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

[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))

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Rubi [A]  time = 0.205141, 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} \[ -\frac{\sqrt{d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^3 e^3 (d+e x)}-\frac{\sqrt{d^2-e^2 x^2} \left (e (2 A e+3 B d)+7 C d^2\right )}{15 d^2 e^3 (d+e x)^2}-\frac{\sqrt{d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{5 d e^3 (d+e x)^3}+\frac{C \sqrt{d^2-e^2 x^2}}{e^3 (d+e x)^2} \]

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 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{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*}

Mathematica [A]  time = 0.19948, 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]

-(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))))/(15*d^3*e^3*(d + e*x)^3)

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Maple [A]  time = 0.049, size = 116, normalized size = 0.6 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( 2\,A{e}^{4}{x}^{2}+3\,Bd{e}^{3}{x}^{2}+7\,C{d}^{2}{e}^{2}{x}^{2}+6\,Ad{e}^{3}x+9\,B{d}^{2}{e}^{2}x+6\,C{d}^{3}ex+7\,A{d}^{2}{e}^{2}+3\,B{d}^{3}e+2\,C{d}^{4} \right ) }{15\,{e}^{3}{d}^{3} \left ( ex+d \right ) ^{2}}{\frac{1}{\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*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 [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*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78098, size = 510, normalized size = 2.83 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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

[Out]

Exception raised: NotImplementedError