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

Optimal. Leaf size=180 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^2 e^3 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{5 d e^3 (d+e x)^3}-\frac {\sqrt {d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^3 e^3 (d+e x)}+\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2} \]

[Out]

-1/5*(A*e^2-B*d*e+C*d^2)*(-e^2*x^2+d^2)^(1/2)/d/e^3/(e*x+d)^3+C*(-e^2*x^2+d^2)^(1/2)/e^3/(e*x+d)^2-1/15*(7*C*d
^2+e*(2*A*e+3*B*d))*(-e^2*x^2+d^2)^(1/2)/d^2/e^3/(e*x+d)^2-1/15*(7*C*d^2+e*(2*A*e+3*B*d))*(-e^2*x^2+d^2)^(1/2)
/d^3/e^3/(e*x+d)

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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} \[ -\frac {\sqrt {d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^3 e^3 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^2 e^3 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{5 d e^3 (d+e x)^3}+\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C*d^2 - B*d*e + A*e^2)*Sqrt[d^2 - e^2*x^2])/(5*d*e^3*(d + e*x)^3) + (C*Sqrt[d^2 - e^2*x^2])/(e^3*(d + e*x)^
2) - ((7*C*d^2 + e*(3*B*d + 2*A*e))*Sqrt[d^2 - e^2*x^2])/(15*d^2*e^3*(d + e*x)^2) - ((7*C*d^2 + e*(3*B*d + 2*A
*e))*Sqrt[d^2 - e^2*x^2])/(15*d^3*e^3*(d + e*x))

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 {A+B x+C x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}+\frac {\int \frac {e^2 \left (2 C d^2+A e^2\right )+e^3 (C d+B e) x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac {\left (C d^2-B d e+A e^2\right ) \sqrt {d^2-e^2 x^2}}{5 d e^3 (d+e x)^3}+\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}+\frac {\left (7 C d^2+e (3 B d+2 A e)\right ) \int \frac {1}{(d+e x)^2 \sqrt {d^2-e^2 x^2}} \, dx}{5 d e^2}\\ &=-\frac {\left (C d^2-B d e+A e^2\right ) \sqrt {d^2-e^2 x^2}}{5 d e^3 (d+e x)^3}+\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}-\frac {\left (7 C d^2+e (3 B d+2 A e)\right ) \sqrt {d^2-e^2 x^2}}{15 d^2 e^3 (d+e x)^2}+\frac {\left (7 C d^2+e (3 B d+2 A e)\right ) \int \frac {1}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx}{15 d^2 e^2}\\ &=-\frac {\left (C d^2-B d e+A e^2\right ) \sqrt {d^2-e^2 x^2}}{5 d e^3 (d+e x)^3}+\frac {C \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}-\frac {\left (7 C d^2+e (3 B d+2 A e)\right ) \sqrt {d^2-e^2 x^2}}{15 d^2 e^3 (d+e x)^2}-\frac {\left (7 C d^2+e (3 B d+2 A e)\right ) \sqrt {d^2-e^2 x^2}}{15 d^3 e^3 (d+e x)}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 103, normalized size = 0.57 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (e \left (A e \left (7 d^2+6 d e x+2 e^2 x^2\right )+3 B d \left (d^2+3 d e x+e^2 x^2\right )\right )+C d^2 \left (2 d^2+6 d e x+7 e^2 x^2\right )\right )}{15 d^3 e^3 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[d^2 - e^2*x^2]*(C*d^2*(2*d^2 + 6*d*e*x + 7*e^2*x^2) + e*(3*B*d*(d^2 + 3*d*e*x + e^2*x^2) + A*e*(7*
d^2 + 6*d*e*x + 2*e^2*x^2))))/(d^3*e^3*(d + e*x)^3)

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fricas [A]  time = 0.79, size = 244, normalized size = 1.36 \[ -\frac {2 \, C d^{5} + 3 \, B d^{4} e + 7 \, A d^{3} e^{2} + {\left (2 \, C d^{2} e^{3} + 3 \, B d e^{4} + 7 \, A e^{5}\right )} x^{3} + 3 \, {\left (2 \, C d^{3} e^{2} + 3 \, B d^{2} e^{3} + 7 \, A d e^{4}\right )} x^{2} + 3 \, {\left (2 \, C d^{4} e + 3 \, B d^{3} e^{2} + 7 \, A d^{2} e^{3}\right )} x + {\left (2 \, C d^{4} + 3 \, B d^{3} e + 7 \, A d^{2} e^{2} + {\left (7 \, C d^{2} e^{2} + 3 \, B d e^{3} + 2 \, A e^{4}\right )} x^{2} + 3 \, {\left (2 \, C d^{3} e + 3 \, B d^{2} e^{2} + 2 \, A d e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{6} x^{3} + 3 \, d^{4} e^{5} x^{2} + 3 \, d^{5} e^{4} x + d^{6} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-2*A*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*
exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^2+7*A*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^
2*exp(1)^6*exp(2)^3-2*A*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)-2*B*d*e
xp(1)*exp(2)^5+2*C*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^2+5*A*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^4*exp(2)^4-11/2*A*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))*exp(1)^4*exp(2)^4/x/exp(2)+A*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^2/x
/exp(2)-5*B*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^5*exp(2)^3-2*B*d*(-1/2*(-2*
d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^9*exp(2)-2*B*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)*exp(2)^5-3*B*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*
exp(1)^3*exp(2)^4+3*C*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^3+C*d^
2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^5-A*exp(1)^6*exp(2)^3-B*d*exp(1)^5*exp(
2)^3+3*C*d^2*exp(1)^4*exp(2)^3+4*A*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^6+4*A*
exp(2)^6+6*C*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)+1/2*C*d^2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^5/x/exp(2)+5/2*B*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*e
xp(1)^3*exp(2)^4/x/exp(2)+2*B*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^7*exp(2)^2/x/exp(2)-5*C*d^2
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^2/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^2/(-d^3*exp(1)^9+2*d^3*
exp(1)^5*exp(2)^2-d^3*exp(1)*exp(2)^4)+1/2*(-2*A*exp(1)^4*exp(2)^3+6*B*d*exp(1)^3*exp(2)^3-2*C*d^2*exp(2)^4-4*
A*exp(2)^5-4*C*d^2*exp(1)^6*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp
(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(d^3*exp(1)^9-2*d^3*exp(1)^5*exp(2)^2+d^3*exp(1)*exp(2)^4)

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maple [A]  time = 0.01, size = 116, normalized size = 0.64 \[ -\frac {\left (-e x +d \right ) \left (2 A \,e^{4} x^{2}+3 B d \,e^{3} x^{2}+7 C \,d^{2} e^{2} x^{2}+6 A d \,e^{3} x +9 B \,d^{2} e^{2} x +6 C \,d^{3} e x +7 A \,d^{2} e^{2}+3 B \,d^{3} e +2 C \,d^{4}\right )}{15 \left (e x +d \right )^{2} \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-e*x+d)*(2*A*e^4*x^2+3*B*d*e^3*x^2+7*C*d^2*e^2*x^2+6*A*d*e^3*x+9*B*d^2*e^2*x+6*C*d^3*e*x+7*A*d^2*e^2+3*
B*d^3*e+2*C*d^4)/(e*x+d)^2/d^3/e^3/(-e^2*x^2+d^2)^(1/2)

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maxima [B]  time = 1.02, size = 608, normalized size = 3.38 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{5 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{15 \, {\left (d^{2} e^{5} x^{2} + 2 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{15 \, {\left (d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d}{5 \, {\left (d e^{5} x^{3} + 3 \, d^{2} e^{4} x^{2} + 3 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{15 \, {\left (d^{2} e^{4} x^{2} + 2 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} B d}{15 \, {\left (d^{3} e^{3} x + d^{4} e^{2}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{3 \, {\left (d e^{5} x^{2} + 2 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d}{3 \, {\left (d^{2} e^{4} x + d^{3} e^{3}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{5 \, {\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{15 \, {\left (d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} A}{15 \, {\left (d^{3} e^{2} x + d^{4} e\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{3 \, {\left (d e^{4} x^{2} + 2 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B}{3 \, {\left (d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} C}{d e^{4} x + d^{2} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/5*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) - 2/15*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) - 2/15*sqrt(-e^2*x^2 + d^2)*C*d^2/(d^3*e^4*x + d^4*e^3) + 1/5*sq
rt(-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) + 2/15*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) + 2/15*sqrt(-e^2*x^2 + d^2)*B*d/(d^3*e^3*x + d^4*e^2) + 2/3*sqrt(-e^2*x^2 +
d^2)*C*d/(d*e^5*x^2 + 2*d^2*e^4*x + d^3*e^3) + 2/3*sqrt(-e^2*x^2 + d^2)*C*d/(d^2*e^4*x + d^3*e^3) - 1/5*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) - 2/15*sqrt(-e^2*x^2 + d^2)*A/(d^2*e^3*x^2
+ 2*d^3*e^2*x + d^4*e) - 2/15*sqrt(-e^2*x^2 + d^2)*A/(d^3*e^2*x + d^4*e) - 1/3*sqrt(-e^2*x^2 + d^2)*B/(d*e^4*
x^2 + 2*d^2*e^3*x + d^3*e^2) - 1/3*sqrt(-e^2*x^2 + d^2)*B/(d^2*e^3*x + d^3*e^2) - sqrt(-e^2*x^2 + d^2)*C/(d*e^
4*x + d^2*e^3)

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mupad [B]  time = 3.80, size = 109, normalized size = 0.61 \[ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,C\,d^4+6\,C\,d^3\,e\,x+3\,B\,d^3\,e+7\,C\,d^2\,e^2\,x^2+9\,B\,d^2\,e^2\,x+7\,A\,d^2\,e^2+3\,B\,d\,e^3\,x^2+6\,A\,d\,e^3\,x+2\,A\,e^4\,x^2\right )}{15\,d^3\,e^3\,{\left (d+e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d^2 - e^2*x^2)^(1/2)*(2*C*d^4 + 7*A*d^2*e^2 + 2*A*e^4*x^2 + 3*B*d^3*e + 7*C*d^2*e^2*x^2 + 6*A*d*e^3*x + 6*C
*d^3*e*x + 9*B*d^2*e^2*x + 3*B*d*e^3*x^2))/(15*d^3*e^3*(d + e*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x + C x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

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

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