3.9 \(\int x^2 \text {sech}^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=279 \[ \frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {2 i a^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 a \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{3 b^3}-\frac {2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {x}{3 b^2} \]

[Out]

-1/3*x/b^2+1/3*a^3*arcsech(b*x+a)^2/b^3+1/3*x^3*arcsech(b*x+a)^2-2/3*arcsech(b*x+a)*arctan(1/(b*x+a)+(1/(b*x+a
)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/b^3-4*a^2*arcsech(b*x+a)*arctan(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1
/2))/b^3+2*a*ln(b*x+a)/b^3+1/3*I*polylog(2,-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b^3+2*I*a^2
*polylog(2,-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b^3-1/3*I*polylog(2,I*(1/(b*x+a)+(1/(b*x+a)
-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b^3-2*I*a^2*polylog(2,I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b
^3+2*a*(b*x+a+1)*arcsech(b*x+a)*((-b*x-a+1)/(b*x+a+1))^(1/2)/b^3-1/3*(b*x+a)*(b*x+a+1)*arcsech(b*x+a)*((-b*x-a
+1)/(b*x+a+1))^(1/2)/b^3

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Rubi [A]  time = 0.24, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6321, 5468, 4190, 4180, 2279, 2391, 4184, 3475, 4185} \[ \frac {2 i a^2 \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 a \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{3 b^3}-\frac {2 \text {sech}^{-1}(a+b x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {x}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSech[a + b*x]^2,x]

[Out]

-x/(3*b^2) + (2*a*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x])/b^3 - ((a + b*x)*Sqrt[(1 -
a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x])/(3*b^3) + (a^3*ArcSech[a + b*x]^2)/(3*b^3) + (x^3*Arc
Sech[a + b*x]^2)/3 - (2*ArcSech[a + b*x]*ArcTan[E^ArcSech[a + b*x]])/(3*b^3) - (4*a^2*ArcSech[a + b*x]*ArcTan[
E^ArcSech[a + b*x]])/b^3 + (2*a*Log[a + b*x])/b^3 + ((I/3)*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b^3 + ((2*I)*a
^2*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b^3 - ((I/3)*PolyLog[2, I*E^ArcSech[a + b*x]])/b^3 - ((2*I)*a^2*PolyLo
g[2, I*E^ArcSech[a + b*x]])/b^3

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 4180

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 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 5468

Int[((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sech[(c_.) + (d_.)*(x_)])^(n_.)*Tanh[(c_
.) + (d_.)*(x_)], x_Symbol] :> -Simp[((e + f*x)^m*(a + b*Sech[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*Sech[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x
] && IGtQ[m, 0] && NeQ[n, -1]

Rule 6321

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 2.00, size = 305, normalized size = 1.09 \[ -\frac {-\left (6 a^2+1\right ) \left (2 i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \text {Li}_2\left (i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \text {sech}^{-1}(a+b x) \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )+\pi \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )+2 i \text {sech}^{-1}(a+b x) \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (\cot \left (\frac {1}{4} \left (\pi +2 i \text {sech}^{-1}(a+b x)\right )\right )\right )\right )+2 \left (-3 a^2 (a+b x) \text {sech}^{-1}(a+b x)^2-6 a \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)+a+b x\right )+12 a \log \left (\frac {1}{a+b x}\right )-2 (a+b x)^3 \text {sech}^{-1}(a+b x)^2+6 a (a+b x)^2 \text {sech}^{-1}(a+b x)^2+2 \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1) (a+b x) \text {sech}^{-1}(a+b x)}{6 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*ArcSech[a + b*x]^2,x]

[Out]

-1/6*(2*(a + b*x)*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)*ArcSech[a + b*x] + 6*a*(a + b*x)^2*ArcSe
ch[a + b*x]^2 - 2*(a + b*x)^3*ArcSech[a + b*x]^2 + 2*(a + b*x - 6*a*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 +
a + b*x)*ArcSech[a + b*x] - 3*a^2*(a + b*x)*ArcSech[a + b*x]^2) + 12*a*Log[(a + b*x)^(-1)] - (1 + 6*a^2)*(Pi*
Log[1 - I*E^ArcSech[a + b*x]] - (2*I)*ArcSech[a + b*x]*Log[1 - I*E^ArcSech[a + b*x]] - Pi*Log[1 + I*E^ArcSech[
a + b*x]] + (2*I)*ArcSech[a + b*x]*Log[1 + I*E^ArcSech[a + b*x]] - Pi*Log[Cot[(Pi + (2*I)*ArcSech[a + b*x])/4]
] + (2*I)*PolyLog[2, (-I)*E^ArcSech[a + b*x]] - (2*I)*PolyLog[2, I*E^ArcSech[a + b*x]]))/b^3

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fricas [F]  time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arsech}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arcsech(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arsech}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(b*x + a)^2, x)

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maple [A]  time = 1.68, size = 655, normalized size = 2.35 \[ -\frac {a}{3 b^{3}}-\frac {x}{3 b^{2}}-\frac {2 a \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )}{b^{3}}+\frac {4 a \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{b^{3}}+\frac {2 i a^{2} \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}+\frac {a^{3} \mathrm {arcsech}\left (b x +a \right )^{2}}{3 b^{3}}+\frac {4 \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \mathrm {arcsech}\left (b x +a \right ) x a}{3 b^{2}}-\frac {2 a \,\mathrm {arcsech}\left (b x +a \right )}{b^{3}}-\frac {2 i a^{2} \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}-\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3 b^{3}}+\frac {2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}+\frac {x^{3} \mathrm {arcsech}\left (b x +a \right )^{2}}{3}-\frac {2 i a^{2} \mathrm {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{3}}+\frac {i \dilog \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3 b^{3}}+\frac {5 \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \mathrm {arcsech}\left (b x +a \right ) a^{2}}{3 b^{3}}-\frac {i \dilog \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3 b^{3}}+\frac {i \mathrm {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3 b^{3}}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \mathrm {arcsech}\left (b x +a \right ) x^{2}}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsech(b*x+a)^2,x)

[Out]

-1/3/b^3*a-1/3*x/b^2-2/b^3*a*ln(1+(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)+4/b^3*a*ln(1/(b*x+a)+
(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))-2*I/b^3*a^2*dilog(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1
/2)))+1/3*a^3*arcsech(b*x+a)^2/b^3+4/3/b^2*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*arcsech(b*x+a)
*x*a-2/b^3*a*arcsech(b*x+a)-1/3*I/b^3*dilog(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-1/3*I/b^3
*arcsech(b*x+a)*ln(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+2*I/b^3*a^2*arcsech(b*x+a)*ln(1+I*
(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+1/3*x^3*arcsech(b*x+a)^2-2*I/b^3*a^2*arcsech(b*x+a)*ln(1-
I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+1/3*I/b^3*dilog(1+I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(
b*x+a)+1)^(1/2)))+5/3/b^3*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*arcsech(b*x+a)*a^2+2*I/b^3*a^2*
dilog(1+I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+1/3*I/b^3*arcsech(b*x+a)*ln(1+I*(1/(b*x+a)+(1/(
b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-1/3/b*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*arcsech(b*x+a
)*x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a)^2 - i
ntegrate(-2/3*(6*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*sqrt(b*x + a + 1)*sqrt(-b*x - a +
1)*log(b*x + a)^2 + 6*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a)^2 - (b^3*x^5 +
2*a*b^2*x^4 + (a^2*b - b)*x^3 + 6*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a) +
(3*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*sqrt(b*x + a + 1)*log(b*x + a) + (2*b^3*x^5 + 4
*a*b^2*x^4 + (2*a^2*b - b)*x^3 + 3*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a))*s
qrt(b*x + a + 1))*sqrt(-b*x - a + 1))*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b
*x - a + 1)*a + b*x + a))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*s
qrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b - b)*x - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acosh(1/(a + b*x))^2,x)

[Out]

int(x^2*acosh(1/(a + b*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asech(b*x+a)**2,x)

[Out]

Integral(x**2*asech(a + b*x)**2, x)

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