\(\int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) [35]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {d^2 (32 d+21 e x) \sqrt {d^2-e^2 x^2}}{24 e^3}+\frac {7 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

7/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-2/3*d*x^2*(-e^2*x^2+d^2)^(1/2)/e-1/4*x^3*(-e^2*x^2+d^2)^(1/2)-1/2
4*d^2*(21*e*x+32*d)*(-e^2*x^2+d^2)^(1/2)/e^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1823, 847, 794, 223, 209} \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {7 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {d^2 (32 d+21 e x) \sqrt {d^2-e^2 x^2}}{24 e^3} \]

[In]

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

[Out]

(-2*d*x^2*Sqrt[d^2 - e^2*x^2])/(3*e) - (x^3*Sqrt[d^2 - e^2*x^2])/4 - (d^2*(32*d + 21*e*x)*Sqrt[d^2 - e^2*x^2])
/(24*e^3) + (7*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*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 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 847

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

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}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^2 \left (-7 d^2 e^2-8 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^2} \\ & = -\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x \left (16 d^3 e^3+21 d^2 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{12 e^4} \\ & = -\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {d^2 (32 d+21 e x) \sqrt {d^2-e^2 x^2}}{24 e^3}+\frac {\left (7 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2} \\ & = -\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {d^2 (32 d+21 e x) \sqrt {d^2-e^2 x^2}}{24 e^3}+\frac {\left (7 d^4\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^2} \\ & = -\frac {2 d x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2}-\frac {d^2 (32 d+21 e x) \sqrt {d^2-e^2 x^2}}{24 e^3}+\frac {7 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (32 d^3+21 d^2 e x+16 d e^2 x^2+6 e^3 x^3\right )+42 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e^3} \]

[In]

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

[Out]

-1/24*(Sqrt[d^2 - e^2*x^2]*(32*d^3 + 21*d^2*e*x + 16*d*e^2*x^2 + 6*e^3*x^3) + 42*d^4*ArcTan[(e*x)/(Sqrt[d^2] -
Sqrt[d^2 - e^2*x^2])])/e^3

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (6 e^{3} x^{3}+16 d \,e^{2} x^{2}+21 d^{2} e x +32 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e^{3}}+\frac {7 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\)	\(86\)
default	\(e^{2} \left (-\frac {x^{3} \sqrt {-e^{2} x^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )+d^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )+2 d e \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )\)	\(202\)

[In]

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

[Out]

-1/24*(6*e^3*x^3+16*d*e^2*x^2+21*d^2*e*x+32*d^3)/e^3*(-e^2*x^2+d^2)^(1/2)+7/8*d^4/e^2/(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 = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {42 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} + 16 \, d e^{2} x^{2} + 21 \, d^{2} e x + 32 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e^{3}} \]

[In]

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

[Out]

-1/24*(42*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 + 16*d*e^2*x^2 + 21*d^2*e*x + 32*d^3)*sqr
t(-e^2*x^2 + d^2))/e^3

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.35 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {7 d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {4 d^{3}}{3 e^{3}} - \frac {7 d^{2} x}{8 e^{2}} - \frac {2 d x^{2}}{3 e} - \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d^{2} x^{3}}{3} + \frac {d e x^{4}}{2} + \frac {e^{2} x^{5}}{5}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((7*d**4*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))/(8*e**2) + sqrt(d**2 - e**2*x**2)*(-4*d**3/(3*e**3) - 7*d**2*x/(8*e**2) - 2
*d*x**2/(3*e) - x**3/4), Ne(e**2, 0)), ((d**2*x**3/3 + d*e*x**4/2 + e**2*x**5/5)/sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{3} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d x^{2}}{3 \, e} + \frac {7 \, d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{2}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63 \[ \int \frac {x^2 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {7 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} - \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, x + \frac {8 \, d}{e}\right )} x + \frac {21 \, d^{2}}{e^{2}}\right )} x + \frac {32 \, d^{3}}{e^{3}}\right )} \]

[In]

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

[Out]

7/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) - 1/24*sqrt(-e^2*x^2 + d^2)*((2*(3*x + 8*d/e)*x + 21*d^2/e^2)
*x + 32*d^3/e^3)

Mupad [F(-1)]

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

[In]

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

[Out]

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