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

Optimal. Leaf size=79 \[ -\frac {\sqrt {1-d^2 x^2} \left (2 \left (3 a d^2+2 c\right )+3 b d^2 x\right )}{6 d^4}+\frac {b \sin ^{-1}(d x)}{2 d^3}-\frac {c x^2 \sqrt {1-d^2 x^2}}{3 d^2} \]

[Out]

1/2*b*arcsin(d*x)/d^3-1/3*c*x^2*(-d^2*x^2+1)^(1/2)/d^2-1/6*(3*b*d^2*x+6*a*d^2+4*c)*(-d^2*x^2+1)^(1/2)/d^4

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Rubi [A]  time = 0.14, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1609, 1809, 780, 216} \[ -\frac {\sqrt {1-d^2 x^2} \left (2 \left (3 a d^2+2 c\right )+3 b d^2 x\right )}{6 d^4}+\frac {b \sin ^{-1}(d x)}{2 d^3}-\frac {c x^2 \sqrt {1-d^2 x^2}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x^2*Sqrt[1 - d^2*x^2])/(3*d^2) - ((2*(2*c + 3*a*d^2) + 3*b*d^2*x)*Sqrt[1 - d^2*x^2])/(6*d^4) + (b*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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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 {x \left (a+b x+c x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx &=\int \frac {x \left (a+b x+c x^2\right )}{\sqrt {1-d^2 x^2}} \, dx\\ &=-\frac {c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {\int \frac {x \left (-2 c-3 a d^2-3 b d^2 x\right )}{\sqrt {1-d^2 x^2}} \, dx}{3 d^2}\\ &=-\frac {c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {\left (2 \left (2 c+3 a d^2\right )+3 b d^2 x\right ) \sqrt {1-d^2 x^2}}{6 d^4}+\frac {b \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=-\frac {c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {\left (2 \left (2 c+3 a d^2\right )+3 b d^2 x\right ) \sqrt {1-d^2 x^2}}{6 d^4}+\frac {b \sin ^{-1}(d x)}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 57, normalized size = 0.72 \[ \frac {3 b d \sin ^{-1}(d x)-\sqrt {1-d^2 x^2} \left (3 d^2 (2 a+b x)+2 c \left (d^2 x^2+2\right )\right )}{6 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[1 - d^2*x^2]*(3*d^2*(2*a + b*x) + 2*c*(2 + d^2*x^2))) + 3*b*d*ArcSin[d*x])/(6*d^4)

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fricas [A]  time = 0.79, size = 78, normalized size = 0.99 \[ -\frac {6 \, b d \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right ) + {\left (2 \, c d^{2} x^{2} + 3 \, b d^{2} x + 6 \, a d^{2} + 4 \, c\right )} \sqrt {d x + 1} \sqrt {-d x + 1}}{6 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*b*d*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)) + (2*c*d^2*x^2 + 3*b*d^2*x + 6*a*d^2 + 4*c)*sqrt(
d*x + 1)*sqrt(-d*x + 1))/d^4

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giac [A]  time = 1.34, size = 101, normalized size = 1.28 \[ -\frac {\sqrt {d x + 1} \sqrt {-d x + 1} {\left ({\left (d x + 1\right )} {\left (\frac {2 \, {\left (d x + 1\right )} c}{d^{3}} + \frac {3 \, b d^{10} - 4 \, c d^{9}}{d^{12}}\right )} + \frac {3 \, {\left (2 \, a d^{11} - b d^{10} + 2 \, c d^{9}\right )}}{d^{12}}\right )} - \frac {6 \, b \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)*c/d^3 + (3*b*d^10 - 4*c*d^9)/d^12) + 3*(2*a*d^11 -
b*d^10 + 2*c*d^9)/d^12) - 6*b*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^2)/d

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maple [C]  time = 0.04, size = 139, normalized size = 1.76 \[ -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (2 \sqrt {-d^{2} x^{2}+1}\, c \,d^{2} x^{2} \mathrm {csgn}\relax (d )+3 \sqrt {-d^{2} x^{2}+1}\, b \,d^{2} x \,\mathrm {csgn}\relax (d )+6 \sqrt {-d^{2} x^{2}+1}\, a \,d^{2} \mathrm {csgn}\relax (d )-3 b d \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+4 \sqrt {-d^{2} x^{2}+1}\, c \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{6 \sqrt {-d^{2} x^{2}+1}\, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(2*csgn(d)*x^2*c*d^2*(-d^2*x^2+1)^(1/2)+3*(-d^2*x^2+1)^(1/2)*csgn(d)*x*b*d^2
+6*(-d^2*x^2+1)^(1/2)*csgn(d)*a*d^2+4*(-d^2*x^2+1)^(1/2)*csgn(d)*c-3*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*b*
d)*csgn(d)/d^4/(-d^2*x^2+1)^(1/2)

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maxima [A]  time = 0.97, size = 87, normalized size = 1.10 \[ -\frac {\sqrt {-d^{2} x^{2} + 1} c x^{2}}{3 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} b x}{2 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} a}{d^{2}} + \frac {b \arcsin \left (d x\right )}{2 \, d^{3}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} c}{3 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(-d^2*x^2 + 1)*c*x^2/d^2 - 1/2*sqrt(-d^2*x^2 + 1)*b*x/d^2 - sqrt(-d^2*x^2 + 1)*a/d^2 + 1/2*b*arcsin(d
*x)/d^3 - 2/3*sqrt(-d^2*x^2 + 1)*c/d^4

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mupad [B]  time = 7.44, size = 244, normalized size = 3.09 \[ -\frac {\sqrt {1-d\,x}\,\left (\frac {a}{d^2}+\frac {a\,x}{d}\right )}{\sqrt {d\,x+1}}-\frac {2\,b\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {\frac {14\,b\,{\left (\sqrt {1-d\,x}-1\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {14\,b\,{\left (\sqrt {1-d\,x}-1\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,b\,{\left (\sqrt {1-d\,x}-1\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {2\,b\,\left (\sqrt {1-d\,x}-1\right )}{\sqrt {d\,x+1}-1}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^4}-\frac {\sqrt {1-d\,x}\,\left (\frac {2\,c}{3\,d^4}+\frac {c\,x^3}{3\,d}+\frac {c\,x^2}{3\,d^2}+\frac {2\,c\,x}{3\,d^3}\right )}{\sqrt {d\,x+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 82.40, size = 313, normalized size = 3.96 \[ - \frac {i a {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {a {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} - \frac {i b {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & - \frac {1}{2}, - \frac {1}{2}, 0, 1 \\-1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} + \frac {b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & - \frac {3}{2}, -1, -1, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} - \frac {i c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} - \frac {c {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*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) - a*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*b*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) + b*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) - I*c*meijerg(((-5/4, -3/4), (-1, -1, -1/2,
1)), ((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**4) - c*meijerg(((-2, -7/4, -3/2, -
5/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4)

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