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]

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

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Rubi [A]  time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 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])

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 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 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 {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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.64, size = 71, normalized size = 0.82 \[ -\frac {2 \, {\left (C d^{2} + 2 \, A e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (C e x + 2 \, B e\right )}}{2 \, e^{3}} \]

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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giac [A]  time = 0.34, size = 52, normalized size = 0.60 \[ \frac {1}{2} \, {\left (C d^{2} + 2 \, A e^{2}\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (C x e^{\left (-2\right )} + 2 \, B e^{\left (-2\right )}\right )} \]

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))

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maple [A]  time = 0.01, size = 108, normalized size = 1.24 \[ \frac {A \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {C \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C x}{2 e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, B}{e^{2}} \]

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

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maxima [A]  time = 0.99, size = 70, normalized size = 0.80 \[ \frac {C d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {A \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C x}{2 \, e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{e^{2}} \]

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]

1/2*C*d^2*arcsin(e*x/d)/e^3 + A*arcsin(e*x/d)/e - 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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mupad [B]  time = 4.40, size = 148, normalized size = 1.70 \[ \left \{\begin {array}{cl} \frac {2\,C\,x^3+3\,B\,x^2+6\,A\,x}{6\,\sqrt {d^2}} & \text {\ if\ \ }e=0\\ \frac {A\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{\sqrt {-e^2}}-\frac {B\,\sqrt {d^2-e^2\,x^2}}{e^2}-\frac {C\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e^2}-\frac {C\,d^2\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}} & \text {\ if\ \ }e\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(e == 0, (6*A*x + 3*B*x^2 + 2*C*x^3)/(6*(d^2)^(1/2)), e ~= 0, (A*log(x*(-e^2)^(1/2) + (d^2 - e^2*x^2)
^(1/2)))/(-e^2)^(1/2) - (B*(d^2 - e^2*x^2)^(1/2))/e^2 - (C*x*(d^2 - e^2*x^2)^(1/2))/(2*e^2) - (C*d^2*log(2*x*(
-e^2)^(1/2) + 2*(d^2 - e^2*x^2)^(1/2)))/(2*(-e^2)^(3/2)))

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sympy [A]  time = 4.56, size = 262, normalized size = 3.01 \[ A \left (\begin {cases} \frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {asin}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac {\sqrt {- \frac {d^{2}}{e^{2}}} \operatorname {asinh}{\left (x \sqrt {- \frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {acosh}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: d^{2} < 0 \wedge e^{2} < 0 \end {cases}\right ) + B \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d^{2}}} & \text {for}\: e^{2} = 0 \\- \frac {\sqrt {d^{2} - e^{2} x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) + C \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {i d x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x}{2 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{3}}{2 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

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/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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