3.142 \(\int \frac{x}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{2 x}{15 d^4 e \sqrt{d^2-e^2 x^2}}+\frac{x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0287794, antiderivative size = 85, 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 = {793, 192, 191} \[ \frac{2 x}{15 d^4 e \sqrt{d^2-e^2 x^2}}+\frac{x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

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

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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[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}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac{1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 e}\\ &=\frac{x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 \int \frac{1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2 e}\\ &=\frac{x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac{1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 x}{15 d^4 e \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0583992, size = 82, normalized size = 0.96 \[ \frac{\sqrt{d^2-e^2 x^2} \left (3 d^2 e^2 x^2+3 d^3 e x+3 d^4-2 d e^3 x^3-2 e^4 x^4\right )}{15 d^4 e^2 (d-e x)^2 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.05, size = 70, normalized size = 0.8 \begin{align*}{\frac{ \left ( -ex+d \right ) \left ( -2\,{e}^{4}{x}^{4}-2\,{e}^{3}{x}^{3}d+3\,{x}^{2}{d}^{2}{e}^{2}+3\,x{d}^{3}e+3\,{d}^{4} \right ) }{15\,{d}^{4}{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

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

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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/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.7433, size = 339, normalized size = 3.99 \begin{align*} \frac{3 \, e^{5} x^{5} + 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} - 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + 3 \, d^{5} -{\left (2 \, e^{4} x^{4} + 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} - 3 \, d^{3} e x - 3 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{4} e^{7} x^{5} + d^{5} e^{6} x^{4} - 2 \, d^{6} e^{5} x^{3} - 2 \, d^{7} e^{4} x^{2} + d^{8} e^{3} x + d^{9} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

1/15*(3*e^5*x^5 + 3*d*e^4*x^4 - 6*d^2*e^3*x^3 - 6*d^3*e^2*x^2 + 3*d^4*e*x + 3*d^5 - (2*e^4*x^4 + 2*d*e^3*x^3 -
3*d^2*e^2*x^2 - 3*d^3*e*x - 3*d^4)*sqrt(-e^2*x^2 + d^2))/(d^4*e^7*x^5 + d^5*e^6*x^4 - 2*d^6*e^5*x^3 - 2*d^7*e
^4*x^2 + d^8*e^3*x + d^9*e^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

[undef, undef, undef, undef, undef, 1]