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

Optimal result

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

[Out]

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*arcsec(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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5330, 4494, 4269, 3800, 2221, 2317, 2438} \[ \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx=-\frac {3 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac {3 b x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac {3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3+\frac {3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2} \]

[In]

Int[x*(a + b*ArcSec[c*x])^3,x]

[Out]

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

Rule 2221

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]
- Dist[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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[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] &&
IGtQ[m, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[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 4494

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] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5330

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+b x)^3 \sec ^2(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^2} \\ & = \frac {1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac {(3 b) \text {Subst}\left (\int (a+b x)^2 \sec ^2(x) \, dx,x,\sec ^{-1}(c x)\right )}{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 {\left (3 b^2\right ) \text {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^2} \\ & = \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 {\left (6 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )}{c^2} \\ & = \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 {\left (3 b^3\right ) \text {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^2} \\ & = \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 {\left (3 i b^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[x*(a + b*ArcSec[c*x])^3,x]

[Out]

(-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*Ar
cSec[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*
Sqrt[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)

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	\(\frac {a^{3} x^{2}}{2}+\frac {b^{3} \left (\frac {\operatorname {arcsec}\left (c x \right )^{2} \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-3 x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-3 i\right )}{2}+3 i \operatorname {arcsec}\left (c x \right )^{2}-3 \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+\frac {3 i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )}{c^{2}}+\frac {3 a \,b^{2} \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )^{2}}{2}-\operatorname {arcsec}\left (c x \right ) c x \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-\ln \left (\frac {1}{c x}\right )\right )}{c^{2}}+\frac {3 a^{2} b \left (\frac {c^{2} x^{2} \operatorname {arcsec}\left (c x \right )}{2}-\frac {c^{2} x^{2}-1}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}\)	\(254\)

[In]

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

[Out]

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)))

Fricas [F]

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

[In]

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

[Out]

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)

Sympy [F]

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

[In]

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

[Out]

Integral(x*(a + b*asec(c*x))**3, x)

Maxima [F]

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

[In]

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

[Out]

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*sq
rt(-1/(c^2*x^2) + 1)*arcsec(c*x)/c - log(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
)*log(x)^2 - ((2*c^2*log(c) + c^2)*x^3 - x*(2*log(c) + 1) + 2*(c^2*x^3 - x)*log(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

Giac [F]

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

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^3*x, x)

Mupad [F(-1)]

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

[In]

int(x*(a + b*acos(1/(c*x)))^3,x)

[Out]

int(x*(a + b*acos(1/(c*x)))^3, x)