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

[Out]

-1/7*(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(3/2)/d/e^3/(e*x+d)^5+C*(-e^2*x^2+d^2)^(3/2)/e^3/(e*x+d)^4-1/35*(23*C*
d^2+e*(2*A*e+5*B*d))*(-e^2*x^2+d^2)^(3/2)/d^2/e^3/(e*x+d)^4-1/105*(23*C*d^2+e*(2*A*e+5*B*d))*(-e^2*x^2+d^2)^(3
/2)/d^3/e^3/(e*x+d)^3

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Rubi [A]  time = 0.21, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 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]

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 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)^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*}

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Mathematica [A]  time = 0.20, 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]

-1/105*((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))))/(d^3*e^3*(d + e*x)^4)

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fricas [A]  time = 0.92, size = 320, normalized size = 1.78 \[ -\frac {2 \, C d^{6} + 5 \, B d^{5} e + 23 \, A d^{4} e^{2} + {\left (2 \, C d^{2} e^{4} + 5 \, B d e^{5} + 23 \, A e^{6}\right )} x^{4} + 4 \, {\left (2 \, C d^{3} e^{3} + 5 \, B d^{2} e^{4} + 23 \, A d e^{5}\right )} x^{3} + 6 \, {\left (2 \, C d^{4} e^{2} + 5 \, B d^{3} e^{3} + 23 \, A d^{2} e^{4}\right )} x^{2} + 4 \, {\left (2 \, C d^{5} e + 5 \, B d^{4} e^{2} + 23 \, A d^{3} e^{3}\right )} x + {\left (2 \, C d^{5} + 5 \, B d^{4} e + 23 \, A d^{3} e^{2} - {\left (23 \, C d^{2} e^{3} + 5 \, B d e^{4} + 2 \, A e^{5}\right )} x^{3} + {\left (13 \, C d^{3} e^{2} - 20 \, B d^{2} e^{3} - 8 \, A d e^{4}\right )} x^{2} + {\left (8 \, C d^{4} e + 20 \, B d^{3} e^{2} - 13 \, A d^{2} e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{105 \, {\left (d^{3} e^{7} x^{4} + 4 \, d^{4} e^{6} x^{3} + 6 \, d^{5} e^{5} x^{2} + 4 \, d^{6} e^{4} x + d^{7} e^{3}\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, choosing root of [1,0,%%%{2,[2,0]%%%},0,%%%{1,[4,0]%%%}+%%%{-4,[2,1]%%%}+%%%{4,[0,2]%%%}]
at parameters values [86,-97]Limit: Max order reached or unable to make series expansion Error: Bad Argument
Value

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maple [A]  time = 0.01, size = 116, normalized size = 0.64 \[ -\frac {\left (-e x +d \right ) \left (2 A \,e^{4} x^{2}+5 B d \,e^{3} x^{2}+23 C \,d^{2} e^{2} x^{2}+10 A d \,e^{3} x +25 B \,d^{2} e^{2} x +10 C \,d^{3} e x +23 A \,d^{2} e^{2}+5 B \,d^{3} e +2 C \,d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{105 \left (e x +d \right )^{4} d^{3} e^{3}} \]

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 [B]  time = 0.54, size = 945, normalized size = 5.25 \[ -\frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{7 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{35 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{105 \, {\left (d^{2} e^{5} x^{2} + 2 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{105 \, {\left (d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{7 \, {\left (e^{6} x^{4} + 4 \, d e^{5} x^{3} + 6 \, d^{2} e^{4} x^{2} + 4 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d}{35 \, {\left (d e^{5} x^{3} + 3 \, d^{2} e^{4} x^{2} + 3 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{105 \, {\left (d^{2} e^{4} x^{2} + 2 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{105 \, {\left (d^{3} e^{3} x + d^{4} e^{2}\right )}} + \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{15 \, {\left (d e^{5} x^{2} + 2 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{15 \, {\left (d^{2} e^{4} x + d^{3} e^{3}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{7 \, {\left (e^{5} x^{4} + 4 \, d e^{4} x^{3} + 6 \, d^{2} e^{3} x^{2} + 4 \, d^{3} e^{2} x + d^{4} e\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{35 \, {\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{105 \, {\left (d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{105 \, {\left (d^{3} e^{2} x + d^{4} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B}{5 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{15 \, {\left (d e^{4} x^{2} + 2 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{15 \, {\left (d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C}{3 \, {\left (d e^{4} x + d^{2} 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)^5,x, algorithm="maxima")

[Out]

-2/7*sqrt(-e^2*x^2 + d^2)*C*d^2/(e^7*x^4 + 4*d*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x + d^4*e^3) + 1/35*sqrt(-e
^2*x^2 + d^2)*C*d^2/(d*e^6*x^3 + 3*d^2*e^5*x^2 + 3*d^3*e^4*x + d^4*e^3) + 2/105*sqrt(-e^2*x^2 + d^2)*C*d^2/(d^
2*e^5*x^2 + 2*d^3*e^4*x + d^4*e^3) + 2/105*sqrt(-e^2*x^2 + d^2)*C*d^2/(d^3*e^4*x + d^4*e^3) + 2/7*sqrt(-e^2*x^
2 + d^2)*B*d/(e^6*x^4 + 4*d*e^5*x^3 + 6*d^2*e^4*x^2 + 4*d^3*e^3*x + d^4*e^2) - 1/35*sqrt(-e^2*x^2 + d^2)*B*d/(
d*e^5*x^3 + 3*d^2*e^4*x^2 + 3*d^3*e^3*x + d^4*e^2) - 2/105*sqrt(-e^2*x^2 + d^2)*B*d/(d^2*e^4*x^2 + 2*d^3*e^3*x
+ d^4*e^2) - 2/105*sqrt(-e^2*x^2 + d^2)*B*d/(d^3*e^3*x + d^4*e^2) + 4/5*sqrt(-e^2*x^2 + d^2)*C*d/(e^6*x^3 + 3
*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) - 2/15*sqrt(-e^2*x^2 + d^2)*C*d/(d*e^5*x^2 + 2*d^2*e^4*x + d^3*e^3) - 2/15
*sqrt(-e^2*x^2 + d^2)*C*d/(d^2*e^4*x + d^3*e^3) - 2/7*sqrt(-e^2*x^2 + d^2)*A/(e^5*x^4 + 4*d*e^4*x^3 + 6*d^2*e^
3*x^2 + 4*d^3*e^2*x + d^4*e) + 1/35*sqrt(-e^2*x^2 + d^2)*A/(d*e^4*x^3 + 3*d^2*e^3*x^2 + 3*d^3*e^2*x + d^4*e) +
2/105*sqrt(-e^2*x^2 + d^2)*A/(d^2*e^3*x^2 + 2*d^3*e^2*x + d^4*e) + 2/105*sqrt(-e^2*x^2 + d^2)*A/(d^3*e^2*x +
d^4*e) - 2/5*sqrt(-e^2*x^2 + d^2)*B/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2) + 1/15*sqrt(-e^2*x^2 + d^2
)*B/(d*e^4*x^2 + 2*d^2*e^3*x + d^3*e^2) + 1/15*sqrt(-e^2*x^2 + d^2)*B/(d^2*e^3*x + d^3*e^2) - 2/3*sqrt(-e^2*x^
2 + d^2)*C/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) + 1/3*sqrt(-e^2*x^2 + d^2)*C/(d*e^4*x + d^2*e^3)

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mupad [B]  time = 4.67, size = 601, normalized size = 3.34 \[ \frac {B\,\sqrt {d^2-e^2\,x^2}}{21\,\left (d^3\,e^2+x\,d^2\,e^3\right )}-\frac {3\,B\,\sqrt {d^2-e^2\,x^2}}{7\,\left (d^3\,e^2+3\,d^2\,e^3\,x+3\,d\,e^4\,x^2+e^5\,x^3\right )}+\frac {2\,A\,\sqrt {d^2-e^2\,x^2}}{105\,\left (d^4\,e+2\,d^3\,e^2\,x+d^2\,e^3\,x^2\right )}+\frac {B\,\sqrt {d^2-e^2\,x^2}}{21\,\left (d^3\,e^2+2\,d^2\,e^3\,x+d\,e^4\,x^2\right )}-\frac {82\,C\,\sqrt {d^2-e^2\,x^2}}{105\,\left (d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2\right )}+\frac {2\,A\,\sqrt {d^2-e^2\,x^2}}{105\,\left (d^4\,e+x\,d^3\,e^2\right )}+\frac {23\,C\,\sqrt {d^2-e^2\,x^2}}{105\,\left (d^2\,e^3+x\,d\,e^4\right )}-\frac {2\,A\,\sqrt {d^2-e^2\,x^2}}{7\,\left (d^4\,e+4\,d^3\,e^2\,x+6\,d^2\,e^3\,x^2+4\,d\,e^4\,x^3+e^5\,x^4\right )}+\frac {A\,\sqrt {d^2-e^2\,x^2}}{35\,\left (d^4\,e+3\,d^3\,e^2\,x+3\,d^2\,e^3\,x^2+d\,e^4\,x^3\right )}-\frac {2\,C\,d^2\,\sqrt {d^2-e^2\,x^2}}{7\,\left (d^4\,e^3+4\,d^3\,e^4\,x+6\,d^2\,e^5\,x^2+4\,d\,e^6\,x^3+e^7\,x^4\right )}+\frac {2\,B\,d\,\sqrt {d^2-e^2\,x^2}}{7\,\left (d^4\,e^2+4\,d^3\,e^3\,x+6\,d^2\,e^4\,x^2+4\,d\,e^5\,x^3+e^6\,x^4\right )}+\frac {29\,C\,d\,\sqrt {d^2-e^2\,x^2}}{35\,\left (d^3\,e^3+3\,d^2\,e^4\,x+3\,d\,e^5\,x^2+e^6\,x^3\right )} \]

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)^5,x)

[Out]

(B*(d^2 - e^2*x^2)^(1/2))/(21*(d^3*e^2 + d^2*e^3*x)) - (3*B*(d^2 - e^2*x^2)^(1/2))/(7*(d^3*e^2 + e^5*x^3 + 3*d
^2*e^3*x + 3*d*e^4*x^2)) + (2*A*(d^2 - e^2*x^2)^(1/2))/(105*(d^4*e + 2*d^3*e^2*x + d^2*e^3*x^2)) + (B*(d^2 - e
^2*x^2)^(1/2))/(21*(d^3*e^2 + 2*d^2*e^3*x + d*e^4*x^2)) - (82*C*(d^2 - e^2*x^2)^(1/2))/(105*(d^2*e^3 + e^5*x^2
+ 2*d*e^4*x)) + (2*A*(d^2 - e^2*x^2)^(1/2))/(105*(d^4*e + d^3*e^2*x)) + (23*C*(d^2 - e^2*x^2)^(1/2))/(105*(d^
2*e^3 + d*e^4*x)) - (2*A*(d^2 - e^2*x^2)^(1/2))/(7*(d^4*e + e^5*x^4 + 4*d^3*e^2*x + 4*d*e^4*x^3 + 6*d^2*e^3*x^
2)) + (A*(d^2 - e^2*x^2)^(1/2))/(35*(d^4*e + 3*d^3*e^2*x + d*e^4*x^3 + 3*d^2*e^3*x^2)) - (2*C*d^2*(d^2 - e^2*x
^2)^(1/2))/(7*(d^4*e^3 + e^7*x^4 + 4*d^3*e^4*x + 4*d*e^6*x^3 + 6*d^2*e^5*x^2)) + (2*B*d*(d^2 - e^2*x^2)^(1/2))
/(7*(d^4*e^2 + e^6*x^4 + 4*d^3*e^3*x + 4*d*e^5*x^3 + 6*d^2*e^4*x^2)) + (29*C*d*(d^2 - e^2*x^2)^(1/2))/(35*(d^3
*e^3 + e^6*x^3 + 3*d^2*e^4*x + 3*d*e^5*x^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (A + B x + C x^{2}\right )}{\left (d + e x\right )^{5}}\, dx \]

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)**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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