\(\int \frac {x^4 (a+b x^2)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [348]
Optimal result
Integrand size = 29, antiderivative size = 125 \[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\left (5 b+6 a c^2\right ) x \sqrt {-1+c x} \sqrt {1+c x}}{16 c^6}+\frac {\left (5 b+6 a c^2\right ) x^3 \sqrt {-1+c x} \sqrt {1+c x}}{24 c^4}+\frac {b x^5 \sqrt {-1+c x} \sqrt {1+c x}}{6 c^2}+\frac {\left (5 b+6 a c^2\right ) \text {arccosh}(c x)}{16 c^7}
\]
[Out]
1/16*(6*a*c^2+5*b)*arccosh(c*x)/c^7+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
Rubi [A] (verified)
Time = 0.06 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {471, 102, 12, 92, 54}
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\left (6 a c^2+5 b\right ) \text {arccosh}(c x)}{16 c^7}+\frac {x \sqrt {c x-1} \sqrt {c x+1} \left (6 a c^2+5 b\right )}{16 c^6}+\frac {x^3 \sqrt {c x-1} \sqrt {c x+1} \left (6 a c^2+5 b\right )}{24 c^4}+\frac {b x^5 \sqrt {c x-1} \sqrt {c x+1}}{6 c^2}
\]
[In]
Int[(x^4*(a + b*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]
[Out]
((5*b + 6*a*c^2)*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(16*c^6) + ((5*b + 6*a*c^2)*x^3*Sqrt[-1 + c*x]*Sqrt[1 + c*x])
/(24*c^4) + (b*x^5*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*c^2) + ((5*b + 6*a*c^2)*ArcCosh[c*x])/(16*c^7)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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^5 \sqrt {-1+c x} \sqrt {1+c x}}{6 c^2}-\frac {1}{6} \left (-6 a-\frac {5 b}{c^2}\right ) \int \frac {x^4}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx \\ & = \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 ) \int \frac {3 x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{24 c^4} \\ & = \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 ) \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{8 c^4} \\ & = \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 ) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{16 c^6} \\ & = \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 ) \cosh ^{-1}(c x)}{16 c^7} \\
\end{align*}
Mathematica [A] (warning: unable to verify)
Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {c x \sqrt {-1+c x} \sqrt {1+c x} \left (6 a c^2 \left (3+2 c^2 x^2\right )+b \left (15+10 c^2 x^2+8 c^4 x^4\right )\right )+6 \left (5 b+6 a c^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )}{48 c^7}
\]
[In]
Integrate[(x^4*(a + b*x^2))/(Sqrt[-1 + c*x]*Sqrt[1 + c*x]),x]
[Out]
(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)
Maple [A] (verified)
Time = 4.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03
| | |
| method | result | size |
| | |
| risch |
\(\frac {x \left (8 b \,x^{4} c^{4}+12 a \,c^{4} x^{2}+10 b \,c^{2} x^{2}+18 c^{2} a +15 b \right ) \sqrt {c x -1}\, \sqrt {c x +1}}{48 c^{6}}+\frac {\left (6 c^{2} a +5 b \right ) \ln \left (\frac {c^{2} x}{\sqrt {c^{2}}}+\sqrt {c^{2} x^{2}-1}\right ) \sqrt {\left (c x -1\right ) \left (c x +1\right )}}{16 c^{6} \sqrt {c^{2}}\, \sqrt {c x -1}\, \sqrt {c x +1}}\) |
\(129\) |
| default |
\(\frac {\sqrt {c x -1}\, \sqrt {c x +1}\, \left (8 \,\operatorname {csgn}\left (c \right ) b \,c^{5} x^{5} \sqrt {c^{2} x^{2}-1}+12 \,\operatorname {csgn}\left (c \right ) a \,c^{5} x^{3} \sqrt {c^{2} x^{2}-1}+10 \,\operatorname {csgn}\left (c \right ) c^{3} \sqrt {c^{2} x^{2}-1}\, b \,x^{3}+18 \,\operatorname {csgn}\left (c \right ) c^{3} \sqrt {c^{2} x^{2}-1}\, a x +15 \,\operatorname {csgn}\left (c \right ) c \sqrt {c^{2} x^{2}-1}\, b x +18 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) a \,c^{2}+15 \ln \left (\left (\sqrt {c^{2} x^{2}-1}\, \operatorname {csgn}\left (c \right )+c x \right ) \operatorname {csgn}\left (c \right )\right ) b \right ) \operatorname {csgn}\left (c \right )}{48 c^{7} \sqrt {c^{2} x^{2}-1}}\) |
\(191\) |
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[In]
int(x^4*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x,method=_RETURNVERBOSE)
[Out]
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)
none
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left (8 \, b c^{5} x^{5} + 2 \, {\left (6 \, a c^{5} + 5 \, b c^{3}\right )} x^{3} + 3 \, {\left (6 \, a c^{3} + 5 \, b c\right )} x\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 3 \, {\left (6 \, a c^{2} + 5 \, b\right )} \log \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{48 \, c^{7}}
\]
[In]
integrate(x^4*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="fricas")
[Out]
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)*sqrt(c*x + 1)*sqrt(c*x - 1) - 3*(6*a*c
^2 + 5*b)*log(-c*x + sqrt(c*x + 1)*sqrt(c*x - 1)))/c^7
Sympy [F(-1)]
Timed out. \[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Timed out}
\]
[In]
integrate(x**4*(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 = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {\sqrt {c^{2} x^{2} - 1} b x^{5}}{6 \, c^{2}} + \frac {\sqrt {c^{2} x^{2} - 1} a x^{3}}{4 \, c^{2}} + \frac {5 \, \sqrt {c^{2} x^{2} - 1} b x^{3}}{24 \, c^{4}} + \frac {3 \, \sqrt {c^{2} x^{2} - 1} a x}{8 \, c^{4}} + \frac {3 \, a \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{8 \, c^{5}} + \frac {5 \, \sqrt {c^{2} x^{2} - 1} b x}{16 \, c^{6}} + \frac {5 \, b \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{16 \, c^{7}}
\]
[In]
integrate(x^4*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="maxima")
[Out]
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*sqrt(c^2*x^2 - 1)*b*x^3/c^4 + 3/8*sqr
t(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 + 5/1
6*b*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^7
Giac [A] (verification not implemented)
none
Time = 0.30 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.38
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\frac {{\left ({\left (2 \, {\left ({\left (c x + 1\right )} {\left (4 \, {\left (c x + 1\right )} {\left (\frac {{\left (c x + 1\right )} b}{c^{6}} - \frac {5 \, b}{c^{6}}\right )} + \frac {3 \, {\left (2 \, a c^{38} + 15 \, b c^{36}\right )}}{c^{42}}\right )} - \frac {18 \, a c^{38} + 55 \, b c^{36}}{c^{42}}\right )} {\left (c x + 1\right )} + \frac {54 \, a c^{38} + 85 \, b c^{36}}{c^{42}}\right )} {\left (c x + 1\right )} - \frac {3 \, {\left (10 \, a c^{38} + 11 \, b c^{36}\right )}}{c^{42}}\right )} \sqrt {c x + 1} \sqrt {c x - 1} - \frac {6 \, {\left (6 \, a c^{2} + 5 \, b\right )} \log \left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}{c^{6}}}{48 \, c}
\]
[In]
integrate(x^4*(b*x^2+a)/(c*x-1)^(1/2)/(c*x+1)^(1/2),x, algorithm="giac")
[Out]
1/48*(((2*((c*x + 1)*(4*(c*x + 1)*((c*x + 1)*b/c^6 - 5*b/c^6) + 3*(2*a*c^38 + 15*b*c^36)/c^42) - (18*a*c^38 +
55*b*c^36)/c^42)*(c*x + 1) + (54*a*c^38 + 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 = 45.44 (sec) , antiderivative size = 1154, normalized size of antiderivative = 9.23
\[
\int \frac {x^4 \left (a+b x^2\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx=\text {Too large to display}
\]
[In]
int((x^4*(a + b*x^2))/((c*x - 1)^(1/2)*(c*x + 1)^(1/2)),x)
[Out]
((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) + (2529
5*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) - 1i)^17)/(4*((c*x + 1)^(1/2) - 1)^17) + (311*b*((c*x - 1)^(1/2) -
1i)^19)/(4*((c*x + 1)^(1/2) - 1)^19) - (175*b*((c*x - 1)^(1/2) - 1i)^21)/(12*((c*x + 1)^(1/2) - 1)^21) + (5*b
*((c*x - 1)^(1/2) - 1i)^23)/(4*((c*x + 1)^(1/2) - 1)^23) + (5*b*((c*x - 1)^(1/2) - 1i))/(4*((c*x + 1)^(1/2) -
1)))/(c^7 - (12*c^7*((c*x - 1)^(1/2) - 1i)^2)/((c*x + 1)^(1/2) - 1)^2 + (66*c^7*((c*x - 1)^(1/2) - 1i)^4)/((c*
x + 1)^(1/2) - 1)^4 - (220*c^7*((c*x - 1)^(1/2) - 1i)^6)/((c*x + 1)^(1/2) - 1)^6 + (495*c^7*((c*x - 1)^(1/2) -
1i)^8)/((c*x + 1)^(1/2) - 1)^8 - (792*c^7*((c*x - 1)^(1/2) - 1i)^10)/((c*x + 1)^(1/2) - 1)^10 + (924*c^7*((c*
x - 1)^(1/2) - 1i)^12)/((c*x + 1)^(1/2) - 1)^12 - (792*c^7*((c*x - 1)^(1/2) - 1i)^14)/((c*x + 1)^(1/2) - 1)^14
+ (495*c^7*((c*x - 1)^(1/2) - 1i)^16)/((c*x + 1)^(1/2) - 1)^16 - (220*c^7*((c*x - 1)^(1/2) - 1i)^18)/((c*x +
1)^(1/2) - 1)^18 + (66*c^7*((c*x - 1)^(1/2) - 1i)^20)/((c*x + 1)^(1/2) - 1)^20 - (12*c^7*((c*x - 1)^(1/2) - 1i
)^22)/((c*x + 1)^(1/2) - 1)^22 + (c^7*((c*x - 1)^(1/2) - 1i)^24)/((c*x + 1)^(1/2) - 1)^24) + (3*a*atanh(((c*x
- 1)^(1/2) - 1i)/((c*x + 1)^(1/2) - 1)))/(2*c^5) + (5*b*atanh(((c*x - 1)^(1/2) - 1i)/((c*x + 1)^(1/2) - 1)))/(
4*c^7)