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

Optimal. Leaf size=170 \[ -\frac {\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e^3 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2} \left (5 C d^2-2 e (2 B d-A e)\right )}{2 d e^3}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (5 C d^2-2 e (2 B d-A e)\right )}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)} \]

[Out]

-(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(3/2)/d/e^3/(e*x+d)^2-1/2*C*(-e^2*x^2+d^2)^(3/2)/e^3/(e*x+d)-1/2*(5*C*d^2-
2*e*(-A*e+2*B*d))*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-1/2*(5*C*d^2-2*e*(-A*e+2*B*d))*(-e^2*x^2+d^2)^(1/2)/d/e
^3

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Rubi [A]  time = 0.20, antiderivative size = 170, 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 = {1639, 793, 665, 217, 203} \[ -\frac {\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{d e^3 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2} \left (5 C d^2-2 e (2 B d-A e)\right )}{2 d e^3}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (5 C d^2-2 e (2 B d-A e)\right )}{2 e^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*C*d^2 - 2*e*(2*B*d - A*e))*Sqrt[d^2 - e^2*x^2])/(2*d*e^3) - ((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2
))/(d*e^3*(d + e*x)^2) - (C*(d^2 - e^2*x^2)^(3/2))/(2*e^3*(d + e*x)) - ((5*C*d^2 - 2*e*(2*B*d - A*e))*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 665

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 793

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

Rule 1639

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)^2} \, dx &=-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\int \frac {\left (e^2 \left (C d^2-2 A e^2\right )+e^3 (3 C d-2 B e) x\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{2 e^4}\\ &=-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{d e^3 (d+e x)^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\left (-3 e^5 \left (C d^2-2 A e^2\right )-2 \left (-d e^5 (3 C d-2 B e)-e^5 \left (C d^2-2 A e^2\right )\right )\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx}{2 d e^7}\\ &=-\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \sqrt {d^2-e^2 x^2}}{2 d e^3}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{d e^3 (d+e x)^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\left (-3 e^5 \left (C d^2-2 A e^2\right )-2 \left (-d e^5 (3 C d-2 B e)-e^5 \left (C d^2-2 A e^2\right )\right )\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^7}\\ &=-\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \sqrt {d^2-e^2 x^2}}{2 d e^3}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{d e^3 (d+e x)^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\left (-3 e^5 \left (C d^2-2 A e^2\right )-2 \left (-d e^5 (3 C d-2 B e)-e^5 \left (C d^2-2 A e^2\right )\right )\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^7}\\ &=-\frac {\left (5 C d^2-2 e (2 B d-A e)\right ) \sqrt {d^2-e^2 x^2}}{2 d e^3}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{d e^3 (d+e x)^2}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)}-\frac {\left (5 C d^2-2 e (2 B d-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 = 0.23, size = 109, normalized size = 0.64 \[ \frac {\frac {\sqrt {d^2-e^2 x^2} \left (2 e (-2 A e+3 B d+B e x)+C \left (-8 d^2-3 d e x+e^2 x^2\right )\right )}{d+e x}-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (2 e (A e-2 B d)+5 C d^2\right )}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(2*e*(3*B*d - 2*A*e + B*e*x) + C*(-8*d^2 - 3*d*e*x + e^2*x^2)))/(d + e*x) - (5*C*d^2 + 2
*e*(-2*B*d + A*e))*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^3)

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fricas [A]  time = 0.68, size = 190, normalized size = 1.12 \[ -\frac {8 \, C d^{3} - 6 \, B d^{2} e + 4 \, A d e^{2} + 2 \, {\left (4 \, C d^{2} e - 3 \, B d e^{2} + 2 \, A e^{3}\right )} x - 2 \, {\left (5 \, C d^{3} - 4 \, B d^{2} e + 2 \, A d e^{2} + {\left (5 \, C d^{2} e - 4 \, B d e^{2} + 2 \, A e^{3}\right )} x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (C e^{2} x^{2} - 8 \, C d^{2} + 6 \, B d e - 4 \, A e^{2} - {\left (3 \, C d e - 2 \, B e^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{4} x + d e^{3}\right )}} \]

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)/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/2*(8*C*d^3 - 6*B*d^2*e + 4*A*d*e^2 + 2*(4*C*d^2*e - 3*B*d*e^2 + 2*A*e^3)*x - 2*(5*C*d^3 - 4*B*d^2*e + 2*A*d
*e^2 + (5*C*d^2*e - 4*B*d*e^2 + 2*A*e^3)*x)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (C*e^2*x^2 - 8*C*d^2 +
6*B*d*e - 4*A*e^2 - (3*C*d*e - 2*B*e^2)*x)*sqrt(-e^2*x^2 + d^2))/(e^4*x + d*e^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.03, size = 439, normalized size = 2.58 \[ -\frac {A \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}}+\frac {2 B d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e}-\frac {3 C \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}+\frac {C \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C x}{2 e^{2}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, A}{d e}+\frac {2 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, B}{e^{2}}-\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, C d}{e^{3}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} A}{\left (x +\frac {d}{e}\right )^{2} d \,e^{3}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} B}{\left (x +\frac {d}{e}\right )^{2} e^{4}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} C d}{\left (x +\frac {d}{e}\right )^{2} e^{5}} \]

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)^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)/(-e^2*x^2+d^2)^(1/2)*x)-1/e^3/d/
(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*A+1/e^4/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*B-1/e^5*d/
(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*C-1/e/d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*A+2/e^2*(2*(x+d/e)*d
*e-(x+d/e)^2*e^2)^(1/2)*B-3/e^3*d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*C-1/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x
+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)*A+2/e/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)*
B*d-3/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)*C*d^2

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maxima [A]  time = 1.01, size = 197, normalized size = 1.16 \[ -\frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{e^{4} x + d e^{3}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{e^{3} x + d e^{2}} - \frac {5 \, C d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {2 \, B d \arcsin \left (\frac {e x}{d}\right )}{e^{2}} - \frac {A \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{e^{2} x + d e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C x}{2 \, e^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{e^{2}} \]

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)/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

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

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)^2,x)

[Out]

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

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

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)**2,x)

[Out]

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

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