3.1 \(\int x^3 \text {sech}^{-1}(a+b x) \, dx\)

Optimal. Leaf size=203 \[ -\frac {a^4 \text {sech}^{-1}(a+b x)}{4 b^4}-\frac {\left (17 a^2+2\right ) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{12 b^4}+\frac {\left (2 a^2+1\right ) a \tan ^{-1}\left (\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{2 b^4}+\frac {a (a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{3 b^4}-\frac {x^2 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{12 b^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x) \]

[Out]

-1/4*a^4*arcsech(b*x+a)/b^4+1/4*x^4*arcsech(b*x+a)+1/2*a*(2*a^2+1)*arctan((b*x+a+1)*((-b*x-a+1)/(b*x+a+1))^(1/
2)/(b*x+a))/b^4-1/12*(17*a^2+2)*(b*x+a+1)*((-b*x-a+1)/(b*x+a+1))^(1/2)/b^4-1/12*x^2*(b*x+a+1)*((-b*x-a+1)/(b*x
+a+1))^(1/2)/b^2+1/3*a*(b*x+a)*(b*x+a+1)*((-b*x-a+1)/(b*x+a+1))^(1/2)/b^4

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Rubi [A]  time = 0.16, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6321, 5468, 3782, 4048, 3770, 3767, 8} \[ -\frac {\left (17 a^2+2\right ) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{12 b^4}+\frac {\left (2 a^2+1\right ) a \tan ^{-1}\left (\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{2 b^4}-\frac {a^4 \text {sech}^{-1}(a+b x)}{4 b^4}-\frac {x^2 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{12 b^2}+\frac {a (a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{3 b^4}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSech[a + b*x],x]

[Out]

-((2 + 17*a^2)*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x))/(12*b^4) - (x^2*Sqrt[(1 - a - b*x)/(1 + a + b*
x)]*(1 + a + b*x))/(12*b^2) + (a*(a + b*x)*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x))/(3*b^4) - (a^4*Arc
Sech[a + b*x])/(4*b^4) + (x^4*ArcSech[a + b*x])/4 + (a*(1 + 2*a^2)*ArcTan[(Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(
1 + a + b*x))/(a + b*x)])/(2*b^4)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

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^3 \text {sech}^{-1}(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int x \text {sech}(x) (-a+\text {sech}(x))^3 \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^4}\\ &=\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int (-a+\text {sech}(x))^4 \, dx,x,\text {sech}^{-1}(a+b x)\right )}{4 b^4}\\ &=-\frac {x^2 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int (-a+\text {sech}(x)) \left (-3 a^3+\left (2+9 a^2\right ) \text {sech}(x)-8 a \text {sech}^2(x)\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{12 b^4}\\ &=-\frac {x^2 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^2}+\frac {a (a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{3 b^4}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int \left (6 a^4-12 a \left (1+2 a^2\right ) \text {sech}(x)+2 \left (2+17 a^2\right ) \text {sech}^2(x)\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{24 b^4}\\ &=-\frac {x^2 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^2}+\frac {a (a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{3 b^4}-\frac {a^4 \text {sech}^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)+\frac {\left (a \left (1+2 a^2\right )\right ) \operatorname {Subst}\left (\int \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{2 b^4}-\frac {\left (2+17 a^2\right ) \operatorname {Subst}\left (\int \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{12 b^4}\\ &=-\frac {x^2 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^2}+\frac {a (a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{3 b^4}-\frac {a^4 \text {sech}^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)+\frac {a \left (1+2 a^2\right ) \tan ^{-1}\left (\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{2 b^4}-\frac {\left (i \left (2+17 a^2\right )\right ) \operatorname {Subst}\left (\int 1 \, dx,x,-i \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)\right )}{12 b^4}\\ &=-\frac {\left (2+17 a^2\right ) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^4}-\frac {x^2 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{12 b^2}+\frac {a (a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{3 b^4}-\frac {a^4 \text {sech}^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {sech}^{-1}(a+b x)+\frac {a \left (1+2 a^2\right ) \tan ^{-1}\left (\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{2 b^4}\\ \end {align*}

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Mathematica [C]  time = 0.54, size = 225, normalized size = 1.11 \[ -\frac {-3 a^4 \log (a+b x)+3 a^4 \log \left (a \sqrt {-\frac {a+b x-1}{a+b x+1}}+b x \sqrt {-\frac {a+b x-1}{a+b x+1}}+\sqrt {-\frac {a+b x-1}{a+b x+1}}+1\right )+6 i \left (2 a^2+1\right ) a \log \left (2 \sqrt {-\frac {a+b x-1}{a+b x+1}} (a+b x+1)-2 i (a+b x)\right )+\sqrt {-\frac {a+b x-1}{a+b x+1}} \left (13 a^3+\left (9 a^2-4 a+2\right ) b x+13 a^2+(1-3 a) b^2 x^2+2 a+b^3 x^3+2\right )-3 b^4 x^4 \text {sech}^{-1}(a+b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSech[a + b*x],x]

[Out]

-1/12*(Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(2 + 2*a + 13*a^2 + 13*a^3 + (2 - 4*a + 9*a^2)*b*x + (1 - 3*a)*b^
2*x^2 + b^3*x^3) - 3*b^4*x^4*ArcSech[a + b*x] - 3*a^4*Log[a + b*x] + 3*a^4*Log[1 + Sqrt[-((-1 + a + b*x)/(1 +
a + b*x))] + a*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))] + b*x*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]] + (6*I)*a*(1
+ 2*a^2)*Log[(-2*I)*(a + b*x) + 2*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)])/b^4

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fricas [A]  time = 0.48, size = 345, normalized size = 1.70 \[ \frac {6 \, b^{4} x^{4} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) - 3 \, a^{4} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) + 3 \, a^{4} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + 12 \, {\left (2 \, a^{3} + a\right )} \arctan \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, {\left (b^{3} x^{3} - 3 \, a b^{2} x^{2} + 13 \, a^{3} + {\left (9 \, a^{2} + 2\right )} b x + 2 \, a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{24 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x+a),x, algorithm="fricas")

[Out]

1/24*(6*b^4*x^4*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/(b*x + a))
- 3*a^4*log(((b*x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/x) + 3*a^4*log(((b*
x + a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2)) - 1)/x) + 12*(2*a^3 + a)*arctan((b^2*x^2
+ 2*a*b*x + a^2)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 + 2*a*b*x + a^2))/(b^2*x^2 + 2*a*b*x + a^2 - 1)
) - 2*(b^3*x^3 - 3*a*b^2*x^2 + 13*a^3 + (9*a^2 + 2)*b*x + 2*a)*sqrt(-(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b^2*x^2 +
2*a*b*x + a^2)))/b^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arsech}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x+a),x, algorithm="giac")

[Out]

integrate(x^3*arcsech(b*x + a), x)

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maple [A]  time = 0.10, size = 250, normalized size = 1.23 \[ \frac {\frac {\mathrm {arcsech}\left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\mathrm {arcsech}\left (b x +a \right ) \left (b x +a \right )^{3} a +\frac {3 \,\mathrm {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2} a^{2}}{2}-\mathrm {arcsech}\left (b x +a \right ) \left (b x +a \right ) a^{3}+\frac {\mathrm {arcsech}\left (b x +a \right ) a^{4}}{4}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (3 a^{4} \arctanh \left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right )+12 a^{3} \arcsin \left (b x +a \right )+\left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}-6 a \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+18 a^{2} \sqrt {1-\left (b x +a \right )^{2}}+6 a \arcsin \left (b x +a \right )+2 \sqrt {1-\left (b x +a \right )^{2}}\right )}{12 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsech(b*x+a),x)

[Out]

1/b^4*(1/4*arcsech(b*x+a)*(b*x+a)^4-arcsech(b*x+a)*(b*x+a)^3*a+3/2*arcsech(b*x+a)*(b*x+a)^2*a^2-arcsech(b*x+a)
*(b*x+a)*a^3+1/4*arcsech(b*x+a)*a^4-1/12*(-(b*x+a-1)/(b*x+a))^(1/2)*(b*x+a)*((b*x+a+1)/(b*x+a))^(1/2)*(3*a^4*a
rctanh(1/(1-(b*x+a)^2)^(1/2))+12*a^3*arcsin(b*x+a)+(b*x+a)^2*(1-(b*x+a)^2)^(1/2)-6*a*(b*x+a)*(1-(b*x+a)^2)^(1/
2)+18*a^2*(1-(b*x+a)^2)^(1/2)+6*a*arcsin(b*x+a)+2*(1-(b*x+a)^2)^(1/2))/(1-(b*x+a)^2)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, b^{4} x^{4} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right ) - 2 \, b^{4} x^{4} \log \left (b x + a\right ) - b^{2} x^{2} + 6 \, a b x - {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) - 2 \, {\left (b^{4} x^{4} - a^{4}\right )} \log \left (b x + a\right ) - {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (-b x - a + 1\right )}{8 \, b^{4}} + \int \frac {b^{2} x^{5} + a b x^{4}}{4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + 1\right ) + \frac {1}{2} \, \log \left (-b x - a + 1\right )\right )} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x+a),x, algorithm="maxima")

[Out]

1/8*(2*b^4*x^4*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*b^4*x^4*log(b*x + a) - b^2*x^2 + 6*a*b*x - (a^4 + 4*a^3 + 6*a^2 + 4*a + 1)*log(b*x + a + 1) - 2*(b^4*x^4
- a^4)*log(b*x + a) - (a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(-b*x - a + 1))/b^4 + integrate(1/4*(b^2*x^5 + a*b*x
^4)/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x + a^2 - 1)*e^(1/2*log(b*x + a + 1) + 1/2*log(-b*x - a + 1))
- 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,\mathrm {acosh}\left (\frac {1}{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acosh(1/(a + b*x)),x)

[Out]

int(x^3*acosh(1/(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {asech}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asech(b*x+a),x)

[Out]

Integral(x**3*asech(a + b*x), x)

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