3.1.17 \(\int \frac {x^2 (d+e x)}{(d^2-e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac {d (d+e x)}{e^3 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e^3}-\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {797, 641, 217, 203, 637} \begin {gather*} \frac {d (d+e x)}{e^3 \sqrt {d^2-e^2 x^2}}+\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))/(d^2 - e^2*x^2)^(3/2),x]

[Out]

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

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 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 641

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 797

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

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Mathematica [A]  time = 0.03, size = 77, normalized size = 1.05 \begin {gather*} \frac {-d \sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+2 d^2+d e x-e^2 x^2}{e^3 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d + e*x))/(d^2 - e^2*x^2)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.47, size = 84, normalized size = 1.15 \begin {gather*} -\frac {d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^4}-\frac {\sqrt {d^2-e^2 x^2} (2 d-e x)}{e^3 (e x-d)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(d + e*x))/(d^2 - e^2*x^2)^(3/2),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 = 0.40, size = 87, normalized size = 1.19 \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)^(3/2),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 [A]  time = 0.25, size = 66, normalized size = 0.90 \begin {gather*} -d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left (2 \, d^{2} e^{\left (-3\right )} - {\left (x e^{\left (-1\right )} - d e^{\left (-2\right )}\right )} x\right )}}{x^{2} e^{2} - d^{2}} \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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 99, normalized size = 1.36 \begin {gather*} -\frac {x^{2}}{\sqrt {-e^{2} x^{2}+d^{2}}\, e}+\frac {d x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{2}}+\frac {2 d^{2}}{\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)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.98, size = 78, normalized size = 1.07 \begin {gather*} -\frac {x^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e} + \frac {d x}{\sqrt {-e^{2} x^{2} + d^{2}} e^{2}} - \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{3}} + \frac {2 \, d^{2}}{\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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.96, size = 87, normalized size = 1.19 \begin {gather*} \frac {2\,d^2-e^2\,x^2}{e^3\,\sqrt {d^2-e^2\,x^2}}+\frac {d\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{{\left (-e^2\right )}^{3/2}}+\frac {d\,x}{e^2\,\sqrt {d^2-e^2\,x^2}} \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)^(3/2),x)

[Out]

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

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sympy [C]  time = 9.71, size = 184, normalized size = 2.52 \begin {gather*} d \left (\begin {cases} \frac {i \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{e^{3}} - \frac {i x}{d e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {\operatorname {asin}{\left (\frac {e x}{d} \right )}}{e^{3}} + \frac {x}{d e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} \tilde {\infty } x^{4} & \text {for}\: \left (d = 0 \vee d = - \sqrt {e^{2} x^{2}} \vee d = \sqrt {e^{2} x^{2}}\right ) \wedge \left (d = - \sqrt {e^{2} x^{2}} \vee d = \sqrt {e^{2} x^{2}} \vee e = 0\right ) \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {3}{2}}} & \text {for}\: e = 0 \\\frac {2 d^{2}}{e^{4} \sqrt {d^{2} - e^{2} x^{2}}} - \frac {x^{2}}{e^{2} \sqrt {d^{2} - e^{2} x^{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)**(3/2),x)

[Out]

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

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