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

Optimal. Leaf size=183 \[ \frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.38, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1805, 1807, 807, 266, 63, 208} \begin {gather*} \frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^2*(d - e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e^2*(25*d - 31*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(90*d - 107*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^5*x^2) + (3*e*Sqrt[d^2 - e^2*x^2])/(
d^6*x) - (13*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^6)

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]

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 {1}{x^3 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {(d-e x)^3}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^3+15 d^2 e x-20 d e^2 x^2+16 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^3-45 d^2 e x+75 d e^2 x^2-62 e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^3+45 d^2 e x-90 d e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {\int \frac {-90 d^4 e+195 d^3 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^5}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {\left (13 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^5}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 \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 )}{2 d^5}\\ &=\frac {4 e^2 (d-e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (25 d-31 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e^2 (90 d-107 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^5 x^2}+\frac {3 e \sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^6}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 107, normalized size = 0.58 \begin {gather*} \frac {-195 e^2 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-15 d^4+45 d^3 e x+479 d^2 e^2 x^2+717 d e^3 x^3+304 e^4 x^4\right )}{x^2 (d+e x)^3}+195 e^2 \log (x)}{30 d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-15*d^4 + 45*d^3*e*x + 479*d^2*e^2*x^2 + 717*d*e^3*x^3 + 304*e^4*x^4))/(x^2*(d + e*x)^3
) + 195*e^2*Log[x] - 195*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(30*d^6)

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IntegrateAlgebraic [A]  time = 0.96, size = 120, normalized size = 0.66 \begin {gather*} \frac {13 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {\sqrt {d^2-e^2 x^2} \left (-15 d^4+45 d^3 e x+479 d^2 e^2 x^2+717 d e^3 x^3+304 e^4 x^4\right )}{30 d^6 x^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-15*d^4 + 45*d^3*e*x + 479*d^2*e^2*x^2 + 717*d*e^3*x^3 + 304*e^4*x^4))/(30*d^6*x^2*(d +
e*x)^3) + (13*e^2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/d^6

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fricas [A]  time = 0.42, size = 202, normalized size = 1.10 \begin {gather*} \frac {254 \, e^{5} x^{5} + 762 \, d e^{4} x^{4} + 762 \, d^{2} e^{3} x^{3} + 254 \, d^{3} e^{2} x^{2} + 195 \, {\left (e^{5} x^{5} + 3 \, d e^{4} x^{4} + 3 \, d^{2} e^{3} x^{3} + d^{3} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (304 \, e^{4} x^{4} + 717 \, d e^{3} x^{3} + 479 \, d^{2} e^{2} x^{2} + 45 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{6} e^{3} x^{5} + 3 \, d^{7} e^{2} x^{4} + 3 \, d^{8} e x^{3} + d^{9} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(254*e^5*x^5 + 762*d*e^4*x^4 + 762*d^2*e^3*x^3 + 254*d^3*e^2*x^2 + 195*(e^5*x^5 + 3*d*e^4*x^4 + 3*d^2*e^3
*x^3 + d^3*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (304*e^4*x^4 + 717*d*e^3*x^3 + 479*d^2*e^2*x^2 + 45*d
^3*e*x - 15*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^3*x^5 + 3*d^7*e^2*x^4 + 3*d^8*e*x^3 + d^9*x^2)

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giac [A]  time = 0.29, 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^3/(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 = 222, normalized size = 1.21 \begin {gather*} -\frac {13 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{5}}+\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^{4} e}+\frac {17 \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^{5}}+\frac {107 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e}{15 \left (x +\frac {d}{e}\right ) d^{6}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e}{d^{6} x}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{5} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*e*(-e^2*x^2+d^2)^(1/2)/d^6/x-1/2*(-e^2*x^2+d^2)^(1/2)/d^5/x^2-13/2/(d^2)^(1/2)/d^5*e^2*ln((2*d^2+2*(d^2)^(1/
2)*(-e^2*x^2+d^2)^(1/2))/x)+1/5/d^4/e/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+17/15/d^5/(x+d/e)^2*(2*(x+
d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+107/15/d^6*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 \begin {gather*} \int \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{3} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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

[Out]

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

[Out]

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

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