Integrand size = 27, antiderivative size = 108 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=-\frac {5 e^2 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}+\frac {5 e^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \]
-1/4*(-e^2*x^2+d^2)^(3/2)/x^4+2/3*e*(-e^2*x^2+d^2)^(3/2)/d/x^3+5/8*e^4*arc tanh((-e^2*x^2+d^2)^(1/2)/d)/d-5/8*e^2*(-e^2*x^2+d^2)^(1/2)/x^2
Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^3+16 d^2 e x-9 d e^2 x^2-16 e^3 x^3\right )}{24 d x^4}+\frac {5 e^4 \log (x)}{8 \sqrt {d^2}}-\frac {5 e^4 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{8 \sqrt {d^2}} \]
(Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 16*d^2*e*x - 9*d*e^2*x^2 - 16*e^3*x^3))/(24 *d*x^4) + (5*e^4*Log[x])/(8*Sqrt[d^2]) - (5*e^4*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(8*Sqrt[d^2])
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {570, 540, 27, 534, 243, 51, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^5}dx\) |
\(\Big \downarrow \) 540 |
\(\displaystyle -\frac {\int \frac {d^2 e (8 d-5 e x) \sqrt {d^2-e^2 x^2}}{x^4}dx}{4 d^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} e \int \frac {(8 d-5 e x) \sqrt {d^2-e^2 x^2}}{x^4}dx-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {1}{4} e \left (-5 e \int \frac {\sqrt {d^2-e^2 x^2}}{x^3}dx-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {1}{4} e \left (-\frac {5}{2} e \int \frac {\sqrt {d^2-e^2 x^2}}{x^4}dx^2-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {1}{4} e \left (-\frac {5}{2} e \left (-\frac {1}{2} e^2 \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {1}{4} e \left (-\frac {5}{2} e \left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {1}{4} e \left (-\frac {5}{2} e \left (\frac {e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {8 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}\) |
-1/4*(d^2 - e^2*x^2)^(3/2)/x^4 - (e*((-8*(d^2 - e^2*x^2)^(3/2))/(3*d*x^3) - (5*e*(-(Sqrt[d^2 - e^2*x^2]/x^2) + (e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/ d))/2))/4
3.2.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) , x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (16 e^{3} x^{3}+9 d \,e^{2} x^{2}-16 d^{2} e x +6 d^{3}\right )}{24 x^{4} d}+\frac {5 e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}\) | \(96\) |
default | \(\text {Expression too large to display}\) | \(1153\) |
-1/24*(-e^2*x^2+d^2)^(1/2)*(16*e^3*x^3+9*d*e^2*x^2-16*d^2*e*x+6*d^3)/x^4/d +5/8*e^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=-\frac {15 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (16 \, e^{3} x^{3} + 9 \, d e^{2} x^{2} - 16 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, d x^{4}} \]
-1/24*(15*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (16*e^3*x^3 + 9*d*e ^2*x^2 - 16*d^2*e*x + 6*d^3)*sqrt(-e^2*x^2 + d^2))/(d*x^4)
Result contains complex when optimal does not.
Time = 5.18 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.91 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=d^{2} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 x} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) \]
d**2*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3* sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*x* *5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1 )) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/ (8*d**3), True)) - 2*d*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), ( -I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), True)) + e**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(2*x ) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(2*e*x* *3*sqrt(-d**2/(e**2*x**2) + 1)) - I*e/(2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**2*asin(d/(e*x))/(2*d), True))
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.20 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {5 \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d^{2}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d^{2} x^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{3 \, d x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{4 \, x^{4}} \]
5/8*e^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 5/8*sqrt(- e^2*x^2 + d^2)*e^4/d^2 - 5/8*(-e^2*x^2 + d^2)^(3/2)*e^2/(d^2*x^2) + 2/3*(- e^2*x^2 + d^2)^(3/2)*e/(d*x^3) - 1/4*(-e^2*x^2 + d^2)^(3/2)/x^4
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.28 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\frac {1}{192} \, {\left (\frac {120 \, e^{3} \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} - \frac {120 \, e^{3} \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} + \frac {4 \, {\left (15 \, e^{3} \log \left (2\right ) - 30 \, e^{3} \log \left (i + 1\right ) + 32 i \, e^{3}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d} - \frac {15 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 73 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 55 \, e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 15 \, e^{3} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d {\left (\frac {d}{e x + d} - 1\right )}^{4}}\right )} {\left | e \right |} \]
1/192*(120*e^3*log(sqrt(2*d/(e*x + d) - 1) + 1)*sgn(1/(e*x + d))*sgn(e)/d - 120*e^3*log(abs(sqrt(2*d/(e*x + d) - 1) - 1))*sgn(1/(e*x + d))*sgn(e)/d + 4*(15*e^3*log(2) - 30*e^3*log(I + 1) + 32*I*e^3)*sgn(1/(e*x + d))*sgn(e) /d - (15*e^3*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) + 73*e^3*(2 *d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) - 55*e^3*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) + 15*e^3*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e *x + d))*sgn(e))/(d*(d/(e*x + d) - 1)^4))*abs(e)
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^5 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^5\,{\left (d+e\,x\right )}^2} \,d x \]