\(\int \frac {x^3 (a+b x^2)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [349]

Optimal result

Integrand size = 29, antiderivative size = 103 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {2 \left (4 b+5 a c^2\right ) \sqrt {-1+c x} \sqrt {1+c x}}{15 c^6}+\frac {\left (4 b+5 a c^2\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{15 c^4}+\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{5 c^2} \]

[Out]

2/15*(5*a*c^2+4*b)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^6+1/15*(5*a*c^2+4*b)*x^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^4+1/5*
b*x^4*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {471, 102, 12, 75} \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {2 \sqrt {c x-1} \sqrt {c x+1} \left (5 a c^2+4 b\right )}{15 c^6}+\frac {x^2 \sqrt {c x-1} \sqrt {c x+1} \left (5 a c^2+4 b\right )}{15 c^4}+\frac {b x^4 \sqrt {c x-1} \sqrt {c x+1}}{5 c^2} \]

[In]

Int[(x^3*(a + b*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]

[Out]

(2*(4*b + 5*a*c^2)*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(15*c^6) + ((4*b + 5*a*c^2)*x^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])
/(15*c^4) + (b*x^4*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(5*c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 75

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

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(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 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^4 \sqrt {-1+c x} \sqrt {1+c x}}{5 c^2}-\frac {1}{5} \left (-5 a-\frac {4 b}{c^2}\right ) \int \frac {x^3}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \frac {\left (4 b+5 a c^2\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{15 c^4}+\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{5 c^2}+\frac {\left (4 b+5 a c^2\right ) \int \frac {2 x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{15 c^4} \\ & = \frac {\left (4 b+5 a c^2\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{15 c^4}+\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{5 c^2}+\frac {\left (2 \left (4 b+5 a c^2\right )\right ) \int \frac {x}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{15 c^4} \\ & = \frac {2 \left (4 b+5 a c^2\right ) \sqrt {-1+c x} \sqrt {1+c x}}{15 c^6}+\frac {\left (4 b+5 a c^2\right ) x^2 \sqrt {-1+c x} \sqrt {1+c x}}{15 c^4}+\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{5 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.59 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (5 a c^2 \left (2+c^2 x^2\right )+b \left (8+4 c^2 x^2+3 c^4 x^4\right )\right )}{15 c^6} \]

[In]

Integrate[(x^3*(a + b*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]

[Out]

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

Maple [A] (verified)

Time = 4.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (3 b \,x^{4} c^{4}+5 a \,c^{4} x^{2}+4 b \,c^{2} x^{2}+10 c^{2} a +8 b \right )}{15 c^{6}}\)	\(57\)
default	\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (3 b \,x^{4} c^{4}+5 a \,c^{4} x^{2}+4 b \,c^{2} x^{2}+10 c^{2} a +8 b \right )}{15 c^{6}}\)	\(57\)
risch	\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (3 b \,x^{4} c^{4}+5 a \,c^{4} x^{2}+4 b \,c^{2} x^{2}+10 c^{2} a +8 b \right )}{15 c^{6}}\)	\(57\)

[In]

int(x^3*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(c*x-1)^(1/2)*(c*x+1)^(1/2)*(3*b*c^4*x^4+5*a*c^4*x^2+4*b*c^2*x^2+10*a*c^2+8*b)/c^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.53 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left (3 \, b c^{4} x^{4} + 10 \, a c^{2} + {\left (5 \, a c^{4} + 4 \, b c^{2}\right )} x^{2} + 8 \, b\right )} \sqrt {c x + 1} \sqrt {c x - 1}}{15 \, c^{6}} \]

[In]

integrate(x^3*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*b*c^4*x^4 + 10*a*c^2 + (5*a*c^4 + 4*b*c^2)*x^2 + 8*b)*sqrt(c*x + 1)*sqrt(c*x - 1)/c^6

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.57 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.10 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {a {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{4}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{4}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{4}, - \frac {7}{4} & -2, -2, - \frac {3}{2}, 1 \\- \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, - \frac {3}{2}, 0 & \end {matrix} \middle | {\frac {1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{6}} + \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {11}{4}, - \frac {5}{2}, - \frac {9}{4}, -2, 1 & \\- \frac {11}{4}, - \frac {9}{4} & -3, - \frac {5}{2}, - \frac {5}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{6}} \]

[In]

integrate(x**3*(b*x**2+a)/(c*x-1)**(1/2)/(c*x+1)**(1/2),x)

[Out]

a*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), 1/(c**2*x**2))/(4*pi**(3/2
)*c**4) + I*a*meijerg(((-2, -7/4, -3/2, -5/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), exp_polar(2*I*
pi)/(c**2*x**2))/(4*pi**(3/2)*c**4) + b*meijerg(((-9/4, -7/4), (-2, -2, -3/2, 1)), ((-5/2, -9/4, -2, -7/4, -3/
2, 0), ()), 1/(c**2*x**2))/(4*pi**(3/2)*c**6) + I*b*meijerg(((-3, -11/4, -5/2, -9/4, -2, 1), ()), ((-11/4, -9/
4), (-3, -5/2, -5/2, 0)), exp_polar(2*I*pi)/(c**2*x**2))/(4*pi**(3/2)*c**6)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {c^{2} x^{2} - 1} b x^{4}}{5 \, c^{2}} + \frac {\sqrt {c^{2} x^{2} - 1} a x^{2}}{3 \, c^{2}} + \frac {4 \, \sqrt {c^{2} x^{2} - 1} b x^{2}}{15 \, c^{4}} + \frac {2 \, \sqrt {c^{2} x^{2} - 1} a}{3 \, c^{4}} + \frac {8 \, \sqrt {c^{2} x^{2} - 1} b}{15 \, c^{6}} \]

[In]

integrate(x^3*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(c^2*x^2 - 1)*b*x^4/c^2 + 1/3*sqrt(c^2*x^2 - 1)*a*x^2/c^2 + 4/15*sqrt(c^2*x^2 - 1)*b*x^2/c^4 + 2/3*sqr
t(c^2*x^2 - 1)*a/c^4 + 8/15*sqrt(c^2*x^2 - 1)*b/c^6

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left ({\left ({\left (c x + 1\right )} {\left (3 \, {\left (c x + 1\right )} {\left (\frac {{\left (c x + 1\right )} b}{c^{5}} - \frac {4 \, b}{c^{5}}\right )} + \frac {5 \, a c^{27} + 22 \, b c^{25}}{c^{30}}\right )} - \frac {10 \, {\left (a c^{27} + 2 \, b c^{25}\right )}}{c^{30}}\right )} {\left (c x + 1\right )} + \frac {15 \, {\left (a c^{27} + b c^{25}\right )}}{c^{30}}\right )} \sqrt {c x + 1} \sqrt {c x - 1}}{15 \, c} \]

[In]

integrate(x^3*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="giac")

[Out]

1/15*(((c*x + 1)*(3*(c*x + 1)*((c*x + 1)*b/c^5 - 4*b/c^5) + (5*a*c^27 + 22*b*c^25)/c^30) - 10*(a*c^27 + 2*b*c^
25)/c^30)*(c*x + 1) + 15*(a*c^27 + b*c^25)/c^30)*sqrt(c*x + 1)*sqrt(c*x - 1)/c

Mupad [B] (verification not implemented)

Time = 6.64 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {c\,x-1}\,\left (\frac {10\,a\,c^2+8\,b}{15\,c^6}+\frac {b\,x^5}{5\,c}+\frac {b\,x^4}{5\,c^2}+\frac {x^2\,\left (5\,a\,c^4+4\,b\,c^2\right )}{15\,c^6}+\frac {x^3\,\left (5\,a\,c^5+4\,b\,c^3\right )}{15\,c^6}+\frac {x\,\left (10\,a\,c^3+8\,b\,c\right )}{15\,c^6}\right )}{\sqrt {c\,x+1}} \]

[In]

int((x^3*(a + b*x^2))/((c*x - 1)^(1/2)*(c*x + 1)^(1/2)),x)

[Out]

((c*x - 1)^(1/2)*((8*b + 10*a*c^2)/(15*c^6) + (b*x^5)/(5*c) + (b*x^4)/(5*c^2) + (x^2*(5*a*c^4 + 4*b*c^2))/(15*
c^6) + (x^3*(5*a*c^5 + 4*b*c^3))/(15*c^6) + (x*(8*b*c + 10*a*c^3))/(15*c^6)))/(c*x + 1)^(1/2)