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

Optimal. Leaf size=91 \[ -\frac {2}{15 d e^2 (d+e x) \sqrt {d^2-e^2 x^2}}+\frac {1}{5 e^2 (d+e x)^2 \sqrt {d^2-e^2 x^2}}+\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {793, 659, 191} \begin {gather*} \frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}}-\frac {2}{15 d e^2 (d+e x) \sqrt {d^2-e^2 x^2}}+\frac {1}{5 e^2 (d+e x)^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (d^3+2 d^2 e x+8 d e^2 x^2+4 e^3 x^3\right )}{15 d^3 e^2 (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(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)*(d + e*x)^3)

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IntegrateAlgebraic [A]  time = 0.48, size = 69, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (d^3+2 d^2 e x+8 d e^2 x^2+4 e^3 x^3\right )}{15 d^3 e^2 (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(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)*(d + e*x)^3)

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fricas [A]  time = 0.40, size = 116, normalized size = 1.27 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.01, size = 64, normalized size = 0.70 \begin {gather*} \frac {\left (-e x +d \right ) \left (4 e^{3} x^{3}+8 d \,e^{2} x^{2}+2 d^{2} e x +d^{3}\right )}{15 \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 138, normalized size = 1.52 \begin {gather*} \frac {1}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{4} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{3} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2}\right )}} - \frac {2}{15 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} d e^{3} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2}\right )}} + \frac {4 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.88, size = 65, normalized size = 0.71 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (d^3+2\,d^2\,e\,x+8\,d\,e^2\,x^2+4\,e^3\,x^3\right )}{15\,d^3\,e^2\,{\left (d+e\,x\right )}^3\,\left (d-e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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