\(\int x^4 (a+b \text {sech}^{-1}(c x)) \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 131 \[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=-\frac {3 b x \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{40 c^4}-\frac {b x^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{20 c^2}+\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {3 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \arcsin (c x)}{40 c^5 \sqrt {1-c^2 x^2}} \] Output:

-3/40*b*x*((-c*x+1)/(c*x+1))^(1/2)*(c*x+1)/c^4-1/20*b*x^3*((-c*x+1)/(c*x+1 
))^(1/2)*(c*x+1)/c^2+1/5*x^5*(a+b*arcsech(c*x))+3/40*b*((-c*x+1)/(c*x+1))^ 
(1/2)*(c*x+1)*arcsin(c*x)/c^5/(-c^2*x^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94 \[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\frac {a x^5}{5}+b \sqrt {\frac {1-c x}{1+c x}} \left (-\frac {3 x}{40 c^4}-\frac {3 x^2}{40 c^3}-\frac {x^3}{20 c^2}-\frac {x^4}{20 c}\right )+\frac {1}{5} b x^5 \text {sech}^{-1}(c x)+\frac {3 i b \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{40 c^5} \] Input:

Integrate[x^4*(a + b*ArcSech[c*x]),x]
 

Output:

(a*x^5)/5 + b*Sqrt[(1 - c*x)/(1 + c*x)]*((-3*x)/(40*c^4) - (3*x^2)/(40*c^3 
) - x^3/(20*c^2) - x^4/(20*c)) + (b*x^5*ArcSech[c*x])/5 + (((3*I)/40)*b*Lo 
g[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)])/c^5
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6837, 111, 27, 101, 25, 39, 223}

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.

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

\(\Big \downarrow \) 6837

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

\(\Big \downarrow \) 111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 39

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

\(\Big \downarrow \) 223

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

Input:

Int[x^4*(a + b*ArcSech[c*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 39

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[( 
a*c + b*d*x^2)^m, x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 0] && ( 
IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 6837

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Si 
mp[(d*x)^(m + 1)*((a + b*ArcSech[c*x])/(d*(m + 1))), x] + Simp[b*(Sqrt[1 + 
c*x]/(m + 1))*Sqrt[1/(1 + c*x)]   Int[(d*x)^m/(Sqrt[1 - c*x]*Sqrt[1 + c*x]) 
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {a \,x^{5}}{5}+\frac {b \left (\frac {c^{5} x^{5} \operatorname {arcsech}\left (c x \right )}{5}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{40 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{5}}\)	\(114\)
derivativedivides	\(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \operatorname {arcsech}\left (c x \right )}{5}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{40 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{5}}\)	\(118\)
default	\(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \operatorname {arcsech}\left (c x \right )}{5}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{40 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{5}}\)	\(118\)

Input:

int(x^4*(a+b*arcsech(c*x)),x,method=_RETURNVERBOSE)
 

Output:

1/5*a*x^5+b/c^5*(1/5*c^5*x^5*arcsech(c*x)+1/40*(-(c*x-1)/c/x)^(1/2)*c*x*(( 
c*x+1)/c/x)^(1/2)*(-2*c^3*x^3*(-c^2*x^2+1)^(1/2)-3*c*x*(-c^2*x^2+1)^(1/2)+ 
3*arcsin(c*x))/(-c^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.33 \[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\frac {8 \, a c^{5} x^{5} - 8 \, b c^{5} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 6 \, b \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) + 8 \, {\left (b c^{5} x^{5} - b c^{5}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (2 \, b c^{4} x^{4} + 3 \, b c^{2} x^{2}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{40 \, c^{5}} \] Input:

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="fricas")
 

Output:

1/40*(8*a*c^5*x^5 - 8*b*c^5*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x 
) - 6*b*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(c*x)) + 8*(b*c^5* 
x^5 - b*c^5)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (2*b*c^ 
4*x^4 + 3*b*c^2*x^2)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/c^5
 

Sympy [F]

\[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int x^{4} \left (a + b \operatorname {asech}{\left (c x \right )}\right )\, dx \] Input:

integrate(x**4*(a+b*asech(c*x)),x)
 

Output:

Integral(x**4*(a + b*asech(c*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{40} \, {\left (8 \, x^{5} \operatorname {arsech}\left (c x\right ) - \frac {\frac {3 \, {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 5 \, \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{4}} + \frac {3 \, \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{4}}}{c}\right )} b \] Input:

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="maxima")
 

Output:

1/5*a*x^5 + 1/40*(8*x^5*arcsech(c*x) - ((3*(1/(c^2*x^2) - 1)^(3/2) + 5*sqr 
t(1/(c^2*x^2) - 1))/(c^4*(1/(c^2*x^2) - 1)^2 + 2*c^4*(1/(c^2*x^2) - 1) + c 
^4) + 3*arctan(sqrt(1/(c^2*x^2) - 1))/c^4)/c)*b
 

Giac [F]

\[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\int { {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{4} \,d x } \] Input:

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="giac")
 

Output:

integrate((b*arcsech(c*x) + a)*x^4, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx=\left (\int \mathit {asech} \left (c x \right ) x^{4}d x \right ) b +\frac {a \,x^{5}}{5} \] Input:

int(x^4*(a+b*asech(c*x)),x)
 

Output:

(5*int(asech(c*x)*x**4,x)*b + a*x**5)/5