\(\int \frac {x^2}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\) [141]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {869, 792, 197} \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {x^2}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{15 d e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^3 e^2 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[x^2/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

-1/5*x^2/(d*e*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (2*(d + e*x))/(15*d*e^3*(d^2 - e^2*x^2)^(3/2)) - (2*x)/(15*d^
3*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 869

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

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^4+2 d^3 e x-3 d^2 e^2 x^2+2 d e^3 x^3+2 e^4 x^4\right )}{15 d^3 e^3 (d-e x)^2 (d+e x)^3} \]

[In]

Integrate[x^2/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (2 e^{4} x^{4}+2 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}+2 d^{3} e x +2 d^{4}\right )}{15 d^{3} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(70\)
trager	\(\frac {\left (2 e^{4} x^{4}+2 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}+2 d^{3} e x +2 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} e^{3} \left (e x +d \right )^{3} \left (-e x +d \right )^{2}}\)	\(79\)
default	\(\frac {1}{3 e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2}}+\frac {d^{2} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{3}}\)	\(234\)

[In]

int(x^2/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (83) = 166\).

Time = 0.29 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.79 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 \, e^{5} x^{5} + 2 \, d e^{4} x^{4} - 4 \, d^{2} e^{3} x^{3} - 4 \, d^{3} e^{2} x^{2} + 2 \, d^{4} e x + 2 \, d^{5} + {\left (2 \, e^{4} x^{4} + 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 2 \, d^{3} e x + 2 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{8} x^{5} + d^{4} e^{7} x^{4} - 2 \, d^{5} e^{6} x^{3} - 2 \, d^{6} e^{5} x^{2} + d^{7} e^{4} x + d^{8} e^{3}\right )}} \]

[In]

integrate(x^2/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate(x**2/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**2/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {d}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{3}\right )}} - \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{2}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e^{2}} \]

[In]

integrate(x^2/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}} \,d x } \]

[In]

integrate(x^2/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

integrate(x^2/((-e^2*x^2 + d^2)^(5/2)*(e*x + d)), x)

Mupad [B] (verification not implemented)

Time = 11.85 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^4+2\,d^3\,e\,x-3\,d^2\,e^2\,x^2+2\,d\,e^3\,x^3+2\,e^4\,x^4\right )}{15\,d^3\,e^3\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \]

[In]

int(x^2/((d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)

[Out]

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