Integrand size = 41, antiderivative size = 81 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=-\frac {5}{8} x \sqrt {-1+c x} \sqrt {1+c x}+\frac {5}{12} x (-1+c x)^{3/2} (1+c x)^{3/2}+\frac {2}{3} x (-1+c x)^{5/2} (1+c x)^{5/2}+\frac {5 \text {arccosh}(c x)}{8 c} \] Output:
-5/8*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)+5/12*x*(c*x-1)^(3/2)*(c*x+1)^(3/2)+2/3* x*(c*x-1)^(5/2)*(c*x+1)^(5/2)+5/8*arccosh(c*x)/c
Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {1}{24} x \sqrt {\frac {-1+c x}{1+c x}} \left (-9-9 c x-22 c^2 x^2-22 c^3 x^3+16 c^4 x^4+16 c^5 x^5\right )+\frac {5 \text {arctanh}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )}{4 c} \] Input:
Integrate[((1 - c^2*x^2)^2*(1 + 4*c^2*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]) ,x]
Output:
(x*Sqrt[(-1 + c*x)/(1 + c*x)]*(-9 - 9*c*x - 22*c^2*x^2 - 22*c^3*x^3 + 16*c ^4*x^4 + 16*c^5*x^5))/24 + (5*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)]])/(4*c)
Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2003, 35, 646, 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.
\(\displaystyle \int \frac {\left (1-c^2 x^2\right )^2 \left (4 c^2 x^2+1\right )}{\sqrt {c x-1} \sqrt {c x+1}} \, dx\) |
\(\Big \downarrow \) 2003 |
\(\displaystyle \int \frac {(-c x-1)^2 (c x-1)^{3/2} \left (4 c^2 x^2+1\right )}{\sqrt {c x+1}}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \int (c x-1)^{3/2} (c x+1)^{3/2} \left (4 c^2 x^2+1\right )dx\) |
\(\Big \downarrow \) 646 |
\(\displaystyle \frac {5}{3} \int (c x-1)^{3/2} (c x+1)^{3/2}dx+\frac {2}{3} x (c x-1)^{5/2} (c x+1)^{5/2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle \frac {5}{3} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \int \sqrt {c x-1} \sqrt {c x+1}dx\right )+\frac {2}{3} x (c x-1)^{5/2} (c x+1)^{5/2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle \frac {5}{3} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \left (\frac {1}{2} x \sqrt {c x-1} \sqrt {c x+1}-\frac {1}{2} \int \frac {1}{\sqrt {c x-1} \sqrt {c x+1}}dx\right )\right )+\frac {2}{3} x (c x-1)^{5/2} (c x+1)^{5/2}\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {5}{3} \left (\frac {1}{4} x (c x-1)^{3/2} (c x+1)^{3/2}-\frac {3}{4} \left (\frac {1}{2} x \sqrt {c x-1} \sqrt {c x+1}-\frac {\text {arccosh}(c x)}{2 c}\right )\right )+\frac {2}{3} x (c x-1)^{5/2} (c x+1)^{5/2}\) |
Input:
Int[((1 - c^2*x^2)^2*(1 + 4*c^2*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]
Output:
(2*x*(-1 + c*x)^(5/2)*(1 + c*x)^(5/2))/3 + (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))/3
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 = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {x \left (16 c^{4} x^{4}-22 c^{2} x^{2}-9\right ) \sqrt {c x -1}\, \sqrt {c x +1}}{24}+\frac {5 \ln \left (\frac {c^{2} x}{\sqrt {c^{2}}}+\sqrt {c^{2} x^{2}-1}\right ) \sqrt {\left (c x -1\right ) \left (c x +1\right )}}{8 \sqrt {c^{2}}\, \sqrt {c x -1}\, \sqrt {c x +1}}\) | \(94\) |
default | \(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (16 \,\operatorname {csgn}\left (c \right ) c^{5} x^{5} \sqrt {c^{2} x^{2}-1}-22 \,\operatorname {csgn}\left (c \right ) c^{3} x^{3} \sqrt {c^{2} x^{2}-1}-9 x \sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right ) c +15 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right )\right ) \operatorname {csgn}\left (c \right )}{24 \sqrt {c^{2} x^{2}-1}\, c}\) | \(117\) |
Input:
int((-c^2*x^2+1)^2*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x,method=_RET URNVERBOSE)
Output:
1/24*x*(16*c^4*x^4-22*c^2*x^2-9)*(c*x-1)^(1/2)*(c*x+1)^(1/2)+5/8*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 = 65, normalized size of antiderivative = 0.80 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left (16 \, c^{5} x^{5} - 22 \, c^{3} x^{3} - 9 \, c x\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 15 \, \log \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{24 \, c} \] Input:
integrate((-c^2*x^2+1)^2*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="fricas")
Output:
1/24*((16*c^5*x^5 - 22*c^3*x^3 - 9*c*x)*sqrt(c*x + 1)*sqrt(c*x - 1) - 15*l og(-c*x + sqrt(c*x + 1)*sqrt(c*x - 1)))/c
Timed out. \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Timed out} \] Input:
integrate((-c**2*x**2+1)**2*(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 = 70, normalized size of antiderivative = 0.86 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {2}{3} \, {\left (c^{2} x^{2} - 1\right )}^{\frac {5}{2}} x + \frac {5}{12} \, {\left (c^{2} x^{2} - 1\right )}^{\frac {3}{2}} x - \frac {5}{8} \, \sqrt {c^{2} x^{2} - 1} x + \frac {5 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{8 \, c} \] Input:
integrate((-c^2*x^2+1)^2*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="maxima")
Output:
2/3*(c^2*x^2 - 1)^(5/2)*x + 5/12*(c^2*x^2 - 1)^(3/2)*x - 5/8*sqrt(c^2*x^2 - 1)*x + 5/8*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left ({\left (2 \, {\left ({\left (8 \, {\left (c x + 1\right )} {\left (c x - 4\right )} + 69\right )} {\left (c x + 1\right )} - 47\right )} {\left (c x + 1\right )} + 5\right )} {\left (c x + 1\right )} + 15\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 30 \, \log \left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}{24 \, c} \] Input:
integrate((-c^2*x^2+1)^2*(4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algo rithm="giac")
Output:
1/24*(((2*((8*(c*x + 1)*(c*x - 4) + 69)*(c*x + 1) - 47)*(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 = 100.22 (sec) , antiderivative size = 1056, normalized size of antiderivative = 13.04 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Too large to display} \] Input:
int(((c^2*x^2 - 1)^2*(4*c^2*x^2 + 1))/((c*x - 1)^(1/2)*(c*x + 1)^(1/2)),x)
Output:
((5*((c*x - 1)^(1/2) - 1i))/(2*((c*x + 1)^(1/2) - 1)) + (47*((c*x - 1)^(1/ 2) - 1i)^3)/(2*((c*x + 1)^(1/2) - 1)^3) + (213*((c*x - 1)^(1/2) - 1i)^5)/( 2*((c*x + 1)^(1/2) - 1)^5) + (759*((c*x - 1)^(1/2) - 1i)^7)/(2*((c*x + 1)^ (1/2) - 1)^7) + (759*((c*x - 1)^(1/2) - 1i)^9)/(2*((c*x + 1)^(1/2) - 1)^9) + (213*((c*x - 1)^(1/2) - 1i)^11)/(2*((c*x + 1)^(1/2) - 1)^11) + (47*((c* x - 1)^(1/2) - 1i)^13)/(2*((c*x + 1)^(1/2) - 1)^13) + (5*((c*x - 1)^(1/2) - 1i)^15)/(2*((c*x + 1)^(1/2) - 1)^15))/(c - (8*c*((c*x - 1)^(1/2) - 1i)^2 )/((c*x + 1)^(1/2) - 1)^2 + (28*c*((c*x - 1)^(1/2) - 1i)^4)/((c*x + 1)^(1/ 2) - 1)^4 - (56*c*((c*x - 1)^(1/2) - 1i)^6)/((c*x + 1)^(1/2) - 1)^6 + (70* c*((c*x - 1)^(1/2) - 1i)^8)/((c*x + 1)^(1/2) - 1)^8 - (56*c*((c*x - 1)^(1/ 2) - 1i)^10)/((c*x + 1)^(1/2) - 1)^10 + (28*c*((c*x - 1)^(1/2) - 1i)^12)/( (c*x + 1)^(1/2) - 1)^12 - (8*c*((c*x - 1)^(1/2) - 1i)^14)/((c*x + 1)^(1/2) - 1)^14 + (c*((c*x - 1)^(1/2) - 1i)^16)/((c*x + 1)^(1/2) - 1)^16) - (((c* x - 1)^(1/2) - 1i)/((c*x + 1)^(1/2) - 1) + (221*((c*x - 1)^(1/2) - 1i)^3)/ (3*((c*x + 1)^(1/2) - 1)^3) + (1035*((c*x - 1)^(1/2) - 1i)^5)/((c*x + 1)^( 1/2) - 1)^5 + (6997*((c*x - 1)^(1/2) - 1i)^7)/((c*x + 1)^(1/2) - 1)^7 + (7 9982*((c*x - 1)^(1/2) - 1i)^9)/(3*((c*x + 1)^(1/2) - 1)^9) + (52614*((c*x - 1)^(1/2) - 1i)^11)/((c*x + 1)^(1/2) - 1)^11 + (52614*((c*x - 1)^(1/2) - 1i)^13)/((c*x + 1)^(1/2) - 1)^13 + (79982*((c*x - 1)^(1/2) - 1i)^15)/(3*(( c*x + 1)^(1/2) - 1)^15) + (6997*((c*x - 1)^(1/2) - 1i)^17)/((c*x + 1)^(...
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1-c^2 x^2\right )^2 \left (1+4 c^2 x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {16 \sqrt {c x +1}\, \sqrt {c x -1}\, c^{5} x^{5}-22 \sqrt {c x +1}\, \sqrt {c x -1}\, c^{3} x^{3}-9 \sqrt {c x +1}\, \sqrt {c x -1}\, c x +30 \,\mathrm {log}\left (\frac {\sqrt {c x -1}+\sqrt {c x +1}}{\sqrt {2}}\right )}{24 c} \] Input:
int((-c^2*x^2+1)^2*(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**5*x**5 - 22*sqrt(c*x + 1)*sqrt(c*x - 1) *c**3*x**3 - 9*sqrt(c*x + 1)*sqrt(c*x - 1)*c*x + 30*log((sqrt(c*x - 1) + s qrt(c*x + 1))/sqrt(2)))/(24*c)