3.28 \(\int \frac {(a+b \sec ^{-1}(c x))^3}{x} \, dx\)

Optimal. Leaf size=128 \[ -\frac {3}{2} b^2 \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {3}{2} i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2+\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^3-\frac {3}{4} i b^3 \text {Li}_4\left (-e^{2 i \sec ^{-1}(c x)}\right ) \]

[Out]

1/4*I*(a+b*arcsec(c*x))^4/b-(a+b*arcsec(c*x))^3*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+3/2*I*b*(a+b*arcsec(c*x)
)^2*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)-3/2*b^2*(a+b*arcsec(c*x))*polylog(3,-(1/c/x+I*(1-1/c^2/x^2)^(1
/2))^2)-3/4*I*b^3*polylog(4,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)

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Rubi [A]  time = 0.14, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5222, 3719, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3}{2} b^2 \text {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {3}{2} i b \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac {3}{4} i b^3 \text {PolyLog}\left (4,-e^{2 i \sec ^{-1}(c x)}\right )+\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^3/x,x]

[Out]

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3719

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 5222

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])

Rule 6589

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

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x} \, dx &=\operatorname {Subst}\left (\int (a+b x)^3 \tan (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^3}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\left (3 i b^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )+\frac {1}{2} \left (3 b^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{4} \left (3 i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )\\ &=\frac {i \left (a+b \sec ^{-1}(c x)\right )^4}{4 b}-\left (a+b \sec ^{-1}(c x)\right )^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+\frac {3}{2} i b \left (a+b \sec ^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {3}{2} b^2 \left (a+b \sec ^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right )-\frac {3}{4} i b^3 \text {Li}_4\left (-e^{2 i \sec ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 204, normalized size = 1.59 \[ \frac {1}{4} \left (4 a^3 \log (c x)+6 i a^2 b \sec ^{-1}(c x)^2-12 a^2 b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-6 b^2 \text {Li}_3\left (-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+4 i a b^2 \sec ^{-1}(c x)^3-12 a b^2 \sec ^{-1}(c x)^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+6 i b \text {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-3 i b^3 \text {Li}_4\left (-e^{2 i \sec ^{-1}(c x)}\right )+i b^3 \sec ^{-1}(c x)^4-4 b^3 \sec ^{-1}(c x)^3 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSec[c*x])^3/x,x]

[Out]

((6*I)*a^2*b*ArcSec[c*x]^2 + (4*I)*a*b^2*ArcSec[c*x]^3 + I*b^3*ArcSec[c*x]^4 - 12*a^2*b*ArcSec[c*x]*Log[1 + E^
((2*I)*ArcSec[c*x])] - 12*a*b^2*ArcSec[c*x]^2*Log[1 + E^((2*I)*ArcSec[c*x])] - 4*b^3*ArcSec[c*x]^3*Log[1 + E^(
(2*I)*ArcSec[c*x])] + 4*a^3*Log[c*x] + (6*I)*b*(a + b*ArcSec[c*x])^2*PolyLog[2, -E^((2*I)*ArcSec[c*x])] - 6*b^
2*(a + b*ArcSec[c*x])*PolyLog[3, -E^((2*I)*ArcSec[c*x])] - (3*I)*b^3*PolyLog[4, -E^((2*I)*ArcSec[c*x])])/4

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arcsec}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname {arcsec}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname {arcsec}\left (c x\right ) + a^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{3}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^3/x, x)

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maple [B]  time = 0.26, size = 390, normalized size = 3.05 \[ a^{3} \ln \left (c x \right )+\frac {i b^{3} \mathrm {arcsec}\left (c x \right )^{4}}{4}-b^{3} \mathrm {arcsec}\left (c x \right )^{3} \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {3 i b^{3} \mathrm {arcsec}\left (c x \right )^{2} \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}-\frac {3 b^{3} \mathrm {arcsec}\left (c x \right ) \polylog \left (3, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}-\frac {3 i b^{3} \polylog \left (4, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{4}+i a \,b^{2} \mathrm {arcsec}\left (c x \right )^{3}-3 a \,b^{2} \mathrm {arcsec}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+3 i a \,b^{2} \mathrm {arcsec}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )-\frac {3 a \,b^{2} \polylog \left (3, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}+\frac {3 i a^{2} b \mathrm {arcsec}\left (c x \right )^{2}}{2}-3 a^{2} b \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {3 i a^{2} b \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*ln(c*x)+1/4*I*b^3*arcsec(c*x)^4-b^3*arcsec(c*x)^3*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+3/2*I*b^3*arcsec(c
*x)^2*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)-3/2*b^3*arcsec(c*x)*polylog(3,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))
^2)-3/4*I*b^3*polylog(4,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+I*a*b^2*arcsec(c*x)^3-3*a*b^2*arcsec(c*x)^2*ln(1+(1/
c/x+I*(1-1/c^2/x^2)^(1/2))^2)+3*I*a*b^2*arcsec(c*x)*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)-3/2*a*b^2*poly
log(3,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+3/2*I*a^2*b*arcsec(c*x)^2-3*a^2*b*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x
^2)^(1/2))^2)+3/2*I*a^2*b*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*a*b^2*c^2*(log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*log(c)^2 - 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x +
1)*sqrt(c*x - 1))/(c^2*x^3 - x), x)*log(c)^2 + 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1
))*log(c^2*x^2)/(c^2*x^3 - x), x)*log(c) - 24*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*lo
g(x)/(c^2*x^3 - x), x)*log(c) + 12*a*b^2*c^2*integrate(1/4*x^2*log(c^2*x^2)/(c^2*x^3 - x), x)*log(c) - 24*a*b^
2*c^2*integrate(1/4*x^2*log(x)/(c^2*x^3 - x), x)*log(c) + b^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3*log(x) - 3
/4*b^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2*log(x) + 24*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*
x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^3 - x), x) - 12*b^3*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1
)*sqrt(c*x - 1))*log(x)^2/(c^2*x^3 - x), x) + 12*a*b^2*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1
))^2/(c^2*x^3 - x), x) - 3*a*b^2*c^2*integrate(1/4*x^2*log(c^2*x^2)^2/(c^2*x^3 - x), x) + 12*a*b^2*c^2*integra
te(1/4*x^2*log(c^2*x^2)*log(x)/(c^2*x^3 - x), x) - 12*a*b^2*c^2*integrate(1/4*x^2*log(x)^2/(c^2*x^3 - x), x) +
12*a^2*b*c^2*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^3 - x), x) + 3/2*a*b^2*(log(c*x + 1
) + log(c*x - 1) - 2*log(x))*log(c)^2 + 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^3 - x)
, x)*log(c)^2 - 12*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^3 - x), x)*log(c)
+ 24*b^3*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^3 - x), x)*log(c) - 12*a*b^2*integra
te(1/4*log(c^2*x^2)/(c^2*x^3 - x), x)*log(c) + 24*a*b^2*integrate(1/4*log(x)/(c^2*x^3 - x), x)*log(c) - 12*b^3
*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2*log(x)/(c^2*x^3 - x), x) + 3*
b^3*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*log(c^2*x^2)^2*log(x)/(c^2*x^3 - x), x) - 24*b^3*integrate(1/4*a
rctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^3 - x), x) + 12*b^3*integrate(1/4*arctan(sqrt(c*
x + 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^3 - x), x) - 12*a*b^2*integrate(1/4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^
2/(c^2*x^3 - x), x) + 3*a*b^2*integrate(1/4*log(c^2*x^2)^2/(c^2*x^3 - x), x) - 12*a*b^2*integrate(1/4*log(c^2*
x^2)*log(x)/(c^2*x^3 - x), x) + 12*a*b^2*integrate(1/4*log(x)^2/(c^2*x^3 - x), x) - 12*a^2*b*integrate(1/4*arc
tan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^3 - x), x) + a^3*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))^3/x,x)

[Out]

int((a + b*acos(1/(c*x)))^3/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))**3/x, x)

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