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]

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

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Rubi [A]  time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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])

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*}

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Mathematica [A]  time = 0.15, 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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fricas [A]  time = 0.93, size = 74, normalized size = 0.69 \[ \frac {3 \, e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (5 \, e^{2} x^{2} + 3 \, d e x + d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, d^{2} x^{3}} \]

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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giac [B]  time = 0.30, size = 239, normalized size = 2.23 \[ \frac {x^{3} {\left (\frac {6 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{6}}{x} + \frac {21 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{4}}{x^{2}} + e^{8}\right )} e}{24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{2}} - \frac {e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{d^{2}} - \frac {{\left (\frac {21 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{4} e^{16}}{x} + \frac {6 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{14}}{x^{2}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{12}}{x^{3}}\right )} e^{\left (-15\right )}}{24 \, d^{6}} \]

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

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maple [A]  time = 0.01, size = 114, normalized size = 1.07 \[ -\frac {e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}{3 d^{2} x}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e}{d \,x^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{3 x^{3}} \]

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]

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

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maxima [A]  time = 0.97, size = 108, normalized size = 1.01 \[ -\frac {e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{2}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2}}{3 \, d^{2} x} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e}{d x^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{3 \, x^{3}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{x^4\,\sqrt {d^2-e^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.09, size = 303, normalized size = 2.83 \[ d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2} x^{2}} - \frac {2 e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{4}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2} x^{2}} - \frac {2 i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{4}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 d^{2} x} - \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{d^{2}} & \text {otherwise} \end {cases}\right ) \]

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/(e**2*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/(e**2*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/(e
**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/d**2, True))

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