\(\int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\) [191]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}+\frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1651, 673, 665, 677, 223, 209} \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}+\frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}-\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)} \]

[In]

Int[(x^2*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]

[Out]

(-2*Sqrt[d^2 - e^2*x^2])/(e^3*(d + e*x)) - (d*(d^2 - e^2*x^2)^(3/2))/(5*e^3*(d + e*x)^4) + (3*(d^2 - e^2*x^2)^
(3/2))/(5*e^3*(d + e*x)^3) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^3

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 677

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^2 (d+e x)^4}-\frac {2 d \sqrt {d^2-e^2 x^2}}{e^2 (d+e x)^3}+\frac {\sqrt {d^2-e^2 x^2}}{e^2 (d+e x)^2}\right ) \, dx \\ & = \frac {\int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^2}-\frac {(2 d) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^2}+\frac {d^2 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx}{e^2} \\ & = -\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}+\frac {2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3 (d+e x)^3}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2}+\frac {d \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e^2} \\ & = -\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}+\frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {\text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \\ & = -\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}+\frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\left (-8 d^2-19 d e x-13 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac {2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[In]

Integrate[(x^2*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]

[Out]

((-8*d^2 - 19*d*e*x - 13*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(5*e^3*(d + e*x)^3) + (2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt
[d^2 - e^2*x^2])])/e^3

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(103)=206\).

Time = 0.41 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.33

method	result	size
default	\(\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}}{e^{4}}+\frac {d^{2} \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{6}}+\frac {2 \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{6} \left (x +\frac {d}{e}\right )^{3}}\)	\(268\)

[In]

int(x^2*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.37 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {8 \, e^{3} x^{3} + 24 \, d e^{2} x^{2} + 24 \, d^{2} e x + 8 \, d^{3} - 10 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (13 \, e^{2} x^{2} + 19 \, d e x + 8 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/5*(8*e^3*x^3 + 24*d*e^2*x^2 + 24*d^2*e*x + 8*d^3 - 10*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(d
- sqrt(-e^2*x^2 + d^2))/(e*x)) + (13*e^2*x^2 + 19*d*e*x + 8*d^2)*sqrt(-e^2*x^2 + d^2))/(e^6*x^3 + 3*d*e^5*x^2
+ 3*d^2*e^4*x + d^3*e^3)

Sympy [F]

\[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate(x**2*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(x**2*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)

Maxima [F]

\[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int { \frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)*x^2/(e*x + d)^4, x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.62 \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{2} {\left | e \right |}} + \frac {2 \, {\left (\frac {35 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {55 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {25 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + 8\right )}}{5 \, e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

-arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + 2/5*(35*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 55*(d*e + sq
rt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 25*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 5*(d*e + sqrt(-e
^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 8)/(e^2*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^2\,\sqrt {d^2-e^2\,x^2}}{{\left (d+e\,x\right )}^4} \,d x \]

[In]

int((x^2*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4,x)

[Out]

int((x^2*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4, x)