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

Optimal. Leaf size=64 \[ \frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {793, 651} \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.78 \begin {gather*} -\frac {\left (d^2+3 d e x-4 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{15 d e^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.59, size = 52, normalized size = 0.81 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-d^2-3 d e x+4 e^2 x^2\right )}{15 d e^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 102, normalized size = 1.59 \begin {gather*} -\frac {e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3} - {\left (4 \, e^{2} x^{2} - 3 \, d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{5} x^{3} + 3 \, d^{2} e^{4} x^{2} + 3 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (6*exp(1)*exp(2)^7+12*(-1/2*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)*exp(2)^7+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1)
)/x/exp(2))^4*exp(1)^11*exp(2)^2+48*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^11*ex
p(2)^2+36*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^9*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)^7*exp(2)^4+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(
1))/x/exp(2))^2*exp(1)^11*exp(2)^2+14*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^9*e
xp(2)^3+4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^13*exp(2)+3*(-1/2*(-2*d*exp(1)-
2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^7*exp(2)^4+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1
))/x/exp(2))^4*exp(1)*exp(2)^7+3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^5*exp(2)
^5+108*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^9*exp(2)^3+96*(-1/2*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^7*exp(2)^4+48*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1
))/x/exp(2))^4*exp(1)^5*exp(2)^5+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^3*exp
(2)^6+24*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^7*exp(2)^4+12*(-1/2*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^5*exp(2)^5+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(
1))/x/exp(2))^4*exp(1)^3*exp(2)^6+60*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^5*ex
p(2)^5+36*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^3*exp(2)^6+exp(1)^7*exp(2)^4+12
*exp(1)^5*exp(2)^5+2*exp(1)^3*exp(2)^6-12*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^3*exp(2)^6/x/exp(
2)-9/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^5*exp(2)^5/x/exp(2)-33*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))*exp(1)^7*exp(2)^4/x/exp(2)-3*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^9*exp(2)^3/x/ex
p(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*exp(1)^11-6*d*exp(1)^7*exp(2)^2-6*d*exp(1)^5*exp(2)^3+3*d*exp(1)^9*exp(2)+6*d*exp(1
)*exp(2)^5)+1/2*(4*exp(1)^7*exp(2)^2+2*exp(1)^5*exp(2)^3+8*exp(1)^3*exp(2)^4)*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*exp(1)^11-2*d*exp(1)^7*
exp(2)^2-2*d*exp(1)^5*exp(2)^3+d*exp(1)^9*exp(2)+2*d*exp(1)*exp(2)^5)

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maple [A]  time = 0.01, size = 42, normalized size = 0.66 \begin {gather*} -\frac {\left (-e x +d \right ) \left (4 e x +d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 \left (e x +d \right )^{3} d \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 125, normalized size = 1.95 \begin {gather*} \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{3} x + d^{2} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-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)*d/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2) - 11/15*sqrt(-e^2*x^2 + d^2)/(e^4*x
^2 + 2*d*e^3*x + d^2*e^2) + 4/15*sqrt(-e^2*x^2 + d^2)/(d*e^3*x + d^2*e^2)

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mupad [B]  time = 2.90, size = 46, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {d^2-e^2\,x^2}\,\left (d^2+3\,d\,e\,x-4\,e^2\,x^2\right )}{15\,d\,e^2\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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(x*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

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

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