3.27 \(\int x^3 \sec ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=381 \[ -\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {2 i a^3 \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {2 i a^3 \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {3 a^2 \log (a+b x)}{b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {(a+b x)^2}{12 b^4}+\frac {\log (a+b x)}{3 b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {2 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {a x}{b^3}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2 \]

[Out]

-a*x/b^3+1/12*(b*x+a)^2/b^4-1/4*a^4*arcsec(b*x+a)^2/b^4+1/4*x^4*arcsec(b*x+a)^2-4*I*a^3*arcsec(b*x+a)*arctan(1
/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/b^4-2*I*a*arcsec(b*x+a)*arctan(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/b^4+1/3*ln
(b*x+a)/b^4+3*a^2*ln(b*x+a)/b^4-2*I*a^3*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4+2*I*a^3*polylog(2
,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4-I*a*polylog(2,I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4+I*a*poly
log(2,-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))/b^4-1/3*(b*x+a)*arcsec(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^4-3*a^2*(b
*x+a)*arcsec(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^4+a*(b*x+a)^2*arcsec(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^4-1/6*(b*x+a)^
3*arcsec(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^4

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Rubi [A]  time = 0.30, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 20, 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} \[ \frac {2 i a^3 \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {2 i a^3 \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {i a \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {i a \text {PolyLog}\left (2,i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {3 a^2 \log (a+b x)}{b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}+\frac {\log (a+b x)}{3 b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {2 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSec[a + b*x]^2,x]

[Out]

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

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^3 \sec ^{-1}(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int x^2 \sec (x) (-a+\sec (x))^3 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int x (-a+\sec (x))^4 \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^4}\\ &=\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int \left (a^4 x-4 a^3 x \sec (x)+6 a^2 x \sec ^2(x)-4 a x \sec ^3(x)+x \sec ^4(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^4}\\ &=-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int x \sec ^4(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^4}+\frac {(2 a) \operatorname {Subst}\left (\int x \sec ^3(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int x \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}\\ &=-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {\operatorname {Subst}\left (\int x \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{3 b^4}+\frac {a \operatorname {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}-\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}\\ &=-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {2 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {3 a^2 \log (a+b x)}{b^4}+\frac {\operatorname {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{3 b^4}-\frac {a \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {a \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (2 i a^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {\left (2 i a^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^4}\\ &=-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {2 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {\log (a+b x)}{3 b^4}+\frac {3 a^2 \log (a+b x)}{b^4}+\frac {2 i a^3 \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {2 i a^3 \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {(i a) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {(i a) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^4}\\ &=-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}-\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{3 b^4}-\frac {3 a^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^4}-\frac {(a+b x)^3 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{6 b^4}-\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2-\frac {2 i a \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \tan ^{-1}\left (e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {\log (a+b x)}{3 b^4}+\frac {3 a^2 \log (a+b x)}{b^4}+\frac {i a \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}+\frac {2 i a^3 \text {Li}_2\left (-i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {i a \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {2 i a^3 \text {Li}_2\left (i e^{i \sec ^{-1}(a+b x)}\right )}{b^4}\\ \end {align*}

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Mathematica [A]  time = 9.42, size = 667, normalized size = 1.75 \[ \frac {\left (1-\frac {a}{a+b x}\right )^3 \left (-24 a \left (2 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 )+16 \left (9 a^2+1\right ) \log \left (\frac {1}{a+b x}\right )+24 a \left (\left (2 a^2+1\right ) \sec ^{-1}(a+b x)^2+2\right )+\frac {3 \left (12 a^2-4 a+1\right ) \sec ^{-1}(a+b x)^2+(24 a-2) \sec ^{-1}(a+b x)+2}{\sqrt {1-\frac {1}{(a+b x)^2}}-1}-\frac {3 \left (12 a^2-4 a+1\right ) \sec ^{-1}(a+b x)^2+(2-24 a) \sec ^{-1}(a+b x)+2}{\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}+\frac {8 \left (6 a^3 \sec ^{-1}(a+b x)^2+18 a^2 \sec ^{-1}(a+b x)+3 a \left (\sec ^{-1}(a+b x)^2+2\right )+2 \sec ^{-1}(a+b x)\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 {8 \left (6 a^3 \sec ^{-1}(a+b x)^2-18 a^2 \sec ^{-1}(a+b x)+3 a \left (\sec ^{-1}(a+b x)^2+2\right )-2 \sec ^{-1}(a+b x)\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 {3 \sec ^{-1}(a+b x)^2}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^4}-\frac {3 \sec ^{-1}(a+b x)^2}{\left (\cos \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )+\sin \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )^4}+\frac {4 \left (6 a \sec ^{-1}(a+b x)+1\right ) \sec ^{-1}(a+b x) \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 {4 \left (1-6 a \sec ^{-1}(a+b x)\right ) \sec ^{-1}(a+b x) \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}\right )}{48 b^4 \left (\frac {a}{a+b x}-1\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*ArcSec[a + b*x]^2,x]

[Out]

((1 - a/(a + b*x))^3*(24*a*(2 + (1 + 2*a^2)*ArcSec[a + b*x]^2) + (2 + (-2 + 24*a)*ArcSec[a + b*x] + 3*(1 - 4*a
+ 12*a^2)*ArcSec[a + b*x]^2)/(-1 + Sqrt[1 - (a + b*x)^(-2)]) + 16*(1 + 9*a^2)*Log[(a + b*x)^(-1)] - 24*a*(1 +
2*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*L
og[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])])) - (3*ArcSec[a + b*x]^2)/(Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2])^4 + (4*ArcSec[a + b*x]
*(1 + 6*a*ArcSec[a + b*x])*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSec[a + b*x]/2] - Sin[ArcSec[a + b*x]/2])^3 + (8*(2
*ArcSec[a + b*x] + 18*a^2*ArcSec[a + b*x] + 6*a^3*ArcSec[a + b*x]^2 + 3*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]) - (3*ArcSec[a + b*x]^2)/(Cos[ArcSec[a + b*x]/2]
+ Sin[ArcSec[a + b*x]/2])^4 + (4*ArcSec[a + b*x]*(1 - 6*a*ArcSec[a + b*x])*Sin[ArcSec[a + b*x]/2])/(Cos[ArcSe
c[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^3 - (2 + (2 - 24*a)*ArcSec[a + b*x] + 3*(1 - 4*a + 12*a^2)*ArcSec[a +
b*x]^2)/(Cos[ArcSec[a + b*x]/2] + Sin[ArcSec[a + b*x]/2])^2 - (8*(-2*ArcSec[a + b*x] - 18*a^2*ArcSec[a + b*x]
+ 6*a^3*ArcSec[a + b*x]^2 + 3*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])))/(48*b^4*(-1 + a/(a + b*x))^3)

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fricas [F]  time = 2.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \operatorname {arcsec}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^3*arcsec(b*x + a)^2, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(b*x+a)^2,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(a+b*x)]Simp
lification assuming t_nostep near 0Simplification assuming t_nostep near 0Evaluation time: 0.59sym2poly/r2sym(
const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 4.00, size = 734, normalized size = 1.93 \[ \frac {x^{4} \mathrm {arcsec}\left (b x +a \right )^{2}}{4}-\frac {a^{4} \mathrm {arcsec}\left (b x +a \right )^{2}}{4 b^{4}}-\frac {5 a x}{6 b^{3}}-\frac {\ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{3 b^{4}}+\frac {2 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{3 b^{4}}-\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x \,a^{2}}{2 b^{3}}-\frac {11 a^{2}}{12 b^{4}}+\frac {x^{2}}{12 b^{2}}-\frac {13 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a^{3}}{6 b^{4}}-\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a}{3 b^{4}}-\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x^{3}}{6 b}-\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) x}{3 b^{3}}+\frac {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 )}{b^{4}}-\frac {2 a^{3} \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^{4}}+\frac {2 a^{3} \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^{4}}-\frac {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 )}{b^{4}}-\frac {3 i \mathrm {arcsec}\left (b x +a \right ) a^{2}}{b^{4}}+\frac {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 )}{b^{4}}+\frac {2 i a^{3} \dilog \left (1+i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{4}}-\frac {2 i a^{3} \dilog \left (1-i \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )\right )}{b^{4}}-\frac {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 )}{b^{4}}+\frac {6 a^{2} \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{4}}-\frac {3 a^{2} \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{b^{4}}-\frac {i \mathrm {arcsec}\left (b x +a \right )}{3 b^{4}}+\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} a}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsec(b*x+a)^2,x)

[Out]

1/4*x^4*arcsec(b*x+a)^2-1/4*a^4*arcsec(b*x+a)^2/b^4-5/6*a*x/b^3+6/b^4*a^2*ln(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)
)-3/b^4*a^2*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)-1/3*I/b^4*arcsec(b*x+a)-3/2/b^3*((-1+(b*x+a)^2)/(b*x+a
)^2)^(1/2)*arcsec(b*x+a)*x*a^2-11/12/b^4*a^2+1/12/b^2*x^2-1/3/b^4*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)+
2/3/b^4*ln(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))+1/b^4*a*arcsec(b*x+a)*ln(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))
)-2/b^4*a^3*arcsec(b*x+a)*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))+2/b^4*a^3*arcsec(b*x+a)*ln(1-I*(1/(b*x+a
)+I*(1-1/(b*x+a)^2)^(1/2)))-1/b^4*a*arcsec(b*x+a)*ln(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-13/6/b^4*((-1+(b
*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*a^3+I/b^4*a*dilog(1+I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-1/3/b^4*((-
1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*a-1/6/b*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*x^3-1/3/b^3
*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*x-3*I/b^4*arcsec(b*x+a)*a^2+2*I/b^4*a^3*dilog(1+I*(1/(b*x+a)+I
*(1-1/(b*x+a)^2)^(1/2)))-2*I/b^4*a^3*dilog(1-I*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2)))-I/b^4*a*dilog(1-I*(1/(b*x+
a)+I*(1-1/(b*x+a)^2)^(1/2)))+1/2/b^2*((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)*arcsec(b*x+a)*x^2*a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, x^{4} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{16} \, x^{4} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x^{4} \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 4 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{5} + {\left (3 \, a^{2} - 1\right )} b x^{4} + {\left (a^{3} - a\right )} x^{3}\right )} \log \left (b x + a\right )^{2} - {\left (b^{3} x^{6} + 2 \, a b^{2} x^{5} + {\left (a^{2} - 1\right )} b x^{4} + 4 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{5} + {\left (3 \, a^{2} - 1\right )} b x^{4} + {\left (a^{3} - a\right )} x^{3}\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsec(b*x+a)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/16*x^4*log(b^2*x^2 + 2*a*b*x + a^2)^2 - integrate(1/
4*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x^4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + 4*(b^3*x^6 + 3*a*
b^2*x^5 + (3*a^2 - 1)*b*x^4 + (a^3 - a)*x^3)*log(b*x + a)^2 - (b^3*x^6 + 2*a*b^2*x^5 + (a^2 - 1)*b*x^4 + 4*(b^
3*x^6 + 3*a*b^2*x^5 + (3*a^2 - 1)*b*x^4 + (a^3 - a)*x^3)*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 \[ \int x^3\,{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(1/(a + b*x))^2,x)

[Out]

int(x^3*acos(1/(a + b*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asec(b*x+a)**2,x)

[Out]

Integral(x**3*asec(a + b*x)**2, x)

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