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

Optimal. Leaf size=288 \[ \frac {a^3 \sec ^{-1}(a+b x)^2}{3 b^3}-\frac {2 i a^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {2 i a^2 \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {4 i a^2 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {i \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 a \log (a+b x)}{b^3}+\frac {2 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^3}+\frac {2 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^2+\frac {x}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5258, 4426, 4190, 4181, 2279, 2391, 4184, 3475, 4185} \[ -\frac {2 i a^2 \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}+\frac {2 i a^2 \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {i \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {i \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {a^3 \sec ^{-1}(a+b x)^2}{3 b^3}+\frac {4 i a^2 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^3}-\frac {2 a \log (a+b x)}{b^3}+\frac {2 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^3}-\frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^3}+\frac {2 i \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{3 b^3}+\frac {1}{3} x^3 \sec ^{-1}(a+b x)^2+\frac {x}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSec[a + b*x]^2,x]

[Out]

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

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 3475

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

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 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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 6.58, size = 473, normalized size = 1.64 \[ \frac {2 \left (-6 a^2-1\right ) \left (2 i \left (\text {Li}_2\left (-i e^{-i \sec ^{-1}(a+b x)}\right )-\text {Li}_2\left (i e^{-i \sec ^{-1}(a+b x)}\right )\right )+\left (\pi -2 \sec ^{-1}(a+b x)\right ) \left (\log \left (1-i e^{-i \sec ^{-1}(a+b x)}\right )-\log \left (1+i e^{-i \sec ^{-1}(a+b x)}\right )\right )-\pi \log \left (\cot \left (\frac {1}{4} \left (2 \sec ^{-1}(a+b x)+\pi \right )\right )\right )\right )+2 \left (6 a^2+1\right ) \sec ^{-1}(a+b x)^2+\frac {2 \left (\left (6 a^2+1\right ) \sec ^{-1}(a+b x)^2+12 a \sec ^{-1}(a+b x)+2\right ) \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}-\frac {2 \left (\left (6 a^2+1\right ) \sec ^{-1}(a+b x)^2-12 a \sec ^{-1}(a+b x)+2\right ) \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}+24 a \log \left (\frac {1}{a+b x}\right )+\frac {\left ((6 a-1) \sec ^{-1}(a+b x)+2\right ) \sec ^{-1}(a+b x)}{\sqrt {1-\frac {1}{(a+b x)^2}}-1}+\frac {2 \sec ^{-1}(a+b x)^2 \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^3}-\frac {2 \sec ^{-1}(a+b x)^2 \sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^3}+\frac {\left ((1-6 a) \sec ^{-1}(a+b x)+2\right ) \sec ^{-1}(a+b x)}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^2}+4}{12 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*ArcSec[a + b*x]^2,x]

[Out]

(4 + 2*(1 + 6*a^2)*ArcSec[a + b*x]^2 + (ArcSec[a + b*x]*(2 + (-1 + 6*a)*ArcSec[a + b*x]))/(-1 + Sqrt[1 - (a +
b*x)^(-2)]) + 24*a*Log[(a + b*x)^(-1)] + 2*(-1 - 6*a^2)*((Pi - 2*ArcSec[a + b*x])*(Log[1 - I/E^(I*ArcSec[a + b
*x])] - Log[1 + I/E^(I*ArcSec[a + b*x])]) - Pi*Log[Cot[(Pi + 2*ArcSec[a + b*x])/4]] + (2*I)*(PolyLog[2, (-I)/E
^(I*ArcSec[a + b*x])] - PolyLog[2, I/E^(I*ArcSec[a + b*x])])) + (2*ArcSec[a + b*x]^2*Sin[ArcSec[a + b*x]/2])/(
Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2])^3 + (2*(2 + 12*a*ArcSec[a + b*x] + (1 + 6*a^2)*ArcSec[a + b*x
]^2)*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2]) - (2*ArcSec[a + b*x]^2*Sin[ArcS
ec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^3 + (ArcSec[a + b*x]*(2 + (1 - 6*a)*ArcSec[a
+ b*x]))/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^2 - (2*(2 - 12*a*ArcSec[a + b*x] + (1 + 6*a^2)*Arc
Sec[a + b*x]^2)*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2]))/(12*b^3)

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arcsec}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arcsec(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsec(b*x + a)^2, x)

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maple [A]  time = 1.69, size = 540, normalized size = 1.88 \[ -\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x^{2}}{3 b}+\frac {4 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x a}{3 b^{2}}+\frac {x^{3} \mathrm {arcsec}\left (b x +a \right )^{2}}{3}+\frac {5 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a^{2}}{3 b^{3}}+\frac {i \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{3 b^{3}}+\frac {2 a^{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^{3}}-\frac {2 a^{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^{3}}-\frac {i \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{3 b^{3}}+\frac {2 i a \,\mathrm {arcsec}\left (b x +a \right )}{b^{3}}+\frac {2 i a^{2} \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}}-\frac {\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 )}{3 b^{3}}+\frac {\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 )}{3 b^{3}}+\frac {a^{3} \mathrm {arcsec}\left (b x +a \right )^{2}}{3 b^{3}}-\frac {2 i a^{2} \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{3}}+\frac {2 a \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{b^{3}}-\frac {4 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right ) a}{b^{3}}+\frac {a}{3 b^{3}}+\frac {x}{3 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/b*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*x^2+4/3/b^2*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+
a)*x*a+1/3*x^3*arcsec(b*x+a)^2+5/3/b^3*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*a^2-2*I/b^3*a^2*dilog(1+
I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2/b^3*a^2*arcsec(b*x+a)*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-2/b
^3*a^2*arcsec(b*x+a)*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2*I/b^3*a*arcsec(b*x+a)+1/3*I/b^3*dilog(1-I*(
1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2*I/b^3*a^2*dilog(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-1/3/b^3*arcsec(
b*x+a)*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+1/3/b^3*arcsec(b*x+a)*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(
1/2)))+1/3*a^3*arcsec(b*x+a)^2/b^3-1/3*I/b^3*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2/b^3*a*ln(1+(1/(b
*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)-4/b^3*ln(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))*a+1/3/b^3*a+1/3*x/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, x^{3} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{12} \, x^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x^{3} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 3 \, {\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} - 1\right )} b x^{3} + {\left (a^{3} - a\right )} x^{2}\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{5} + 2 \, a b^{2} x^{4} + {\left (a^{2} - 1\right )} b x^{3} + 3 \, {\left (b^{3} x^{5} + 3 \, a b^{2} x^{4} + {\left (3 \, a^{2} - 1\right )} b x^{3} + {\left (a^{3} - a\right )} x^{2}\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{3 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/12*x^3*log(b^2*x^2 + 2*a*b*x + a^2)^2 - integrate(1/
3*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 3*(b^3*x^5 + 3*a*
b^2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2)*log(b*x + a)^2 - (b^3*x^5 + 2*a*b^2*x^4 + (a^2 - 1)*b*x^3 + 3*(b^
3*x^5 + 3*a*b^2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)*x^2)*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 = -0.00 \[ \int x^2\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(1/(a + b*x))^2,x)

[Out]

int(x^2*acos(1/(a + b*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*asec(a + b*x)**2, x)

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