\(\int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [182]

Optimal result

Integrand size = 27, antiderivative size = 95 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d \sqrt {d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac {8 \sqrt {d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac {7 \sqrt {d^2-e^2 x^2}}{15 d e^3 (d+e x)} \]

[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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1653, 807, 673, 665} \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d \sqrt {d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac {8 \sqrt {d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac {7 \sqrt {d^2-e^2 x^2}}{15 d e^3 (d+e x)} \]

[In]

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

[Out]

-1/5*(d*Sqrt[d^2 - e^2*x^2])/(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 665

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 673

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 807

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 1653

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*} \text {integral}& = \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] (verified)

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.55 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-2 d^2-6 d e x-7 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{15 d e^3 (d+e x)^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.52

method	result	size
trager	\(-\frac {\left (7 e^{2} x^{2}+6 d e x +2 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d \,e^{3} \left (e x +d \right )^{3}}\)	\(49\)
gosper	\(-\frac {\left (-e x +d \right ) \left (7 e^{2} x^{2}+6 d e x +2 d^{2}\right )}{15 \left (e x +d \right )^{2} d \,e^{3} \sqrt {-e^{2} x^{2}+d^{2}}}\)	\(55\)
default	\(-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} d \left (x +\frac {d}{e}\right )}+\frac {d^{2} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{5}}-\frac {2 d \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{4}}\)	\(288\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \, e^{3} x^{3} + 6 \, d e^{2} x^{2} + 6 \, d^{2} e x + 2 \, d^{3} + {\left (7 \, e^{2} x^{2} + 6 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{4} x + d^{2} e^{3}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {4 \, {\left (\frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + 1\right )}}{15 \, d e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

4/15*(5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 10*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 1)/
(d*e^2*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [B] (verification not implemented)

Time = 11.69 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d+e\,x\right )}^3} \]

[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)