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

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

[Out]

-7/8*e^4*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^3-1/4*(-e^2*x^2+d^2)^(1/2)/x^4-2/3*e*(-e^2*x^2+d^2)^(1/2)/d/x^3-7/8
*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^2-4/3*e^3*(-e^2*x^2+d^2)^(1/2)/d^3/x

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Rubi [A]  time = 0.17, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 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^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*}

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Mathematica [C]  time = 0.15, size = 155, normalized size = 1.11 \[ -\frac {e \sqrt {d^2-e^2 x^2} \left (d \left (4 d^2+3 d e x+8 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}+6 e^3 x^3 \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+3 e^3 x^3 \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )\right )}{6 d^4 x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 87, normalized size = 0.62 \[ \frac {21 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (32 \, e^{3} x^{3} + 21 \, d e^{2} x^{2} + 16 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, d^{3} x^{4}} \]

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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giac [B]  time = 0.29, size = 305, normalized size = 2.18 \[ \frac {x^{4} {\left (\frac {16 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{8}}{x} + \frac {48 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{6}}{x^{2}} + \frac {144 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{4}}{x^{3}} + 3 \, e^{10}\right )} e^{2}}{192 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{3}} - \frac {7 \, e^{4} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{8 \, d^{3}} - \frac {{\left (\frac {144 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{9} e^{26}}{x} + \frac {48 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{9} e^{24}}{x^{2}} + \frac {16 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{9} e^{22}}{x^{3}} + \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{9} e^{20}}{x^{4}}\right )} e^{\left (-24\right )}}{192 \, d^{12}} \]

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

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maple [A]  time = 0.02, size = 139, normalized size = 0.99 \[ -\frac {7 e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}\, d^{2}}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{3 d^{3} x}-\frac {7 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}{8 d^{2} x^{2}}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, e}{3 d \,x^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{4 x^{4}} \]

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-7/8*e^2*(-e^2*x^2+d^2)^(1/2)/d^2/x^2-7/8*
e^4/d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/4*(-e^2*x^2+d^2)^(1/2)/x^4

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maxima [A]  time = 0.97, size = 133, normalized size = 0.95 \[ -\frac {7 \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d^{3}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{3 \, d^{3} x} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2}}{8 \, d^{2} x^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} e}{3 \, d x^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{4 \, x^{4}} \]

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]

-7/8*e^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^3 - 4/3*sqrt(-e^2*x^2 + d^2)*e^3/(d^3*x) - 7/8*
sqrt(-e^2*x^2 + d^2)*e^2/(d^2*x^2) - 2/3*sqrt(-e^2*x^2 + d^2)*e/(d*x^3) - 1/4*sqrt(-e^2*x^2 + d^2)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{x^5\,\sqrt {d^2-e^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.32, size = 449, normalized size = 3.21 \[ d^{2} \left (\begin {cases} - \frac {1}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e}{8 d^{2} x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e^{3}}{8 d^{4} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {3 e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{5}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e}{8 d^{2} x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e^{3}}{8 d^{4} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {3 i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{5}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2} x^{2}} - \frac {2 e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{4}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2} x^{2}} - \frac {2 i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{4}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 d^{2} x} - \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {otherwise} \end {cases}\right ) \]

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/(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)) + 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/(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)) + 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/(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))

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