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

Optimal. Leaf size=115 \[ -\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}+\frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[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]
time = 0.10, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1651, 673, 665, 677, 223, 209} \begin {gather*} -\frac {\text {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)} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx &=\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]
time = 0.35, size = 91, normalized size = 0.79 \begin {gather*} \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 {\log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{\left (-e^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(103)=206\).
time = 0.08, size = 268, normalized size = 2.33

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.64, size = 148, normalized size = 1.29 \begin {gather*} -\frac {8 \, x^{3} e^{3} + 24 \, d x^{2} e^{2} + 24 \, d^{2} x e + 8 \, d^{3} - 10 \, {\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (13 \, x^{2} e^{2} + 19 \, d x e + 8 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{3} e^{6} + 3 \, d x^{2} e^{5} + 3 \, d^{2} x e^{4} + d^{3} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]
time = 1.05, size = 170, normalized size = 1.48 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\left (d\right ) + \frac {2 \, {\left (\frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + \frac {55 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} + \frac {25 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-6\right )}}{x^{3}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-8\right )}}{x^{4}} + 8\right )} e^{\left (-3\right )}}{5 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\sqrt {d^2-e^2\,x^2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

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)^4,x)

[Out]

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

________________________________________________________________________________________