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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.13137, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5258, 4426, 4190, 4181, 2279, 2391, 4184, 3475} \[ \frac{2 i a \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{2 i a \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \sec ^{-1}(a+b x)^2}{2 b^2}+\frac{\log (a+b x)}{b^2}-\frac{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2}-\frac{4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \sec ^{-1}(a+b x)^2 \]

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 5258

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]

Rule 4426

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 4190

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 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 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 4184

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 3475

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

Rubi steps

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

Mathematica [A]  time = 0.145175, size = 142, normalized size = 0.92 \[ \frac{2 i a \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )-2 i a \text{PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )-\frac{1}{2} a^2 \sec ^{-1}(a+b x)^2+\frac{1}{2} b^2 x^2 \sec ^{-1}(a+b x)^2+\log (a+b x)-(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}} \sec ^{-1}(a+b x)-4 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^2} \]

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

________________________________________________________________________________________

Maple [A]  time = 0.524, size = 327, normalized size = 2.1 \begin{align*}{\frac{{x}^{2} \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{2}}-{\frac{{a}^{2} \left ({\rm arcsec} \left (bx+a\right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{x{\rm arcsec} \left (bx+a\right )}{b}\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}-{\frac{{\rm arcsec} \left (bx+a\right )a}{{b}^{2}}\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}-{\frac{i{\rm arcsec} \left (bx+a\right )}{{b}^{2}}}-{\frac{1}{{b}^{2}}\ln \left ( 1+ \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) ^{2} \right ) }+2\,{\frac{1}{{b}^{2}}\ln \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) }-2\,{\frac{{\rm arcsec} \left (bx+a\right )a}{{b}^{2}}\ln \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }+2\,{\frac{{\rm arcsec} \left (bx+a\right )a}{{b}^{2}}\ln \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }+{\frac{2\,ia}{{b}^{2}}{\it dilog} \left ( 1+i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) }-{\frac{2\,ia}{{b}^{2}}{\it dilog} \left ( 1-i \left ( \left ( bx+a \right ) ^{-1}+i\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )^{2} - \frac{1}{8} \, x^{2} \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^{2} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) + 2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} - 1\right )} b x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} -{\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} +{\left (a^{2} - 1\right )} b x^{2} + 2 \,{\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} - 1\right )} b x^{2} +{\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x \operatorname{arcsec}\left (b x + a\right )^{2}, x\right ) \end{align*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arcsec}\left (b x + a\right )^{2}\,{d x} \end{align*}

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)