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

Optimal. Leaf size=83 \[ -\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[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

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Rubi [A]
time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1823, 794, 223, 209} \begin {gather*} \frac {d^3 \text {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {\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*}

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Mathematica [A]
time = 0.16, size = 89, normalized size = 1.07 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-5 d^2-3 d e x-e^2 x^2\right )}{3 e^2}+\frac {d^3 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 133, normalized size = 1.60

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	\(e^{2} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+2 d e \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e^{2}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 72, normalized size = 0.87 \begin {gather*} d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} - \sqrt {-x^{2} e^{2} + d^{2}} d x e^{\left (-1\right )} - \frac {5}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-2\right )} - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.22, size = 67, normalized size = 0.81 \begin {gather*} -\frac {1}{3} \, {\left (6 \, d^{3} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (x^{2} e^{2} + 3 \, d x e + 5 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [A]
time = 2.42, size = 218, normalized size = 2.63 \begin {gather*} d^{2} \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d^{2}}} & \text {for}\: e^{2} = 0 \\- \frac {\sqrt {d^{2} - e^{2} x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) + 2 d e \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^{2} \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*(e*x+d)**2/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e**2, True)) + 2*d*e*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**2
*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 = 1.14, size = 49, normalized size = 0.59 \begin {gather*} d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (5 \, d^{2} e^{\left (-2\right )} + {\left (3 \, d e^{\left (-1\right )} + x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (d+e\,x\right )}^2}{\sqrt {d^2-e^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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