3.2.75 \(\int \frac {1}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\)

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

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {659, 191} \begin {gather*} \frac {2 x}{5 d^4 \sqrt {d^2-e^2 x^2}}-\frac {1}{5 d^2 e (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {1}{5 d e (d+e x)^2 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 70, normalized size = 0.77 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-2 d^3+d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )}{5 d^4 e (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.52, size = 70, normalized size = 0.77 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-2 d^3+d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )}{5 d^4 e (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.39, size = 115, normalized size = 1.26 \begin {gather*} -\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} + 4 \, d e^{2} x^{2} + d^{2} e x - 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{4} e^{5} x^{4} + 2 \, d^{5} e^{4} x^{3} - 2 \, d^{7} e^{2} x - d^{8} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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(1/(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 = 66, normalized size = 0.73 \begin {gather*} -\frac {\left (-e x +d \right ) \left (-2 e^{3} x^{3}-4 d \,e^{2} x^{2}-d^{2} e x +2 d^{3}\right )}{5 \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(-e^2*x^2+d^2)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.44, size = 136, normalized size = 1.49 \begin {gather*} -\frac {1}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} d e^{3} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e\right )}} - \frac {1}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e\right )}} + \frac {2 \, x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 2.85, size = 66, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (-2\,d^3+d^2\,e\,x+4\,d\,e^2\,x^2+2\,e^3\,x^3\right )}{5\,d^4\,e\,{\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(1/((d^2 - e^2*x^2)^(3/2)*(d + e*x)^2),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(2*e^3*x^3 - 2*d^3 + 4*d*e^2*x^2 + d^2*e*x))/(5*d^4*e*(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 {1}{\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(1/(e*x+d)**2/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)**2), x)

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