\(\int \frac {x^4 (a+b x^2)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [12]

Optimal result

Integrand size = 29, antiderivative size = 125 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\left (5 b+6 a c^2\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{16 c^6}+\frac {\left (5 b+6 a c^2\right ) x^3 \sqrt {-1+c x} \sqrt {1+c x}}{24 c^4}+\frac {b x^5 \sqrt {-1+c x} \sqrt {1+c x}}{6 c^2}+\frac {\left (5 b+6 a c^2\right ) \text {arccosh}(c x)}{16 c^7} \] Output:

1/16*(6*a*c^2+5*b)*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^6+1/24*(6*a*c^2+5*b)*x^ 
3*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^4+1/6*b*x^5*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^ 
2+1/16*(6*a*c^2+5*b)*arccosh(c*x)/c^7
 

Mathematica [A] (warning: unable to verify)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {c x \sqrt {-1+c x} \sqrt {1+c x} \left (6 a c^2 \left (3+2 c^2 x^2\right )+b \left (15+10 c^2 x^2+8 c^4 x^4\right )\right )+6 \left (5 b+6 a c^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )}{48 c^7} \] Input:

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

Output:

(c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(6*a*c^2*(3 + 2*c^2*x^2) + b*(15 + 10*c^ 
2*x^2 + 8*c^4*x^4)) + 6*(5*b + 6*a*c^2)*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)] 
])/(48*c^7)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {960, 111, 27, 101, 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.

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

\(\Big \downarrow \) 960

\(\displaystyle \frac {1}{6} \left (6 a+\frac {5 b}{c^2}\right ) \int \frac {x^4}{\sqrt {c x-1} \sqrt {c x+1}}dx+\frac {b x^5 \sqrt {c x-1} \sqrt {c x+1}}{6 c^2}\)

\(\Big \downarrow \) 111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 43

\(\displaystyle \frac {1}{6} \left (6 a+\frac {5 b}{c^2}\right ) \left (\frac {3 \left (\frac {\text {arccosh}(c x)}{2 c^3}+\frac {x \sqrt {c x-1} \sqrt {c x+1}}{2 c^2}\right )}{4 c^2}+\frac {x^3 \sqrt {c x-1} \sqrt {c x+1}}{4 c^2}\right )+\frac {b x^5 \sqrt {c x-1} \sqrt {c x+1}}{6 c^2}\)

Input:

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

Output:

(b*x^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*c^2) + ((6*a + (5*b)/c^2)*((x^3*Sq 
rt[-1 + c*x]*Sqrt[1 + c*x])/(4*c^2) + (3*((x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) 
/(2*c^2) + ArcCosh[c*x]/(2*c^3)))/(4*c^2)))/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 43

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]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
 

rule 960

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) 
*(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^( 
m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(m + n 
*(p + 1) + 1))), x] - Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/ 
(b1*b2*(m + n*(p + 1) + 1))   Int[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n 
/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] && EqQ[non2, 
 n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {x \left (8 b \,x^{4} c^{4}+12 a \,c^{4} x^{2}+10 b \,c^{2} x^{2}+18 a \,c^{2}+15 b \right ) \sqrt {c x +1}\, \sqrt {c x -1}}{48 c^{6}}+\frac {\left (6 a \,c^{2}+5 b \right ) \ln \left (\frac {c^{2} x}{\sqrt {c^{2}}}+\sqrt {c^{2} x^{2}-1}\right ) \sqrt {\left (c x +1\right ) \left (c x -1\right )}}{16 c^{6} \sqrt {c^{2}}\, \sqrt {c x -1}\, \sqrt {c x +1}}\)	\(129\)
default	\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (8 \,\operatorname {csgn}\left (c \right ) b \,c^{5} x^{5} \sqrt {c^{2} x^{2}-1}+12 \,\operatorname {csgn}\left (c \right ) a \,c^{5} x^{3} \sqrt {c^{2} x^{2}-1}+10 \sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right ) c^{3} b \,x^{3}+18 \sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right ) c^{3} a x +15 \sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right ) c b x +18 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) a \,c^{2}+15 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) b \right ) \operatorname {csgn}\left (c \right )}{48 c^{7} \sqrt {c^{2} x^{2}-1}}\)	\(191\)

Input:

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

Output:

1/48*x*(8*b*c^4*x^4+12*a*c^4*x^2+10*b*c^2*x^2+18*a*c^2+15*b)*(c*x+1)^(1/2) 
*(c*x-1)^(1/2)/c^6+1/16*(6*a*c^2+5*b)/c^6*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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left (8 \, b c^{5} x^{5} + 2 \, {\left (6 \, a c^{5} + 5 \, b c^{3}\right )} x^{3} + 3 \, {\left (6 \, a c^{3} + 5 \, b c\right )} x\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 3 \, {\left (6 \, a c^{2} + 5 \, b\right )} \log \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{48 \, c^{7}} \] Input:

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

Output:

1/48*((8*b*c^5*x^5 + 2*(6*a*c^5 + 5*b*c^3)*x^3 + 3*(6*a*c^3 + 5*b*c)*x)*sq 
rt(c*x + 1)*sqrt(c*x - 1) - 3*(6*a*c^2 + 5*b)*log(-c*x + sqrt(c*x + 1)*sqr 
t(c*x - 1)))/c^7
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {c^{2} x^{2} - 1} b x^{5}}{6 \, c^{2}} + \frac {\sqrt {c^{2} x^{2} - 1} a x^{3}}{4 \, c^{2}} + \frac {5 \, \sqrt {c^{2} x^{2} - 1} b x^{3}}{24 \, c^{4}} + \frac {3 \, \sqrt {c^{2} x^{2} - 1} a x}{8 \, c^{4}} + \frac {3 \, a \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{8 \, c^{5}} + \frac {5 \, \sqrt {c^{2} x^{2} - 1} b x}{16 \, c^{6}} + \frac {5 \, b \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{16 \, c^{7}} \] Input:

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

Output:

1/6*sqrt(c^2*x^2 - 1)*b*x^5/c^2 + 1/4*sqrt(c^2*x^2 - 1)*a*x^3/c^2 + 5/24*s 
qrt(c^2*x^2 - 1)*b*x^3/c^4 + 3/8*sqrt(c^2*x^2 - 1)*a*x/c^4 + 3/8*a*log(2*c 
^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^5 + 5/16*sqrt(c^2*x^2 - 1)*b*x/c^6 + 5/16* 
b*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^7
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.38 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left ({\left (2 \, {\left ({\left (c x + 1\right )} {\left (4 \, {\left (c x + 1\right )} {\left (\frac {{\left (c x + 1\right )} b}{c^{6}} - \frac {5 \, b}{c^{6}}\right )} + \frac {3 \, {\left (2 \, a c^{38} + 15 \, b c^{36}\right )}}{c^{42}}\right )} - \frac {18 \, a c^{38} + 55 \, b c^{36}}{c^{42}}\right )} {\left (c x + 1\right )} + \frac {54 \, a c^{38} + 85 \, b c^{36}}{c^{42}}\right )} {\left (c x + 1\right )} - \frac {3 \, {\left (10 \, a c^{38} + 11 \, b c^{36}\right )}}{c^{42}}\right )} \sqrt {c x + 1} \sqrt {c x - 1} - \frac {6 \, {\left (6 \, a c^{2} + 5 \, b\right )} \log \left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}{c^{6}}}{48 \, c} \] Input:

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

Output:

1/48*(((2*((c*x + 1)*(4*(c*x + 1)*((c*x + 1)*b/c^6 - 5*b/c^6) + 3*(2*a*c^3 
8 + 15*b*c^36)/c^42) - (18*a*c^38 + 55*b*c^36)/c^42)*(c*x + 1) + (54*a*c^3 
8 + 85*b*c^36)/c^42)*(c*x + 1) - 3*(10*a*c^38 + 11*b*c^36)/c^42)*sqrt(c*x 
+ 1)*sqrt(c*x - 1) - 6*(6*a*c^2 + 5*b)*log(sqrt(c*x + 1) - sqrt(c*x - 1))/ 
c^6)/c
 

Mupad [B] (verification not implemented)

Time = 41.24 (sec) , antiderivative size = 1154, normalized size of antiderivative = 9.23 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Too large to display} \] Input:

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

Output:

((23*a*((c*x - 1)^(1/2) - 1i)^3)/(2*((c*x + 1)^(1/2) - 1)^3) + (333*a*((c* 
x - 1)^(1/2) - 1i)^5)/(2*((c*x + 1)^(1/2) - 1)^5) + (671*a*((c*x - 1)^(1/2 
) - 1i)^7)/(2*((c*x + 1)^(1/2) - 1)^7) + (671*a*((c*x - 1)^(1/2) - 1i)^9)/ 
(2*((c*x + 1)^(1/2) - 1)^9) + (333*a*((c*x - 1)^(1/2) - 1i)^11)/(2*((c*x + 
 1)^(1/2) - 1)^11) + (23*a*((c*x - 1)^(1/2) - 1i)^13)/(2*((c*x + 1)^(1/2) 
- 1)^13) - (3*a*((c*x - 1)^(1/2) - 1i)^15)/(2*((c*x + 1)^(1/2) - 1)^15) - 
(3*a*((c*x - 1)^(1/2) - 1i))/(2*((c*x + 1)^(1/2) - 1)))/(c^5 - (8*c^5*((c* 
x - 1)^(1/2) - 1i)^2)/((c*x + 1)^(1/2) - 1)^2 + (28*c^5*((c*x - 1)^(1/2) - 
 1i)^4)/((c*x + 1)^(1/2) - 1)^4 - (56*c^5*((c*x - 1)^(1/2) - 1i)^6)/((c*x 
+ 1)^(1/2) - 1)^6 + (70*c^5*((c*x - 1)^(1/2) - 1i)^8)/((c*x + 1)^(1/2) - 1 
)^8 - (56*c^5*((c*x - 1)^(1/2) - 1i)^10)/((c*x + 1)^(1/2) - 1)^10 + (28*c^ 
5*((c*x - 1)^(1/2) - 1i)^12)/((c*x + 1)^(1/2) - 1)^12 - (8*c^5*((c*x - 1)^ 
(1/2) - 1i)^14)/((c*x + 1)^(1/2) - 1)^14 + (c^5*((c*x - 1)^(1/2) - 1i)^16) 
/((c*x + 1)^(1/2) - 1)^16) - ((311*b*((c*x - 1)^(1/2) - 1i)^5)/(4*((c*x + 
1)^(1/2) - 1)^5) - (175*b*((c*x - 1)^(1/2) - 1i)^3)/(12*((c*x + 1)^(1/2) - 
 1)^3) + (8361*b*((c*x - 1)^(1/2) - 1i)^7)/(4*((c*x + 1)^(1/2) - 1)^7) + ( 
42259*b*((c*x - 1)^(1/2) - 1i)^9)/(6*((c*x + 1)^(1/2) - 1)^9) + (25295*b*( 
(c*x - 1)^(1/2) - 1i)^11)/(2*((c*x + 1)^(1/2) - 1)^11) + (25295*b*((c*x - 
1)^(1/2) - 1i)^13)/(2*((c*x + 1)^(1/2) - 1)^13) + (42259*b*((c*x - 1)^(1/2 
) - 1i)^15)/(6*((c*x + 1)^(1/2) - 1)^15) + (8361*b*((c*x - 1)^(1/2) - 1...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.22 \[ \int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {12 \sqrt {c x +1}\, \sqrt {c x -1}\, a \,c^{5} x^{3}+18 \sqrt {c x +1}\, \sqrt {c x -1}\, a \,c^{3} x +8 \sqrt {c x +1}\, \sqrt {c x -1}\, b \,c^{5} x^{5}+10 \sqrt {c x +1}\, \sqrt {c x -1}\, b \,c^{3} x^{3}+15 \sqrt {c x +1}\, \sqrt {c x -1}\, b c x +36 \,\mathrm {log}\left (\frac {\sqrt {c x -1}+\sqrt {c x +1}}{\sqrt {2}}\right ) a \,c^{2}+30 \,\mathrm {log}\left (\frac {\sqrt {c x -1}+\sqrt {c x +1}}{\sqrt {2}}\right ) b}{48 c^{7}} \] Input:

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

Output:

(12*sqrt(c*x + 1)*sqrt(c*x - 1)*a*c**5*x**3 + 18*sqrt(c*x + 1)*sqrt(c*x - 
1)*a*c**3*x + 8*sqrt(c*x + 1)*sqrt(c*x - 1)*b*c**5*x**5 + 10*sqrt(c*x + 1) 
*sqrt(c*x - 1)*b*c**3*x**3 + 15*sqrt(c*x + 1)*sqrt(c*x - 1)*b*c*x + 36*log 
((sqrt(c*x - 1) + sqrt(c*x + 1))/sqrt(2))*a*c**2 + 30*log((sqrt(c*x - 1) + 
 sqrt(c*x + 1))/sqrt(2))*b)/(48*c**7)