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

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]

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

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Rubi [B]  time = 0.184061, antiderivative size = 135, normalized size of antiderivative = 2.45, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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])

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 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 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 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 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]

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*}

Mathematica [B]  time = 0.399322, 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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Maple [C]  time = 0.019, size = 95, normalized size = 1.7 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{{d}^{2}} \left ( -{\it csgn} \left ( d \right ) \arctan \left ({\frac{1}{\sqrt{{d}^{2}{x}^{2}-1}}} \right ) a{d}^{2}+{\it csgn} \left ( d \right ) \sqrt{{d}^{2}{x}^{2}-1}c+\ln \left ( \left ({\it csgn} \left ( d \right ) \sqrt{ \left ( dx+1 \right ) \left ( dx-1 \right ) }+dx \right ){\it csgn} \left ( d \right ) \right ) bd \right ) \sqrt{dx-1}\sqrt{dx+1}{\frac{1}{\sqrt{{d}^{2}{x}^{2}-1}}}} \end{align*}

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]

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

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Maxima [A]  time = 3.91204, size = 86, normalized size = 1.56 \begin{align*} -a \arcsin \left (\frac{1}{\sqrt{d^{2}}{\left | x \right |}}\right ) + \frac{b \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - 1} \sqrt{d^{2}}\right )}{\sqrt{d^{2}}} + \frac{\sqrt{d^{2} x^{2} - 1} c}{d^{2}} \end{align*}

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

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Fricas [A]  time = 1.08568, size = 184, normalized size = 3.35 \begin{align*} \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}} \end{align*}

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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Sympy [C]  time = 26.2475, size = 240, normalized size = 4.36 \begin{align*} - \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}} \end{align*}

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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Giac [A]  time = 1.98647, size = 96, normalized size = 1.75 \begin{align*} -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}} \end{align*}

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