3.4 \(\int \sqrt {1-d x} \sqrt {1+d x} (A+B x+C x^2) \, dx\)

Optimal. Leaf size=95 \[ \frac {x \sqrt {1-d^2 x^2} \left (4 A d^2+C\right )}{8 d^2}+\frac {\left (4 A d^2+C\right ) \sin ^{-1}(d x)}{8 d^3}-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2} \]

[Out]

-1/3*B*(-d^2*x^2+1)^(3/2)/d^2-1/4*C*x*(-d^2*x^2+1)^(3/2)/d^2+1/8*(4*A*d^2+C)*arcsin(d*x)/d^3+1/8*(4*A*d^2+C)*x
*(-d^2*x^2+1)^(1/2)/d^2

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {899, 1815, 641, 195, 216} \[ \frac {x \sqrt {1-d^2 x^2} \left (4 A d^2+C\right )}{8 d^2}+\frac {\left (4 A d^2+C\right ) \sin ^{-1}(d x)}{8 d^3}-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - d*x]*Sqrt[1 + d*x]*(A + B*x + C*x^2),x]

[Out]

((C + 4*A*d^2)*x*Sqrt[1 - d^2*x^2])/(8*d^2) - (B*(1 - d^2*x^2)^(3/2))/(3*d^2) - (C*x*(1 - d^2*x^2)^(3/2))/(4*d
^2) + ((C + 4*A*d^2)*ArcSin[d*x])/(8*d^3)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx &=\int \left (A+B x+C x^2\right ) \sqrt {1-d^2 x^2} \, dx\\ &=-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2}-\frac {\int \left (-C-4 A d^2-4 B d^2 x\right ) \sqrt {1-d^2 x^2} \, dx}{4 d^2}\\ &=-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2}-\frac {\left (-C-4 A d^2\right ) \int \sqrt {1-d^2 x^2} \, dx}{4 d^2}\\ &=\frac {\left (C+4 A d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^2}-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2}+\frac {\left (C+4 A d^2\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{8 d^2}\\ &=\frac {\left (C+4 A d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^2}-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2}+\frac {\left (C+4 A d^2\right ) \sin ^{-1}(d x)}{8 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 71, normalized size = 0.75 \[ \frac {d \sqrt {1-d^2 x^2} \left (12 A d^2 x+8 B d^2 x^2-8 B+6 C d^2 x^3-3 C x\right )+3 \left (4 A d^2+C\right ) \sin ^{-1}(d x)}{24 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - d*x]*Sqrt[1 + d*x]*(A + B*x + C*x^2),x]

[Out]

(d*Sqrt[1 - d^2*x^2]*(-8*B - 3*C*x + 12*A*d^2*x + 8*B*d^2*x^2 + 6*C*d^2*x^3) + 3*(C + 4*A*d^2)*ArcSin[d*x])/(2
4*d^3)

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fricas [A]  time = 0.95, size = 95, normalized size = 1.00 \[ \frac {{\left (6 \, C d^{3} x^{3} + 8 \, B d^{3} x^{2} - 8 \, B d + 3 \, {\left (4 \, A d^{3} - C d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 6 \, {\left (4 \, A d^{2} + C\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{3}} \]

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]

1/24*((6*C*d^3*x^3 + 8*B*d^3*x^2 - 8*B*d + 3*(4*A*d^3 - C*d)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(4*A*d^2 + C)
*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^3

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giac [B]  time = 1.54, size = 336, normalized size = 3.54 \[ \frac {4 \, {\left (\sqrt {d x + 1} \sqrt {-d x + 1} {\left ({\left (d x + 1\right )} {\left (\frac {2 \, {\left (d x + 1\right )}}{d^{2}} - \frac {7}{d^{2}}\right )} + \frac {9}{d^{2}}\right )} + \frac {6 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{2}}\right )} B d + {\left ({\left ({\left (d x + 1\right )} {\left (2 \, {\left (d x + 1\right )} {\left (\frac {3 \, {\left (d x + 1\right )}}{d^{3}} - \frac {13}{d^{3}}\right )} + \frac {43}{d^{3}}\right )} - \frac {39}{d^{3}}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - \frac {18 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{3}}\right )} C d + 12 \, {\left (\sqrt {d x + 1} {\left (d x - 2\right )} \sqrt {-d x + 1} - 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} A + 24 \, {\left (\sqrt {d x + 1} \sqrt {-d x + 1} + 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} A + 4 \, {\left (\sqrt {d x + 1} \sqrt {-d x + 1} {\left ({\left (d x + 1\right )} {\left (\frac {2 \, {\left (d x + 1\right )}}{d^{2}} - \frac {7}{d^{2}}\right )} + \frac {9}{d^{2}}\right )} + \frac {6 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{2}}\right )} C + \frac {12 \, {\left (\sqrt {d x + 1} {\left (d x - 2\right )} \sqrt {-d x + 1} - 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} B}{d}}{24 \, d} \]

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]

1/24*(4*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2) + 6*arcsin(1/2*sqrt(2)*sqr
t(d*x + 1))/d^2)*B*d + (((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)/d^3 - 13/d^3) + 43/d^3) - 39/d^3)*sqrt(d*x + 1)*s
qrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^3)*C*d + 12*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2
*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*A + 24*(sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))
)*A + 4*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2) + 6*arcsin(1/2*sqrt(2)*sqr
t(d*x + 1))/d^2)*C + 12*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B/d)/d

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maple [C]  time = 0.01, size = 185, normalized size = 1.95 \[ \frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (6 \sqrt {-d^{2} x^{2}+1}\, C \,d^{3} x^{3} \mathrm {csgn}\relax (d )+8 \sqrt {-d^{2} x^{2}+1}\, B \,d^{3} x^{2} \mathrm {csgn}\relax (d )+12 \sqrt {-d^{2} x^{2}+1}\, A \,d^{3} x \,\mathrm {csgn}\relax (d )+12 A \,d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-3 \sqrt {-d^{2} x^{2}+1}\, C d x \,\mathrm {csgn}\relax (d )-8 \sqrt {-d^{2} x^{2}+1}\, B d \,\mathrm {csgn}\relax (d )+3 C \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (d )}{24 \sqrt {-d^{2} x^{2}+1}\, d^{3}} \]

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]

1/24*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(6*C*csgn(d)*x^3*d^3*(-d^2*x^2+1)^(1/2)+8*B*csgn(d)*x^2*d^3*(-d^2*x^2+1)^(1/
2)+12*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x-3*C*csgn(d)*d*(-d^2*x^2+1)^(1/2)*x+12*A*arctan(1/(-d^2*x^2+1)^(1/2)*d
*x*csgn(d))*d^2-8*B*(-d^2*x^2+1)^(1/2)*csgn(d)*d+3*C*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d)))*csgn(d)/(-d^2*x
^2+1)^(1/2)/d^3

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maxima [A]  time = 0.98, size = 93, normalized size = 0.98 \[ \frac {1}{2} \, \sqrt {-d^{2} x^{2} + 1} A x - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C x}{4 \, d^{2}} + \frac {A \arcsin \left (d x\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} B}{3 \, d^{2}} + \frac {\sqrt {-d^{2} x^{2} + 1} C x}{8 \, d^{2}} + \frac {C \arcsin \left (d x\right )}{8 \, d^{3}} \]

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]

1/2*sqrt(-d^2*x^2 + 1)*A*x - 1/4*(-d^2*x^2 + 1)^(3/2)*C*x/d^2 + 1/2*A*arcsin(d*x)/d - 1/3*(-d^2*x^2 + 1)^(3/2)
*B/d^2 + 1/8*sqrt(-d^2*x^2 + 1)*C*x/d^2 + 1/8*C*arcsin(d*x)/d^3

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mupad [B]  time = 7.21, size = 361, normalized size = 3.80 \[ \frac {A\,x\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{2}-\frac {\frac {35\,C\,{\left (\sqrt {1-d\,x}-1\right )}^3}{2\,{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {273\,C\,{\left (\sqrt {1-d\,x}-1\right )}^5}{2\,{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {715\,C\,{\left (\sqrt {1-d\,x}-1\right )}^7}{2\,{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {715\,C\,{\left (\sqrt {1-d\,x}-1\right )}^9}{2\,{\left (\sqrt {d\,x+1}-1\right )}^9}+\frac {273\,C\,{\left (\sqrt {1-d\,x}-1\right )}^{11}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{11}}-\frac {35\,C\,{\left (\sqrt {1-d\,x}-1\right )}^{13}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{13}}+\frac {C\,{\left (\sqrt {1-d\,x}-1\right )}^{15}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{15}}-\frac {C\,\left (\sqrt {1-d\,x}-1\right )}{2\,\left (\sqrt {d\,x+1}-1\right )}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^8}-\frac {C\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2\,d^3}-\frac {A\,\sqrt {d}\,\ln \left (\sqrt {-d}\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}-d^{3/2}\,x\right )}{2\,{\left (-d\right )}^{3/2}}+\frac {B\,\left (d^2\,x^2-1\right )\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{3\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - d*x)^(1/2)*(d*x + 1)^(1/2)*(A + B*x + C*x^2),x)

[Out]

(A*x*(1 - d*x)^(1/2)*(d*x + 1)^(1/2))/2 - ((35*C*((1 - d*x)^(1/2) - 1)^3)/(2*((d*x + 1)^(1/2) - 1)^3) - (273*C
*((1 - d*x)^(1/2) - 1)^5)/(2*((d*x + 1)^(1/2) - 1)^5) + (715*C*((1 - d*x)^(1/2) - 1)^7)/(2*((d*x + 1)^(1/2) -
1)^7) - (715*C*((1 - d*x)^(1/2) - 1)^9)/(2*((d*x + 1)^(1/2) - 1)^9) + (273*C*((1 - d*x)^(1/2) - 1)^11)/(2*((d*
x + 1)^(1/2) - 1)^11) - (35*C*((1 - d*x)^(1/2) - 1)^13)/(2*((d*x + 1)^(1/2) - 1)^13) + (C*((1 - d*x)^(1/2) - 1
)^15)/(2*((d*x + 1)^(1/2) - 1)^15) - (C*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1)^(1/2) - 1)))/(d^3*(((1 - d*x)^(1/
2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^8) - (C*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/(2*d^3) - (A
*d^(1/2)*log((-d)^(1/2)*(1 - d*x)^(1/2)*(d*x + 1)^(1/2) - d^(3/2)*x))/(2*(-d)^(3/2)) + (B*(d^2*x^2 - 1)*(1 - d
*x)^(1/2)*(d*x + 1)^(1/2))/(3*d^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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