 + d*x]*Sqrt[1 + d*x]) + (c*Sqrt[-1 + d^2*x^2]*ArcTanh[(d*x)/Sqrt[-1 + d^2*x^2]])/(d*Sqrt[-1 + d*x]*Sqrt[1
+ d*x])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1624

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Dist[(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] &&  !Intege
rQ[m]

Rule 1821

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]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[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] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-1+d^2 x^2} \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {a \left (1-d^2 x^2\right )}{x \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\sqrt {-1+d^2 x^2} \int \frac {b+c x}{x \sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {a \left (1-d^2 x^2\right )}{x \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (b \sqrt {-1+d^2 x^2}\right ) \int \frac {1}{x \sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (c \sqrt {-1+d^2 x^2}\right ) \int \frac {1}{\sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {a \left (1-d^2 x^2\right )}{x \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (b \sqrt {-1+d^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {-1+d^2 x}} \, dx,x,x^2\right )}{2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (c \sqrt {-1+d^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{1-d^2 x^2} \, dx,x,\frac {x}{\sqrt {-1+d^2 x^2}}\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {a \left (1-d^2 x^2\right )}{x \sqrt {-1+d x} \sqrt {1+d x}}+\frac {c \sqrt {-1+d^2 x^2} \tanh ^{-1}\left (\frac {d x}{\sqrt {-1+d^2 x^2}}\right )}{d \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (b \sqrt {-1+d^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{d^2}+\frac {x^2}{d^2}} \, dx,x,\sqrt {-1+d^2 x^2}\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}} \\ & = -\frac {a \left (1-d^2 x^2\right )}{x \sqrt {-1+d x} \sqrt {1+d x}}+\frac {b \sqrt {-1+d^2 x^2} \tan ^{-1}\left (\sqrt {-1+d^2 x^2}\right )}{\sqrt {-1+d x} \sqrt {1+d x}}+\frac {c \sqrt {-1+d^2 x^2} \tanh ^{-1}\left (\frac {d x}{\sqrt {-1+d^2 x^2}}\right )}{d \sqrt {-1+d x} \sqrt {1+d x}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

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

[In]

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

[Out]

(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

Maple [A] (verified)

Time = 1.62 (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	\(\frac {\left (-\arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \operatorname {csgn}\left (d \right ) d b x +\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right ) d a +\ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) c x \right ) \sqrt {d x -1}\, \sqrt {d x +1}\, \operatorname {csgn}\left (d \right )}{\sqrt {d^{2} x^{2}-1}\, x d}\)	\(96\)

[In]

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

[Out]

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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a d^{2} x + 2 \, b d x \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + \sqrt {d x + 1} \sqrt {d x - 1} a d - c x \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right )}{d x} \]

[In]

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

[Out]

(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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.24 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.93 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=- \frac {a d {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {3}{2}, \frac {3}{2}, 2 \\1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2 & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {i a d {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 1 & \\\frac {3}{4}, \frac {5}{4} & \frac {1}{2}, 1, 1, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {b {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} \]

[In]

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

[Out]

-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*p
i**(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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-b \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {c \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} a}{x} \]

[In]

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

[Out]

-b*arcsin(1/(d*abs(x))) + c*log(2*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d + sqrt(d^2*x^2 - 1)*a/x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {2 \, b d \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) - \frac {8 \, a d^{2}}{{\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} + 4} + c \log \left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right )}{d} \]

[In]

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

[Out]

-(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*lo
g((sqrt(d*x + 1) - sqrt(d*x - 1))^2))/d

Mupad [B] (verification not implemented)

Time = 4.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.15 \[ \int \frac {a+b x+c x^2}{x^2 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{x}-\frac {4\,c\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-b\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i} \]

[In]

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

[Out]

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