Integrand size = 43, antiderivative size = 103 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=-\frac {5 x \sqrt {-1+c x}}{4 \sqrt {1-c x} \left (1-c^2 x^2\right )^2}+\frac {x \sqrt {-1+c x}}{8 \sqrt {1-c x} \left (1-c^2 x^2\right )}+\frac {\sqrt {-1+c x} \text {arctanh}(c x)}{8 c \sqrt {1-c x}} \] Output:
-5/4*x*(c*x-1)^(1/2)/(-c*x+1)^(1/2)/(-c^2*x^2+1)^2+1/8*x*(c*x-1)^(1/2)/(-c *x+1)^(1/2)/(-c^2*x^2+1)+1/8*(c*x-1)^(1/2)*arctanh(c*x)/c/(-c*x+1)^(1/2)
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (-2 c x \left (9+c^2 x^2\right )-\left (-1+c^2 x^2\right )^2 \log (-1+c x)+\left (-1+c^2 x^2\right )^2 \log (1+c x)\right )}{16 c \left (1-c^2 x^2\right )^{5/2}} \] Input:
Integrate[(1 + 4*c^2*x^2)/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(5/2 )),x]
Output:
(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-2*c*x*(9 + c^2*x^2) - (-1 + c^2*x^2)^2*Log [-1 + c*x] + (-1 + c^2*x^2)^2*Log[1 + c*x]))/(16*c*(1 - c^2*x^2)^(5/2))
Time = 0.41 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2003, 37, 25, 643, 298, 215, 219}
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.
| \(\displaystyle \int \frac {4 c^2 x^2+1}{\sqrt {c x-1} \sqrt {c x+1} \left (1-c^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle \int \frac {4 c^2 x^2+1}{(1-c x)^{5/2} \sqrt {c x-1} (c x+1)^3}dx\) |
| \(\Big \downarrow \) 37 |
| \(\displaystyle \frac {(c x-1)^{5/2} \int -\frac {4 c^2 x^2+1}{(1-c x)^3 (c x+1)^3}dx}{(1-c x)^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {(c x-1)^{5/2} \int \frac {4 c^2 x^2+1}{(1-c x)^3 (c x+1)^3}dx}{(1-c x)^{5/2}}\) |
| \(\Big \downarrow \) 643 |
| \(\displaystyle -\frac {(c x-1)^{5/2} \int \frac {4 c^2 x^2+1}{\left (1-c^2 x^2\right )^3}dx}{(1-c x)^{5/2}}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle -\frac {(c x-1)^{5/2} \left (\frac {5 x}{4 \left (1-c^2 x^2\right )^2}-\frac {1}{4} \int \frac {1}{\left (1-c^2 x^2\right )^2}dx\right )}{(1-c x)^{5/2}}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle -\frac {(c x-1)^{5/2} \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {1}{1-c^2 x^2}dx-\frac {x}{2 \left (1-c^2 x^2\right )}\right )+\frac {5 x}{4 \left (1-c^2 x^2\right )^2}\right )}{(1-c x)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {(c x-1)^{5/2} \left (\frac {1}{4} \left (-\frac {\text {arctanh}(c x)}{2 c}-\frac {x}{2 \left (1-c^2 x^2\right )}\right )+\frac {5 x}{4 \left (1-c^2 x^2\right )^2}\right )}{(1-c x)^{5/2}}\) |
Input:
Int[(1 + 4*c^2*x^2)/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(5/2)),x]
Output:
-(((-1 + c*x)^(5/2)*((5*x)/(4*(1 - c^2*x^2)^2) + (-1/2*x/(1 - c^2*x^2) - A rcTanh[c*x]/(2*c))/4))/(1 - c*x)^(5/2))
Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S imp[(a + b*x)^m/(c + d*x)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] && !SimplerQ[a + b*x, c + d*x]
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) ^2)^(p_.), x_Symbol] :> Int[(c*e + d*f*x^2)^m*(a + b*x^2)^p, x] /; FreeQ[{a , b, c, d, e, f, m, n, p}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && (Integer Q[m] || (GtQ[c, 0] && GtQ[e, 0]))
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.11
| method | result | size |
| default | \(-\frac {\ln \left (c x -1\right ) c^{4} x^{4}-\ln \left (c x +1\right ) c^{4} x^{4}+2 c^{3} x^{3}-2 \ln \left (c x -1\right ) c^{2} x^{2}+2 \ln \left (c x +1\right ) c^{2} x^{2}+18 c x +\ln \left (c x -1\right )-\ln \left (c x +1\right )}{16 \left (c x -1\right )^{\frac {3}{2}} \left (c x +1\right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}\, c}\) | \(114\) |
| risch | \(\frac {\sqrt {\frac {-c^{2} x^{2}+1}{\left (c x -1\right ) \left (c x +1\right )}}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \left (-\frac {i \left (-\frac {1}{8} c^{2} x^{3}-\frac {9}{8} x \right )}{\left (c^{2} x^{2}-1\right ) \left (c x -1\right ) \left (c x +1\right )}-\frac {i \ln \left (-c x -1\right )}{16 c}+\frac {i \ln \left (c x -1\right )}{16 c}\right )}{\sqrt {-c^{2} x^{2}+1}}\) | \(121\) |
Input:
int((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(5/2),x,method= _RETURNVERBOSE)
Output:
-1/16/(c*x-1)^(3/2)/(c*x+1)^(3/2)*(ln(c*x-1)*c^4*x^4-ln(c*x+1)*c^4*x^4+2*c ^3*x^3-2*ln(c*x-1)*c^2*x^2+2*ln(c*x+1)*c^2*x^2+18*c*x+ln(c*x-1)-ln(c*x+1)) /(-c^2*x^2+1)^(1/2)/c
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.37 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (c^{3} x^{3} + 9 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c x + 1} \sqrt {c x - 1} - {\left (c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1\right )} \arctan \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} \sqrt {c x + 1} \sqrt {c x - 1} c x}{c^{4} x^{4} - 1}\right )}{16 \, {\left (c^{7} x^{6} - 3 \, c^{5} x^{4} + 3 \, c^{3} x^{2} - c\right )}} \] Input:
integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(5/2),x, algorithm="fricas")
Output:
1/16*(2*(c^3*x^3 + 9*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*x + 1)*sqrt(c*x - 1) - (c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - 1)*arctan(2*sqrt(-c^2*x^2 + 1)*sqrt(c* x + 1)*sqrt(c*x - 1)*c*x/(c^4*x^4 - 1)))/(c^7*x^6 - 3*c^5*x^4 + 3*c^3*x^2 - c)
\[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\int \frac {4 c^{2} x^{2} + 1}{\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \sqrt {c x - 1} \sqrt {c x + 1}}\, dx \] Input:
integrate((4*c**2*x**2+1)/(c*x-1)**(1/2)/(c*x+1)**(1/2)/(-c**2*x**2+1)**(5 /2),x)
Output:
Integral((4*c**2*x**2 + 1)/((-(c*x - 1)*(c*x + 1))**(5/2)*sqrt(c*x - 1)*sq rt(c*x + 1)), x)
\[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\int { \frac {4 \, c^{2} x^{2} + 1}{{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}} \sqrt {c x + 1} \sqrt {c x - 1}} \,d x } \] Input:
integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(5/2),x, algorithm="maxima")
Output:
integrate((4*c^2*x^2 + 1)/((-c^2*x^2 + 1)^(5/2)*sqrt(c*x + 1)*sqrt(c*x - 1 )), x)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.68 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=-\frac {i \, \log \left (c x + 1\right )}{16 \, c} + \frac {i \, \log \left ({\left | c x - 1 \right |}\right )}{16 \, c} - \frac {i \, {\left (c x + 1\right )}^{3} - 3 i \, {\left (c x + 1\right )}^{2} + 12 i \, c x + 2 i}{8 \, {\left (i \, {\left (c x + 1\right )}^{2} - 2 i \, c x - 2 i\right )}^{2} c} \] Input:
integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(5/2),x, algorithm="giac")
Output:
-1/16*I*log(c*x + 1)/c + 1/16*I*log(abs(c*x - 1))/c - 1/8*(I*(c*x + 1)^3 - 3*I*(c*x + 1)^2 + 12*I*c*x + 2*I)/((I*(c*x + 1)^2 - 2*I*c*x - 2*I)^2*c)
Timed out. \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\int \frac {4\,c^2\,x^2+1}{{\left (1-c^2\,x^2\right )}^{5/2}\,\sqrt {c\,x-1}\,\sqrt {c\,x+1}} \,d x \] Input:
int((4*c^2*x^2 + 1)/((1 - c^2*x^2)^(5/2)*(c*x - 1)^(1/2)*(c*x + 1)^(1/2)), x)
Output:
int((4*c^2*x^2 + 1)/((1 - c^2*x^2)^(5/2)*(c*x - 1)^(1/2)*(c*x + 1)^(1/2)), x)
Time = 0.18 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.98 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{5/2}} \, dx=\frac {i \left (2 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right ) c^{4} x^{4}-4 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right ) c^{2} x^{2}+2 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right )+2 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right ) c^{4} x^{4}-4 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right ) c^{2} x^{2}+2 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right )-4 \,\mathrm {log}\left (\sqrt {-c x +1}\right ) c^{4} x^{4}+8 \,\mathrm {log}\left (\sqrt {-c x +1}\right ) c^{2} x^{2}-4 \,\mathrm {log}\left (\sqrt {-c x +1}\right )+c^{4} x^{4}-4 c^{3} x^{3}-2 c^{2} x^{2}-36 c x +1\right )}{32 c \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )} \] Input:
int((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(5/2),x)
Output:
(i*(2*log(sqrt( - c*x + 1) - sqrt(2))*c**4*x**4 - 4*log(sqrt( - c*x + 1) - sqrt(2))*c**2*x**2 + 2*log(sqrt( - c*x + 1) - sqrt(2)) + 2*log(sqrt( - c* x + 1) + sqrt(2))*c**4*x**4 - 4*log(sqrt( - c*x + 1) + sqrt(2))*c**2*x**2 + 2*log(sqrt( - c*x + 1) + sqrt(2)) - 4*log(sqrt( - c*x + 1))*c**4*x**4 + 8*log(sqrt( - c*x + 1))*c**2*x**2 - 4*log(sqrt( - c*x + 1)) + c**4*x**4 - 4*c**3*x**3 - 2*c**2*x**2 - 36*c*x + 1))/(32*c*(c**4*x**4 - 2*c**2*x**2 + 1))