3.17 \(\int \frac{A+B x+C x^2}{(d+e x)^4 \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=234 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{105 d^4 e^3 (d+e x)}-\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{105 d^3 e^3 (d+e x)^2}-\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{70 d^2 e^3 (d+e x)^3}-\frac{\sqrt{d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3} \]

[Out]

-((C*d^2 - B*d*e + A*e^2)*Sqrt[d^2 - e^2*x^2])/(7*d*e^3*(d + e*x)^4) + (C*Sqrt[d^2 - e^2*x^2])/(2*e^3*(d + e*x
)^3) - ((13*C*d^2 + 8*B*d*e + 6*A*e^2)*Sqrt[d^2 - e^2*x^2])/(70*d^2*e^3*(d + e*x)^3) - ((13*C*d^2 + 8*B*d*e +
6*A*e^2)*Sqrt[d^2 - e^2*x^2])/(105*d^3*e^3*(d + e*x)^2) - ((13*C*d^2 + 8*B*d*e + 6*A*e^2)*Sqrt[d^2 - e^2*x^2])
/(105*d^4*e^3*(d + e*x))

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Rubi [A]  time = 0.248522, 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} \[ -\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{105 d^4 e^3 (d+e x)}-\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{105 d^3 e^3 (d+e x)^2}-\frac{\sqrt{d^2-e^2 x^2} \left (6 A e^2+8 B d e+13 C d^2\right )}{70 d^2 e^3 (d+e x)^3}-\frac{\sqrt{d^2-e^2 x^2} \left (A e^2-B d e+C d^2\right )}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/((d + e*x)^4*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])/(7*d*e^3*(d + e*x)^4) + (C*Sqrt[d^2 - e^2*x^2])/(2*e^3*(d + e*x
)^3) - ((13*C*d^2 + 8*B*d*e + 6*A*e^2)*Sqrt[d^2 - e^2*x^2])/(70*d^2*e^3*(d + e*x)^3) - ((13*C*d^2 + 8*B*d*e +
6*A*e^2)*Sqrt[d^2 - e^2*x^2])/(105*d^3*e^3*(d + e*x)^2) - ((13*C*d^2 + 8*B*d*e + 6*A*e^2)*Sqrt[d^2 - e^2*x^2])
/(105*d^4*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)^4 \sqrt{d^2-e^2 x^2}} \, dx &=\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3}+\frac{\int \frac{e^2 \left (3 C d^2+2 A e^2\right )+e^3 (C d+2 B e) x}{(d+e x)^4 \sqrt{d^2-e^2 x^2}} \, dx}{2 e^4}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \sqrt{d^2-e^2 x^2}}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3}+\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \int \frac{1}{(d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx}{14 d e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \sqrt{d^2-e^2 x^2}}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{70 d^2 e^3 (d+e x)^3}+\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \int \frac{1}{(d+e x)^2 \sqrt{d^2-e^2 x^2}} \, dx}{35 d^2 e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \sqrt{d^2-e^2 x^2}}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{70 d^2 e^3 (d+e x)^3}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{105 d^3 e^3 (d+e x)^2}+\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \int \frac{1}{(d+e x) \sqrt{d^2-e^2 x^2}} \, dx}{105 d^3 e^2}\\ &=-\frac{\left (C d^2-B d e+A e^2\right ) \sqrt{d^2-e^2 x^2}}{7 d e^3 (d+e x)^4}+\frac{C \sqrt{d^2-e^2 x^2}}{2 e^3 (d+e x)^3}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{70 d^2 e^3 (d+e x)^3}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{105 d^3 e^3 (d+e x)^2}-\frac{\left (13 C d^2+8 B d e+6 A e^2\right ) \sqrt{d^2-e^2 x^2}}{105 d^4 e^3 (d+e x)}\\ \end{align*}

Mathematica [A]  time = 0.224481, size = 139, normalized size = 0.59 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (e \left (3 A e \left (13 d^2 e x+12 d^3+8 d e^2 x^2+2 e^3 x^3\right )+B d \left (52 d^2 e x+13 d^3+32 d e^2 x^2+8 e^3 x^3\right )\right )+C d^2 \left (32 d^2 e x+8 d^3+52 d e^2 x^2+13 e^3 x^3\right )\right )}{105 d^4 e^3 (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/((d + e*x)^4*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-(Sqrt[d^2 - e^2*x^2]*(C*d^2*(8*d^3 + 32*d^2*e*x + 52*d*e^2*x^2 + 13*e^3*x^3) + e*(3*A*e*(12*d^3 + 13*d^2*e*x
+ 8*d*e^2*x^2 + 2*e^3*x^3) + B*d*(13*d^3 + 52*d^2*e*x + 32*d*e^2*x^2 + 8*e^3*x^3))))/(105*d^4*e^3*(d + e*x)^4)

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Maple [A]  time = 0.049, size = 152, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( 6\,A{e}^{5}{x}^{3}+8\,Bd{e}^{4}{x}^{3}+13\,C{d}^{2}{e}^{3}{x}^{3}+24\,Ad{e}^{4}{x}^{2}+32\,B{d}^{2}{e}^{3}{x}^{2}+52\,C{d}^{3}{e}^{2}{x}^{2}+39\,A{d}^{2}{e}^{3}x+52\,B{d}^{3}{e}^{2}x+32\,C{d}^{4}ex+36\,A{d}^{3}{e}^{2}+13\,B{d}^{4}e+8\,C{d}^{5} \right ) }{105\,{e}^{3}{d}^{4} \left ( ex+d \right ) ^{3}}{\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)^4/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/105*(-e*x+d)*(6*A*e^5*x^3+8*B*d*e^4*x^3+13*C*d^2*e^3*x^3+24*A*d*e^4*x^2+32*B*d^2*e^3*x^2+52*C*d^3*e^2*x^2+3
9*A*d^2*e^3*x+52*B*d^3*e^2*x+32*C*d^4*e*x+36*A*d^3*e^2+13*B*d^4*e+8*C*d^5)/(e*x+d)^3/d^4/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)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05572, size = 689, normalized size = 2.94 \begin{align*} -\frac{8 \, C d^{6} + 13 \, B d^{5} e + 36 \, A d^{4} e^{2} +{\left (8 \, C d^{2} e^{4} + 13 \, B d e^{5} + 36 \, A e^{6}\right )} x^{4} + 4 \,{\left (8 \, C d^{3} e^{3} + 13 \, B d^{2} e^{4} + 36 \, A d e^{5}\right )} x^{3} + 6 \,{\left (8 \, C d^{4} e^{2} + 13 \, B d^{3} e^{3} + 36 \, A d^{2} e^{4}\right )} x^{2} + 4 \,{\left (8 \, C d^{5} e + 13 \, B d^{4} e^{2} + 36 \, A d^{3} e^{3}\right )} x +{\left (8 \, C d^{5} + 13 \, B d^{4} e + 36 \, A d^{3} e^{2} +{\left (13 \, C d^{2} e^{3} + 8 \, B d e^{4} + 6 \, A e^{5}\right )} x^{3} + 4 \,{\left (13 \, C d^{3} e^{2} + 8 \, B d^{2} e^{3} + 6 \, A d e^{4}\right )} x^{2} +{\left (32 \, C d^{4} e + 52 \, B d^{3} e^{2} + 39 \, A d^{2} e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{105 \,{\left (d^{4} e^{7} x^{4} + 4 \, d^{5} e^{6} x^{3} + 6 \, d^{6} e^{5} x^{2} + 4 \, d^{7} e^{4} x + d^{8} 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)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/105*(8*C*d^6 + 13*B*d^5*e + 36*A*d^4*e^2 + (8*C*d^2*e^4 + 13*B*d*e^5 + 36*A*e^6)*x^4 + 4*(8*C*d^3*e^3 + 13*
B*d^2*e^4 + 36*A*d*e^5)*x^3 + 6*(8*C*d^4*e^2 + 13*B*d^3*e^3 + 36*A*d^2*e^4)*x^2 + 4*(8*C*d^5*e + 13*B*d^4*e^2
+ 36*A*d^3*e^3)*x + (8*C*d^5 + 13*B*d^4*e + 36*A*d^3*e^2 + (13*C*d^2*e^3 + 8*B*d*e^4 + 6*A*e^5)*x^3 + 4*(13*C*
d^3*e^2 + 8*B*d^2*e^3 + 6*A*d*e^4)*x^2 + (32*C*d^4*e + 52*B*d^3*e^2 + 39*A*d^2*e^3)*x)*sqrt(-e^2*x^2 + d^2))/(
d^4*e^7*x^4 + 4*d^5*e^6*x^3 + 6*d^6*e^5*x^2 + 4*d^7*e^4*x + d^8*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 )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(e*x+d)**4/(-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)**4), 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)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError