3.1.16 \(\int x^2 (a+b \sec ^{-1}(c x))^2 \, dx\) [16]

Optimal. Leaf size=147 \[ \frac {b^2 x}{3 c^2}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )}{3 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {2 i b \left (a+b \sec ^{-1}(c x)\right ) \text {ArcTan}\left (e^{i \sec ^{-1}(c x)}\right )}{3 c^3}-\frac {i b^2 \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )}{3 c^3}+\frac {i b^2 \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )}{3 c^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5330, 4494, 4270, 4266, 2317, 2438} \begin {gather*} \frac {2 i b \text {ArcTan}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{3 c^3}-\frac {b x^2 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{3 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^2-\frac {i b^2 \text {Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{3 c^3}+\frac {i b^2 \text {Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{3 c^3}+\frac {b^2 x}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSec[c*x])^2,x]

[Out]

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

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 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 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 4494

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 5330

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rubi steps

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

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Mathematica [A]
time = 0.89, size = 225, normalized size = 1.53 \begin {gather*} \frac {1}{3} \left (a^2 x^3+\frac {a b \left (2 x^4 \sec ^{-1}(c x)-\frac {-c x+c^3 x^3+\sqrt {-1+c^2 x^2} \tanh ^{-1}\left (\frac {c x}{\sqrt {-1+c^2 x^2}}\right )}{c^4 \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{x}+\frac {b^2 \left (c x-c^2 \sqrt {1-\frac {1}{c^2 x^2}} x^2 \sec ^{-1}(c x)+c^3 x^3 \sec ^{-1}(c x)^2-\sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+\sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )-i \text {PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right )+i \text {PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right )\right )}{c^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSec[c*x])^2,x]

[Out]

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

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Maple [A]
time = 0.50, size = 320, normalized size = 2.18

method	result	size
derivativedivides	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+\frac {b^{2} \mathrm {arcsec}\left (c x \right )^{2} c^{3} x^{3}}{3}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{2} x^{2}}{3}+\frac {b^{2} c x}{3}+\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}-\frac {i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}+\frac {i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}+\frac {2 a b \,c^{3} x^{3} \mathrm {arcsec}\left (c x \right )}{3}-\frac {a b \left (c^{2} x^{2}-1\right )}{3 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {a b \sqrt {c^{2} x^{2}-1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{3 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}}{c^{3}}\)	\(320\)
default	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+\frac {b^{2} \mathrm {arcsec}\left (c x \right )^{2} c^{3} x^{3}}{3}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{2} x^{2}}{3}+\frac {b^{2} c x}{3}+\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}-\frac {b^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}-\frac {i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}+\frac {i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{3}+\frac {2 a b \,c^{3} x^{3} \mathrm {arcsec}\left (c x \right )}{3}-\frac {a b \left (c^{2} x^{2}-1\right )}{3 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {a b \sqrt {c^{2} x^{2}-1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{3 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}}{c^{3}}\)	\(320\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(1/3*c^3*x^3*a^2+1/3*b^2*arcsec(c*x)^2*c^3*x^3-1/3*b^2*arcsec(c*x)*((c^2*x^2-1)/c^2/x^2)^(1/2)*c^2*x^2+1
/3*b^2*c*x+1/3*b^2*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-1/3*b^2*arcsec(c*x)*ln(1-I*(1/c/x+I*(1-1/
c^2/x^2)^(1/2)))-1/3*I*b^2*dilog(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+1/3*I*b^2*dilog(1-I*(1/c/x+I*(1-1/c^2/x^2)
^(1/2)))+2/3*a*b*c^3*x^3*arcsec(c*x)-1/3*a*b*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)-1/3*a*b*(c^2*x^2-1)^(1/2)
/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x*ln(c*x+(c^2*x^2-1)^(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^2*(a+b*arcsec(c*x))^2,x, algorithm="maxima")

[Out]

1/3*a^2*x^3 + 1/6*(4*x^3*arcsec(c*x) - (2*sqrt(-1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + log(sqrt(-1/(
c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) - 1)/c^2)/c)*a*b + 1/12*(4*x^3*arctan(sqrt(c*x + 1)*sqrt(c
*x - 1))^2 - x^3*log(c^2*x^2)^2 - 2*c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)*log(
c)^2 + 36*c^2*integrate(1/3*x^4*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) - 72*c^2*integrate(1/3*x^4*log(x)/(c^2*x
^2 - 1), x)*log(c) + 36*c^2*integrate(1/3*x^4*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 36*c^2*integrate(1/3*x^4
*log(x)^2/(c^2*x^2 - 1), x) + 12*c^2*integrate(1/3*x^4*log(c^2*x^2)/(c^2*x^2 - 1), x) + 6*(2*x/c^2 - log(c*x +
1)/c^3 + log(c*x - 1)/c^3)*log(c)^2 - 36*integrate(1/3*x^2*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) + 72*integra
te(1/3*x^2*log(x)/(c^2*x^2 - 1), x)*log(c) - 24*integrate(1/3*sqrt(c*x + 1)*sqrt(c*x - 1)*x^2*arctan(sqrt(c*x
+ 1)*sqrt(c*x - 1))/(c^2*x^2 - 1), x) - 36*integrate(1/3*x^2*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) + 36*integr
ate(1/3*x^2*log(x)^2/(c^2*x^2 - 1), x) - 12*integrate(1/3*x^2*log(c^2*x^2)/(c^2*x^2 - 1), x))*b^2

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

[Out]

integral(b^2*x^2*arcsec(c*x)^2 + 2*a*b*x^2*arcsec(c*x) + a^2*x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*asec(c*x))**2, x)

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Giac [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^2*(a+b*arcsec(c*x))^2,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)^2*x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*acos(1/(c*x)))^2,x)

[Out]

int(x^2*(a + b*acos(1/(c*x)))^2, x)

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