3.13 \(\int \frac{A+B x+C x^2}{\sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=87 \[ \frac{\left (2 A e^2+C d^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3}-\frac{B \sqrt{d^2-e^2 x^2}}{e^2}-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2} \]

[Out]

-((B*Sqrt[d^2 - e^2*x^2])/e^2) - (C*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + ((C*d^2 + 2*A*e^2)*ArcTan[(e*x)/Sqrt[d^2
- e^2*x^2]])/(2*e^3)

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Rubi [A]  time = 0.0510074, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1815, 641, 217, 203} \[ \frac{\left (2 A e^2+C d^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3}-\frac{B \sqrt{d^2-e^2 x^2}}{e^2}-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*Sqrt[d^2 - e^2*x^2])/e^2) - (C*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + ((C*d^2 + 2*A*e^2)*ArcTan[(e*x)/Sqrt[d^2
- e^2*x^2]])/(2*e^3)

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]

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 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 203

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

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{\sqrt{d^2-e^2 x^2}} \, dx &=-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2}-\frac{\int \frac{-C d^2-2 A e^2-2 B e^2 x}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^2}\\ &=-\frac{B \sqrt{d^2-e^2 x^2}}{e^2}-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2}-\frac{\left (-C d^2-2 A e^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^2}\\ &=-\frac{B \sqrt{d^2-e^2 x^2}}{e^2}-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2}-\frac{\left (-C d^2-2 A e^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^2}\\ &=-\frac{B \sqrt{d^2-e^2 x^2}}{e^2}-\frac{C x \sqrt{d^2-e^2 x^2}}{2 e^2}+\frac{\left (C d^2+2 A e^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3}\\ \end{align*}

Mathematica [A]  time = 0.0417679, size = 67, normalized size = 0.77 \[ \frac{\left (2 A e^2+C d^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-e (2 B+C x) \sqrt{d^2-e^2 x^2}}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(e*(2*B + C*x)*Sqrt[d^2 - e^2*x^2]) + (C*d^2 + 2*A*e^2)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^3)

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Maple [A]  time = 0.105, size = 108, normalized size = 1.2 \begin{align*} -{\frac{Cx}{2\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{C{d}^{2}}{2\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{B}{{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{A\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*C*x*(-e^2*x^2+d^2)^(1/2)/e^2+1/2*C/e^2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-B*(-e^2
*x^2+d^2)^(1/2)/e^2+A/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [A]  time = 1.52561, size = 126, normalized size = 1.45 \begin{align*} \frac{A \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}}} + \frac{C d^{2} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}} e^{2}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} C x}{2 \, e^{2}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} B}{e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77747, size = 153, normalized size = 1.76 \begin{align*} -\frac{2 \,{\left (C d^{2} + 2 \, A e^{2}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt{-e^{2} x^{2} + d^{2}}{\left (C e x + 2 \, B e\right )}}{2 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(2*(C*d^2 + 2*A*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + sqrt(-e^2*x^2 + d^2)*(C*e*x + 2*B*e))/e^
3

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Sympy [A]  time = 3.22441, size = 264, normalized size = 3.03 \begin{align*} A \left (\begin{cases} \frac{\sqrt{\frac{d^{2}}{e^{2}}} \operatorname{asin}{\left (x \sqrt{\frac{e^{2}}{d^{2}}} \right )}}{\sqrt{d^{2}}} & \text{for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac{\sqrt{- \frac{d^{2}}{e^{2}}} \operatorname{asinh}{\left (x \sqrt{- \frac{e^{2}}{d^{2}}} \right )}}{\sqrt{d^{2}}} & \text{for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac{\sqrt{\frac{d^{2}}{e^{2}}} \operatorname{acosh}{\left (x \sqrt{\frac{e^{2}}{d^{2}}} \right )}}{\sqrt{- d^{2}}} & \text{for}\: d^{2} < 0 \wedge e^{2} < 0 \end{cases}\right ) + B \left (\begin{cases} \frac{x^{2}}{2 \sqrt{d^{2}}} & \text{for}\: e^{2} = 0 \\- \frac{\sqrt{d^{2} - e^{2} x^{2}}}{e^{2}} & \text{otherwise} \end{cases}\right ) + C \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{i d x \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{d x}{2 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{x^{3}}{2 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*Piecewise((sqrt(d**2/e**2)*asin(x*sqrt(e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 > 0)), (sqrt(-d**2/e**2)*a
sinh(x*sqrt(-e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 < 0)), (sqrt(d**2/e**2)*acosh(x*sqrt(e**2/d**2))/sqrt(
-d**2), (d**2 < 0) & (e**2 < 0))) + B*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e
**2, True)) + C*Piecewise((-I*d**2*acosh(e*x/d)/(2*e**3) - I*d*x*sqrt(-1 + e**2*x**2/d**2)/(2*e**2), Abs(e**2*
x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2*d*sqrt(1 -
e**2*x**2/d**2)), True))

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Giac [A]  time = 1.21307, size = 70, normalized size = 0.8 \begin{align*} \frac{1}{2} \,{\left (C d^{2} + 2 \, A e^{2}\right )} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{2} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left (C x e^{\left (-2\right )} + 2 \, B e^{\left (-2\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

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