\(\int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx\) [16]

Optimal result

Integrand size = 25, antiderivative size = 103 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 655, 201, 223, 209} \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3} \]

[In]

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

[Out]

-((d^2*Sqrt[d^2 - e^2*x^2])/e^3) - (d*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + (d^2 - e^2*x^2)^(3/2)/(3*e^3) + (d^3*Ar
cTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*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) \sqrt {d^2-e^2 x^2} \, dx}{e^2}+\frac {d^2 \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2} \\ & = -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d \int \sqrt {d^2-e^2 x^2} \, dx}{e^2}+\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2} \\ & = -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 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^2} \\ & = -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\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 )}{2 e^2} \\ & = -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-4 d^2-3 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 e^3}-\frac {d^3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[In]

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

[Out]

((-4*d^2 - 3*d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(6*e^3) - (d^3*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x
^2])])/e^3

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {\left (2 e^{2} x^{2}+3 d e x +4 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{3}}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\)	\(75\)
default	\(e \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+d \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\)	\(107\)

[In]

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

[Out]

-1/6*(2*e^2*x^2+3*d*e*x+4*d^2)/e^3*(-e^2*x^2+d^2)^(1/2)+1/2*d^3/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 = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{2} x^{2} + 3 \, d e x + 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

[In]

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

[Out]

-1/6*(6*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^2*x^2 + 3*d*e*x + 4*d^2)*sqrt(-e^2*x^2 + d^2))/e^
3

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.25 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {d^{3} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{2}}{3 e^{3}} - \frac {d x}{2 e^{2}} - \frac {x^{2}}{3 e}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\frac {d x^{3}}{3} + \frac {e x^{4}}{4}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d**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))/(2*e**2) + sqrt(d**2 - e**2*x**2)*(-2*d**2/(3*e**3) - d*x/(2*e**2) - x**2/(3*
e)), 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 = 93, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e} + \frac {d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}} e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{2} {\left | e \right |}} - \frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x {\left (\frac {2 \, x}{e} + \frac {3 \, d}{e^{2}}\right )} + \frac {4 \, d^{2}}{e^{3}}\right )} \]

[In]

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

[Out]

1/2*d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) - 1/6*sqrt(-e^2*x^2 + d^2)*(x*(2*x/e + 3*d/e^2) + 4*d^2/e^3)

Mupad [B] (verification not implemented)

Time = 11.93 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx=\left \{\begin {array}{cl} \frac {d\,x^3}{3\,\sqrt {d^2}} & \text {\ if\ \ }e=0\\ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+e^2\,x^2\right )}{3\,e^3}-\frac {d^3\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}}-\frac {d\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e^2} & \text {\ if\ \ }e\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(e == 0, (d*x^3)/(3*(d^2)^(1/2)), e ~= 0, - ((d^2 - e^2*x^2)^(1/2)*(2*d^2 + e^2*x^2))/(3*e^3) - (d^3*
log(2*x*(-e^2)^(1/2) + 2*(d^2 - e^2*x^2)^(1/2)))/(2*(-e^2)^(3/2)) - (d*x*(d^2 - e^2*x^2)^(1/2))/(2*e^2))