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]

-1/5*d*(-e^2*x^2+d^2)^(1/2)/e^3/(e*x+d)^3+8/15*(-e^2*x^2+d^2)^(1/2)/e^3/(e*x+d)^2-7/15*(-e^2*x^2+d^2)^(1/2)/d/
e^3/(e*x+d)

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Rubi [A]  time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, 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]

-1/15*(Sqrt[d^2 - e^2*x^2]*(2*d^2 + 6*d*e*x + 7*e^2*x^2))/(d*e^3*(d + e*x)^3)

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fricas [A]  time = 0.98, size = 104, normalized size = 1.09 \[ -\frac {2 \, e^{3} x^{3} + 6 \, d e^{2} x^{2} + 6 \, d^{2} e x + 2 \, d^{3} + {\left (7 \, e^{2} x^{2} + 6 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

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

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 >> Unable to parse Giac output: (3*(-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)^3+6*(-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)+2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^2+(-1/2*(-2*d*exp
(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^5+3*exp(1)^4*exp(2)^3+1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))*exp(2)^5/x/exp(2)-5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^2/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))^2/(-d*exp(1)^9+2*d*exp(1)^5*exp(2)^2-d*exp(1)*exp(2)^4)+1/2*(-2*exp(2)^4-4*exp(1)^6*exp(2))*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)^9-2*d*exp(1)^5*exp(2)^2+d*exp(1)*exp(2)^4)

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maple [A]  time = 0.01, size = 55, normalized size = 0.58 \[ -\frac {\left (-e x +d \right ) \left (7 e^{2} x^{2}+6 d e x +2 d^{2}\right )}{15 \left (e x +d \right )^{2} \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{3}} \]

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 [A]  time = 0.98, size = 125, normalized size = 1.32 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{4} x + d^{2} e^{3}\right )}} \]

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]

-1/5*sqrt(-e^2*x^2 + d^2)*d/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) + 8/15*sqrt(-e^2*x^2 + d^2)/(e^5*x
^2 + 2*d*e^4*x + d^2*e^3) - 7/15*sqrt(-e^2*x^2 + d^2)/(d*e^4*x + d^2*e^3)

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mupad [B]  time = 2.76, size = 48, normalized size = 0.51 \[ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d+e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d^2 - e^2*x^2)^(1/2)*(2*d^2 + 7*e^2*x^2 + 6*d*e*x))/(15*d*e^3*(d + e*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

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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