Integrand size = 41, antiderivative size = 104 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {15}{32} x \sqrt {-1+c x} \sqrt {1+c x}+\frac {5}{16} x (-1+c x)^{3/2} (1+c x)^{3/2}-\frac {1}{4} x (-1+c x)^{5/2} (1+c x)^{5/2}-\frac {1}{2} x (-1+c x)^{7/2} (1+c x)^{7/2}+\frac {15 \text {arccosh}(c x)}{32 c} \] Output:
-15/32*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)+5/16*x*(c*x-1)^(3/2)*(c*x+1)^(3/2)-1/ 4*x*(c*x-1)^(5/2)*(c*x+1)^(5/2)-1/2*x*(c*x-1)^(7/2)*(c*x+1)^(7/2)+15/32*ar ccosh(c*x)/c
Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {c x \sqrt {\frac {-1+c x}{1+c x}} \left (17+17 c x+22 c^2 x^2+22 c^3 x^3-40 c^4 x^4-40 c^5 x^5+16 c^6 x^6+16 c^7 x^7\right )+15 \log \left (c \left (-1+\sqrt {\frac {-1+c x}{1+c x}}\right )\right )-15 \log \left (1+\sqrt {\frac {-1+c x}{1+c x}}\right )}{32 c} \] Input:
Integrate[((1 - c^2*x^2)^3*(1 + 4*c^2*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) ,x]
Output:
-1/32*(c*x*Sqrt[(-1 + c*x)/(1 + c*x)]*(17 + 17*c*x + 22*c^2*x^2 + 22*c^3*x ^3 - 40*c^4*x^4 - 40*c^5*x^5 + 16*c^6*x^6 + 16*c^7*x^7) + 15*Log[c*(-1 + S qrt[(-1 + c*x)/(1 + c*x)])] - 15*Log[1 + Sqrt[(-1 + c*x)/(1 + c*x)]])/c
Time = 0.42 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2003, 35, 646, 40, 40, 40, 43}
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.
\(\displaystyle \int \frac {\left (1-c^2 x^2\right )^3 \left (4 c^2 x^2+1\right )}{\sqrt {c x-1} \sqrt {c x+1}} \, dx\) |
\(\Big \downarrow \) 2003 |
\(\displaystyle \int \frac {(-c x-1)^3 (c x-1)^{5/2} \left (4 c^2 x^2+1\right )}{\sqrt {c x+1}}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle -\int (c x-1)^{5/2} (c x+1)^{5/2} \left (4 c^2 x^2+1\right )dx\) |
\(\Big \downarrow \) 646 |
\(\displaystyle -\frac {3}{2} \int (c x-1)^{5/2} (c x+1)^{5/2}dx-\frac {1}{2} x (c x-1)^{7/2} (c x+1)^{7/2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} x (c x-1)^{5/2} (c x+1)^{5/2}-\frac {5}{6} \int (c x-1)^{3/2} (c x+1)^{3/2}dx\right )-\frac {1}{2} x (c x-1)^{7/2} (c x+1)^{7/2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} x (c x-1)^{5/2} (c x+1)^{5/2}-\frac {5}{6} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \int \sqrt {c x-1} \sqrt {c x+1}dx\right )\right )-\frac {1}{2} x (c x-1)^{7/2} (c x+1)^{7/2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} x (c x-1)^{5/2} (c x+1)^{5/2}-\frac {5}{6} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \left (\frac {1}{2} x \sqrt {c x-1} \sqrt {c x+1}-\frac {1}{2} \int \frac {1}{\sqrt {c x-1} \sqrt {c x+1}}dx\right )\right )\right )-\frac {1}{2} x (c x-1)^{7/2} (c x+1)^{7/2}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} x (c x-1)^{5/2} (c x+1)^{5/2}-\frac {5}{6} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \left (\frac {1}{2} x \sqrt {c x-1} \sqrt {c x+1}-\frac {\text {arccosh}(c x)}{2 c}\right )\right )\right )-\frac {1}{2} x (c x-1)^{7/2} (c x+1)^{7/2}\) |
Input:
Int[((1 - c^2*x^2)^3*(1 + 4*c^2*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]
Output:
-1/2*(x*(-1 + c*x)^(7/2)*(1 + c*x)^(7/2)) - (3*((x*(-1 + c*x)^(5/2)*(1 + c *x)^(5/2))/6 - (5*((x*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2))/4 - (3*((x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/2 - ArcCosh[c*x]/(2*c)))/4))/6))/2
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 0]
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) ^2), x_Symbol] :> Simp[b*x*(c + d*x)^(m + 1)*((e + f*x)^(n + 1)/(d*f*(2*m + 3))), x] - Simp[(b*c*e - a*d*f*(2*m + 3))/(d*f*(2*m + 3)) Int[(c + d*x)^ m*(e + f*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && !LtQ[m, -1]
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 = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98
method | result | size |
risch | \(-\frac {x \left (16 c^{6} x^{6}-40 c^{4} x^{4}+22 c^{2} x^{2}+17\right ) \sqrt {c x -1}\, \sqrt {c x +1}}{32}+\frac {15 \ln \left (\frac {c^{2} x}{\sqrt {c^{2}}}+\sqrt {c^{2} x^{2}-1}\right ) \sqrt {\left (c x -1\right ) \left (c x +1\right )}}{32 \sqrt {c^{2}}\, \sqrt {c x -1}\, \sqrt {c x +1}}\) | \(102\) |
default | \(-\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (16 \,\operatorname {csgn}\left (c \right ) c^{7} x^{7} \sqrt {c^{2} x^{2}-1}-40 \,\operatorname {csgn}\left (c \right ) c^{5} x^{5} \sqrt {c^{2} x^{2}-1}+22 \,\operatorname {csgn}\left (c \right ) c^{3} x^{3} \sqrt {c^{2} x^{2}-1}+17 x \sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right ) c -15 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right )\right ) \operatorname {csgn}\left (c \right )}{32 \sqrt {c^{2} x^{2}-1}\, c}\) | \(138\) |
Input:
int((-c^2*x^2+1)^3*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x,method=_RET URNVERBOSE)
Output:
-1/32*x*(16*c^6*x^6-40*c^4*x^4+22*c^2*x^2+17)*(c*x-1)^(1/2)*(c*x+1)^(1/2)+ 15/32*ln(c^2*x/(c^2)^(1/2)+(c^2*x^2-1)^(1/2))/(c^2)^(1/2)*((c*x-1)*(c*x+1) )^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {{\left (16 \, c^{7} x^{7} - 40 \, c^{5} x^{5} + 22 \, c^{3} x^{3} + 17 \, c x\right )} \sqrt {c x + 1} \sqrt {c x - 1} + 15 \, \log \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{32 \, c} \] Input:
integrate((-c^2*x^2+1)^3*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="fricas")
Output:
-1/32*((16*c^7*x^7 - 40*c^5*x^5 + 22*c^3*x^3 + 17*c*x)*sqrt(c*x + 1)*sqrt( c*x - 1) + 15*log(-c*x + sqrt(c*x + 1)*sqrt(c*x - 1)))/c
Timed out. \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Timed out} \] Input:
integrate((-c**2*x**2+1)**3*(4*c**2*x**2+1)/(c*x-1)**(1/2)/(c*x+1)**(1/2), x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.81 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {1}{2} \, {\left (c^{2} x^{2} - 1\right )}^{\frac {7}{2}} x - \frac {1}{4} \, {\left (c^{2} x^{2} - 1\right )}^{\frac {5}{2}} x + \frac {5}{16} \, {\left (c^{2} x^{2} - 1\right )}^{\frac {3}{2}} x - \frac {15}{32} \, \sqrt {c^{2} x^{2} - 1} x + \frac {15 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{32 \, c} \] Input:
integrate((-c^2*x^2+1)^3*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="maxima")
Output:
-1/2*(c^2*x^2 - 1)^(7/2)*x - 1/4*(c^2*x^2 - 1)^(5/2)*x + 5/16*(c^2*x^2 - 1 )^(3/2)*x - 15/32*sqrt(c^2*x^2 - 1)*x + 15/32*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {{\left ({\left (2 \, {\left ({\left (4 \, {\left ({\left (2 \, {\left (c x + 1\right )} {\left (c x - 6\right )} + 37\right )} {\left (c x + 1\right )} - 45\right )} {\left (c x + 1\right )} + 91\right )} {\left (c x + 1\right )} - 1\right )} {\left (c x + 1\right )} - 5\right )} {\left (c x + 1\right )} - 15\right )} \sqrt {c x + 1} \sqrt {c x - 1} + 30 \, \log \left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}{32 \, c} \] Input:
integrate((-c^2*x^2+1)^3*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="giac")
Output:
-1/32*(((2*((4*((2*(c*x + 1)*(c*x - 6) + 37)*(c*x + 1) - 45)*(c*x + 1) + 9 1)*(c*x + 1) - 1)*(c*x + 1) - 5)*(c*x + 1) - 15)*sqrt(c*x + 1)*sqrt(c*x - 1) + 30*log(sqrt(c*x + 1) - sqrt(c*x - 1)))/c
Time = 162.44 (sec) , antiderivative size = 1468, normalized size of antiderivative = 14.12 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Too large to display} \] Input:
int(-((c^2*x^2 - 1)^3*(4*c^2*x^2 + 1))/((c*x - 1)^(1/2)*(c*x + 1)^(1/2)),x )
Output:
((11*((c*x - 1)^(1/2) - 1i))/(4*((c*x + 1)^(1/2) - 1)) + (383*((c*x - 1)^( 1/2) - 1i)^3)/(12*((c*x + 1)^(1/2) - 1)^3) + (1145*((c*x - 1)^(1/2) - 1i)^ 5)/(4*((c*x + 1)^(1/2) - 1)^5) + (7335*((c*x - 1)^(1/2) - 1i)^7)/(4*((c*x + 1)^(1/2) - 1)^7) + (36445*((c*x - 1)^(1/2) - 1i)^9)/(6*((c*x + 1)^(1/2) - 1)^9) + (27233*((c*x - 1)^(1/2) - 1i)^11)/(2*((c*x + 1)^(1/2) - 1)^11) + (27233*((c*x - 1)^(1/2) - 1i)^13)/(2*((c*x + 1)^(1/2) - 1)^13) + (36445*( (c*x - 1)^(1/2) - 1i)^15)/(6*((c*x + 1)^(1/2) - 1)^15) + (7335*((c*x - 1)^ (1/2) - 1i)^17)/(4*((c*x + 1)^(1/2) - 1)^17) + (1145*((c*x - 1)^(1/2) - 1i )^19)/(4*((c*x + 1)^(1/2) - 1)^19) + (383*((c*x - 1)^(1/2) - 1i)^21)/(12*( (c*x + 1)^(1/2) - 1)^21) + (11*((c*x - 1)^(1/2) - 1i)^23)/(4*((c*x + 1)^(1 /2) - 1)^23))/(c - (12*c*((c*x - 1)^(1/2) - 1i)^2)/((c*x + 1)^(1/2) - 1)^2 + (66*c*((c*x - 1)^(1/2) - 1i)^4)/((c*x + 1)^(1/2) - 1)^4 - (220*c*((c*x - 1)^(1/2) - 1i)^6)/((c*x + 1)^(1/2) - 1)^6 + (495*c*((c*x - 1)^(1/2) - 1i )^8)/((c*x + 1)^(1/2) - 1)^8 - (792*c*((c*x - 1)^(1/2) - 1i)^10)/((c*x + 1 )^(1/2) - 1)^10 + (924*c*((c*x - 1)^(1/2) - 1i)^12)/((c*x + 1)^(1/2) - 1)^ 12 - (792*c*((c*x - 1)^(1/2) - 1i)^14)/((c*x + 1)^(1/2) - 1)^14 + (495*c*( (c*x - 1)^(1/2) - 1i)^16)/((c*x + 1)^(1/2) - 1)^16 - (220*c*((c*x - 1)^(1/ 2) - 1i)^18)/((c*x + 1)^(1/2) - 1)^18 + (66*c*((c*x - 1)^(1/2) - 1i)^20)/( (c*x + 1)^(1/2) - 1)^20 - (12*c*((c*x - 1)^(1/2) - 1i)^22)/((c*x + 1)^(1/2 ) - 1)^22 + (c*((c*x - 1)^(1/2) - 1i)^24)/((c*x + 1)^(1/2) - 1)^24) - (...
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1-c^2 x^2\right )^3 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {-16 \sqrt {c x +1}\, \sqrt {c x -1}\, c^{7} x^{7}+40 \sqrt {c x +1}\, \sqrt {c x -1}\, c^{5} x^{5}-22 \sqrt {c x +1}\, \sqrt {c x -1}\, c^{3} x^{3}-17 \sqrt {c x +1}\, \sqrt {c x -1}\, c x +30 \,\mathrm {log}\left (\frac {\sqrt {c x -1}+\sqrt {c x +1}}{\sqrt {2}}\right )}{32 c} \] Input:
int((-c^2*x^2+1)^3*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x)
Output:
( - 16*sqrt(c*x + 1)*sqrt(c*x - 1)*c**7*x**7 + 40*sqrt(c*x + 1)*sqrt(c*x - 1)*c**5*x**5 - 22*sqrt(c*x + 1)*sqrt(c*x - 1)*c**3*x**3 - 17*sqrt(c*x + 1 )*sqrt(c*x - 1)*c*x + 30*log((sqrt(c*x - 1) + sqrt(c*x + 1))/sqrt(2)))/(32 *c)