3.6 \(\int \frac{(A+B x+C x^2) \sqrt{d^2-e^2 x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=149 \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{3 d e^3 (d+e x)^3}+\frac{2 \sqrt{d^2-e^2 x^2} (3 C d-B e)}{e^3 (d+e x)}+\frac{(3 C d-B e) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3}-\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.188542, antiderivative size = 149, normalized size of antiderivative = 1., 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, 663, 217, 203} \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{3 d e^3 (d+e x)^3}+\frac{2 \sqrt{d^2-e^2 x^2} (3 C d-B e)}{e^3 (d+e x)}+\frac{(3 C d-B e) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3}-\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 663

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

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

Mathematica [A]  time = 0.23575, size = 114, normalized size = 0.77 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (e (A e (e x-d)-B d (5 d+7 e x))+C d \left (14 d^2+19 d e x+3 e^2 x^2\right )\right )}{d (d+e x)^2}+3 (3 C d-B e) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(C*d*(14*d^2 + 19*d*e*x + 3*e^2*x^2) + e*(A*e*(-d + e*x) - B*d*(5*d + 7*e*x))))/(d*(d +
e*x)^2) + 3*(3*C*d - B*e)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(3*e^3)

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Maple [B]  time = 0.062, size = 318, normalized size = 2.1 \begin{align*} 3\,{\frac{C}{{e}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+3\,{\frac{Cd}{{e}^{2}\sqrt{{e}^{2}}}\arctan \left ({\sqrt{{e}^{2}}x{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ) }-{\frac{B}{{e}^{4}d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-2}}+2\,{\frac{C}{{e}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{B}{d{e}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{B}{e}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{A{e}^{2}-Bde+C{d}^{2}}{3\,{e}^{6}d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}} \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)/(e*x+d)^3,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82373, size = 533, normalized size = 3.58 \begin{align*} \frac{14 \, C d^{4} - 5 \, B d^{3} e - A d^{2} e^{2} +{\left (14 \, C d^{2} e^{2} - 5 \, B d e^{3} - A e^{4}\right )} x^{2} + 2 \,{\left (14 \, C d^{3} e - 5 \, B d^{2} e^{2} - A d e^{3}\right )} x - 6 \,{\left (3 \, C d^{4} - B d^{3} e +{\left (3 \, C d^{2} e^{2} - B d e^{3}\right )} x^{2} + 2 \,{\left (3 \, C d^{3} e - B d^{2} e^{2}\right )} x\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (3 \, C d e^{2} x^{2} + 14 \, C d^{3} - 5 \, B d^{2} e - A d e^{2} +{\left (19 \, C d^{2} e - 7 \, B d e^{2} + A e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \,{\left (d e^{5} x^{2} + 2 \, d^{2} e^{4} x + d^{3} e^{3}\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)/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/3*(14*C*d^4 - 5*B*d^3*e - A*d^2*e^2 + (14*C*d^2*e^2 - 5*B*d*e^3 - A*e^4)*x^2 + 2*(14*C*d^3*e - 5*B*d^2*e^2 -
A*d*e^3)*x - 6*(3*C*d^4 - B*d^3*e + (3*C*d^2*e^2 - B*d*e^3)*x^2 + 2*(3*C*d^3*e - B*d^2*e^2)*x)*arctan(-(d - s
qrt(-e^2*x^2 + d^2))/(e*x)) + (3*C*d*e^2*x^2 + 14*C*d^3 - 5*B*d^2*e - A*d*e^2 + (19*C*d^2*e - 7*B*d*e^2 + A*e^
3)*x)*sqrt(-e^2*x^2 + d^2))/(d*e^5*x^2 + 2*d^2*e^4*x + d^3*e^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )^{3}}\, dx \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)/(e*x+d)**3,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))*(A + B*x + C*x**2)/(d + e*x)**3, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)/(e*x+d)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError