Integrand size = 29, antiderivative size = 76 \[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {x \sqrt {-1+c^2 x^2}}{2 \left (d-c^2 d x^2\right )^{3/2}}-\frac {\sqrt {-1+c^2 x^2} \text {arctanh}(c x)}{2 c d \sqrt {d-c^2 d x^2}} \] Output:
-1/2*x*(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2)-1/2*(c^2*x^2-1)^(1/2)*arctan h(c*x)/c/d/(-c^2*d*x^2+d)^(1/2)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {c x+\left (1-c^2 x^2\right ) \text {arctanh}(c x)}{2 c d \sqrt {-1+c^2 x^2} \sqrt {d-c^2 d x^2}} \] Input:
Integrate[1/(Sqrt[-1 + c^2*x^2]*(d - c^2*d*x^2)^(3/2)),x]
Output:
(c*x + (1 - c^2*x^2)*ArcTanh[c*x])/(2*c*d*Sqrt[-1 + c^2*x^2]*Sqrt[d - c^2* d*x^2])
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {283, 215, 220}
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.
\(\displaystyle \int \frac {1}{\sqrt {c^2 x^2-1} \left (d-c^2 d x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 283 |
\(\displaystyle \frac {\left (c^2 x^2-1\right )^{3/2} \int \frac {1}{\left (c^2 x^2-1\right )^2}dx}{\left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\left (c^2 x^2-1\right )^{3/2} \left (\frac {x}{2 \left (1-c^2 x^2\right )}-\frac {1}{2} \int \frac {1}{c^2 x^2-1}dx\right )}{\left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {\left (c^2 x^2-1\right )^{3/2} \left (\frac {\text {arctanh}(c x)}{2 c}+\frac {x}{2 \left (1-c^2 x^2\right )}\right )}{\left (d-c^2 d x^2\right )^{3/2}}\) |
Input:
Int[1/(Sqrt[-1 + c^2*x^2]*(d - c^2*d*x^2)^(3/2)),x]
Output:
((-1 + c^2*x^2)^(3/2)*(x/(2*(1 - c^2*x^2)) + ArcTanh[c*x]/(2*c)))/(d - c^2 *d*x^2)^(3/2)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Sy mbol] :> Simp[(a + b*x^n)^p/(c + d*x^n)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && !SimplerQ[a + b*x^n, c + d*x^n]
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\sqrt {-\left (c^{2} x^{2}-1\right ) d}\, \left (\ln \left (c x +1\right ) c^{2} x^{2}-\ln \left (c x -1\right ) c^{2} x^{2}-2 c x -\ln \left (c x +1\right )+\ln \left (c x -1\right )\right )}{4 \sqrt {c^{2} x^{2}-1}\, d^{2} c \left (c x +1\right ) \left (c x -1\right )}\) | \(94\) |
risch | \(\frac {x}{2 d \sqrt {c^{2} x^{2}-1}\, \sqrt {-\left (c^{2} x^{2}-1\right ) d}}+\frac {\sqrt {c^{2} x^{2}-1}\, \ln \left (c x -1\right )}{4 d \sqrt {-\left (c^{2} x^{2}-1\right ) d}\, c}-\frac {\sqrt {c^{2} x^{2}-1}\, \ln \left (-c x -1\right )}{4 d \sqrt {-\left (c^{2} x^{2}-1\right ) d}\, c}\) | \(112\) |
Input:
int(1/(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4/(c^2*x^2-1)^(1/2)*(-(c^2*x^2-1)*d)^(1/2)*(ln(c*x+1)*c^2*x^2-ln(c*x-1)* c^2*x^2-2*c*x-ln(c*x+1)+ln(c*x-1))/d^2/c/(c*x+1)/(c*x-1)
Time = 0.12 (sec) , antiderivative size = 314, normalized size of antiderivative = 4.13 \[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c x + {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} \sqrt {-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right )}{8 \, {\left (c^{5} d^{2} x^{4} - 2 \, c^{3} d^{2} x^{2} + c d^{2}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c x - {\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right )}{4 \, {\left (c^{5} d^{2} x^{4} - 2 \, c^{3} d^{2} x^{2} + c d^{2}\right )}}\right ] \] Input:
integrate(1/(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")
Output:
[-1/8*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*x + (c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 + 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c^4 *x^4 + 3*c^2*x^2 - 1)))/(c^5*d^2*x^4 - 2*c^3*d^2*x^2 + c*d^2), -1/4*(2*sqr t(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*x - (c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(d )*arctan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)))/(c^5*d^2*x^4 - 2*c^3*d^2*x^2 + c*d^2)]
\[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {\left (c x - 1\right ) \left (c x + 1\right )} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(c**2*x**2-1)**(1/2)/(-c**2*d*x**2+d)**(3/2),x)
Output:
Integral(1/(sqrt((c*x - 1)*(c*x + 1))*(-d*(c*x - 1)*(c*x + 1))**(3/2)), x)
\[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} \sqrt {c^{2} x^{2} - 1}} \,d x } \] Input:
integrate(1/(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((-c^2*d*x^2 + d)^(3/2)*sqrt(c^2*x^2 - 1)), x)
\[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} \sqrt {c^{2} x^{2} - 1}} \,d x } \] Input:
integrate(1/(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")
Output:
integrate(1/((-c^2*d*x^2 + d)^(3/2)*sqrt(c^2*x^2 - 1)), x)
Timed out. \[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (d-c^2\,d\,x^2\right )}^{3/2}\,\sqrt {c^2\,x^2-1}} \,d x \] Input:
int(1/((d - c^2*d*x^2)^(3/2)*(c^2*x^2 - 1)^(1/2)),x)
Output:
int(1/((d - c^2*d*x^2)^(3/2)*(c^2*x^2 - 1)^(1/2)), x)
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\sqrt {-1+c^2 x^2} \left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d}\, i \left (-\mathrm {log}\left (c^{2} x -c \right ) c^{2} x^{2}+\mathrm {log}\left (c^{2} x -c \right )+\mathrm {log}\left (c^{2} x +c \right ) c^{2} x^{2}-\mathrm {log}\left (c^{2} x +c \right )-2 c x \right )}{4 c \,d^{2} \left (c^{2} x^{2}-1\right )} \] Input:
int(1/(c^2*x^2-1)^(1/2)/(-c^2*d*x^2+d)^(3/2),x)
Output:
(sqrt(d)*i*( - log(c**2*x - c)*c**2*x**2 + log(c**2*x - c) + log(c**2*x + c)*c**2*x**2 - log(c**2*x + c) - 2*c*x))/(4*c*d**2*(c**2*x**2 - 1))