3.1.17 \(\int \frac {x^2 (d+e x)}{(d^2-e^2 x^2)^{3/2}} \, dx\) [17]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {811, 655, 223, 209, 651} \begin {gather*} -\frac {d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}+\frac {d (d+e x)}{e^3 \sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{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 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 651

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 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)}{\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 \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 (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 = 0.20, size = 79, normalized size = 1.08 \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))/(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*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(-e^2)^(3
/2)

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Maple [A]
time = 0.06, size = 103, normalized size = 1.41

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}-\frac {d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{e^{4} \left (x -\frac {d}{e}\right )}\)	\(99\)
default	\(e \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )+d \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )\)	\(103\)

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,method=_RETURNVERBOSE)

[Out]

e*(-x^2/e^2/(-e^2*x^2+d^2)^(1/2)+2*d^2/e^4/(-e^2*x^2+d^2)^(1/2))+d*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/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 = 0.48, size = 72, normalized size = 0.99 \begin {gather*} -d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \frac {x^{2} e^{\left (-1\right )}}{\sqrt {-x^{2} e^{2} + d^{2}}} + \frac {d x e^{\left (-2\right )}}{\sqrt {-x^{2} e^{2} + d^{2}}} + \frac {2 \, d^{2} e^{\left (-3\right )}}{\sqrt {-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="maxima")

[Out]

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

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Fricas [A]
time = 2.34, size = 85, normalized size = 1.16 \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)^(3/2),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 [C] Result contains complex when optimal does not.
time = 4.06, size = 163, normalized size = 2.23 \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}\: d = 0 \wedge e = 0 \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {3}{2}}} & \text {for}\: e = 0 \\\tilde {\infty } x^{4} & \text {for}\: d = - \sqrt {e^{2} x^{2}} \vee d = \sqrt {e^{2} x^{2}} \\\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(e, 0)), (x**4/(
4*(d**2)**(3/2)), Eq(e, 0)), (zoo*x**4, Eq(d, sqrt(e**2*x**2)) | Eq(d, -sqrt(e**2*x**2))), (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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Giac [A]
time = 1.08, size = 68, normalized size = 0.93 \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)^(3/2),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 [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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