\(\int \frac {x^2 (d+e x)}{(d^2-e^2 x^2)^{3/2}} \, dx\) [17]

Optimal result

Integrand size = 25, antiderivative size = 73 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d (d+e x)}{e^3 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 655, 223, 209, 651} \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}+\frac {d (d+e x)}{e^3 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^3} \]

[In]

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

[Out]

(d*(d + e*x))/(e^3*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^3 - (d*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 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 655

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

Rule 811

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {(-2 d+e x) \sqrt {d^2-e^2 x^2}}{e^3 (-d+e x)}+\frac {2 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{e^{4} \left (x -\frac {d}{e}\right )}\)	\(99\)
default	\(e \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )+d \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )\)	\(103\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.19 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2 \, d e x - 2 \, d^{2} + 2 \, {\left (d e x - d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x - 2 \, d\right )}}{e^{4} x - d e^{3}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.84 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=d \left (\begin {cases} \frac {i \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{e^{3}} - \frac {i x}{d e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {\operatorname {asin}{\left (\frac {e x}{d} \right )}}{e^{3}} + \frac {x}{d e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} \frac {2 d^{2}}{e^{4} \sqrt {d^{2} - e^{2} x^{2}}} - \frac {x^{2}}{e^{2} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d*Piecewise((I*acosh(e*x/d)/e**3 - I*x/(d*e**2*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (-asin(e*
x/d)/e**3 + x/(d*e**2*sqrt(1 - e**2*x**2/d**2)), True)) + e*Piecewise((2*d**2/(e**4*sqrt(d**2 - e**2*x**2)) -
x**2/(e**2*sqrt(d**2 - e**2*x**2)), Ne(e, 0)), (x**4/(4*(d**2)**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {x^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e} + \frac {d x}{\sqrt {-e^{2} x^{2} + d^{2}} e^{2}} - \frac {d \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{2}} + \frac {2 \, d^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{3}} \]

[In]

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

[Out]

-x^2/(sqrt(-e^2*x^2 + d^2)*e) + d*x/(sqrt(-e^2*x^2 + d^2)*e^2) - d*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2)
+ 2*d^2/(sqrt(-e^2*x^2 + d^2)*e^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{2} {\left | e \right |}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{3}} + \frac {2 \, d}{e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

-d*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + sqrt(-e^2*x^2 + d^2)/e^3 + 2*d/(e^2*((d*e + sqrt(-e^2*x^2 + d^2)
*abs(e))/(e^2*x) - 1)*abs(e))

Mupad [B] (verification not implemented)

Time = 11.78 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.19 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2\,d^2-e^2\,x^2}{e^3\,\sqrt {d^2-e^2\,x^2}}+\frac {d\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{{\left (-e^2\right )}^{3/2}}+\frac {d\,x}{e^2\,\sqrt {d^2-e^2\,x^2}} \]

[In]

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

[Out]

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