3.1.34 \(\int x^2 (a+b \text {sech}^{-1}(c x))^2 \, dx\) [34]

Optimal. Leaf size=140 \[ -\frac {b^2 x}{3 c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {2 b \left (a+b \text {sech}^{-1}(c x)\right ) \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}+\frac {i b^2 \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}-\frac {i b^2 \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(c x)}\right )}{3 c^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 140, 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 = {6420, 5559, 4270, 4265, 2317, 2438} \begin {gather*} -\frac {2 b \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^3}-\frac {b x \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2+\frac {i b^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}-\frac {i b^2 \text {Li}_2\left (i e^{\text {sech}^{-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*ArcSech[c*x])^2,x]

[Out]

-1/3*(b^2*x)/c^2 - (b*x*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(3*c^2) + (x^3*(a + b*ArcSec
h[c*x])^2)/3 - (2*b*(a + b*ArcSech[c*x])*ArcTan[E^ArcSech[c*x]])/(3*c^3) + ((I/3)*b^2*PolyLog[2, (-I)*E^ArcSec
h[c*x]])/c^3 - ((I/3)*b^2*PolyLog[2, I*E^ArcSech[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 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, 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 5559

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

Rule 6420

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[-(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[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 \text {sech}^{-1}(c x)\right )^2 \, dx &=-\frac {\text {Subst}\left (\int (a+b x)^2 \text {sech}^3(x) \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )}{c^3}\\ &=\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {(2 b) \text {Subst}\left (\int (a+b x) \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac {b^2 x}{3 c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {b \text {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\text {sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac {b^2 x}{3 c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {2 b \left (a+b \text {sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{3 c^3}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(c x)\right )}{3 c^3}\\ &=-\frac {b^2 x}{3 c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {2 b \left (a+b \text {sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text {sech}^{-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^{\text {sech}^{-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^{\text {sech}^{-1}(c x)}\right )}{3 c^3}\\ &=-\frac {b^2 x}{3 c^2}-\frac {b x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {2 b \left (a+b \text {sech}^{-1}(c x)\right ) \tan ^{-1}\left (e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}+\frac {i b^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}-\frac {i b^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{3 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 241, normalized size = 1.72 \begin {gather*} \frac {1}{3} \left (a^2 x^3+a b \left (2 x^3 \text {sech}^{-1}(c x)+\frac {\sqrt {\frac {1-c x}{1+c x}} \left (c x-c^3 x^3+2 \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c^3 (-1+c x)}\right )+\frac {b^2 \left (-c x-c x \sqrt {\frac {1-c x}{1+c x}} (1+c x) \text {sech}^{-1}(c x)+c^3 x^3 \text {sech}^{-1}(c x)^2+i \text {sech}^{-1}(c x) \log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-i \text {sech}^{-1}(c x) \log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )+i \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(c x)}\right )-i \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(c x)}\right )\right )}{c^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.55, size = 349, normalized size = 2.49

method	result	size
derivativedivides	\(\frac {\frac {a^{2} c^{3} x^{3}}{3}+\frac {b^{2} \mathrm {arcsech}\left (c x \right )^{2} c^{3} x^{3}}{3}-\frac {b^{2} \mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c^{2} x^{2}}{3}-\frac {b^{2} c x}{3}+\frac {i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}-\frac {i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}+\frac {i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}-\frac {i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}+2 a b \left (\frac {c^{3} x^{3} \mathrm {arcsech}\left (c x \right )}{3}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (c x \sqrt {-c^{2} x^{2}+1}-\arcsin \left (c x \right )\right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{3}}\)	\(349\)
default	\(\frac {\frac {a^{2} c^{3} x^{3}}{3}+\frac {b^{2} \mathrm {arcsech}\left (c x \right )^{2} c^{3} x^{3}}{3}-\frac {b^{2} \mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c^{2} x^{2}}{3}-\frac {b^{2} c x}{3}+\frac {i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}-\frac {i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}+\frac {i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}-\frac {i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{3}+2 a b \left (\frac {c^{3} x^{3} \mathrm {arcsech}\left (c x \right )}{3}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (c x \sqrt {-c^{2} x^{2}+1}-\arcsin \left (c x \right )\right )}{6 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{3}}\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(1/3*a^2*c^3*x^3+1/3*b^2*arcsech(c*x)^2*c^3*x^3-1/3*b^2*arcsech(c*x)*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^
(1/2)*c^2*x^2-1/3*b^2*c*x+1/3*I*b^2*arcsech(c*x)*ln(1+I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))-1/3*I*b^2*ar
csech(c*x)*ln(1-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))+1/3*I*b^2*dilog(1+I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c
/x)^(1/2)))-1/3*I*b^2*dilog(1-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))+2*a*b*(1/3*c^3*x^3*arcsech(c*x)-1/6*
(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(c*x*(-c^2*x^2+1)^(1/2)-arcsin(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*arcsech(c*x))^2,x, algorithm="maxima")

[Out]

1/3*a^2*x^3 + 1/3*(2*x^3*arcsech(c*x) - (sqrt(1/(c^2*x^2) - 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + arctan(sqrt(1/(
c^2*x^2) - 1))/c^2)/c)*a*b + b^2*integrate(x^2*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2, 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^2*(a+b*arcsech(c*x))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^2*arcsech(c*x)^2 + 2*a*b*x^2*arcsech(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 {asech}{\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*asech(c*x))**2,x)

[Out]

Integral(x**2*(a + b*asech(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*arcsech(c*x))^2,x, algorithm="giac")

[Out]

integrate((b*arcsech(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 {acosh}\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*acosh(1/(c*x)))^2,x)

[Out]

int(x^2*(a + b*acosh(1/(c*x)))^2, x)

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