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

Optimal. Leaf size=103 \[ -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

[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

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Rubi [A]
time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {811, 655, 201, 223, 209} \begin {gather*} \frac {d^3 \text {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx &=-\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*}

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Mathematica [A]
time = 0.14, size = 92, normalized size = 0.89 \begin {gather*} \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 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[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*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2
- e^2*x^2]])/(2*e^4)

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Maple [A]
time = 0.06, size = 107, normalized size = 1.04

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	\(e \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+d \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 75, normalized size = 0.73 \begin {gather*} \frac {1}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{\left (-1\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{\left (-2\right )} - \frac {2}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.35, size = 68, normalized size = 0.66 \begin {gather*} -\frac {1}{6} \, {\left (6 \, d^{3} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (2 \, x^{2} e^{2} + 3 \, d x e + 4 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [C] Result contains complex when optimal does not.
time = 2.11, size = 177, normalized size = 1.72 \begin {gather*} d \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} + \frac {i d x}{2 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.28, size = 54, normalized size = 0.52 \begin {gather*} \frac {1}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (4 \, d^{2} e^{\left (-3\right )} + {\left (2 \, x e^{\left (-1\right )} + 3 \, d e^{\left (-2\right )}\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.14, size = 112, normalized size = 1.09 \begin {gather*} \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 . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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