\(\int \frac {x^2 (a+b x^2)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [350]

Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\left (3 b+4 a c^2\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{8 c^4}+\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{4 c^2}+\frac {\left (3 b+4 a c^2\right ) \text {arccosh}(c x)}{8 c^5} \]

[Out]

1/8*(4*a*c^2+3*b)*arccosh(c*x)/c^5+1/8*(4*a*c^2+3*b)*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^4+1/4*b*x^3*(c*x-1)^(1/2)
*(c*x+1)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {471, 92, 54} \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\left (4 a c^2+3 b\right ) \text {arccosh}(c x)}{8 c^5}+\frac {x \sqrt {c x-1} \sqrt {c x+1} \left (4 a c^2+3 b\right )}{8 c^4}+\frac {b x^3 \sqrt {c x-1} \sqrt {c x+1}}{4 c^2} \]

[In]

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

[Out]

((3*b + 4*a*c^2)*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(8*c^4) + (b*x^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*c^2) + ((3*
b + 4*a*c^2)*ArcCosh[c*x])/(8*c^5)

Rule 54

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[b*(x/a)]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[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 471

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] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{4 c^2}-\frac {1}{4} \left (-4 a-\frac {3 b}{c^2}\right ) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \frac {\left (3 b+4 a c^2\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{8 c^4}+\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{4 c^2}+\frac {\left (3 b+4 a c^2\right ) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{8 c^4} \\ & = \frac {\left (3 b+4 a c^2\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{8 c^4}+\frac {b x^3 \sqrt {-1+c x} \sqrt {1+c x}}{4 c^2}+\frac {\left (3 b+4 a c^2\right ) \cosh ^{-1}(c x)}{8 c^5} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {c x \sqrt {-1+c x} \sqrt {1+c x} \left (4 a c^2+b \left (3+2 c^2 x^2\right )\right )+\left (6 b+8 a c^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )}{8 c^5} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.28

method	result	size
risch	\(\frac {x \left (2 b \,c^{2} x^{2}+4 c^{2} a +3 b \right ) \sqrt {c x -1}\, \sqrt {c x +1}}{8 c^{4}}+\frac {\left (4 c^{2} a +3 b \right ) \ln \left (\frac {c^{2} x}{\sqrt {c^{2}}}+\sqrt {c^{2} x^{2}-1}\right ) \sqrt {\left (c x -1\right ) \left (c x +1\right )}}{8 c^{4} \sqrt {c^{2}}\, \sqrt {c x -1}\, \sqrt {c x +1}}\)	\(111\)
default	\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (2 \,\operatorname {csgn}\left (c \right ) c^{3} \sqrt {c^{2} x^{2}-1}\, b \,x^{3}+4 \,\operatorname {csgn}\left (c \right ) c^{3} \sqrt {c^{2} x^{2}-1}\, a x +3 \,\operatorname {csgn}\left (c \right ) c \sqrt {c^{2} x^{2}-1}\, b x +4 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) a \,c^{2}+3 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) b \right ) \operatorname {csgn}\left (c \right )}{8 c^{5} \sqrt {c^{2} x^{2}-1}}\)	\(147\)

[In]

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

[Out]

1/8*x*(2*b*c^2*x^2+4*a*c^2+3*b)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^4+1/8*(4*a*c^2+3*b)/c^4*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)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left (2 \, b c^{3} x^{3} + {\left (4 \, a c^{3} + 3 \, b c\right )} x\right )} \sqrt {c x + 1} \sqrt {c x - 1} - {\left (4 \, a c^{2} + 3 \, b\right )} \log \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{8 \, c^{5}} \]

[In]

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

[Out]

1/8*((2*b*c^3*x^3 + (4*a*c^3 + 3*b*c)*x)*sqrt(c*x + 1)*sqrt(c*x - 1) - (4*a*c^2 + 3*b)*log(-c*x + sqrt(c*x + 1
)*sqrt(c*x - 1)))/c^5

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {c^{2} x^{2} - 1} b x^{3}}{4 \, c^{2}} + \frac {\sqrt {c^{2} x^{2} - 1} a x}{2 \, c^{2}} + \frac {a \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{2 \, c^{3}} + \frac {3 \, \sqrt {c^{2} x^{2} - 1} b x}{8 \, c^{4}} + \frac {3 \, b \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{8 \, c^{5}} \]

[In]

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

[Out]

1/4*sqrt(c^2*x^2 - 1)*b*x^3/c^2 + 1/2*sqrt(c^2*x^2 - 1)*a*x/c^2 + 1/2*a*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c
^3 + 3/8*sqrt(c^2*x^2 - 1)*b*x/c^4 + 3/8*b*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left ({\left (c x + 1\right )} {\left (2 \, {\left (c x + 1\right )} {\left (\frac {{\left (c x + 1\right )} b}{c^{4}} - \frac {3 \, b}{c^{4}}\right )} + \frac {4 \, a c^{18} + 9 \, b c^{16}}{c^{20}}\right )} - \frac {4 \, a c^{18} + 5 \, b c^{16}}{c^{20}}\right )} \sqrt {c x + 1} \sqrt {c x - 1} - \frac {2 \, {\left (4 \, a c^{2} + 3 \, b\right )} \log \left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}{c^{4}}}{8 \, c} \]

[In]

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

[Out]

1/8*(((c*x + 1)*(2*(c*x + 1)*((c*x + 1)*b/c^4 - 3*b/c^4) + (4*a*c^18 + 9*b*c^16)/c^20) - (4*a*c^18 + 5*b*c^16)
/c^20)*sqrt(c*x + 1)*sqrt(c*x - 1) - 2*(4*a*c^2 + 3*b)*log(sqrt(c*x + 1) - sqrt(c*x - 1))/c^4)/c

Mupad [B] (verification not implemented)

Time = 29.41 (sec) , antiderivative size = 720, normalized size of antiderivative = 8.28 \[ \int \frac {x^2 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\frac {23\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^3}{2\,{\left (\sqrt {c\,x+1}-1\right )}^3}+\frac {333\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^5}{2\,{\left (\sqrt {c\,x+1}-1\right )}^5}+\frac {671\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^7}{2\,{\left (\sqrt {c\,x+1}-1\right )}^7}+\frac {671\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^9}{2\,{\left (\sqrt {c\,x+1}-1\right )}^9}+\frac {333\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{11}}{2\,{\left (\sqrt {c\,x+1}-1\right )}^{11}}+\frac {23\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{13}}{2\,{\left (\sqrt {c\,x+1}-1\right )}^{13}}-\frac {3\,b\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{15}}{2\,{\left (\sqrt {c\,x+1}-1\right )}^{15}}-\frac {3\,b\,\left (\sqrt {c\,x-1}-\mathrm {i}\right )}{2\,\left (\sqrt {c\,x+1}-1\right )}}{c^5-\frac {8\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+\frac {28\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {c\,x+1}-1\right )}^4}-\frac {56\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {c\,x+1}-1\right )}^6}+\frac {70\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {c\,x+1}-1\right )}^8}-\frac {56\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {c\,x+1}-1\right )}^{10}}+\frac {28\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {c\,x+1}-1\right )}^{12}}-\frac {8\,c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{14}}{{\left (\sqrt {c\,x+1}-1\right )}^{14}}+\frac {c^5\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^{16}}{{\left (\sqrt {c\,x+1}-1\right )}^{16}}}-\frac {\frac {14\,a\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {c\,x+1}-1\right )}^3}+\frac {14\,a\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {c\,x+1}-1\right )}^5}+\frac {2\,a\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {c\,x+1}-1\right )}^7}+\frac {2\,a\,\left (\sqrt {c\,x-1}-\mathrm {i}\right )}{\sqrt {c\,x+1}-1}}{c^3-\frac {4\,c^3\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+\frac {6\,c^3\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {c\,x+1}-1\right )}^4}-\frac {4\,c^3\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {c\,x+1}-1\right )}^6}+\frac {c^3\,{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {c\,x+1}-1\right )}^8}}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )}{c^3}+\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )}{2\,c^5} \]

[In]

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

[Out]

((23*b*((c*x - 1)^(1/2) - 1i)^3)/(2*((c*x + 1)^(1/2) - 1)^3) + (333*b*((c*x - 1)^(1/2) - 1i)^5)/(2*((c*x + 1)^
(1/2) - 1)^5) + (671*b*((c*x - 1)^(1/2) - 1i)^7)/(2*((c*x + 1)^(1/2) - 1)^7) + (671*b*((c*x - 1)^(1/2) - 1i)^9
)/(2*((c*x + 1)^(1/2) - 1)^9) + (333*b*((c*x - 1)^(1/2) - 1i)^11)/(2*((c*x + 1)^(1/2) - 1)^11) + (23*b*((c*x -
1)^(1/2) - 1i)^13)/(2*((c*x + 1)^(1/2) - 1)^13) - (3*b*((c*x - 1)^(1/2) - 1i)^15)/(2*((c*x + 1)^(1/2) - 1)^15
) - (3*b*((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) -
((14*a*((c*x - 1)^(1/2) - 1i)^3)/((c*x + 1)^(1/2) - 1)^3 + (14*a*((c*x - 1)^(1/2) - 1i)^5)/((c*x + 1)^(1/2) -
1)^5 + (2*a*((c*x - 1)^(1/2) - 1i)^7)/((c*x + 1)^(1/2) - 1)^7 + (2*a*((c*x - 1)^(1/2) - 1i))/((c*x + 1)^(1/2)
- 1))/(c^3 - (4*c^3*((c*x - 1)^(1/2) - 1i)^2)/((c*x + 1)^(1/2) - 1)^2 + (6*c^3*((c*x - 1)^(1/2) - 1i)^4)/((c*
x + 1)^(1/2) - 1)^4 - (4*c^3*((c*x - 1)^(1/2) - 1i)^6)/((c*x + 1)^(1/2) - 1)^6 + (c^3*((c*x - 1)^(1/2) - 1i)^8
)/((c*x + 1)^(1/2) - 1)^8) + (2*a*atanh(((c*x - 1)^(1/2) - 1i)/((c*x + 1)^(1/2) - 1)))/c^3 + (3*b*atanh(((c*x
- 1)^(1/2) - 1i)/((c*x + 1)^(1/2) - 1)))/(2*c^5)