3.1.4 \(\int \frac {(A+B x+C x^2) \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [4]

Optimal. Leaf size=148 \[ \frac {\left (C d^2-e (B d-2 A e)\right ) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {(C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}+\frac {d \left (C d^2-e (B d-2 A e)\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {1653, 809, 679, 223, 209} \begin {gather*} \frac {d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {\sqrt {d^2-e^2 x^2} \left (C d^2-e (B d-2 A e)\right )}{2 e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2} (C d-B e)}{2 e^3 (d+e x)}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 223

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 679

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] - Dist[2*c*d*(p/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 809

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*
((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 1653

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{d+e x} \, dx &=-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {\int \frac {\left (-3 A e^4+3 e^3 (C d-B e) x\right ) \sqrt {d^2-e^2 x^2}}{d+e x} \, dx}{3 e^4}\\ &=-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {(C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}+\frac {\left (C d^2-e (B d-2 A e)\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx}{2 e^2}\\ &=\frac {\left (C d^2-e (B d-2 A e)\right ) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {(C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}+\frac {\left (d \left (C d^2-e (B d-2 A e)\right )\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^2}\\ &=\frac {\left (C d^2-e (B d-2 A e)\right ) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {(C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}+\frac {\left (d \left (C d^2-e (B d-2 A e)\right )\right ) \text {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 {\left (C d^2-e (B d-2 A e)\right ) \sqrt {d^2-e^2 x^2}}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {(C d-B e) \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}+\frac {d \left (C d^2-e (B d-2 A e)\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 = 0.45, size = 123, normalized size = 0.83 \begin {gather*} \frac {e \sqrt {d^2-e^2 x^2} \left (3 e (-2 B d+2 A e+B e x)+C \left (4 d^2-3 d e x+2 e^2 x^2\right )\right )+3 \sqrt {-e^2} \left (C d^3+d e (-B d+2 A e)\right ) \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{6 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[d^2 - e^2*x^2]*(3*e*(-2*B*d + 2*A*e + B*e*x) + C*(4*d^2 - 3*d*e*x + 2*e^2*x^2)) + 3*Sqrt[-e^2]*(C*d^3
+ d*e*(-(B*d) + 2*A*e))*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(6*e^4)

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Maple [A]
time = 0.08, size = 227, normalized size = 1.53

method	result	size
risch	\(\frac {\left (2 C \,e^{2} x^{2}+3 B \,e^{2} x -3 C d e x +6 A \,e^{2}-6 B d e +4 C \,d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{3}}+\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) A}{\sqrt {e^{2}}}-\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) B}{2 e \sqrt {e^{2}}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) C}{2 e^{2} \sqrt {e^{2}}}\)	\(163\)
default	\(\frac {-\frac {C \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e}+B e \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )-C d \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{2}}+\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{3}}\)	\(227\)

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)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 159, normalized size = 1.07 \begin {gather*} \frac {1}{2} \, C d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \frac {1}{2} \, B d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} C d x e^{\left (-2\right )} + \sqrt {-x^{2} e^{2} + d^{2}} C d^{2} e^{\left (-3\right )} + A d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} B x e^{\left (-1\right )} - \sqrt {-x^{2} e^{2} + d^{2}} B d e^{\left (-2\right )} - \frac {1}{3} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} C e^{\left (-3\right )} + \sqrt {-x^{2} e^{2} + d^{2}} A e^{\left (-1\right )} \end {gather*}

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)/(e*x+d),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 104, normalized size = 0.70 \begin {gather*} -\frac {1}{6} \, {\left (6 \, {\left (C d^{3} - B d^{2} e + 2 \, A d e^{2}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (4 \, C d^{2} + {\left (2 \, C x^{2} + 3 \, B x + 6 \, A\right )} e^{2} - 3 \, {\left (C d x + 2 \, B d\right )} e\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-3\right )} \end {gather*}

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)/(e*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (A + B x + C x^{2}\right )}{d + e x}\, dx \end {gather*}

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)/(e*x+d),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))*(A + B*x + C*x**2)/(d + e*x), x)

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Giac [A]
time = 2.29, size = 99, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, {\left (C d^{3} - B d^{2} e + 2 \, A d e^{2}\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\left (d\right ) + \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left (2 \, C x e^{\left (-1\right )} - 3 \, {\left (C d e^{3} - B e^{4}\right )} e^{\left (-5\right )}\right )} x + 2 \, {\left (2 \, C d^{2} e^{2} - 3 \, B d e^{3} + 3 \, A e^{4}\right )} e^{\left (-5\right )}\right )} \end {gather*}

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)/(e*x+d),x, algorithm="giac")

[Out]

1/2*(C*d^3 - B*d^2*e + 2*A*d*e^2)*arcsin(x*e/d)*e^(-3)*sgn(d) + 1/6*sqrt(-x^2*e^2 + d^2)*((2*C*x*e^(-1) - 3*(C
*d*e^3 - B*e^4)*e^(-5))*x + 2*(2*C*d^2*e^2 - 3*B*d*e^3 + 3*A*e^4)*e^(-5))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}\,\left (C\,x^2+B\,x+A\right )}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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