\(\int \frac {x^2}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}}-\frac {d}{5 e^3 (d+e x)^2 \sqrt {d^2-e^2 x^2}}+\frac {7}{15 e^3 (d+e x) \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, 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 = {866, 1649, 792, 197} \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}}-\frac {d (d-e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 (d-e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1649

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = -\frac {d (d-e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {\left (\frac {2 d^2}{e^2}-\frac {5 d x}{e}\right ) (d-e x)}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = -\frac {d (d-e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 (d-e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 e^2} \\ & = -\frac {d (d-e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 (d-e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (4 d^3+8 d^2 e x+2 d e^2 x^2+e^3 x^3\right )}{15 d^2 e^3 (d-e x) (d+e x)^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(4*d^3 + 8*d^2*e*x + 2*d*e^2*x^2 + e^3*x^3))/(15*d^2*e^3*(d - e*x)*(d + e*x)^3)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (e^{3} x^{3}+2 d \,e^{2} x^{2}+8 d^{2} e x +4 d^{3}\right )}{15 \left (e x +d \right ) d^{2} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(65\)
trager	\(\frac {\left (e^{3} x^{3}+2 d \,e^{2} x^{2}+8 d^{2} e x +4 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{2} e^{3} \left (e x +d \right )^{3} \left (-e x +d \right )}\)	\(67\)
default	\(\frac {x}{d^{2} e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {d^{2} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{4}}-\frac {2 d \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{3}}\)	\(287\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.33 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {4 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 8 \, d^{3} e x - 4 \, d^{4} - {\left (e^{3} x^{3} + 2 \, d e^{2} x^{2} + 8 \, d^{2} e x + 4 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{7} x^{4} + 2 \, d^{3} e^{6} x^{3} - 2 \, d^{5} e^{4} x - d^{6} e^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.53 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{5} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{3}\right )}} + \frac {7}{15 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{4} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{3}\right )}} + \frac {x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.15 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {-\frac {8 i \, \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2} e^{2}} - \frac {15}{d^{2} e^{2} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )} + \frac {3 \, d^{8} e^{8} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} - 5 \, d^{8} e^{8} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} - 15 \, d^{8} e^{8} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4}}{d^{10} e^{10} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{5} \mathrm {sgn}\left (e\right )^{5}}}{120 \, {\left | e \right |}} \]

[In]

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

[Out]

-1/120*(-8*I*sgn(1/(e*x + d))*sgn(e)/(d^2*e^2) - 15/(d^2*e^2*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))
+ (3*d^8*e^8*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))^4*sgn(e)^4 - 5*d^8*e^8*(2*d/(e*x + d) - 1)^(3/2)*sgn(1
/(e*x + d))^4*sgn(e)^4 - 15*d^8*e^8*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))^4*sgn(e)^4)/(d^10*e^10*sgn(1/(e*x
+ d))^5*sgn(e)^5))/abs(e)

Mupad [B] (verification not implemented)

Time = 11.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (4\,d^3+8\,d^2\,e\,x+2\,d\,e^2\,x^2+e^3\,x^3\right )}{15\,d^2\,e^3\,{\left (d+e\,x\right )}^3\,\left (d-e\,x\right )} \]

[In]

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

[Out]

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