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

Optimal. Leaf size=77 \[ \frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )-\frac {b \sqrt {1-c x}}{6 c^4 \sqrt {\frac {1}{c x+1}}}-\frac {b x^2 \sqrt {1-c x}}{12 c^2 \sqrt {\frac {1}{c x+1}}} \]

[Out]

1/4*x^4*(a+b*arcsech(c*x))-1/6*b*(-c*x+1)^(1/2)/c^4/(1/(c*x+1))^(1/2)-1/12*b*x^2*(-c*x+1)^(1/2)/c^2/(1/(c*x+1)
)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6283, 100, 12, 74} \[ \frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )-\frac {b x^2 \sqrt {1-c x}}{12 c^2 \sqrt {\frac {1}{c x+1}}}-\frac {b \sqrt {1-c x}}{6 c^4 \sqrt {\frac {1}{c x+1}}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSech[c*x]),x]

[Out]

-(b*Sqrt[1 - c*x])/(6*c^4*Sqrt[(1 + c*x)^(-1)]) - (b*x^2*Sqrt[1 - c*x])/(12*c^2*Sqrt[(1 + c*x)^(-1)]) + (x^4*(
a + b*ArcSech[c*x]))/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 6283

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSech[c*
x]))/(d*(m + 1)), x] + Dist[(b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)])/(m + 1), Int[(d*x)^m/(Sqrt[1 - c*x]*Sqrt[1 + c
*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^3 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{4} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^3}{\sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b x^2 \sqrt {1-c x}}{12 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {2 x}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{12 c^2}\\ &=-\frac {b x^2 \sqrt {1-c x}}{12 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{6 c^2}\\ &=-\frac {b \sqrt {1-c x}}{6 c^4 \sqrt {\frac {1}{1+c x}}}-\frac {b x^2 \sqrt {1-c x}}{12 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{4} x^4 \left (a+b \text {sech}^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 77, normalized size = 1.00 \[ \frac {a x^4}{4}+b \sqrt {\frac {1-c x}{c x+1}} \left (-\frac {1}{6 c^4}-\frac {x}{6 c^3}-\frac {x^2}{12 c^2}-\frac {x^3}{12 c}\right )+\frac {1}{4} b x^4 \text {sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSech[c*x]),x]

[Out]

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

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fricas [A]  time = 0.47, size = 90, normalized size = 1.17 \[ \frac {3 \, b c^{3} x^{4} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 3 \, a c^{3} x^{4} - {\left (b c^{2} x^{3} + 2 \, b x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{12 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*c^3*x^4*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 3*a*c^3*x^4 - (b*c^2*x^3 + 2*b*x)*sqrt
(-(c^2*x^2 - 1)/(c^2*x^2)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsech(c*x) + a)*x^3, x)

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maple [A]  time = 0.06, size = 72, normalized size = 0.94 \[ \frac {\frac {c^{4} x^{4} a}{4}+b \left (\frac {c^{4} x^{4} \mathrm {arcsech}\left (c x \right )}{4}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (c^{2} x^{2}+2\right )}{12}\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(1/4*c^4*x^4*a+b*(1/4*c^4*x^4*arcsech(c*x)-1/12*(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(c^2*x^2+2)
))

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maxima [A]  time = 0.31, size = 57, normalized size = 0.74 \[ \frac {1}{4} \, a x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arsech}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{3}}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*acosh(1/(c*x))),x)

[Out]

int(x^3*(a + b*acosh(1/(c*x))), x)

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sympy [A]  time = 1.97, size = 68, normalized size = 0.88 \[ \begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {asech}{\left (c x \right )}}{4} - \frac {b x^{2} \sqrt {- c^{2} x^{2} + 1}}{12 c^{2}} - \frac {b \sqrt {- c^{2} x^{2} + 1}}{6 c^{4}} & \text {for}\: c \neq 0 \\\frac {x^{4} \left (a + \infty b\right )}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*asech(c*x)/4 - b*x**2*sqrt(-c**2*x**2 + 1)/(12*c**2) - b*sqrt(-c**2*x**2 + 1)/(6*
c**4), Ne(c, 0)), (x**4*(a + oo*b)/4, True))

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