3.2.93 \(\int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=67 \[ -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 e (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 d e (d+e x)^4} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {659, 651} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 e (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 d e (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d^2 - e^2*x^2]/(d + e*x)^4,x]

[Out]

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

Rule 651

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*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 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 {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 d e (d+e x)^4}+\frac {\int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 d}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 d e (d+e x)^4}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 e (d+e x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-4 d^2+3 d e x+e^2 x^2\right )}{15 d^2 e (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d^2 - e^2*x^2]/(d + e*x)^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.62, size = 51, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-4 d^2+3 d e x+e^2 x^2\right )}{15 d^2 e (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[d^2 - e^2*x^2]/(d + e*x)^4,x]

[Out]

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

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fricas [A]  time = 0.41, size = 104, normalized size = 1.55 \begin {gather*} -\frac {4 \, e^{3} x^{3} + 12 \, d e^{2} x^{2} + 12 \, d^{2} e x + 4 \, d^{3} - {\left (e^{2} x^{2} + 3 \, d e x - 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{4} x^{3} + 3 \, d^{3} e^{3} x^{2} + 3 \, d^{4} e^{2} x + d^{5} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/15*(4*e^3*x^3 + 12*d*e^2*x^2 + 12*d^2*e*x + 4*d^3 - (e^2*x^2 + 3*d*e*x - 4*d^2)*sqrt(-e^2*x^2 + d^2))/(d^2*
e^4*x^3 + 3*d^3*e^3*x^2 + 3*d^4*e^2*x + d^5*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^2+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*e
xp(1)^10*exp(2)^3+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^10*exp(2)^3+6*(-1/2*
(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^4+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^2-8*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^
3*exp(1)^10*exp(2)^3-24*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^8*exp(2)^4-12*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(2)^5-12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^3-72*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(
2))^3*exp(1)^8*exp(2)^4-84*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp(2)^5-24*
(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^4*exp(2)^6-30*(-1/2*(-2*d*exp(1)-2*sqrt(d
^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^4-30*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^3*exp(1)^6*exp(2)^5-3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^4*exp(2)^6+3*
(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^8-132*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^5-108*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3
*exp(1)^4*exp(2)^6+2*exp(1)^8*exp(2)^4-18*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)
^8-12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^6-5*exp(1)^4*exp(2)^6-36*(
-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8-18*exp(2)^8+8*(-1/2*(-2*d*exp(1)-2*sqrt(
d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^14*exp(2)+3/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/
x/exp(2)+42*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/exp(2)+9*(-2*d*exp(1)-2*sqrt(d^2-x
^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/
x/exp(2)-3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*exp(2)^3/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt
(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^3/(3*d^2*exp
(1)^11-6*d^2*exp(1)^7*exp(2)^2-6*d^2*exp(1)^5*exp(2)^3+3*d^2*exp(1)^9*exp(2)+6*d^2*exp(1)*exp(2)^5)+1/2*(8*exp
(1)^4*exp(2)^4+6*exp(2)^6)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp
(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(-d^2*exp(1)^11+2*d^2*exp(1)^7*exp(2)^2+2*d^2*exp(1)^5*exp(2)^3-d^2*exp(1)^9*
exp(2)-2*d^2*exp(1)*exp(2)^5)

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maple [A]  time = 0.01, size = 43, normalized size = 0.64 \begin {gather*} -\frac {\left (-e x +d \right ) \left (e x +4 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 \left (e x +d \right )^{3} d^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x)

[Out]

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

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maxima [B]  time = 0.44, size = 123, normalized size = 1.84 \begin {gather*} -\frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{3} x^{2} + 2 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{2} x + d^{3} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.78, size = 47, normalized size = 0.70 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (-4\,d^2+3\,d\,e\,x+e^2\,x^2\right )}{15\,d^2\,e\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(1/2)/(d + e*x)^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)

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