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

Optimal. Leaf size=118 \[ \frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.18, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1805, 823, 12, 266, 63, 208} \begin {gather*} \frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\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*(d + e*x)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

(2*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (5*d - 8*e*x)/(15*d^3*(d^2 - e^2*x^2)^(3/2)) + (15*d - 16*e*x)/(15
*d^5*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {(d-e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2+8 d e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^4 e^2+16 d^3 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^6 e^4}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^4}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\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^2}\\ &=\frac {2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 95, normalized size = 0.81 \begin {gather*} \frac {-15 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (26 d^3+22 d^2 e x-17 d e^2 x^2-16 e^3 x^3\right )}{(d-e x) (d+e x)^3}+15 \log (x)}{15 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(26*d^3 + 22*d^2*e*x - 17*d*e^2*x^2 - 16*e^3*x^3))/((d - e*x)*(d + e*x)^3) + 15*Log[x] -
15*Log[d + Sqrt[d^2 - e^2*x^2]])/(15*d^5)

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IntegrateAlgebraic [A]  time = 0.76, size = 111, normalized size = 0.94 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {\sqrt {d^2-e^2 x^2} \left (26 d^3+22 d^2 e x-17 d e^2 x^2-16 e^3 x^3\right )}{15 d^5 (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(26*d^3 + 22*d^2*e*x - 17*d*e^2*x^2 - 16*e^3*x^3))/(15*d^5*(d - e*x)*(d + e*x)^3) + (2*Ar
cTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^5

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fricas [A]  time = 0.40, size = 168, normalized size = 1.42 \begin {gather*} \frac {26 \, e^{4} x^{4} + 52 \, d e^{3} x^{3} - 52 \, d^{3} e x - 26 \, d^{4} + 15 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (16 \, e^{3} x^{3} + 17 \, d e^{2} x^{2} - 22 \, d^{2} e x - 26 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{5} e^{4} x^{4} + 2 \, d^{6} e^{3} x^{3} - 2 \, d^{8} e x - d^{9}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(26*e^4*x^4 + 52*d*e^3*x^3 - 52*d^3*e*x - 26*d^4 + 15*(e^4*x^4 + 2*d*e^3*x^3 - 2*d^3*e*x - d^4)*log(-(d -
sqrt(-e^2*x^2 + d^2))/x) + (16*e^3*x^3 + 17*d*e^2*x^2 - 22*d^2*e*x - 26*d^3)*sqrt(-e^2*x^2 + d^2))/(d^5*e^4*x
^4 + 2*d^6*e^3*x^3 - 2*d^8*e*x - d^9)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.01, size = 187, normalized size = 1.58 \begin {gather*} -\frac {\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 {16 e x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{5}}+\frac {1}{5 \left (x +\frac {d}{e}\right )^{2} \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2} e^{2}}+\frac {8}{15 \left (x +\frac {d}{e}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3} e}+\frac {1}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-e^2*x^2+d^2)^(1/2)/d^4-1/(d^2)^(1/2)/d^4*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+8/15/d^3/e/(x+d/
e)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-16/15/d^5*e/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+1/5/d^2/e^2/(x+d/e)^2
/(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}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{2} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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