### 3.9 $$\int \frac{(A+B x+C x^2) \sqrt{d^2-e^2 x^2}}{(d+e x)^6} \, dx$$

Optimal. Leaf size=234 $-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{315 d^4 e^3 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{105 d^3 e^3 (d+e x)^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{42 d^2 e^3 (d+e x)^5}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}$

[Out]

-((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2))/(9*d*e^3*(d + e*x)^6) + (C*(d^2 - e^2*x^2)^(3/2))/(2*e^3*(d +
e*x)^5) - ((11*C*d^2 + 2*e*(2*B*d + A*e))*(d^2 - e^2*x^2)^(3/2))/(42*d^2*e^3*(d + e*x)^5) - ((11*C*d^2 + 2*e*
(2*B*d + A*e))*(d^2 - e^2*x^2)^(3/2))/(105*d^3*e^3*(d + e*x)^4) - ((11*C*d^2 + 2*e*(2*B*d + A*e))*(d^2 - e^2*x
^2)^(3/2))/(315*d^4*e^3*(d + e*x)^3)

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Rubi [A]  time = 0.264216, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {1639, 793, 659, 651} $-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{315 d^4 e^3 (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{105 d^3 e^3 (d+e x)^4}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+11 C d^2\right )}{42 d^2 e^3 (d+e x)^5}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Int[((A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2])/(d + e*x)^6,x]

[Out]

-((C*d^2 - B*d*e + A*e^2)*(d^2 - e^2*x^2)^(3/2))/(9*d*e^3*(d + e*x)^6) + (C*(d^2 - e^2*x^2)^(3/2))/(2*e^3*(d +
e*x)^5) - ((11*C*d^2 + 2*e*(2*B*d + A*e))*(d^2 - e^2*x^2)^(3/2))/(42*d^2*e^3*(d + e*x)^5) - ((11*C*d^2 + 2*e*
(2*B*d + A*e))*(d^2 - e^2*x^2)^(3/2))/(105*d^3*e^3*(d + e*x)^4) - ((11*C*d^2 + 2*e*(2*B*d + A*e))*(d^2 - e^2*x
^2)^(3/2))/(315*d^4*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)^6} \, dx &=\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}+\frac{\int \frac{\left (e^2 \left (5 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)^6} \, 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}}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}+\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^5} \, dx}{6 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}}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{42 d^2 e^3 (d+e x)^5}+\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^4} \, dx}{21 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}}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{42 d^2 e^3 (d+e x)^5}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{105 d^3 e^3 (d+e x)^4}+\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \int \frac{\sqrt{d^2-e^2 x^2}}{(d+e x)^3} \, dx}{105 d^3 e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{9 d e^3 (d+e x)^6}+\frac{C \left (d^2-e^2 x^2\right )^{3/2}}{2 e^3 (d+e x)^5}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{42 d^2 e^3 (d+e x)^5}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{105 d^3 e^3 (d+e x)^4}-\frac{\left (11 C d^2+2 e (2 B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{315 d^4 e^3 (d+e x)^3}\\ \end{align*}

Mathematica [A]  time = 0.227784, size = 144, normalized size = 0.62 $-\frac{(d-e x) \sqrt{d^2-e^2 x^2} \left (e \left (A e \left (33 d^2 e x+58 d^3+12 d e^2 x^2+2 e^3 x^3\right )+B d \left (66 d^2 e x+11 d^3+24 d e^2 x^2+4 e^3 x^3\right )\right )+C d^2 \left (24 d^2 e x+4 d^3+66 d e^2 x^2+11 e^3 x^3\right )\right )}{315 d^4 e^3 (d+e x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2])/(d + e*x)^6,x]

[Out]

-((d - e*x)*Sqrt[d^2 - e^2*x^2]*(C*d^2*(4*d^3 + 24*d^2*e*x + 66*d*e^2*x^2 + 11*e^3*x^3) + e*(A*e*(58*d^3 + 33*
d^2*e*x + 12*d*e^2*x^2 + 2*e^3*x^3) + B*d*(11*d^3 + 66*d^2*e*x + 24*d*e^2*x^2 + 4*e^3*x^3))))/(315*d^4*e^3*(d
+ e*x)^5)

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Maple [A]  time = 0.049, size = 152, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,A{e}^{5}{x}^{3}+4\,Bd{e}^{4}{x}^{3}+11\,C{d}^{2}{e}^{3}{x}^{3}+12\,Ad{e}^{4}{x}^{2}+24\,B{d}^{2}{e}^{3}{x}^{2}+66\,C{d}^{3}{e}^{2}{x}^{2}+33\,A{d}^{2}{e}^{3}x+66\,B{d}^{3}{e}^{2}x+24\,C{d}^{4}ex+58\,A{d}^{3}{e}^{2}+11\,B{d}^{4}e+4\,C{d}^{5} \right ) \left ( -ex+d \right ) }{315\, \left ( ex+d \right ) ^{5}{d}^{4}{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)^6,x)

[Out]

-1/315*(-e*x+d)*(2*A*e^5*x^3+4*B*d*e^4*x^3+11*C*d^2*e^3*x^3+12*A*d*e^4*x^2+24*B*d^2*e^3*x^2+66*C*d^3*e^2*x^2+3
3*A*d^2*e^3*x+66*B*d^3*e^2*x+24*C*d^4*e*x+58*A*d^3*e^2+11*B*d^4*e+4*C*d^5)*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^5/d^4/
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)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5496, size = 852, normalized size = 3.64 \begin{align*} -\frac{4 \, C d^{7} + 11 \, B d^{6} e + 58 \, A d^{5} e^{2} +{\left (4 \, C d^{2} e^{5} + 11 \, B d e^{6} + 58 \, A e^{7}\right )} x^{5} + 5 \,{\left (4 \, C d^{3} e^{4} + 11 \, B d^{2} e^{5} + 58 \, A d e^{6}\right )} x^{4} + 10 \,{\left (4 \, C d^{4} e^{3} + 11 \, B d^{3} e^{4} + 58 \, A d^{2} e^{5}\right )} x^{3} + 10 \,{\left (4 \, C d^{5} e^{2} + 11 \, B d^{4} e^{3} + 58 \, A d^{3} e^{4}\right )} x^{2} + 5 \,{\left (4 \, C d^{6} e + 11 \, B d^{5} e^{2} + 58 \, A d^{4} e^{3}\right )} x +{\left (4 \, C d^{6} + 11 \, B d^{5} e + 58 \, A d^{4} e^{2} -{\left (11 \, C d^{2} e^{4} + 4 \, B d e^{5} + 2 \, A e^{6}\right )} x^{4} - 5 \,{\left (11 \, C d^{3} e^{3} + 4 \, B d^{2} e^{4} + 2 \, A d e^{5}\right )} x^{3} + 21 \,{\left (2 \, C d^{4} e^{2} - 2 \, B d^{3} e^{3} - A d^{2} e^{4}\right )} x^{2} + 5 \,{\left (4 \, C d^{5} e + 11 \, B d^{4} e^{2} - 5 \, A d^{3} e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{315 \,{\left (d^{4} e^{8} x^{5} + 5 \, d^{5} e^{7} x^{4} + 10 \, d^{6} e^{6} x^{3} + 10 \, d^{7} e^{5} x^{2} + 5 \, d^{8} e^{4} x + d^{9} 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)^6,x, algorithm="fricas")

[Out]

-1/315*(4*C*d^7 + 11*B*d^6*e + 58*A*d^5*e^2 + (4*C*d^2*e^5 + 11*B*d*e^6 + 58*A*e^7)*x^5 + 5*(4*C*d^3*e^4 + 11*
B*d^2*e^5 + 58*A*d*e^6)*x^4 + 10*(4*C*d^4*e^3 + 11*B*d^3*e^4 + 58*A*d^2*e^5)*x^3 + 10*(4*C*d^5*e^2 + 11*B*d^4*
e^3 + 58*A*d^3*e^4)*x^2 + 5*(4*C*d^6*e + 11*B*d^5*e^2 + 58*A*d^4*e^3)*x + (4*C*d^6 + 11*B*d^5*e + 58*A*d^4*e^2
- (11*C*d^2*e^4 + 4*B*d*e^5 + 2*A*e^6)*x^4 - 5*(11*C*d^3*e^3 + 4*B*d^2*e^4 + 2*A*d*e^5)*x^3 + 21*(2*C*d^4*e^2
- 2*B*d^3*e^3 - A*d^2*e^4)*x^2 + 5*(4*C*d^5*e + 11*B*d^4*e^2 - 5*A*d^3*e^3)*x)*sqrt(-e^2*x^2 + d^2))/(d^4*e^8
*x^5 + 5*d^5*e^7*x^4 + 10*d^6*e^6*x^3 + 10*d^7*e^5*x^2 + 5*d^8*e^4*x + d^9*e^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**6,x)

[Out]

Timed out

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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)^6,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError