Integrand size = 27, antiderivative size = 111 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {c^2 d \sqrt {c^2-d^2 x^2}}{4 x^4}-\frac {5 d^3 \sqrt {c^2-d^2 x^2}}{8 x^2}-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}+\frac {3 d^5 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{8 c} \] Output:
1/4*c^2*d*(-d^2*x^2+c^2)^(1/2)/x^4-5/8*d^3*(-d^2*x^2+c^2)^(1/2)/x^2-1/5*(- d^2*x^2+c^2)^(5/2)/c/x^5+3/8*d^5*arctanh((-d^2*x^2+c^2)^(1/2)/c)/c
Time = 0.44 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {1}{40} \left (\frac {\sqrt {c^2-d^2 x^2} \left (-8 c^4+10 c^3 d x+16 c^2 d^2 x^2-25 c d^3 x^3-8 d^4 x^4\right )}{c x^5}+\frac {15 d^5 \log (x)}{\sqrt {c^2}}-\frac {15 d^5 \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right )}{\sqrt {c^2}}\right ) \] Input:
Integrate[(c^2 - d^2*x^2)^(5/2)/(x^6*(c + d*x)),x]
Output:
((Sqrt[c^2 - d^2*x^2]*(-8*c^4 + 10*c^3*d*x + 16*c^2*d^2*x^2 - 25*c*d^3*x^3 - 8*d^4*x^4))/(c*x^5) + (15*d^5*Log[x])/Sqrt[c^2] - (15*d^5*Log[Sqrt[c^2] - Sqrt[c^2 - d^2*x^2]])/Sqrt[c^2])/40
Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {566, 534, 243, 51, 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.
\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx\) |
\(\Big \downarrow \) 566 |
\(\displaystyle \int \frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{x^6}dx\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -d \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^5}dx-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {1}{2} d \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^6}dx^2-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {1}{2} d \left (-\frac {3}{4} d^2 \int \frac {\sqrt {c^2-d^2 x^2}}{x^4}dx^2-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {1}{2} d \left (-\frac {3}{4} d^2 \left (-\frac {1}{2} d^2 \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2-\frac {\sqrt {c^2-d^2 x^2}}{x^2}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {1}{2} d \left (-\frac {3}{4} d^2 \left (\int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}-\frac {\sqrt {c^2-d^2 x^2}}{x^2}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {1}{2} d \left (-\frac {3}{4} d^2 \left (\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c}-\frac {\sqrt {c^2-d^2 x^2}}{x^2}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{5 c x^5}\) |
Input:
Int[(c^2 - d^2*x^2)^(5/2)/(x^6*(c + d*x)),x]
Output:
-1/5*(c^2 - d^2*x^2)^(5/2)/(c*x^5) - (d*(-1/2*(c^2 - d^2*x^2)^(3/2)/x^4 - (3*d^2*(-(Sqrt[c^2 - d^2*x^2]/x^2) + (d^2*ArcTanh[Sqrt[c^2 - d^2*x^2]/c])/ c))/4))/2
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_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Int[x^m*(a/c + b*(x/d))*(a + b*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0]
Time = 0.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (8 d^{4} x^{4}+25 c \,d^{3} x^{3}-16 d^{2} c^{2} x^{2}-10 c^{3} d x +8 c^{4}\right )}{40 x^{5} c}+\frac {3 d^{5} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{8 \sqrt {c^{2}}}\) | \(107\) |
default | \(\text {Expression too large to display}\) | \(1100\) |
Input:
int((-d^2*x^2+c^2)^(5/2)/x^6/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-1/40*(-d^2*x^2+c^2)^(1/2)*(8*d^4*x^4+25*c*d^3*x^3-16*c^2*d^2*x^2-10*c^3*d *x+8*c^4)/x^5/c+3/8*d^5/(c^2)^(1/2)*ln((2*c^2+2*(c^2)^(1/2)*(-d^2*x^2+c^2) ^(1/2))/x)
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=-\frac {15 \, d^{5} x^{5} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + {\left (8 \, d^{4} x^{4} + 25 \, c d^{3} x^{3} - 16 \, c^{2} d^{2} x^{2} - 10 \, c^{3} d x + 8 \, c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{40 \, c x^{5}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/x^6/(d*x+c),x, algorithm="fricas")
Output:
-1/40*(15*d^5*x^5*log(-(c - sqrt(-d^2*x^2 + c^2))/x) + (8*d^4*x^4 + 25*c*d ^3*x^3 - 16*c^2*d^2*x^2 - 10*c^3*d*x + 8*c^4)*sqrt(-d^2*x^2 + c^2))/(c*x^5 )
Result contains complex when optimal does not.
Time = 5.25 (sec) , antiderivative size = 774, normalized size of antiderivative = 6.97 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((-d**2*x**2+c**2)**(5/2)/x**6/(d*x+c),x)
Output:
c**3*Piecewise((3*I*c**3*sqrt(-1 + d**2*x**2/c**2)/(-15*c**2*x**5 + 15*d** 2*x**7) - 4*I*c*d**2*x**2*sqrt(-1 + d**2*x**2/c**2)/(-15*c**2*x**5 + 15*d* *2*x**7) + 2*I*d**6*x**6*sqrt(-1 + d**2*x**2/c**2)/(-15*c**5*x**5 + 15*c** 3*d**2*x**7) - I*d**4*x**4*sqrt(-1 + d**2*x**2/c**2)/(-15*c**3*x**5 + 15*c *d**2*x**7), Abs(d**2*x**2/c**2) > 1), (3*c**3*sqrt(1 - d**2*x**2/c**2)/(- 15*c**2*x**5 + 15*d**2*x**7) - 4*c*d**2*x**2*sqrt(1 - d**2*x**2/c**2)/(-15 *c**2*x**5 + 15*d**2*x**7) + 2*d**6*x**6*sqrt(1 - d**2*x**2/c**2)/(-15*c** 5*x**5 + 15*c**3*d**2*x**7) - d**4*x**4*sqrt(1 - d**2*x**2/c**2)/(-15*c**3 *x**5 + 15*c*d**2*x**7), True)) - c**2*d*Piecewise((-c**2/(4*d*x**5*sqrt(c **2/(d**2*x**2) - 1)) + 3*d/(8*x**3*sqrt(c**2/(d**2*x**2) - 1)) - d**3/(8* c**2*x*sqrt(c**2/(d**2*x**2) - 1)) + d**4*acosh(c/(d*x))/(8*c**3), Abs(c** 2/(d**2*x**2)) > 1), (I*c**2/(4*d*x**5*sqrt(-c**2/(d**2*x**2) + 1)) - 3*I* d/(8*x**3*sqrt(-c**2/(d**2*x**2) + 1)) + I*d**3/(8*c**2*x*sqrt(-c**2/(d**2 *x**2) + 1)) - I*d**4*asin(c/(d*x))/(8*c**3), True)) - c*d**2*Piecewise((- d*sqrt(c**2/(d**2*x**2) - 1)/(3*x**2) + d**3*sqrt(c**2/(d**2*x**2) - 1)/(3 *c**2), Abs(c**2/(d**2*x**2)) > 1), (-I*d*sqrt(-c**2/(d**2*x**2) + 1)/(3*x **2) + I*d**3*sqrt(-c**2/(d**2*x**2) + 1)/(3*c**2), True)) + d**3*Piecewis e((-d*sqrt(c**2/(d**2*x**2) - 1)/(2*x) + d**2*acosh(c/(d*x))/(2*c), Abs(c* *2/(d**2*x**2)) > 1), (I*c**2/(2*d*x**3*sqrt(-c**2/(d**2*x**2) + 1)) - I*d /(2*x*sqrt(-c**2/(d**2*x**2) + 1)) - I*d**2*asin(c/(d*x))/(2*c), True))
Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.38 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {3 \, d^{5} \log \left (\frac {2 \, c^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} c}{{\left | x \right |}}\right )}{8 \, c} - \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{5}}{8 \, c^{2}} - \frac {3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{3}}{8 \, c^{2} x^{2}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}}{5 \, c x^{3}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d}{4 \, x^{4}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c}{5 \, x^{5}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/x^6/(d*x+c),x, algorithm="maxima")
Output:
3/8*d^5*log(2*c^2/abs(x) + 2*sqrt(-d^2*x^2 + c^2)*c/abs(x))/c - 3/8*sqrt(- d^2*x^2 + c^2)*d^5/c^2 - 3/8*(-d^2*x^2 + c^2)^(3/2)*d^3/(c^2*x^2) + 1/5*(- d^2*x^2 + c^2)^(3/2)*d^2/(c*x^3) + 1/4*(-d^2*x^2 + c^2)^(3/2)*d/x^4 - 1/5* (-d^2*x^2 + c^2)^(3/2)*c/x^5
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (95) = 190\).
Time = 0.14 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.50 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {{\left (2 \, d^{6} - \frac {5 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} d^{4}}{x} - \frac {10 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} d^{2}}{x^{2}} + \frac {40 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3}}{x^{3}} + \frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4}}{d^{2} x^{4}}\right )} d^{10} x^{5}}{320 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} c {\left | d \right |}} + \frac {3 \, d^{6} \log \left (\frac {{\left | -2 \, c d - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |} \right |}}{2 \, d^{2} {\left | x \right |}}\right )}{8 \, c {\left | d \right |}} - \frac {\frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} c^{4} d^{8}}{x} + \frac {40 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} c^{4} d^{6}}{x^{2}} - \frac {10 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} c^{4} d^{4}}{x^{3}} - \frac {5 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} c^{4} d^{2}}{x^{4}} + \frac {2 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} c^{4}}{x^{5}}}{320 \, c^{5} d^{4} {\left | d \right |}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/x^6/(d*x+c),x, algorithm="giac")
Output:
1/320*(2*d^6 - 5*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*d^4/x - 10*(c*d + sqr t(-d^2*x^2 + c^2)*abs(d))^2*d^2/x^2 + 40*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d ))^3/x^3 + 20*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4/(d^2*x^4))*d^10*x^5/(( c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*c*abs(d)) + 3/8*d^6*log(1/2*abs(-2*c* d - 2*sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*abs(x)))/(c*abs(d)) - 1/320*(20*(c *d + sqrt(-d^2*x^2 + c^2)*abs(d))*c^4*d^8/x + 40*(c*d + sqrt(-d^2*x^2 + c^ 2)*abs(d))^2*c^4*d^6/x^2 - 10*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*c^4*d^ 4/x^3 - 5*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*c^4*d^2/x^4 + 2*(c*d + sqr t(-d^2*x^2 + c^2)*abs(d))^5*c^4/x^5)/(c^5*d^4*abs(d))
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}}{x^6\,\left (c+d\,x\right )} \,d x \] Input:
int((c^2 - d^2*x^2)^(5/2)/(x^6*(c + d*x)),x)
Output:
int((c^2 - d^2*x^2)^(5/2)/(x^6*(c + d*x)), x)
Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.20 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {-8 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+10 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +16 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}-25 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}-15 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) d^{5} x^{5}}{40 c \,x^{5}} \] Input:
int((-d^2*x^2+c^2)^(5/2)/x^6/(d*x+c),x)
Output:
( - 8*sqrt(c**2 - d**2*x**2)*c**4 + 10*sqrt(c**2 - d**2*x**2)*c**3*d*x + 1 6*sqrt(c**2 - d**2*x**2)*c**2*d**2*x**2 - 25*sqrt(c**2 - d**2*x**2)*c*d**3 *x**3 - 8*sqrt(c**2 - d**2*x**2)*d**4*x**4 - 15*log(tan(asin((d*x)/c)/2))* d**5*x**5)/(40*c*x**5)