3.1.43 \(\int \frac {(d+e x)^2}{x^6 \sqrt {d^2-e^2 x^2}} \, dx\) [43]

Optimal. Leaf size=169 \[ -\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4} \]

[Out]

-3/4*e^5*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^4-1/5*(-e^2*x^2+d^2)^(1/2)/x^5-1/2*e*(-e^2*x^2+d^2)^(1/2)/d/x^4-3/5
*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^3-3/4*e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x^2-6/5*e^4*(-e^2*x^2+d^2)^(1/2)/d^4/x

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Rubi [A]
time = 0.12, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1821, 849, 821, 272, 65, 214} \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^6*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-1/5*Sqrt[d^2 - e^2*x^2]/x^5 - (e*Sqrt[d^2 - e^2*x^2])/(2*d*x^4) - (3*e^2*Sqrt[d^2 - e^2*x^2])/(5*d^2*x^3) - (
3*e^3*Sqrt[d^2 - e^2*x^2])/(4*d^3*x^2) - (6*e^4*Sqrt[d^2 - e^2*x^2])/(5*d^4*x) - (3*e^5*ArcTanh[Sqrt[d^2 - e^2
*x^2]/d])/(4*d^4)

Rule 65

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 214

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

Rule 272

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 821

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 849

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 1821

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^6 \sqrt {d^2-e^2 x^2}} \, dx &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {\int \frac {-10 d^3 e-9 d^2 e^2 x}{x^5 \sqrt {d^2-e^2 x^2}} \, dx}{5 d^2}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}+\frac {\int \frac {36 d^4 e^2+30 d^3 e^3 x}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{20 d^4}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {\int \frac {-90 d^5 e^3-72 d^4 e^4 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{60 d^6}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}+\frac {\int \frac {144 d^6 e^4+90 d^5 e^5 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{120 d^8}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {\left (3 e^5\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{4 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}+\frac {\left (3 e^5\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{8 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {\left (3 e^3\right ) \text {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 )}{4 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{5 x^5}-\frac {e \sqrt {d^2-e^2 x^2}}{2 d x^4}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{5 d^2 x^3}-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{4 d^3 x^2}-\frac {6 e^4 \sqrt {d^2-e^2 x^2}}{5 d^4 x}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^4}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 115, normalized size = 0.68 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-4 d^4-10 d^3 e x-12 d^2 e^2 x^2-15 d e^3 x^3-24 e^4 x^4\right )}{20 d^4 x^5}+\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^6*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-4*d^4 - 10*d^3*e*x - 12*d^2*e^2*x^2 - 15*d*e^3*x^3 - 24*e^4*x^4))/(20*d^4*x^5) + (3*e^5
*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/(2*d^4)

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Maple [A]
time = 0.09, size = 240, normalized size = 1.42

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (24 e^{4} x^{4}+15 d \,e^{3} x^{3}+12 d^{2} x^{2} e^{2}+10 d^{3} e x +4 d^{4}\right )}{20 x^{5} d^{4}}-\frac {3 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{4 d^{3} \sqrt {d^{2}}}\)	\(110\)
default	\(d^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{5 d^{2} x^{5}}+\frac {4 e^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{2} x^{3}}-\frac {2 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{4} x}\right )}{5 d^{2}}\right )+2 d e \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{4 d^{2} x^{4}}+\frac {3 e^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{2} \sqrt {d^{2}}}\right )}{4 d^{2}}\right )+e^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{2} x^{3}}-\frac {2 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{4} x}\right )\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

d^2*(-1/5/d^2/x^5*(-e^2*x^2+d^2)^(1/2)+4/5*e^2/d^2*(-1/3/d^2/x^3*(-e^2*x^2+d^2)^(1/2)-2/3*e^2/d^4/x*(-e^2*x^2+
d^2)^(1/2)))+2*d*e*(-1/4/d^2/x^4*(-e^2*x^2+d^2)^(1/2)+3/4*e^2/d^2*(-1/2/d^2/x^2*(-e^2*x^2+d^2)^(1/2)-1/2*e^2/d
^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)))+e^2*(-1/3/d^2/x^3*(-e^2*x^2+d^2)^(1/2)-2/3*e
^2/d^4/x*(-e^2*x^2+d^2)^(1/2))

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Maxima [A]
time = 0.48, size = 149, normalized size = 0.88 \begin {gather*} -\frac {3 \, e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{4 \, d^{4}} - \frac {6 \, \sqrt {-x^{2} e^{2} + d^{2}} e^{4}}{5 \, d^{4} x} - \frac {3 \, \sqrt {-x^{2} e^{2} + d^{2}} e^{3}}{4 \, d^{3} x^{2}} - \frac {3 \, \sqrt {-x^{2} e^{2} + d^{2}} e^{2}}{5 \, d^{2} x^{3}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} e}{2 \, d x^{4}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-3/4*e^5*log(2*d^2/abs(x) + 2*sqrt(-x^2*e^2 + d^2)*d/abs(x))/d^4 - 6/5*sqrt(-x^2*e^2 + d^2)*e^4/(d^4*x) - 3/4*
sqrt(-x^2*e^2 + d^2)*e^3/(d^3*x^2) - 3/5*sqrt(-x^2*e^2 + d^2)*e^2/(d^2*x^3) - 1/2*sqrt(-x^2*e^2 + d^2)*e/(d*x^
4) - 1/5*sqrt(-x^2*e^2 + d^2)/x^5

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Fricas [A]
time = 1.82, size = 93, normalized size = 0.55 \begin {gather*} \frac {15 \, x^{5} e^{5} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) - {\left (24 \, x^{4} e^{4} + 15 \, d x^{3} e^{3} + 12 \, d^{2} x^{2} e^{2} + 10 \, d^{3} x e + 4 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{20 \, d^{4} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/20*(15*x^5*e^5*log(-(d - sqrt(-x^2*e^2 + d^2))/x) - (24*x^4*e^4 + 15*d*x^3*e^3 + 12*d^2*x^2*e^2 + 10*d^3*x*e
+ 4*d^4)*sqrt(-x^2*e^2 + d^2))/(d^4*x^5)

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Sympy [C] Result contains complex when optimal does not.
time = 4.44, size = 510, normalized size = 3.02 \begin {gather*} d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{5 d^{2} x^{4}} - \frac {4 e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{15 d^{4} x^{2}} - \frac {8 e^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{15 d^{6}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{5 d^{2} x^{4}} - \frac {4 i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{15 d^{4} x^{2}} - \frac {8 i e^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{15 d^{6}} & \text {otherwise} \end {cases}\right ) + 2 d e \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}\: \left |{\frac {d^{2}}{e^{2} 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 ) + e^{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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/x**6/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(5*d**2*x**4) - 4*e**3*sqrt(d**2/(e**2*x**2) - 1)/(15*d**4*x**2)
- 8*e**5*sqrt(d**2/(e**2*x**2) - 1)/(15*d**6), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/
(5*d**2*x**4) - 4*I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(15*d**4*x**2) - 8*I*e**5*sqrt(-d**2/(e**2*x**2) + 1)/(15
*d**6), True)) + 2*d*e*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/(e**2*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)) + e**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))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (136) = 272\).
time = 1.57, size = 363, normalized size = 2.15 \begin {gather*} \frac {x^{5} {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{3}}{x} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e}{x^{2}} + \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-1\right )}}{x^{3}} + \frac {110 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-3\right )}}{x^{4}} + e^{5}\right )} e^{10}}{160 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{4}} - \frac {3 \, e^{5} \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 )}{4 \, d^{4}} - \frac {\frac {110 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{16} e^{3}}{x} + \frac {40 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{16} e}{x^{2}} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{16} e^{\left (-1\right )}}{x^{3}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{16} e^{\left (-3\right )}}{x^{4}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{16} e^{\left (-5\right )}}{x^{5}}}{160 \, d^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^6/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/160*x^5*(5*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^3/x + 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e/x^2 + 40*(d*e + sqrt
(-x^2*e^2 + d^2)*e)^3*e^(-1)/x^3 + 110*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(-3)/x^4 + e^5)*e^10/((d*e + sqrt(-x
^2*e^2 + d^2)*e)^5*d^4) - 3/4*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/160*(1
10*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^16*e^3/x + 40*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^16*e/x^2 + 15*(d*e + sqrt
(-x^2*e^2 + d^2)*e)^3*d^16*e^(-1)/x^3 + 5*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^16*e^(-3)/x^4 + (d*e + sqrt(-x^2*
e^2 + d^2)*e)^5*d^16*e^(-5)/x^5)/d^20

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^2}{x^6\,\sqrt {d^2-e^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(x^6*(d^2 - e^2*x^2)^(1/2)),x)

[Out]

int((d + e*x)^2/(x^6*(d^2 - e^2*x^2)^(1/2)), x)

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