3.42 \(\int \frac{(d+e x)^2}{x^5 \sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{7 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d^3} \]

[Out]

-Sqrt[d^2 - e^2*x^2]/(4*x^4) - (2*e*Sqrt[d^2 - e^2*x^2])/(3*d*x^3) - (7*e^2*Sqrt[d^2 - e^2*x^2])/(8*d^2*x^2) -
(4*e^3*Sqrt[d^2 - e^2*x^2])/(3*d^3*x) - (7*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(8*d^3)

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Rubi [A]  time = 0.171534, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{7 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^5*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-Sqrt[d^2 - e^2*x^2]/(4*x^4) - (2*e*Sqrt[d^2 - e^2*x^2])/(3*d*x^3) - (7*e^2*Sqrt[d^2 - e^2*x^2])/(8*d^2*x^2) -
(4*e^3*Sqrt[d^2 - e^2*x^2])/(3*d^3*x) - (7*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(8*d^3)

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^5 \sqrt{d^2-e^2 x^2}} \, dx &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{\int \frac{-8 d^3 e-7 d^2 e^2 x}{x^4 \sqrt{d^2-e^2 x^2}} \, dx}{4 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}+\frac{\int \frac{21 d^4 e^2+16 d^3 e^3 x}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{12 d^4}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{\int \frac{-32 d^5 e^3-21 d^4 e^4 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{24 d^6}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{\left (7 e^4\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{8 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}+\frac{\left (7 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{16 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{\left (7 e^2\right ) \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 )}{8 d^2}\\ &=-\frac{\sqrt{d^2-e^2 x^2}}{4 x^4}-\frac{2 e \sqrt{d^2-e^2 x^2}}{3 d x^3}-\frac{7 e^2 \sqrt{d^2-e^2 x^2}}{8 d^2 x^2}-\frac{4 e^3 \sqrt{d^2-e^2 x^2}}{3 d^3 x}-\frac{7 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{8 d^3}\\ \end{align*}

Mathematica [C]  time = 0.160776, size = 155, normalized size = 1.11 \[ -\frac{e \sqrt{d^2-e^2 x^2} \left (6 e^3 x^3 \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )+d \left (4 d^2+3 d e x+8 e^2 x^2\right ) \sqrt{1-\frac{e^2 x^2}{d^2}}+3 e^3 x^3 \tanh ^{-1}\left (\sqrt{1-\frac{e^2 x^2}{d^2}}\right )\right )}{6 d^4 x^3 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^5*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-(e*Sqrt[d^2 - e^2*x^2]*(d*(4*d^2 + 3*d*e*x + 8*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] + 3*e^3*x^3*ArcTanh[Sqrt[1 -
(e^2*x^2)/d^2]] + 6*e^3*x^3*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[1/2, 3, 3/2, 1 - (e^2*x^2)/d^2]))/(6*d^4
*x^3*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [A]  time = 0.059, size = 139, normalized size = 1. \begin{align*} -{\frac{2\,e}{3\,d{x}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{4\,{e}^{3}}{3\,{d}^{3}x}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{1}{4\,{x}^{4}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{7\,{e}^{2}}{8\,{d}^{2}{x}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{7\,{e}^{4}}{8\,{d}^{2}}\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^5/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-2/3*e*(-e^2*x^2+d^2)^(1/2)/d/x^3-4/3*e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x-1/4*(-e^2*x^2+d^2)^(1/2)/x^4-7/8*e^2*(-e^
2*x^2+d^2)^(1/2)/d^2/x^2-7/8/d^2*e^4/(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^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83666, size = 184, normalized size = 1.31 \begin{align*} \frac{21 \, e^{4} x^{4} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (32 \, e^{3} x^{3} + 21 \, d e^{2} x^{2} + 16 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \, d^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/24*(21*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (32*e^3*x^3 + 21*d*e^2*x^2 + 16*d^2*e*x + 6*d^3)*sqrt(-e
^2*x^2 + d^2))/(d^3*x^4)

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Sympy [C]  time = 9.21469, size = 459, normalized size = 3.28 \begin{align*} d^{2} \left (\begin{cases} - \frac{1}{4 e x^{5} \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} - \frac{e}{8 d^{2} x^{3} \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} + \frac{3 e^{3}}{8 d^{4} x \sqrt{\frac{d^{2}}{e^{2} x^{2}} - 1}} - \frac{3 e^{4} \operatorname{acosh}{\left (\frac{d}{e x} \right )}}{8 d^{5}} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\\frac{i}{4 e x^{5} \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} + \frac{i e}{8 d^{2} x^{3} \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} - \frac{3 i e^{3}}{8 d^{4} x \sqrt{- \frac{d^{2}}{e^{2} x^{2}} + 1}} + \frac{3 i e^{4} \operatorname{asin}{\left (\frac{d}{e x} \right )}}{8 d^{5}} & \text{otherwise} \end{cases}\right ) + 2 d e \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 ) + e^{2} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/x**5/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((-1/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) - e/(8*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) + 3*e**3
/(8*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) - 3*e**4*acosh(d/(e*x))/(8*d**5), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1),
(I/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)) + I*e/(8*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e**3/(8*d**4*
x*sqrt(-d**2/(e**2*x**2) + 1)) + 3*I*e**4*asin(d/(e*x))/(8*d**5), True)) + 2*d*e*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)) + e**2
*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*a
sin(d/(e*x))/(2*d**3), True))

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Giac [B]  time = 1.15415, size = 412, normalized size = 2.94 \begin{align*} \frac{x^{4}{\left (\frac{16 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{8}}{x} + \frac{48 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{6}}{x^{2}} + \frac{144 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{4}}{x^{3}} + 3 \, e^{10}\right )} e^{2}}{192 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{3}} - \frac{7 \, e^{4} \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 )}{8 \, d^{3}} - \frac{{\left (\frac{144 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{9} e^{26}}{x} + \frac{48 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{9} e^{24}}{x^{2}} + \frac{16 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{9} e^{22}}{x^{3}} + \frac{3 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{9} e^{20}}{x^{4}}\right )} e^{\left (-24\right )}}{192 \, d^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/192*x^4*(16*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^8/x + 48*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^6/x^2 + 144*(d*e +
sqrt(-x^2*e^2 + d^2)*e)^3*e^4/x^3 + 3*e^10)*e^2/((d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^3) - 7/8*e^4*log(1/2*abs(-
2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^3 - 1/192*(144*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^9*e^26/x +
48*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^9*e^24/x^2 + 16*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^9*e^22/x^3 + 3*(d*e +
sqrt(-x^2*e^2 + d^2)*e)^4*d^9*e^20/x^4)*e^(-24)/d^12