Integrand size = 27, antiderivative size = 80 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d x}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d} \]
-3/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d-1/2*(-e^2*x^2+d^2)^(1/2)/x^2-2* e*(-e^2*x^2+d^2)^(1/2)/d/x
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {(d+4 e x) \sqrt {d^2-e^2 x^2}-3 e^2 x^2 \log \left (d \left (-d-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )\right )+3 e^2 x^2 \log \left (d-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 d x^2} \]
-1/2*((d + 4*e*x)*Sqrt[d^2 - e^2*x^2] - 3*e^2*x^2*Log[d*(-d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*x^2])] + 3*e^2*x^2*Log[d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2 *x^2]])/(d*x^2)
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {540, 25, 27, 534, 243, 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.
\(\displaystyle \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 540 |
\(\displaystyle -\frac {\int -\frac {d^2 e (4 d+3 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {d^2 e (4 d+3 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} e \int \frac {4 d+3 e x}{x^2 \sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {1}{2} e \left (3 e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {4 \sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} e \left (\frac {3}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {4 \sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} e \left (-\frac {3 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {4 \sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} e \left (-\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {4 \sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {\sqrt {d^2-e^2 x^2}}{2 x^2}\) |
-1/2*Sqrt[d^2 - e^2*x^2]/x^2 + (e*((-4*Sqrt[d^2 - e^2*x^2])/(d*x) - (3*e*A rcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/2
3.1.40.3.1 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] :> 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]
Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (4 e x +d \right )}{2 d \,x^{2}}-\frac {3 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}\) | \(72\) |
default | \(-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+d^{2} \left (-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{2 d^{2} x^{2}}-\frac {e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{2} \sqrt {d^{2}}}\right )-\frac {2 e \sqrt {-e^{2} x^{2}+d^{2}}}{d x}\) | \(139\) |
-1/2*(-e^2*x^2+d^2)^(1/2)*(4*e*x+d)/d/x^2-3/2*e^2/(d^2)^(1/2)*ln((2*d^2+2* (d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.79 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 \, e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - \sqrt {-e^{2} x^{2} + d^{2}} {\left (4 \, e x + d\right )}}{2 \, d x^{2}} \]
1/2*(3*e^2*x^2*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - sqrt(-e^2*x^2 + d^2)*( 4*e*x + d))/(d*x^2)
Result contains complex when optimal does not.
Time = 2.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.68 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=d^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 d^{2} x} - \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d^{3}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{d^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {d}{e x} \right )}}{d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i \operatorname {asin}{\left (\frac {d}{e x} \right )}}{d} & \text {otherwise} \end {cases}\right ) \]
d**2*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(2*d**2*x) - e**2*acosh(d/(e *x))/(2*d**3), Abs(d**2/(e**2*x**2)) > 1), (I/(2*e*x**3*sqrt(-d**2/(e**2*x **2) + 1)) - I*e/(2*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**2*asin(d/(e *x))/(2*d**3), True)) + 2*d*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/d** 2, Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/d**2, Tru e)) + e**2*Piecewise((-acosh(d/(e*x))/d, Abs(d**2/(e**2*x**2)) > 1), (I*as in(d/(e*x))/d, True))
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} e}{d x} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{2 \, x^{2}} \]
-3/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 2*sqrt(-e ^2*x^2 + d^2)*e/(d*x) - 1/2*sqrt(-e^2*x^2 + d^2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (70) = 140\).
Time = 0.28 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.35 \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {{\left (e^{3} + \frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e}{x}\right )} e^{4} x^{2}}{8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left | e \right |}} - \frac {3 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{2 \, d {\left | e \right |}} - \frac {\frac {8 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d e {\left | e \right |}}{x} + \frac {{\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d {\left | e \right |}}{e x^{2}}}{8 \, d^{2} e^{2}} \]
1/8*(e^3 + 8*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e/x)*e^4*x^2/((d*e + sqrt (-e^2*x^2 + d^2)*abs(e))^2*d*abs(e)) - 3/2*e^3*log(1/2*abs(-2*d*e - 2*sqrt (-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d*abs(e)) - 1/8*(8*(d*e + sqrt(-e^ 2*x^2 + d^2)*abs(e))*d*e*abs(e)/x + (d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2* d*abs(e)/(e*x^2))/(d^2*e^2)
Timed out. \[ \int \frac {(d+e x)^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{x^3\,\sqrt {d^2-e^2\,x^2}} \,d x \]