3.41 \(\int \frac{(d+e x)^2}{x^4 \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=107 \[ -\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.137031, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1807, 835, 807, 266, 63, 208} \[ -\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^4*Sqrt[d^2 - e^2*x^2]),x]

[Out]

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{x^4 \sqrt{d^2-e^2 x^2}} \, dx &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{\int \frac{-6 d^3 e-5 d^2 e^2 x}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{3 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}+\frac{\int \frac{10 d^4 e^2+6 d^3 e^3 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{6 d^4}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{e^3 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{e \operatorname{Subst}\left (\int \frac{1}{\frac{d^2}{e^2}-\frac{x^2}{e^2}} \, dx,x,\sqrt{d^2-e^2 x^2}\right )}{d}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{3 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d x^2}-\frac{5 e^2 \sqrt{d^2-e^2 x^2}}{3 d^2 x}-\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.16066, size = 87, normalized size = 0.81 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-\frac{d \left (d^2+3 d e x+5 e^2 x^2\right )}{x^3}-\frac{3 e^3 \tanh ^{-1}\left (\sqrt{1-\frac{e^2 x^2}{d^2}}\right )}{\sqrt{1-\frac{e^2 x^2}{d^2}}}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^4*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-((d*(d^2 + 3*d*e*x + 5*e^2*x^2))/x^3) - (3*e^3*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])/Sqrt[1
- (e^2*x^2)/d^2]))/(3*d^3)

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Maple [A]  time = 0.058, size = 114, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{5\,{e}^{2}}{3\,{d}^{2}x}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{e}{d{x}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{e}^{3}}{d}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88616, size = 153, normalized size = 1.43 \begin{align*} \frac{3 \, e^{3} x^{3} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (5 \, e^{2} x^{2} + 3 \, d e x + d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \, d^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 5.23226, size = 313, normalized size = 2.93 \begin{align*} d^{2} \left (\begin{cases} - \frac{e \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2} x^{2}} - \frac{2 e^{3} \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}}{3 d^{4}} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\- \frac{i e \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2} x^{2}} - \frac{2 i e^{3} \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}}{3 d^{4}} & \text{otherwise} \end{cases}\right ) + 2 d e \left (\begin{cases} - \frac{e \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}}{2 d^{2} x} - \frac{e^{2} \operatorname{acosh}{\left (\frac{d}{e x} \right )}}{2 d^{3}} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\\frac{i}{2 e x^{3} \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} - \frac{i e}{2 d^{2} x \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} + \frac{i e^{2} \operatorname{asin}{\left (\frac{d}{e x} \right )}}{2 d^{3}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{e \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}}{d^{2}} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\- \frac{i e \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}}{d^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/x**4/(-e**2*x**2+d**2)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.20519, size = 323, normalized size = 3.02 \begin{align*} \frac{x^{3}{\left (\frac{6 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{6}}{x} + \frac{21 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{4}}{x^{2}} + e^{8}\right )} e}{24 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{2}} - \frac{e^{3} \log \left (\frac{{\left | -2 \, d e - 2 \, \sqrt{-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \,{\left | x \right |}}\right )}{d^{2}} - \frac{{\left (\frac{21 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{4} e^{16}}{x} + \frac{6 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{14}}{x^{2}} + \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{12}}{x^{3}}\right )} e^{\left (-15\right )}}{24 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/24*x^3*(6*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^6/x + 21*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^4/x^2 + e^8)*e/((d*e
+ sqrt(-x^2*e^2 + d^2)*e)^3*d^2) - e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^2 - 1/2
4*(21*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^4*e^16/x + 6*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^14/x^2 + (d*e + sqr
t(-x^2*e^2 + d^2)*e)^3*d^4*e^12/x^3)*e^(-15)/d^6