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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.301953, antiderivative size = 381, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, 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 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]

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]

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*}

Mathematica [A]  time = 9.37366, 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{PolyLog}\left (2,-i e^{-i \sec ^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,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{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 \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{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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Maple [A]  time = 0.97, size = 734, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \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^3*arcsec(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^3*arcsec(b*x + a)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{asec}^{2}{\left (a + b x \right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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^3*arcsec(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^3*arcsec(b*x + a)^2, x)