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

Optimal. Leaf size=48 \[ -a \tanh ^{-1}\left (\sqrt {1-d^2 x^2}\right )+\frac {b \sin ^{-1}(d x)}{d}-\frac {c \sqrt {1-d^2 x^2}}{d^2} \]

[Out]

b*arcsin(d*x)/d-a*arctanh((-d^2*x^2+1)^(1/2))-c*(-d^2*x^2+1)^(1/2)/d^2

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Rubi [A]  time = 0.18, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1609, 1809, 844, 216, 266, 63, 208} \[ -a \tanh ^{-1}\left (\sqrt {1-d^2 x^2}\right )+\frac {b \sin ^{-1}(d x)}{d}-\frac {c \sqrt {1-d^2 x^2}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

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

Rule 216

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

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 1609

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[P
x*(a*c + b*d*x^2)^m*(e + f*x)^p, 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] || (GtQ[a, 0] && GtQ[c, 0]))

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

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.00 \[ -a \tanh ^{-1}\left (\sqrt {1-d^2 x^2}\right )+\frac {b \sin ^{-1}(d x)}{d}-\frac {c \sqrt {1-d^2 x^2}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 81, normalized size = 1.69 \[ \frac {a d^{2} \log \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{x}\right ) - 2 \, b d \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right ) - \sqrt {d x + 1} \sqrt {-d x + 1} c}{d^{2}} \]

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]

(a*d^2*log((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/x) - 2*b*d*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)) - sq
rt(d*x + 1)*sqrt(-d*x + 1)*c)/d^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2,2]%%%}] at parameters values [70,22]Warning, choosing root of [1,0,-4,0,%%%{4,[2,2]%%%}] at parameters va
lues [42,56]-a*ln(abs(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))+2-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1))
)+a*ln(abs(2*sqrt(d*x+1)/(-2*sqrt(-d*x+1)+2*sqrt(2))-2-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1)))-2*b*(-1/2
*pi-atan(sqrt(d*x+1)*((-1/2*(-2*sqrt(-d*x+1)+2*sqrt(2))/sqrt(d*x+1))^2-1)/(-2*sqrt(-d*x+1)+2*sqrt(2))))/d-2*c*
d^2/2/d^4*sqrt(d*x+1)*sqrt(-d*x+1)

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maple [C]  time = 0.03, size = 96, normalized size = 2.00 \[ \frac {\left (-a \,d^{2} \arctanh \left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) \mathrm {csgn}\relax (d )+b d \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-\left (d x +1\right ) \left (d x -1\right )}}\right )-\sqrt {-d^{2} x^{2}+1}\, c \,\mathrm {csgn}\relax (d )\right ) \sqrt {-d x +1}\, \sqrt {d x +1}\, \mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}\, d^{2}} \]

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]

(-csgn(d)*arctanh(1/(-d^2*x^2+1)^(1/2))*a*d^2-(-d^2*x^2+1)^(1/2)*c*csgn(d)+arctan(csgn(d)*d*x/(-(d*x+1)*(d*x-1
))^(1/2))*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 = 0.97, size = 57, normalized size = 1.19 \[ -a \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {b \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} c}{d^{2}} \]

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

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mupad [B]  time = 3.92, size = 122, normalized size = 2.54 \[ a\,\left (\ln \left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}-1\right )-\ln \left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\right )-\frac {\sqrt {1-d\,x}\,\left (\frac {c}{d^2}+\frac {c\,x}{d}\right )}{\sqrt {d\,x+1}}-\frac {4\,b\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {1-d\,x}-1\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {d^2}}\right )}{\sqrt {d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 55.20, size = 245, normalized size = 5.10 \[ \frac {i a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} - \frac {i b {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} \]

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]

I*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)) - a*me
ijerg(((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)) - I*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) + 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) - I*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) - 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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