3.1.26 \(\int x (a+b \sec ^{-1}(c x))^3 \, dx\) [26]

3.1.26.1 Optimal result

Integrand size = 12, antiderivative size = 126 \[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\frac {3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3 b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^2}+\frac {3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2} \]

output

3/2*I*b*(a+b*arcsec(c*x))^2/c^2+1/2*x^2*(a+b*arcsec(c*x))^3-3*b^2*(a+b*arc 
sec(c*x))*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)/c^2+3/2*I*b^3*polylog(2,-( 
1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)/c^2-3/2*b*x*(a+b*arcsec(c*x))^2*(1-1/c^2/x 
^2)^(1/2)/c
 

3.1.26.2 Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.46 \[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\frac {-3 b^2 \left (-a c^2 x^2+b \left (-i+c \sqrt {1-\frac {1}{c^2 x^2}} x\right )\right ) \sec ^{-1}(c x)^2+b^3 c^2 x^2 \sec ^{-1}(c x)^3-3 b \sec ^{-1}(c x) \left (a c x \left (2 b \sqrt {1-\frac {1}{c^2 x^2}}-a c x\right )+2 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )+a \left (a c x \left (-3 b \sqrt {1-\frac {1}{c^2 x^2}}+a c x\right )-6 b^2 \log \left (\frac {1}{c x}\right )\right )+3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2} \]

input

Integrate[x*(a + b*ArcSec[c*x])^3,x]
 

output

(-3*b^2*(-(a*c^2*x^2) + b*(-I + c*Sqrt[1 - 1/(c^2*x^2)]*x))*ArcSec[c*x]^2 
+ b^3*c^2*x^2*ArcSec[c*x]^3 - 3*b*ArcSec[c*x]*(a*c*x*(2*b*Sqrt[1 - 1/(c^2* 
x^2)] - a*c*x) + 2*b^2*Log[1 + E^((2*I)*ArcSec[c*x])]) + a*(a*c*x*(-3*b*Sq 
rt[1 - 1/(c^2*x^2)] + a*c*x) - 6*b^2*Log[1/(c*x)]) + (3*I)*b^3*PolyLog[2, 
-E^((2*I)*ArcSec[c*x])])/(2*c^2)
 

3.1.26.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5745, 4909, 3042, 4672, 25, 3042, 4202, 2620, 2715, 2838}

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.

\(\displaystyle \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx\)

\(\Big \downarrow \) 5745

\(\displaystyle \frac {\int c^3 \sqrt {1-\frac {1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^3d\sec ^{-1}(c x)}{c^2}\)

\(\Big \downarrow \) 4909

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \int c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^2d\sec ^{-1}(c x)}{c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \int \left (a+b \sec ^{-1}(c x)\right )^2 \csc \left (\sec ^{-1}(c x)+\frac {\pi }{2}\right )^2d\sec ^{-1}(c x)}{c^2}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (2 b \int -c \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )d\sec ^{-1}(c x)+c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2\right )}{c^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \int c \sqrt {1-\frac {1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )d\sec ^{-1}(c x)\right )}{c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \int \left (a+b \sec ^{-1}(c x)\right ) \tan \left (\sec ^{-1}(c x)\right )d\sec ^{-1}(c x)\right )}{c^2}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \left (\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \int \frac {e^{2 i \sec ^{-1}(c x)} \left (a+b \sec ^{-1}(c x)\right )}{1+e^{2 i \sec ^{-1}(c x)}}d\sec ^{-1}(c x)\right )\right )}{c^2}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \left (\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \left (\frac {1}{2} i b \int \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )d\sec ^{-1}(c x)-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )\right )\right )\right )}{c^2}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \left (\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \left (\frac {1}{4} b \int e^{-2 i \sec ^{-1}(c x)} \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )de^{2 i \sec ^{-1}(c x)}-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )\right )\right )\right )}{c^2}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {1}{2} c^2 x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{2} b \left (c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-2 b \left (\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b}-2 i \left (-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )\right )\right )\right )}{c^2}\)

input

Int[x*(a + b*ArcSec[c*x])^3,x]
 

output

((c^2*x^2*(a + b*ArcSec[c*x])^3)/2 - (3*b*(c*Sqrt[1 - 1/(c^2*x^2)]*x*(a + 
b*ArcSec[c*x])^2 - 2*b*(((I/2)*(a + b*ArcSec[c*x])^2)/b - (2*I)*((-1/2*I)* 
(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])] - (b*PolyLog[2, -E^((2* 
I)*ArcSec[c*x])])/4))))/2)/c^2
 

3.1.26.3.1 Defintions of rubi rules used

rule 25

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

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

rule 5745

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[1 
/c^(m + 1)   Subst[Int[(a + b*x)^n*Sec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x 
]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n, 0] | 
| LtQ[m, -1])
 

3.1.26.4 Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.00

method	result	size
derivativedivides	\(\frac {\frac {a^{3} c^{2} x^{2}}{2}+b^{3} \left (\frac {\operatorname {arcsec}\left (c x \right )^{2} \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-3 x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-3 i\right )}{2}+3 i \operatorname {arcsec}\left (c x \right )^{2}-3 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )+3 a \,b^{2} \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )^{2}}{2}-\operatorname {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-\ln \left (\frac {1}{c x}\right )\right )+3 a^{2} b \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )}{2}-\frac {c^{2} x^{2}-1}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}\)	\(252\)
default	\(\frac {\frac {a^{3} c^{2} x^{2}}{2}+b^{3} \left (\frac {\operatorname {arcsec}\left (c x \right )^{2} \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-3 x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-3 i\right )}{2}+3 i \operatorname {arcsec}\left (c x \right )^{2}-3 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )+3 a \,b^{2} \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )^{2}}{2}-\operatorname {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-\ln \left (\frac {1}{c x}\right )\right )+3 a^{2} b \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )}{2}-\frac {c^{2} x^{2}-1}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}\)	\(252\)
parts	\(\frac {a^{3} x^{2}}{2}+\frac {b^{3} \left (\frac {\operatorname {arcsec}\left (c x \right )^{2} \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-3 x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-3 i\right )}{2}+3 i \operatorname {arcsec}\left (c x \right )^{2}-3 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )}{c^{2}}+\frac {3 a \,b^{2} \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )^{2}}{2}-\operatorname {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-\ln \left (\frac {1}{c x}\right )\right )}{c^{2}}+\frac {3 a^{2} b \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )}{2}-\frac {c^{2} x^{2}-1}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}\)	\(254\)

input

int(x*(a+b*arcsec(c*x))^3,x,method=_RETURNVERBOSE)
 

output

1/c^2*(1/2*a^3*c^2*x^2+b^3*(1/2*arcsec(c*x)^2*(c^2*x^2*arcsec(c*x)-3*x*c*( 
(c^2*x^2-1)/c^2/x^2)^(1/2)-3*I)+3*I*arcsec(c*x)^2-3*arcsec(c*x)*ln(1+(1/c/ 
x+I*(1-1/c^2/x^2)^(1/2))^2)+3/2*I*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2)) 
^2))+3*a*b^2*(1/2*c^2*x^2*arcsec(c*x)^2-arcsec(c*x)*c*x*((c^2*x^2-1)/c^2/x 
^2)^(1/2)-ln(1/c/x))+3*a^2*b*(1/2*c^2*x^2*arcsec(c*x)-1/2/((c^2*x^2-1)/c^2 
/x^2)^(1/2)/c/x*(c^2*x^2-1)))
 

3.1.26.5 Fricas [F]

\[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x \,d x } \]

input

integrate(x*(a+b*arcsec(c*x))^3,x, algorithm="fricas")
 

output

integral(b^3*x*arcsec(c*x)^3 + 3*a*b^2*x*arcsec(c*x)^2 + 3*a^2*b*x*arcsec( 
c*x) + a^3*x, x)
 

3.1.26.6 Sympy [F]

\[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int x \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}\, dx \]

input

integrate(x*(a+b*asec(c*x))**3,x)
 

output

Integral(x*(a + b*asec(c*x))**3, x)
 

3.1.26.7 Maxima [F]

\[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x \,d x } \]

input

integrate(x*(a+b*arcsec(c*x))^3,x, algorithm="maxima")
 

output

3/2*a*b^2*x^2*arcsec(c*x)^2 + 1/2*a^3*x^2 + 3/2*(x^2*arcsec(c*x) - x*sqrt( 
-1/(c^2*x^2) + 1)/c)*a^2*b - 3*(x*sqrt(-1/(c^2*x^2) + 1)*arcsec(c*x)/c - l 
og(x)/c^2)*a*b^2 + 1/8*(4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3*x^ 
2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 - 8*integrate(3/8*((4 
*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 - x*log(c^2*x^2)^2)*sqrt(c*x + 1) 
*sqrt(c*x - 1) + 4*(2*c^2*x^3*log(c)^2 - 2*x*log(c)^2 + 2*(c^2*x^3 - x)*lo 
g(x)^2 - ((2*c^2*log(c) + c^2)*x^3 - x*(2*log(c) + 1) + 2*(c^2*x^3 - x)*lo 
g(x))*log(c^2*x^2) + 4*(c^2*x^3*log(c) - x*log(c))*log(x))*arctan(sqrt(c*x 
 + 1)*sqrt(c*x - 1)))/(c^2*x^2 - 1), x))*b^3
 

3.1.26.8 Giac [F]

\[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int { {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3} x \,d x } \]

input

integrate(x*(a+b*arcsec(c*x))^3,x, algorithm="giac")
 

output

integrate((b*arcsec(c*x) + a)^3*x, x)
 

3.1.26.9 Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=\int x\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3 \,d x \]

input

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

output

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