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

Optimal. Leaf size=63 \[ \frac {\left (2 a d^2+c\right ) \sin ^{-1}(d x)}{2 d^3}-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2} \]

[Out]

1/2*(2*a*d^2+c)*arcsin(d*x)/d^3-b*(-d^2*x^2+1)^(1/2)/d^2-1/2*c*x*(-d^2*x^2+1)^(1/2)/d^2

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Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {899, 1815, 641, 216} \[ \frac {\left (2 a d^2+c\right ) \sin ^{-1}(d x)}{2 d^3}-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*Sqrt[1 - d^2*x^2])/d^2) - (c*x*Sqrt[1 - d^2*x^2])/(2*d^2) + ((c + 2*a*d^2)*ArcSin[d*x])/(2*d^3)

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 899

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Int[(d*f + e*g*x^2)^m*(a + b*x + c*x^2)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] &&
EqQ[e*f + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx &=\int \frac {a+b x+c x^2}{\sqrt {1-d^2 x^2}} \, dx\\ &=-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}-\frac {\int \frac {-c-2 a d^2-2 b d^2 x}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}-\frac {\left (-c-2 a d^2\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=-\frac {b \sqrt {1-d^2 x^2}}{d^2}-\frac {c x \sqrt {1-d^2 x^2}}{2 d^2}+\frac {\left (c+2 a d^2\right ) \sin ^{-1}(d x)}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 45, normalized size = 0.71 \[ \frac {\left (2 a d^2+c\right ) \sin ^{-1}(d x)-d \sqrt {1-d^2 x^2} (2 b+c x)}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*(2*b + c*x)*Sqrt[1 - d^2*x^2]) + (c + 2*a*d^2)*ArcSin[d*x])/(2*d^3)

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fricas [A]  time = 1.00, size = 67, normalized size = 1.06 \[ -\frac {{\left (c d x + 2 \, b d\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 2 \, {\left (2 \, a d^{2} + c\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((c*d*x + 2*b*d)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*(2*a*d^2 + c)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)
/(d*x)))/d^3

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giac [A]  time = 0.27, size = 76, normalized size = 1.21 \[ -\frac {\sqrt {d x + 1} \sqrt {-d x + 1} {\left (\frac {{\left (d x + 1\right )} c}{d^{2}} + \frac {2 \, b d^{5} - c d^{4}}{d^{6}}\right )} - \frac {2 \, {\left (2 \, a d^{2} + c\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*c/d^2 + (2*b*d^5 - c*d^4)/d^6) - 2*(2*a*d^2 + c)*arcsin(1/2*sqrt
(2)*sqrt(d*x + 1))/d^2)/d

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maple [C]  time = 0.02, size = 117, normalized size = 1.86 \[ -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (-2 a \,d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+\sqrt {-d^{2} x^{2}+1}\, c d x \,\mathrm {csgn}\relax (d )+2 \sqrt {-d^{2} x^{2}+1}\, b d \,\mathrm {csgn}\relax (d )-c \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (d )}{2 \sqrt {-d^{2} x^{2}+1}\, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-d*x+1)^(1/2)*(d*x+1)^(1/2)/d^3*(csgn(d)*d*(-d^2*x^2+1)^(1/2)*x*c-2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn
(d))*a*d^2+2*csgn(d)*d*(-d^2*x^2+1)^(1/2)*b-arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))*c)/(-d^2*x^2+1)^(1/2)*csg
n(d)

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maxima [A]  time = 0.97, size = 57, normalized size = 0.90 \[ \frac {a \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} c x}{2 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} b}{d^{2}} + \frac {c \arcsin \left (d x\right )}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*arcsin(d*x)/d - 1/2*sqrt(-d^2*x^2 + 1)*c*x/d^2 - sqrt(-d^2*x^2 + 1)*b/d^2 + 1/2*c*arcsin(d*x)/d^3

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mupad [B]  time = 7.76, size = 232, normalized size = 3.68 \[ -\frac {\sqrt {1-d\,x}\,\left (\frac {b}{d^2}+\frac {b\,x}{d}\right )}{\sqrt {d\,x+1}}-\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {1-d\,x}-1\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {d^2}}\right )}{\sqrt {d^2}}-\frac {2\,c\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {\frac {14\,c\,{\left (\sqrt {1-d\,x}-1\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {14\,c\,{\left (\sqrt {1-d\,x}-1\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,c\,{\left (\sqrt {1-d\,x}-1\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {2\,c\,\left (\sqrt {1-d\,x}-1\right )}{\sqrt {d\,x+1}-1}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((1 - d*x)^(1/2)*(b/d^2 + (b*x)/d))/(d*x + 1)^(1/2) - (4*a*atan((d*((1 - d*x)^(1/2) - 1))/(((d*x + 1)^(1/2)
- 1)*(d^2)^(1/2))))/(d^2)^(1/2) - (2*c*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/d^3 - ((14*c*((1 - d
*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 - (14*c*((1 - d*x)^(1/2) - 1)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*c*((1
- d*x)^(1/2) - 1)^7)/((d*x + 1)^(1/2) - 1)^7 - (2*c*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1))/(d^3*(((1 -
d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^4)

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sympy [C]  time = 49.71, size = 282, normalized size = 4.48 \[ - \frac {i a {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {a {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i b {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {b {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {i c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & - \frac {1}{2}, - \frac {1}{2}, 0, 1 \\-1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} + \frac {c {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & - \frac {3}{2}, -1, -1, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*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) +
a*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*b*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) - b*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) - I*c*meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -
1/2, -1/4, 0, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**3) + c*meijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((
-5/4, -3/4), (-3/2, -1, -1, 0)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**3)

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