3.1.29 \(\int x \sec ^{-1}(a+b x)^2 \, dx\) [29]

Optimal. Leaf size=154 \[ -\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2-\frac {4 i a \sec ^{-1}(a+b x) \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {2 i a \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5366, 4511, 4275, 4266, 2317, 2438, 4269, 3556} \begin {gather*} -\frac {a^2 \sec ^{-1}(a+b x)^2}{2 b^2}-\frac {4 i a \sec ^{-1}(a+b x) \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {2 i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}+\frac {1}{2} x^2 \sec ^{-1}(a+b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcSec[a + b*x]^2,x]

[Out]

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

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4266

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 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 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4511

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

Rule 5366

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d^(m + 1),
Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 142, normalized size = 0.92 \begin {gather*} \frac {-\left ((a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)\right )-\frac {1}{2} a^2 \sec ^{-1}(a+b x)^2+\frac {1}{2} b^2 x^2 \sec ^{-1}(a+b x)^2-4 i a \sec ^{-1}(a+b x) \text {ArcTan}\left (e^{i \sec ^{-1}(a+b x)}\right )+\log (a+b x)+2 i a \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSec[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 0.47, size = 227, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {-a \mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )-2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i a \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i a \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{2}}{2}-\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )-\ln \left (\frac {1}{b x +a}\right )}{b^{2}}\)	\(227\)
default	\(\frac {-a \mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )-2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 a \,\mathrm {arcsec}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+2 i a \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )-2 i a \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )+\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{2}}{2}-\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) \left (b x +a \right )-\ln \left (\frac {1}{b x +a}\right )}{b^{2}}\)	\(227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/8*x^2*log(b^2*x^2 + 2*a*b*x + a^2)^2 - integrate(1/2
*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x^2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 2*(b^3*x^4 + 3*a*b
^2*x^3 + (3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*log(b*x + a)^2 - (b^3*x^4 + 2*a*b^2*x^3 + (a^2 - 1)*b*x^2 + 2*(b^3*x
^4 + 3*a*b^2*x^3 + (3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arcsec(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asec(a + b*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsec(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acos(1/(a + b*x))^2,x)

[Out]

int(x*acos(1/(a + b*x))^2, x)

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