3.2.85 \(\int \frac {1}{x (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [185]

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

[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 = 0.11, 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 = {866, 1819, 837, 12, 272, 65, 214} \begin {gather*} \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} \end {gather*}

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 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 837

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] +
Dist[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 866

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 1819

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 {\text {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 {\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 )}{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 = 0.45, size = 88, normalized size = 0.77 \begin {gather*} \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}+30 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^4} \end {gather*}

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 + 30*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^
2*x^2])/d])/(15*d^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(101)=202\).
time = 0.07, size = 332, normalized size = 2.89

method	result	size
default	\(\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e \,d^{4} \left (x +\frac {d}{e}\right )}-\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}}{e^{2} d}-\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{e \,d^{2}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}\)	\(332\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(-x^2*e^2 + d^2)*(x*e + d)^3*x), x)

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Fricas [A]
time = 2.67, size = 148, normalized size = 1.29 \begin {gather*} \frac {32 \, x^{3} e^{3} + 96 \, d x^{2} e^{2} + 96 \, d^{2} x e + 32 \, d^{3} + 15 \, {\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) + {\left (22 \, x^{2} e^{2} + 51 \, d x e + 32 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{4} x^{3} e^{3} + 3 \, d^{5} x^{2} e^{2} + 3 \, d^{6} x e + d^{7}\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \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/(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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Giac [A]
time = 1.83, size = 195, normalized size = 1.70 \begin {gather*} -\frac {\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 )}{d^{4}} - \frac {2 \, {\left (\frac {115 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + \frac {185 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} + \frac {135 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-6\right )}}{x^{3}} + \frac {45 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{\left (-8\right )}}{x^{4}} + 32\right )}}{15 \, d^{4} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

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]

-log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 2/15*(115*(d*e + sqrt(-x^2*e^2 + d^2)*e)*
e^(-2)/x + 185*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^(-4)/x^2 + 135*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^(-6)/x^3 +
45*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^(-8)/x^4 + 32)/(d^4*((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^5)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,\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*(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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