3.173 \(\int \frac{x^2}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac{d}{5 e^3 (d+e x)^2 \sqrt{d^2-e^2 x^2}}+\frac{7}{15 e^3 (d+e x) \sqrt{d^2-e^2 x^2}}+\frac{x}{15 d^2 e^2 \sqrt{d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.141929, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {852, 1635, 778, 191} \[ -\frac{d (d-e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{7 (d-e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{x}{15 d^2 e^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0710843, size = 70, normalized size = 0.79 \[ \frac{\sqrt{d^2-e^2 x^2} \left (8 d^2 e x+4 d^3+2 d e^2 x^2+e^3 x^3\right )}{15 d^2 e^3 (d-e x) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(4*d^3 + 8*d^2*e*x + 2*d*e^2*x^2 + e^3*x^3))/(15*d^2*e^3*(d - e*x)*(d + e*x)^3)

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Maple [A]  time = 0.05, size = 65, normalized size = 0.7 \begin{align*}{\frac{ \left ( -ex+d \right ) \left ({e}^{3}{x}^{3}+2\,d{e}^{2}{x}^{2}+8\,x{d}^{2}e+4\,{d}^{3} \right ) }{ \left ( 15\,ex+15\,d \right ){d}^{2}{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x)

[Out]

1/15*(-e*x+d)*(e^3*x^3+2*d*e^2*x^2+8*d^2*e*x+4*d^3)/(e*x+d)/d^2/e^3/(-e^2*x^2+d^2)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68399, size = 234, normalized size = 2.63 \begin{align*} \frac{4 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 8 \, d^{3} e x - 4 \, d^{4} -{\left (e^{3} x^{3} + 2 \, d e^{2} x^{2} + 8 \, d^{2} e x + 4 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{2} e^{7} x^{4} + 2 \, d^{3} e^{6} x^{3} - 2 \, d^{5} e^{4} x - d^{6} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(4*e^4*x^4 + 8*d*e^3*x^3 - 8*d^3*e*x - 4*d^4 - (e^3*x^3 + 2*d*e^2*x^2 + 8*d^2*e*x + 4*d^3)*sqrt(-e^2*x^2
+ d^2))/(d^2*e^7*x^4 + 2*d^3*e^6*x^3 - 2*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{x^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{3}{2}} \left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(e*x+d)**2/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(x**2/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x