3.2.86 \(\int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.30, antiderivative size = 146, 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 = {852, 1805, 807, 266, 63, 208} \begin {gather*} -\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) - (e*(5*d - 7*e*x))/(5*d^3*(d^2 - e^2*x^2)^(3/2)) - (e*(15*d - 19
*e*x))/(5*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^5*x) + (3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^5

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 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {(d-e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3+15 d^2 e x-16 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3-45 d^2 e x+42 d e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3+45 d^2 e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {(3 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}-\frac {(3 e) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 \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^4 e}\\ &=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 92, normalized size = 0.63 \begin {gather*} -\frac {-15 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (5 d^3+39 d^2 e x+57 d e^2 x^2+24 e^3 x^3\right )}{x (d+e x)^3}+15 e \log (x)}{5 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*((Sqrt[d^2 - e^2*x^2]*(5*d^3 + 39*d^2*e*x + 57*d*e^2*x^2 + 24*e^3*x^3))/(x*(d + e*x)^3) + 15*e*Log[x] - 1
5*e*Log[d + Sqrt[d^2 - e^2*x^2]])/d^5

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IntegrateAlgebraic [A]  time = 0.70, size = 107, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-5 d^3-39 d^2 e x-57 d e^2 x^2-24 e^3 x^3\right )}{5 d^5 x (d+e x)^3}-\frac {6 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-5*d^3 - 39*d^2*e*x - 57*d*e^2*x^2 - 24*e^3*x^3))/(5*d^5*x*(d + e*x)^3) - (6*e*ArcTanh[(
Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^5

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fricas [A]  time = 0.41, size = 181, normalized size = 1.24 \begin {gather*} -\frac {24 \, e^{4} x^{4} + 72 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 24 \, d^{3} e x + 15 \, {\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (24 \, e^{3} x^{3} + 57 \, d e^{2} x^{2} + 39 \, d^{2} e x + 5 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{5} e^{3} x^{4} + 3 \, d^{6} e^{2} x^{3} + 3 \, d^{7} e x^{2} + d^{8} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(24*e^4*x^4 + 72*d*e^3*x^3 + 72*d^2*e^2*x^2 + 24*d^3*e*x + 15*(e^4*x^4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2 + d^
3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (24*e^3*x^3 + 57*d*e^2*x^2 + 39*d^2*e*x + 5*d^3)*sqrt(-e^2*x^2 + d
^2))/(d^5*e^3*x^4 + 3*d^6*e^2*x^3 + 3*d^7*e*x^2 + d^8*x)

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giac [A]  time = 0.30, size = 1, normalized size = 0.01 \begin {gather*} +\infty \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

+Infinity

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maple [A]  time = 0.01, size = 199, normalized size = 1.36 \begin {gather*} \frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{4}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{5 \left (x +\frac {d}{e}\right )^{3} d^{3} e^{2}}-\frac {4 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{5 \left (x +\frac {d}{e}\right )^{2} d^{4} e}-\frac {19 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{5 \left (x +\frac {d}{e}\right ) d^{5}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-e^2*x^2+d^2)^(1/2)/d^5/x+3/(d^2)^(1/2)/d^4*e*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/5/d^3/e^2/(
x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-4/5/d^4/e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-19/5/d^5/
(x+d/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{3} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

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