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

Optimal. Leaf size=95 \[ -\frac{d \sqrt{d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac{8 \sqrt{d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac{7 \sqrt{d^2-e^2 x^2}}{15 d e^3 (d+e x)} \]

[Out]

-(d*Sqrt[d^2 - e^2*x^2])/(5*e^3*(d + e*x)^3) + (8*Sqrt[d^2 - e^2*x^2])/(15*e^3*(d + e*x)^2) - (7*Sqrt[d^2 - e^
2*x^2])/(15*d*e^3*(d + e*x))

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Rubi [A]  time = 0.128102, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1639, 793, 659, 651} \[ -\frac{d \sqrt{d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac{8 \sqrt{d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac{7 \sqrt{d^2-e^2 x^2}}{15 d e^3 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*Sqrt[d^2 - e^2*x^2])/(5*e^3*(d + e*x)^3) + (8*Sqrt[d^2 - e^2*x^2])/(15*e^3*(d + e*x)^2) - (7*Sqrt[d^2 - e^
2*x^2])/(15*d*e^3*(d + e*x))

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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]

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]

Rubi steps

\begin{align*} \int \frac{x^2}{(d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx &=\frac{\sqrt{d^2-e^2 x^2}}{e^3 (d+e x)^2}+\frac{\int \frac{2 d^2 e^2+d e^3 x}{(d+e x)^3 \sqrt{d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac{d \sqrt{d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac{\sqrt{d^2-e^2 x^2}}{e^3 (d+e x)^2}+\frac{(7 d) \int \frac{1}{(d+e x)^2 \sqrt{d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac{d \sqrt{d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac{8 \sqrt{d^2-e^2 x^2}}{15 e^3 (d+e x)^2}+\frac{7 \int \frac{1}{(d+e x) \sqrt{d^2-e^2 x^2}} \, dx}{15 e^2}\\ &=-\frac{d \sqrt{d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac{8 \sqrt{d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac{7 \sqrt{d^2-e^2 x^2}}{15 d e^3 (d+e x)}\\ \end{align*}

Mathematica [A]  time = 0.0670098, size = 52, normalized size = 0.55 \[ -\frac{\sqrt{d^2-e^2 x^2} \left (2 d^2+6 d e x+7 e^2 x^2\right )}{15 d e^3 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[d^2 - e^2*x^2]*(2*d^2 + 6*d*e*x + 7*e^2*x^2))/(15*d*e^3*(d + e*x)^3)

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Maple [A]  time = 0.05, size = 55, normalized size = 0.6 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( 7\,{x}^{2}{e}^{2}+6\,dex+2\,{d}^{2} \right ) }{15\,{e}^{3}d \left ( ex+d \right ) ^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-e*x+d)*(7*e^2*x^2+6*d*e*x+2*d^2)/(e*x+d)^2/d/e^3/(-e^2*x^2+d^2)^(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*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51837, size = 213, normalized size = 2.24 \begin{align*} -\frac{2 \, e^{3} x^{3} + 6 \, d e^{2} x^{2} + 6 \, d^{2} e x + 2 \, d^{3} +{\left (7 \, e^{2} x^{2} + 6 \, d e x + 2 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(2*e^3*x^3 + 6*d*e^2*x^2 + 6*d^2*e*x + 2*d^3 + (7*e^2*x^2 + 6*d*e*x + 2*d^2)*sqrt(-e^2*x^2 + d^2))/(d*e^
6*x^3 + 3*d^2*e^5*x^2 + 3*d^3*e^4*x + d^4*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 )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), 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*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError