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

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

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Rubi [A]  time = 0.10, 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 = {1639, 12, 793, 217, 203} \begin {gather*} -\frac {d \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

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 793

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 1639

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 \operatorname {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 = 0.07, size = 59, normalized size = 0.77 \begin {gather*} -\frac {\frac {\sqrt {d^2-e^2 x^2} (2 d+e x)}{d+e x}+d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \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])/(d + e*x) + d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^3)

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IntegrateAlgebraic [A]  time = 0.34, size = 81, normalized size = 1.05 \begin {gather*} \frac {(-2 d-e x) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.41, size = 85, normalized size = 1.10 \begin {gather*} -\frac {2 \, d e x + 2 \, d^{2} - 2 \, {\left (d e x + d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + 2 \, d\right )}}{e^{4} x + 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*e*x + 2*d^2 - 2*(d*e*x + d^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + sqrt(-e^2*x^2 + d^2)*(e*x + 2*
d))/(e^4*x + d*e^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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]

Exception raised: NotImplementedError >> Unable to parse Giac output: -d*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1
)/exp(2)-2*d*exp(2)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))
/sqrt(-exp(1)^4+exp(2)^2)/exp(1)/exp(2)-4*exp(1)^2*1/4/exp(1)^5*sqrt(-exp(2)*x^2+d^2)

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maple [A]  time = 0.01, size = 97, normalized size = 1.26 \begin {gather*} -\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d}{\left (x +\frac {d}{e}\right ) e^{4}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}} \end {gather*}

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)

[Out]

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

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maxima [A]  time = 0.98, size = 63, normalized size = 0.82 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{4} x + d e^{3}} - \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{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]

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

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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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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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