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

Optimal. Leaf size=55 \[ a \tan ^{-1}\left (\sqrt {d x-1} \sqrt {d x+1}\right )+\frac {b \cosh ^{-1}(d x)}{d}+\frac {c \sqrt {d x-1} \sqrt {d x+1}}{d^2} \]

[Out]

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

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Rubi [B]  time = 0.18, antiderivative size = 135, normalized size of antiderivative = 2.45, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1610, 1809, 844, 217, 206, 266, 63, 205} \[ \frac {a \sqrt {d^2 x^2-1} \tan ^{-1}\left (\sqrt {d^2 x^2-1}\right )}{\sqrt {d x-1} \sqrt {d x+1}}+\frac {b \sqrt {d^2 x^2-1} \tanh ^{-1}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{d \sqrt {d x-1} \sqrt {d x+1}}-\frac {c \left (1-d^2 x^2\right )}{d^2 \sqrt {d x-1} \sqrt {d x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(1 - d^2*x^2))/(d^2*Sqrt[-1 + d*x]*Sqrt[1 + d*x])) + (a*Sqrt[-1 + d^2*x^2]*ArcTan[Sqrt[-1 + d^2*x^2]])/(S
qrt[-1 + d*x]*Sqrt[1 + d*x]) + (b*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 63

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 205

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 266

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 844

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 1610

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 1809

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] + Dist[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])

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{x \sqrt {-1+d x} \sqrt {1+d x}} \, dx &=\frac {\sqrt {-1+d^2 x^2} \int \frac {a+b x+c x^2}{x \sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\sqrt {-1+d^2 x^2} \int \frac {a d^2+b d^2 x}{x \sqrt {-1+d^2 x^2}} \, dx}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (a \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 (b \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 {c \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (a \sqrt {-1+d^2 x^2}\right ) \operatorname {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 (b \sqrt {-1+d^2 x^2}\right ) \operatorname {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 {c \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {b \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 (a \sqrt {-1+d^2 x^2}\right ) \operatorname {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 {c \left (1-d^2 x^2\right )}{d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {a \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 {b \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*}

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Mathematica [B]  time = 0.42, size = 128, normalized size = 2.33 \[ \frac {\frac {a d^2 \sqrt {d^2 x^2-1} \tan ^{-1}\left (\sqrt {d^2 x^2-1}\right )+c d^2 x^2-2 c \sqrt {1-d^2 x^2} \sin ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {2}}\right )-c}{\sqrt {d x-1} \sqrt {d x+1}}-2 (c-b d) \tanh ^{-1}\left (\sqrt {\frac {d x-1}{d x+1}}\right )}{d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.67, size = 73, normalized size = 1.33 \[ \frac {2 \, a d^{2} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) - b d \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + \sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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

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giac [A]  time = 1.37, size = 71, normalized size = 1.29 \[ -2 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) - \frac {b \log \left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right )}{d} + \frac {\sqrt {d x + 1} \sqrt {d x - 1} c}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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maple [C]  time = 0.00, size = 95, normalized size = 1.73 \[ \frac {\left (-a \,d^{2} \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \mathrm {csgn}\relax (d )+b d \ln \left (\left (d x +\sqrt {\left (d x +1\right ) \left (d x -1\right )}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+\sqrt {d^{2} x^{2}-1}\, c \,\mathrm {csgn}\relax (d )\right ) \sqrt {d x -1}\, \sqrt {d x +1}\, \mathrm {csgn}\relax (d )}{\sqrt {d^{2} x^{2}-1}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.34, size = 56, normalized size = 1.02 \[ -a \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - 1} c}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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mupad [B]  time = 5.39, size = 118, normalized size = 2.15 \[ \frac {c\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2}-\frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {-d^2}}\right )}{\sqrt {-d^2}}-a\,\left (\ln \left (\frac {{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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) - 1
i)/((d*x + 1)^(1/2) - 1)))*1i

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sympy [C]  time = 47.37, size = 240, normalized size = 4.36 \[ - \frac {a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*m
eijerg(((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)

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