\(\int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx\) [1]

Optimal result

Integrand size = 25, antiderivative size = 132 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

-1/3*d^2*(-e^2*x^2+d^2)^(3/2)/e^3-1/4*d*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/5*(-e^2*x^2+d^2)^(5/2)/e^3+1/8*d^5*arctan
(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+1/8*d^3*x*(-e^2*x^2+d^2)^(1/2)/e^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 655, 201, 223, 209} \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2} \]

[In]

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

[Out]

(d^3*x*Sqrt[d^2 - e^2*x^2])/(8*e^2) - (d^2*(d^2 - e^2*x^2)^(3/2))/(3*e^3) - (d*x*(d^2 - e^2*x^2)^(3/2))/(4*e^2
) + (d^2 - e^2*x^2)^(5/2)/(5*e^3) + (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}+\frac {d^2 \int (d+e x) \sqrt {d^2-e^2 x^2} \, dx}{e^2} \\ & = -\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}+\frac {d^3 \int \sqrt {d^2-e^2 x^2} \, dx}{e^2} \\ & = \frac {d^3 x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {\left (3 d^3\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{4 e^2}+\frac {d^5 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^2} \\ & = \frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {\left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}+\frac {d^5 \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2} \\ & = \frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}-\frac {\left (3 d^5\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 {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.78 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^4-15 d^3 e x-8 d^2 e^2 x^2+30 d e^3 x^3+24 e^4 x^4\right )-30 d^5 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{120 e^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^4 - 15*d^3*e*x - 8*d^2*e^2*x^2 + 30*d*e^3*x^3 + 24*e^4*x^4) - 30*d^5*ArcTan[(e*x)/
(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(120*e^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {\left (-24 e^{4} x^{4}-30 d \,e^{3} x^{3}+8 d^{2} e^{2} x^{2}+15 d^{3} e x +16 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{120 e^{3}}+\frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}\)	\(97\)
default	\(e \left (-\frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 e^{2}}-\frac {2 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 e^{4}}\right )+d \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}\right )\)	\(130\)

[In]

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

[Out]

-1/120*(-24*e^4*x^4-30*d*e^3*x^3+8*d^2*e^2*x^2+15*d^3*e*x+16*d^4)/e^3*(-e^2*x^2+d^2)^(1/2)+1/8*d^5/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 = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.72 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=-\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, e^{4} x^{4} + 30 \, d e^{3} x^{3} - 8 \, d^{2} e^{2} x^{2} - 15 \, d^{3} e x - 16 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]

[In]

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

[Out]

-1/120*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*e^4*x^4 + 30*d*e^3*x^3 - 8*d^2*e^2*x^2 - 15*d^3
*e*x - 16*d^4)*sqrt(-e^2*x^2 + d^2))/e^3

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\begin {cases} \frac {d^{5} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{3}} - \frac {d^{3} x}{8 e^{2}} - \frac {d^{2} x^{2}}{15 e} + \frac {d x^{3}}{4} + \frac {e x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{3}}{3} + \frac {e x^{4}}{4}\right ) \sqrt {d^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d**5*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)*(-2*d**4/(15*e**3) - d**3*x/(8*e**2) - d**2
*x**2/(15*e) + d*x**3/4 + e*x**4/5), Ne(e**2, 0)), ((d*x**3/3 + e*x**4/4)*sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^{5} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{15 \, e^{3}} \]

[In]

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

[Out]

1/8*d^5*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e^2) + 1/8*sqrt(-e^2*x^2 + d^2)*d^3*x/e^2 - 1/5*(-e^2*x^2 + d^2
)^(3/2)*x^2/e - 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*d^2/e^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.63 \[ \int x^2 (d+e x) \sqrt {d^2-e^2 x^2} \, dx=\frac {d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} + \frac {1}{120} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, e x + 5 \, d\right )} x - \frac {4 \, d^{2}}{e}\right )} x - \frac {15 \, d^{3}}{e^{2}}\right )} x - \frac {16 \, d^{4}}{e^{3}}\right )} \]

[In]

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

[Out]

1/8*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + 1/120*sqrt(-e^2*x^2 + d^2)*((2*(3*(4*e*x + 5*d)*x - 4*d^2/e
)*x - 15*d^3/e^2)*x - 16*d^4/e^3)

Mupad [F(-1)]

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

[In]

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

[Out]

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