\(\int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) [36]

Optimal result

Integrand size = 25, antiderivative size = 83 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1823, 794, 223, 209} \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2} \]

[In]

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

[Out]

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

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 794

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

Rule 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x \left (-5 d^2 e^2-6 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 e^2} \\ & = -\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e} \\ & = -\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \\ & = -\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (5 d^2+3 d e x+e^2 x^2\right )+6 d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{3 e^2} \]

[In]

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

[Out]

-1/3*(Sqrt[d^2 - e^2*x^2]*(5*d^2 + 3*d*e*x + e^2*x^2) + 6*d^3*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])
/e^2

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\left (e^{2} x^{2}+3 d e x +5 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}\)	\(73\)
default	\(e^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )-\frac {d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e^{2}}+2 d e \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\)	\(133\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (e^{2} x^{2} + 3 \, d e x + 5 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, e^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {d^{3} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{e} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {5 d^{2}}{3 e^{2}} - \frac {d x}{e} - \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d^{2} x^{2}}{2} + \frac {2 d e x^{3}}{3} + \frac {e^{2} x^{4}}{4}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d**3*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)), (x
*log(x)/sqrt(-e**2*x**2), True))/e + sqrt(d**2 - e**2*x**2)*(-5*d**2/(3*e**2) - d*x/e - x**2/3), Ne(e**2, 0)),
((d**2*x**2/2 + 2*d*e*x**3/3 + e**2*x**4/4)/sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{2} + \frac {d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{e} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {x (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e {\left | e \right |}} - \frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (x + \frac {3 \, d}{e}\right )} x + \frac {5 \, d^{2}}{e^{2}}\right )} \]

[In]

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

[Out]

d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/(e*abs(e)) - 1/3*sqrt(-e^2*x^2 + d^2)*((x + 3*d/e)*x + 5*d^2/e^2)

Mupad [F(-1)]

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

[In]

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

[Out]

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