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

Optimal. Leaf size=115 \[ \frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

4/5*(-e*x+d)/(-e^2*x^2+d^2)^(5/2)+1/15*(-11*e*x+5*d)/d^2/(-e^2*x^2+d^2)^(3/2)-arctanh((-e^2*x^2+d^2)^(1/2)/d)/
d^4+1/15*(-22*e*x+15*d)/d^4/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 115, 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} \[ \frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) + (5*d - 11*e*x)/(15*d^2*(d^2 - e^2*x^2)^(3/2)) + (15*d - 22*e*x)/(15*
d^4*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^4

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)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {(d-e x)^3}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3+11 d^2 e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^5 e^2+22 d^4 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^7 e^4}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^4}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^3}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \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^3}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \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^3 e^2}\\ &=\frac {4 (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d-11 e x}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-22 e x}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 76, normalized size = 0.66 \[ \frac {\frac {\sqrt {d^2-e^2 x^2} \left (32 d^2+51 d e x+22 e^2 x^2\right )}{(d+e x)^3}-15 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+15 \log (x)}{15 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(32*d^2 + 51*d*e*x + 22*e^2*x^2))/(d + e*x)^3 + 15*Log[x] - 15*Log[d + Sqrt[d^2 - e^2*x^
2]])/(15*d^4)

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fricas [A]  time = 0.95, size = 153, normalized size = 1.33 \[ \frac {32 \, e^{3} x^{3} + 96 \, d e^{2} x^{2} + 96 \, d^{2} e x + 32 \, d^{3} + 15 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (22 \, e^{2} x^{2} + 51 \, d e x + 32 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{3} x^{3} + 3 \, d^{5} e^{2} x^{2} + 3 \, d^{6} e x + d^{7}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(32*e^3*x^3 + 96*d*e^2*x^2 + 96*d^2*e*x + 32*d^3 + 15*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*log(-(d -
sqrt(-e^2*x^2 + d^2))/x) + (22*e^2*x^2 + 51*d*e*x + 32*d^2)*sqrt(-e^2*x^2 + d^2))/(d^4*e^3*x^3 + 3*d^5*e^2*x^
2 + 3*d^6*e*x + d^7)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-9*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)^7*exp(2)^3+6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*ex
p(1)^11*exp(2)+4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^9*exp(2)^2-7*(-1/2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^5*exp(2)^4+3*exp(1)^7*exp(2)^3-6*(-1/2*(-2*d*exp(1)-
2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^3*exp(2)^5-6*exp(1)^3*exp(2)^5+17/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))*exp(1)^5*exp(2)^4/x/exp(2)-4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^9*exp(2)^2
/x/exp(2))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x+exp(2))^2/(-d^4*exp(1)^9+2*d^4*exp(1)^5*exp(2)^2-d^4*exp(1)*exp(2)^4)+1/2*(-10*exp(1)^5*exp(2
)^3+4*exp(1)^9*exp(2)+12*exp(1)*exp(2)^5)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqr
t(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(d^4*exp(1)^9-2*d^4*exp(1)^5*exp(2)^2+d^4*exp(1)*exp(2)^4)-exp
(2)*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^4/exp(1)^2

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maple [A]  time = 0.01, size = 179, normalized size = 1.56 \[ -\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{3}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{5 \left (x +\frac {d}{e}\right )^{3} d^{2} e^{3}}+\frac {7 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{15 \left (x +\frac {d}{e}\right )^{2} d^{3} e^{2}}+\frac {22 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{15 \left (x +\frac {d}{e}\right ) d^{4} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d^2)^(1/2)/d^3*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+1/5/d^2/e^3/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/
e)^2*e^2)^(1/2)+7/15/d^3/e^2/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+22/15/d^4/e/(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 \[ \int \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{3} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(x*(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 \[ \int \frac {1}{x \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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