3.8 \(\int \frac{(A+B x+C x^2) \sqrt{d^2-e^2 x^2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=180 \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (e (2 A e+5 B d)+23 C d^2\right )}{105 d^3 e^3 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (e (2 A e+5 B d)+23 C d^2\right )}{35 d^2 e^3 (d+e x)^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{7 d e^3 (d+e x)^5}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4} \]

[Out]

-((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2))/(7*d*e^3*(d + e*x)^5) + (C*(d^2 - e^2*x^2)^(3/2))/(e^3*(d + e
*x)^4) - ((23*C*d^2 + e*(5*B*d + 2*A*e))*(d^2 - e^2*x^2)^(3/2))/(35*d^2*e^3*(d + e*x)^4) - ((23*C*d^2 + e*(5*B
*d + 2*A*e))*(d^2 - e^2*x^2)^(3/2))/(105*d^3*e^3*(d + e*x)^3)

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Rubi [A]  time = 0.210378, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {1639, 793, 659, 651} \[ -\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (e (2 A e+5 B d)+23 C d^2\right )}{105 d^3 e^3 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (e (2 A e+5 B d)+23 C d^2\right )}{35 d^2 e^3 (d+e x)^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{7 d e^3 (d+e x)^5}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2))/(7*d*e^3*(d + e*x)^5) + (C*(d^2 - e^2*x^2)^(3/2))/(e^3*(d + e
*x)^4) - ((23*C*d^2 + e*(5*B*d + 2*A*e))*(d^2 - e^2*x^2)^(3/2))/(35*d^2*e^3*(d + e*x)^4) - ((23*C*d^2 + e*(5*B
*d + 2*A*e))*(d^2 - e^2*x^2)^(3/2))/(105*d^3*e^3*(d + e*x)^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 659

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

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 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)^5} \, dx &=\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4}+\frac{\int \frac{\left (e^2 \left (4 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)^5} \, 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}}{7 d e^3 (d+e x)^5}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4}+\frac{\left (23 C d^2+e (5 B d+2 A e)\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^4} \, dx}{7 d e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{7 d e^3 (d+e x)^5}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4}-\frac{\left (23 C d^2+e (5 B d+2 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{35 d^2 e^3 (d+e x)^4}+\frac{\left (23 C d^2+e (5 B d+2 A e)\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^3} \, dx}{35 d^2 e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{7 d e^3 (d+e x)^5}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^4}-\frac{\left (23 C d^2+e (5 B d+2 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{35 d^2 e^3 (d+e x)^4}-\frac{\left (23 C d^2+e (5 B d+2 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{105 d^3 e^3 (d+e x)^3}\\ \end{align*}

Mathematica [A]  time = 0.206715, size = 109, normalized size = 0.61 \[ -\frac{(d-e x) \sqrt{d^2-e^2 x^2} \left (e \left (A e \left (23 d^2+10 d e x+2 e^2 x^2\right )+5 B d \left (d^2+5 d e x+e^2 x^2\right )\right )+C d^2 \left (2 d^2+10 d e x+23 e^2 x^2\right )\right )}{105 d^3 e^3 (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d - e*x)*Sqrt[d^2 - e^2*x^2]*(C*d^2*(2*d^2 + 10*d*e*x + 23*e^2*x^2) + e*(5*B*d*(d^2 + 5*d*e*x + e^2*x^2) +
A*e*(23*d^2 + 10*d*e*x + 2*e^2*x^2))))/(105*d^3*e^3*(d + e*x)^4)

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Maple [A]  time = 0.048, size = 116, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,A{e}^{4}{x}^{2}+5\,Bd{e}^{3}{x}^{2}+23\,C{d}^{2}{e}^{2}{x}^{2}+10\,Ad{e}^{3}x+25\,B{d}^{2}{e}^{2}x+10\,C{d}^{3}ex+23\,A{d}^{2}{e}^{2}+5\,B{d}^{3}e+2\,C{d}^{4} \right ) \left ( -ex+d \right ) }{105\, \left ( ex+d \right ) ^{4}{d}^{3}{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{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)/(e*x+d)^5,x)

[Out]

-1/105*(-e*x+d)*(2*A*e^4*x^2+5*B*d*e^3*x^2+23*C*d^2*e^2*x^2+10*A*d*e^3*x+25*B*d^2*e^2*x+10*C*d^3*e*x+23*A*d^2*
e^2+5*B*d^3*e+2*C*d^4)*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4/d^3/e^3

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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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96755, size = 678, normalized size = 3.77 \begin{align*} -\frac{2 \, C d^{6} + 5 \, B d^{5} e + 23 \, A d^{4} e^{2} +{\left (2 \, C d^{2} e^{4} + 5 \, B d e^{5} + 23 \, A e^{6}\right )} x^{4} + 4 \,{\left (2 \, C d^{3} e^{3} + 5 \, B d^{2} e^{4} + 23 \, A d e^{5}\right )} x^{3} + 6 \,{\left (2 \, C d^{4} e^{2} + 5 \, B d^{3} e^{3} + 23 \, A d^{2} e^{4}\right )} x^{2} + 4 \,{\left (2 \, C d^{5} e + 5 \, B d^{4} e^{2} + 23 \, A d^{3} e^{3}\right )} x +{\left (2 \, C d^{5} + 5 \, B d^{4} e + 23 \, A d^{3} e^{2} -{\left (23 \, C d^{2} e^{3} + 5 \, B d e^{4} + 2 \, A e^{5}\right )} x^{3} +{\left (13 \, C d^{3} e^{2} - 20 \, B d^{2} e^{3} - 8 \, A d e^{4}\right )} x^{2} +{\left (8 \, C d^{4} e + 20 \, B d^{3} e^{2} - 13 \, A d^{2} e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{105 \,{\left (d^{3} e^{7} x^{4} + 4 \, d^{4} e^{6} x^{3} + 6 \, d^{5} e^{5} x^{2} + 4 \, d^{6} e^{4} x + d^{7} 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)^5,x, algorithm="fricas")

[Out]

-1/105*(2*C*d^6 + 5*B*d^5*e + 23*A*d^4*e^2 + (2*C*d^2*e^4 + 5*B*d*e^5 + 23*A*e^6)*x^4 + 4*(2*C*d^3*e^3 + 5*B*d
^2*e^4 + 23*A*d*e^5)*x^3 + 6*(2*C*d^4*e^2 + 5*B*d^3*e^3 + 23*A*d^2*e^4)*x^2 + 4*(2*C*d^5*e + 5*B*d^4*e^2 + 23*
A*d^3*e^3)*x + (2*C*d^5 + 5*B*d^4*e + 23*A*d^3*e^2 - (23*C*d^2*e^3 + 5*B*d*e^4 + 2*A*e^5)*x^3 + (13*C*d^3*e^2
- 20*B*d^2*e^3 - 8*A*d*e^4)*x^2 + (8*C*d^4*e + 20*B*d^3*e^2 - 13*A*d^2*e^3)*x)*sqrt(-e^2*x^2 + d^2))/(d^3*e^7*
x^4 + 4*d^4*e^6*x^3 + 6*d^5*e^5*x^2 + 4*d^6*e^4*x + d^7*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 )^{5}}\, 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)**5,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))*(A + B*x + C*x**2)/(d + e*x)**5, 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)^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError