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

Optimal. Leaf size=115 \[ \frac{3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}-\frac{2 \sqrt{d^2-e^2 x^2}}{e^3 (d+e x)}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.146687, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1637, 659, 651, 663, 217, 203} \[ \frac{3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}-\frac{2 \sqrt{d^2-e^2 x^2}}{e^3 (d+e x)}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 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 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 663

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.129219, size = 73, normalized size = 0.63 \[ -\frac{\frac{\sqrt{d^2-e^2 x^2} \left (8 d^2+19 d e x+13 e^2 x^2\right )}{(d+e x)^3}+5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{5 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d^2 - e^2*x^2]*(8*d^2 + 19*d*e*x + 13*e^2*x^2))/(d + e*x)^3 + 5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(5*
e^3)

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Maple [B]  time = 0.066, size = 214, normalized size = 1.9 \begin{align*} -{\frac{d}{5\,{e}^{7}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}+{\frac{3}{5\,{e}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}-{\frac{1}{{e}^{5}d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{1}{{e}^{3}d}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{1}{{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*d/e^7/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)+3/5/e^6/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/
2)-1/e^5/d/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)-1/e^3/d*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-1/e^2/(
e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77875, size = 332, normalized size = 2.89 \begin{align*} -\frac{8 \, e^{3} x^{3} + 24 \, d e^{2} x^{2} + 24 \, d^{2} e x + 8 \, d^{3} - 10 \,{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (13 \, e^{2} x^{2} + 19 \, d e x + 8 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(8*e^3*x^3 + 24*d*e^2*x^2 + 24*d^2*e*x + 8*d^3 - 10*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(d
- sqrt(-e^2*x^2 + d^2))/(e*x)) + (13*e^2*x^2 + 19*d*e*x + 8*d^2)*sqrt(-e^2*x^2 + d^2))/(e^6*x^3 + 3*d*e^5*x^2
+ 3*d^2*e^4*x + d^3*e^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError