3.1.27 \(\int x^3 \sec ^{-1}(a+b x)^2 \, dx\) [27]

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

[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)^2-2*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+1/3*ln(b*x+a)/b^4+3*a^
2*ln(b*x+a)/b^4-4*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+2*I*a^3*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 = 0.23, 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 = {5366, 4511, 4275, 4266, 2317, 2438, 4269, 3556, 4270} \begin {gather*} -\frac {a^4 \sec ^{-1}(a+b x)^2}{4 b^4}-\frac {4 i a^3 \sec ^{-1}(a+b x) \text {ArcTan}\left (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 {2 i a^3 \text {Li}_2\left (i 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 {2 i a \sec ^{-1}(a+b x) \text {ArcTan}\left (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 {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 {a x}{b^3}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)^2 \end {gather*}

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 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 4270

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] &&
GtQ[n, 1] && NeQ[n, 2]

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^3 \sec ^{-1}(a+b x)^2 \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {Subst}\left (\int x \sec ^4(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^4}+\frac {(2 a) \text {Subst}\left (\int x \sec ^3(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int x \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (2 a^3\right ) \text {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 {\text {Subst}\left (\int x \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{3 b^4}+\frac {a \text {Subst}\left (\int x \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}-\frac {\left (2 a^3\right ) \text {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 ) \text {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 {\text {Subst}\left (\int \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{3 b^4}-\frac {a \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}+\frac {a \text {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 ) \text {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 ) \text {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) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sec ^{-1}(a+b x)}\right )}{b^4}-\frac {(i a) \text {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 = 8.28, size = 667, normalized size = 1.75 \begin {gather*} \frac {\left (1-\frac {a}{a+b x}\right )^3 \left (24 a \left (2+\left (1+2 a^2\right ) \sec ^{-1}(a+b x)^2\right )+\frac {2+(-2+24 a) \sec ^{-1}(a+b x)+3 \left (1-4 a+12 a^2\right ) \sec ^{-1}(a+b x)^2}{-1+\sqrt {1-\frac {1}{(a+b x)^2}}}+16 \left (1+9 a^2\right ) \log \left (\frac {1}{a+b x}\right )-24 a \left (1+2 a^2\right ) \left (\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 (\pi +2 \sec ^{-1}(a+b x)\right )\right )\right )+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 )\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 {4 \sec ^{-1}(a+b x) \left (1+6 a \sec ^{-1}(a+b x)\right ) \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 {8 \left (2 \sec ^{-1}(a+b x)+18 a^2 \sec ^{-1}(a+b x)+6 a^3 \sec ^{-1}(a+b x)^2+3 a \left (2+\sec ^{-1}(a+b x)^2\right )\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 {4 \sec ^{-1}(a+b x) \left (1-6 a \sec ^{-1}(a+b x)\right ) \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+(2-24 a) \sec ^{-1}(a+b x)+3 \left (1-4 a+12 a^2\right ) \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 (-2 \sec ^{-1}(a+b x)-18 a^2 \sec ^{-1}(a+b x)+6 a^3 \sec ^{-1}(a+b x)^2+3 a \left (2+\sec ^{-1}(a+b x)^2\right )\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 )}\right )}{48 b^4 \left (-1+\frac {a}{a+b x}\right )^3} \end {gather*}

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 = 5.17, size = 673, normalized size = 1.77

method	result	size
derivativedivides	\(\frac {-\mathrm {arcsec}\left (b x +a \right )^{2} a^{3} \left (b x +a \right )+\frac {3 \mathrm {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )^{2}}{2}-\mathrm {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{3}+\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{4}}{4}-3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a^{2} \left (b x +a \right )+\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )^{2}-\frac {\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 )^{3}}{6}+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 {\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 )}{3}+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 )-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{12}-\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}+\frac {2 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{3}-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 )+6 a^{2} \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )-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 )+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 )-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 )-3 i a^{2} \mathrm {arcsec}\left (b x +a \right )-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 )+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 )-\frac {i \mathrm {arcsec}\left (b x +a \right )}{3}-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}}\)	\(673\)
default	\(\frac {-\mathrm {arcsec}\left (b x +a \right )^{2} a^{3} \left (b x +a \right )+\frac {3 \mathrm {arcsec}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )^{2}}{2}-\mathrm {arcsec}\left (b x +a \right )^{2} a \left (b x +a \right )^{3}+\frac {\mathrm {arcsec}\left (b x +a \right )^{2} \left (b x +a \right )^{4}}{4}-3 \sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a^{2} \left (b x +a \right )+\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \mathrm {arcsec}\left (b x +a \right ) a \left (b x +a \right )^{2}-\frac {\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 )^{3}}{6}+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 {\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 )}{3}+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 )-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{12}-\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}+\frac {2 \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{3}-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 )+6 a^{2} \ln \left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )-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 )+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 )-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 )-3 i a^{2} \mathrm {arcsec}\left (b x +a \right )-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 )+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 )-\frac {i \mathrm {arcsec}\left (b x +a \right )}{3}-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}}\)	\(673\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^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 = 0.00, 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^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {asec}^{2}{\left (a + b x \right )}\, dx \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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(sageVARa+sa
geVARb*sageVAR

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^3\,{\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^3*acos(1/(a + b*x))^2,x)

[Out]

int(x^3*acos(1/(a + b*x))^2, x)

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