3.2.21 \(\int \frac {x^2}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [121]

Optimal. Leaf size=77 \[ -\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1653, 12, 807, 223, 209} \begin {gather*} -\frac {d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 807

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

Rule 1653

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 78, normalized size = 1.01 \begin {gather*} \frac {(-2 d-e x) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{\left (-e^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 97, normalized size = 1.26

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} \left (x +\frac {d}{e}\right )}\)	\(97\)
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} \left (x +\frac {d}{e}\right )}\)	\(97\)

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

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Maxima [A]
time = 0.48, size = 58, normalized size = 0.75 \begin {gather*} -d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-3\right )} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} d}{x e^{4} + d e^{3}} \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]

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

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Fricas [A]
time = 2.40, size = 83, normalized size = 1.08 \begin {gather*} -\frac {2 \, d x e + 2 \, d^{2} - 2 \, {\left (d x e + d^{2}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e + 2 \, d\right )}}{x e^{4} + d e^{3}} \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]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \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]

Integral(x**2/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)), x)

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Giac [A]
time = 1.17, size = 69, normalized size = 0.90 \begin {gather*} -d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\left (d\right ) - \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-3\right )} + \frac {2 \, d e^{\left (-3\right )}}{\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1} \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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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