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

Optimal. Leaf size=89 \[ \frac{(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{4 x}{15 d^3 e \sqrt{d^2-e^2 x^2}} \]

[Out]

(d + e*x)^2/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x))/(15*d*e^2*(d^2 - e^2*x^2)^(3/2)) - (4*x)/(15*d^3*e*S
qrt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.033264, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {789, 639, 191} \[ \frac{(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{4 x}{15 d^3 e \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)^2/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x))/(15*d*e^2*(d^2 - e^2*x^2)^(3/2)) - (4*x)/(15*d^3*e*S
qrt[d^2 - e^2*x^2])

Rule 789

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*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

Rule 639

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

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

Mathematica [A]  time = 0.0561763, size = 62, normalized size = 0.7 \[ \frac{-2 d^2 e x+d^3+8 d e^2 x^2-4 e^3 x^3}{15 d^3 e^2 (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 64, normalized size = 0.7 \begin{align*}{\frac{ \left ( -ex+d \right ) \left ( ex+d \right ) ^{3} \left ( -4\,{e}^{3}{x}^{3}+8\,d{e}^{2}{x}^{2}-2\,{d}^{2}ex+{d}^{3} \right ) }{15\,{d}^{3}{e}^{2}} \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*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

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

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Maxima [A]  time = 1.01071, size = 147, normalized size = 1.65 \begin{align*} \frac{x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{2 \, d x}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{d^{2}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{2 \, x}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d e} - \frac{4 \, x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{3} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.87344, size = 228, normalized size = 2.56 \begin{align*} \frac{e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4} +{\left (4 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 2 \, d^{2} e x - d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{3} e^{6} x^{4} - 2 \, d^{4} e^{5} x^{3} + 2 \, d^{6} e^{3} x - d^{7} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(e^4*x^4 - 2*d*e^3*x^3 + 2*d^3*e*x - d^4 + (4*e^3*x^3 - 8*d*e^2*x^2 + 2*d^2*e*x - d^3)*sqrt(-e^2*x^2 + d^
2))/(d^3*e^6*x^4 - 2*d^4*e^5*x^3 + 2*d^6*e^3*x - d^7*e^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

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

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Giac [A]  time = 1.17491, size = 86, normalized size = 0.97 \begin{align*} \frac{{\left ({\left (2 \, x{\left (\frac{2 \, x^{2} e^{3}}{d^{3}} - \frac{5 \, e}{d}\right )} - 5\right )} x^{2} - d^{2} e^{\left (-2\right )}\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*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

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