3.30 \(\int \sec ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=94 \[ -\frac {2 i \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5252, 5216, 3757, 4181, 2279, 2391} \[ -\frac {2 i \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sec ^{-1}(a+b x)^2}{b}+\frac {4 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]^2,x]

[Out]

((a + b*x)*ArcSec[a + b*x]^2)/b + ((4*I)*ArcSec[a + b*x]*ArcTan[E^(I*ArcSec[a + b*x])])/b - ((2*I)*PolyLog[2,
(-I)*E^(I*ArcSec[a + b*x])])/b + ((2*I)*PolyLog[2, I*E^(I*ArcSec[a + b*x])])/b

Rule 2279

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 2391

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 3757

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(x^(
m - n + 1)*Sec[a + b*x^n]^p)/(b*n*p), x] - Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /;
FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5216

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/c, Subst[Int[(a + b*x)^n*Sec[x]*Tan[x], x], x
, ArcSec[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

Rule 5252

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSec[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 111, normalized size = 1.18 \[ \frac {-2 i \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )+2 i \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \left ((a+b x) \sec ^{-1}(a+b x)-2 \log \left (1-i e^{i \sec ^{-1}(a+b x)}\right )+2 \log \left (1+i e^{i \sec ^{-1}(a+b x)}\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSec[a + b*x]^2,x]

[Out]

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

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fricas [F]  time = 1.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arcsec}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arcsec(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcsec}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(arcsec(b*x + a)^2, x)

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maple [A]  time = 0.28, size = 179, normalized size = 1.90 \[ x \mathrm {arcsec}\left (b x +a \right )^{2}+\frac {\mathrm {arcsec}\left (b x +a \right )^{2} a}{b}+\frac {2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}-\frac {2 \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}+\frac {2 i \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b}-\frac {2 i \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(b*x+a)^2,x)

[Out]

x*arcsec(b*x+a)^2+1/b*arcsec(b*x+a)^2*a+2/b*arcsec(b*x+a)*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-2/b*arcs
ec(b*x+a)*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2*I/b*dilog(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-2*I
/b*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{4} \, x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} - 1\right )} b x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)^2,x, algorithm="maxima")

[Out]

x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/4*x*log(b^2*x^2 + 2*a*b*x + a^2)^2 - integrate((2*sqrt(b*x
+ a + 1)*sqrt(b*x + a - 1)*b*x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (
3*a^2 - 1)*b*x - a)*log(b*x + a)^2 - (b^3*x^3 + 2*a*b^2*x^2 + (a^2 - 1)*b*x + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (
3*a^2 - 1)*b*x - a)*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x
- a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a + b*x))^2,x)

[Out]

int(acos(1/(a + b*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(b*x+a)**2,x)

[Out]

Integral(asec(a + b*x)**2, x)

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