\(\int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} (1-c^2 x^2)^{7/2}} \, dx\) [173]

Optimal result

Integrand size = 43, antiderivative size = 139 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=-\frac {5 x \sqrt {-1+c x}}{6 \sqrt {1-c x} \left (1-c^2 x^2\right )^3}-\frac {x \sqrt {-1+c x}}{24 \sqrt {1-c x} \left (1-c^2 x^2\right )^2}-\frac {x \sqrt {-1+c x}}{16 \sqrt {1-c x} \left (1-c^2 x^2\right )}-\frac {\sqrt {-1+c x} \text {arctanh}(c x)}{16 c \sqrt {1-c x}} \] Output:

-5/6*x*(c*x-1)^(1/2)/(-c*x+1)^(1/2)/(-c^2*x^2+1)^3-1/24*x*(c*x-1)^(1/2)/(- 
c*x+1)^(1/2)/(-c^2*x^2+1)^2-1/16*x*(c*x-1)^(1/2)/(-c*x+1)^(1/2)/(-c^2*x^2+ 
1)-1/16*(c*x-1)^(1/2)*arctanh(c*x)/c/(-c*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.72 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\frac {2 c x \left (45-8 c^2 x^2+3 c^4 x^4\right )+3 \left (-1+c^2 x^2\right )^3 \log (-1+c x)-3 \left (-1+c^2 x^2\right )^3 \log (1+c x)}{96 c (-1+c x)^{5/2} (1+c x)^{5/2} \sqrt {1-c^2 x^2}} \] Input:

Integrate[(1 + 4*c^2*x^2)/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(7/2 
)),x]
 

Output:

(2*c*x*(45 - 8*c^2*x^2 + 3*c^4*x^4) + 3*(-1 + c^2*x^2)^3*Log[-1 + c*x] - 3 
*(-1 + c^2*x^2)^3*Log[1 + c*x])/(96*c*(-1 + c*x)^(5/2)*(1 + c*x)^(5/2)*Sqr 
t[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2003, 37, 643, 298, 215, 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.

\(\displaystyle \int \frac {4 c^2 x^2+1}{\sqrt {c x-1} \sqrt {c x+1} \left (1-c^2 x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 2003

\(\displaystyle \int \frac {4 c^2 x^2+1}{(1-c x)^{7/2} \sqrt {c x-1} (c x+1)^4}dx\)

\(\Big \downarrow \) 37

\(\displaystyle \frac {(c x-1)^{7/2} \int \frac {4 c^2 x^2+1}{(1-c x)^4 (c x+1)^4}dx}{(1-c x)^{7/2}}\)

\(\Big \downarrow \) 643

\(\displaystyle \frac {(c x-1)^{7/2} \int \frac {4 c^2 x^2+1}{\left (1-c^2 x^2\right )^4}dx}{(1-c x)^{7/2}}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {(c x-1)^{7/2} \left (\frac {1}{6} \int \frac {1}{\left (1-c^2 x^2\right )^3}dx+\frac {5 x}{6 \left (1-c^2 x^2\right )^3}\right )}{(1-c x)^{7/2}}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {(c x-1)^{7/2} \left (\frac {1}{6} \left (\frac {3}{4} \int \frac {1}{\left (1-c^2 x^2\right )^2}dx+\frac {x}{4 \left (1-c^2 x^2\right )^2}\right )+\frac {5 x}{6 \left (1-c^2 x^2\right )^3}\right )}{(1-c x)^{7/2}}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {(c x-1)^{7/2} \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-c^2 x^2}dx+\frac {x}{2 \left (1-c^2 x^2\right )}\right )+\frac {x}{4 \left (1-c^2 x^2\right )^2}\right )+\frac {5 x}{6 \left (1-c^2 x^2\right )^3}\right )}{(1-c x)^{7/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(c x-1)^{7/2} \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {\text {arctanh}(c x)}{2 c}+\frac {x}{2 \left (1-c^2 x^2\right )}\right )+\frac {x}{4 \left (1-c^2 x^2\right )^2}\right )+\frac {5 x}{6 \left (1-c^2 x^2\right )^3}\right )}{(1-c x)^{7/2}}\)

Input:

Int[(1 + 4*c^2*x^2)/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(1 - c^2*x^2)^(7/2)),x]
 

Output:

((-1 + c*x)^(7/2)*((5*x)/(6*(1 - c^2*x^2)^3) + (x/(4*(1 - c^2*x^2)^2) + (3 
*(x/(2*(1 - c^2*x^2)) + ArcTanh[c*x]/(2*c)))/4)/6))/(1 - c*x)^(7/2)
 

Defintions of rubi rules used

rule 37

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]
 

rule 215

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])
 

rule 219

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])
 

rule 298

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])
 

rule 643

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]))
 

rule 2003

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]))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {\sqrt {\frac {-c^{2} x^{2}+1}{\left (c x -1\right ) \left (c x +1\right )}}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \left (-\frac {i \left (\frac {1}{16} c^{4} x^{5}-\frac {1}{6} c^{2} x^{3}+\frac {15}{16} x \right )}{\left (c^{2} x^{2}-1\right )^{2} \left (c x -1\right ) \left (c x +1\right )}+\frac {i \ln \left (c x +1\right )}{32 c}-\frac {i \ln \left (-c x +1\right )}{32 c}\right )}{\sqrt {-c^{2} x^{2}+1}}\)	\(129\)
default	\(\frac {3 \ln \left (c x -1\right ) c^{6} x^{6}-3 \ln \left (c x +1\right ) c^{6} x^{6}+6 c^{5} x^{5}-9 \ln \left (c x -1\right ) c^{4} x^{4}+9 \ln \left (c x +1\right ) c^{4} x^{4}-16 c^{3} x^{3}+9 \ln \left (c x -1\right ) c^{2} x^{2}-9 \ln \left (c x +1\right ) c^{2} x^{2}+90 c x -3 \ln \left (c x -1\right )+3 \ln \left (c x +1\right )}{96 \left (c x -1\right )^{\frac {5}{2}} \left (c x +1\right )^{\frac {5}{2}} \sqrt {-c^{2} x^{2}+1}\, c}\)	\(153\)

Input:

int((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(7/2),x,method= 
_RETURNVERBOSE)
 

Output:

((-c^2*x^2+1)/(c*x-1)/(c*x+1))^(1/2)/(-c^2*x^2+1)^(1/2)*(c*x-1)^(1/2)*(c*x 
+1)^(1/2)*(-I*(1/16*c^4*x^5-1/6*c^2*x^3+15/16*x)/(c^2*x^2-1)^2/(c*x-1)/(c* 
x+1)+1/32*I/c*ln(c*x+1)-1/32*I/c*ln(-c*x+1))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.18 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (3 \, c^{5} x^{5} - 8 \, c^{3} x^{3} + 45 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c x + 1} \sqrt {c x - 1} - 3 \, {\left (c^{8} x^{8} - 4 \, c^{6} x^{6} + 6 \, c^{4} x^{4} - 4 \, c^{2} x^{2} + 1\right )} \arctan \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} \sqrt {c x + 1} \sqrt {c x - 1} c x}{c^{4} x^{4} - 1}\right )}{96 \, {\left (c^{9} x^{8} - 4 \, c^{7} x^{6} + 6 \, c^{5} x^{4} - 4 \, c^{3} x^{2} + c\right )}} \] Input:

integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(7/2),x, 
algorithm="fricas")
 

Output:

-1/96*(2*(3*c^5*x^5 - 8*c^3*x^3 + 45*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*x + 1) 
*sqrt(c*x - 1) - 3*(c^8*x^8 - 4*c^6*x^6 + 6*c^4*x^4 - 4*c^2*x^2 + 1)*arcta 
n(2*sqrt(-c^2*x^2 + 1)*sqrt(c*x + 1)*sqrt(c*x - 1)*c*x/(c^4*x^4 - 1)))/(c^ 
9*x^8 - 4*c^7*x^6 + 6*c^5*x^4 - 4*c^3*x^2 + c)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((4*c**2*x**2+1)/(c*x-1)**(1/2)/(c*x+1)**(1/2)/(-c**2*x**2+1)**(7 
/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\int { \frac {4 \, c^{2} x^{2} + 1}{{\left (-c^{2} x^{2} + 1\right )}^{\frac {7}{2}} \sqrt {c x + 1} \sqrt {c x - 1}} \,d x } \] Input:

integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(7/2),x, 
algorithm="maxima")
 

Output:

integrate((4*c^2*x^2 + 1)/((-c^2*x^2 + 1)^(7/2)*sqrt(c*x + 1)*sqrt(c*x - 1 
)), x)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.63 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\frac {i \, \log \left (c x + 1\right )}{32 \, c} - \frac {i \, \log \left ({\left | c x - 1 \right |}\right )}{32 \, c} + \frac {3 \, {\left (c x + 1\right )}^{5} - 15 \, {\left (c x + 1\right )}^{4} + 22 \, {\left (c x + 1\right )}^{3} - 6 \, {\left (c x + 1\right )}^{2} + 36 \, c x - 4}{48 \, {\left (-i \, {\left (c x + 1\right )}^{2} + 2 i \, c x + 2 i\right )}^{3} c} \] Input:

integrate((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(7/2),x, 
algorithm="giac")
 

Output:

1/32*I*log(c*x + 1)/c - 1/32*I*log(abs(c*x - 1))/c + 1/48*(3*(c*x + 1)^5 - 
 15*(c*x + 1)^4 + 22*(c*x + 1)^3 - 6*(c*x + 1)^2 + 36*c*x - 4)/((-I*(c*x + 
 1)^2 + 2*I*c*x + 2*I)^3*c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\int \frac {4\,c^2\,x^2+1}{{\left (1-c^2\,x^2\right )}^{7/2}\,\sqrt {c\,x-1}\,\sqrt {c\,x+1}} \,d x \] Input:

int((4*c^2*x^2 + 1)/((1 - c^2*x^2)^(7/2)*(c*x - 1)^(1/2)*(c*x + 1)^(1/2)), 
x)
 

Output:

int((4*c^2*x^2 + 1)/((1 - c^2*x^2)^(7/2)*(c*x - 1)^(1/2)*(c*x + 1)^(1/2)), 
 x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.05 \[ \int \frac {1+4 c^2 x^2}{\sqrt {-1+c x} \sqrt {1+c x} \left (1-c^2 x^2\right )^{7/2}} \, dx=\frac {i \left (-3 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right ) c^{6} x^{6}+9 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right ) c^{4} x^{4}-9 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right ) c^{2} x^{2}+3 \,\mathrm {log}\left (\sqrt {-c x +1}-\sqrt {2}\right )-3 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right ) c^{6} x^{6}+9 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right ) c^{4} x^{4}-9 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right ) c^{2} x^{2}+3 \,\mathrm {log}\left (\sqrt {-c x +1}+\sqrt {2}\right )+6 \,\mathrm {log}\left (\sqrt {-c x +1}\right ) c^{6} x^{6}-18 \,\mathrm {log}\left (\sqrt {-c x +1}\right ) c^{4} x^{4}+18 \,\mathrm {log}\left (\sqrt {-c x +1}\right ) c^{2} x^{2}-6 \,\mathrm {log}\left (\sqrt {-c x +1}\right )-c^{6} x^{6}+6 c^{5} x^{5}+3 c^{4} x^{4}-16 c^{3} x^{3}-3 c^{2} x^{2}+90 c x +1\right )}{96 c \left (c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1\right )} \] Input:

int((4*c^2*x^2+1)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/(-c^2*x^2+1)^(7/2),x)
 

Output:

(i*( - 3*log(sqrt( - c*x + 1) - sqrt(2))*c**6*x**6 + 9*log(sqrt( - c*x + 1 
) - sqrt(2))*c**4*x**4 - 9*log(sqrt( - c*x + 1) - sqrt(2))*c**2*x**2 + 3*l 
og(sqrt( - c*x + 1) - sqrt(2)) - 3*log(sqrt( - c*x + 1) + sqrt(2))*c**6*x* 
*6 + 9*log(sqrt( - c*x + 1) + sqrt(2))*c**4*x**4 - 9*log(sqrt( - c*x + 1) 
+ sqrt(2))*c**2*x**2 + 3*log(sqrt( - c*x + 1) + sqrt(2)) + 6*log(sqrt( - c 
*x + 1))*c**6*x**6 - 18*log(sqrt( - c*x + 1))*c**4*x**4 + 18*log(sqrt( - c 
*x + 1))*c**2*x**2 - 6*log(sqrt( - c*x + 1)) - c**6*x**6 + 6*c**5*x**5 + 3 
*c**4*x**4 - 16*c**3*x**3 - 3*c**2*x**2 + 90*c*x + 1))/(96*c*(c**6*x**6 - 
3*c**4*x**4 + 3*c**2*x**2 - 1))