3.47 \(\int \frac{x^2 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=87 \[ \frac{d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{7 (d+e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{x}{15 d^2 e^2 \sqrt{d^2-e^2 x^2}} \]

[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*S
qrt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.120037, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {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)^(7/2),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*S
qrt[d^2 - e^2*x^2])

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 )^{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 = 0.0600976, size = 63, normalized size = 0.72 \[ \frac{8 d^2 e x-4 d^3-2 d e^2 x^2+e^3 x^3}{15 d^2 e^3 (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(-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)^2*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.05, size = 66, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( ex+d \right ) ^{3} \left ( -{e}^{3}{x}^{3}+2\,d{e}^{2}{x}^{2}-8\,{d}^{2}ex+4\,{d}^{3} \right ) }{15\,{d}^{2}{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{7}{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)^(7/2),x)

[Out]

-1/15*(-e*x+d)*(e*x+d)^3*(-e^3*x^3+2*d*e^2*x^2-8*d^2*e*x+4*d^3)/d^2/e^3/(-e^2*x^2+d^2)^(7/2)

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Maxima [A]  time = 1.00427, size = 177, normalized size = 2.03 \begin{align*} \frac{x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{2 \, d x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} - \frac{d^{2} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{4 \, d^{3}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{3}} + \frac{x}{30 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{2}} + \frac{x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{2} e^{2}} \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)^(7/2),x, algorithm="maxima")

[Out]

1/2*x^3/(-e^2*x^2 + d^2)^(5/2) + 2/3*d*x^2/((-e^2*x^2 + d^2)^(5/2)*e) - 1/10*d^2*x/((-e^2*x^2 + d^2)^(5/2)*e^2
) - 4/15*d^3/((-e^2*x^2 + d^2)^(5/2)*e^3) + 1/30*x/((-e^2*x^2 + d^2)^(3/2)*e^2) + 1/15*x/(sqrt(-e^2*x^2 + d^2)
*d^2*e^2)

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Fricas [A]  time = 1.78998, size = 235, normalized size = 2.7 \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)^(7/2),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 (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{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)**(7/2),x)

[Out]

Integral(x**2*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.15725, size = 82, normalized size = 0.94 \begin{align*} \frac{{\left (4 \, d^{3} e^{\left (-3\right )} -{\left (x{\left (\frac{x^{2} e^{2}}{d^{2}} + 5\right )} + 10 \, d e^{\left (-1\right )}\right )} x^{2}\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \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)^(7/2),x, algorithm="giac")

[Out]

1/15*(4*d^3*e^(-3) - (x*(x^2*e^2/d^2 + 5) + 10*d*e^(-1))*x^2)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3