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

Optimal. Leaf size=99 \[ \frac{d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-2 e x}{5 d e^4 \sqrt{d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.204107, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {852, 1635, 637} \[ \frac{d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-2 e x}{5 d e^4 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(d - e*x)^2)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (4*d*(d - e*x))/(5*e^4*(d^2 - e^2*x^2)^(3/2)) + (5*d - 2*e*x
)/(5*d*e^4*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 637

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

Rubi steps

\begin{align*} \int \frac{x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac{x^3 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d-e x) \left (-\frac{2 d^3}{e^3}+\frac{5 d^2 x}{e^2}-\frac{5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{-\frac{6 d^3}{e^3}+\frac{15 d^2 x}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{5 d-2 e x}{5 d e^4 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0867032, size = 70, normalized size = 0.71 \[ \frac{\sqrt{d^2-e^2 x^2} \left (4 d^2 e x+2 d^3+d e^2 x^2-2 e^3 x^3\right )}{5 d e^4 (d-e x) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 65, normalized size = 0.7 \begin{align*}{\frac{ \left ( -ex+d \right ) \left ( -2\,{e}^{3}{x}^{3}+d{e}^{2}{x}^{2}+4\,{d}^{2}ex+2\,{d}^{3} \right ) }{ \left ( 5\,ex+5\,d \right ) d{e}^{4}} \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^3/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x)

[Out]

1/5*(-e*x+d)*(-2*e^3*x^3+d*e^2*x^2+4*d^2*e*x+2*d^3)/(e*x+d)/d/e^4/(-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^3/(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 = 1.58563, size = 230, normalized size = 2.32 \begin{align*} \frac{2 \, e^{4} x^{4} + 4 \, d e^{3} x^{3} - 4 \, d^{3} e x - 2 \, d^{4} +{\left (2 \, e^{3} x^{3} - d e^{2} x^{2} - 4 \, d^{2} e x - 2 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (d e^{8} x^{4} + 2 \, d^{2} e^{7} x^{3} - 2 \, d^{4} e^{5} x - d^{5} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\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**3/(e*x+d)**2/(-e**2*x**2+d**2)**(3/2),x)

[Out]

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

[Out]

sage0*x