3.21 \(\int x^4 (a+b \text {sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=110 \[ \frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {3 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{40 c^5}-\frac {3 b x \sqrt {1-c x}}{40 c^4 \sqrt {\frac {1}{c x+1}}}-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{c x+1}}} \]

[Out]

1/5*x^5*(a+b*arcsech(c*x))-3/40*b*x*(-c*x+1)^(1/2)/c^4/(1/(c*x+1))^(1/2)-1/20*b*x^3*(-c*x+1)^(1/2)/c^2/(1/(c*x
+1))^(1/2)+3/40*b*arcsin(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^5

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Rubi [A]  time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6283, 100, 12, 90, 41, 216} \[ \frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{c x+1}}}-\frac {3 b x \sqrt {1-c x}}{40 c^4 \sqrt {\frac {1}{c x+1}}}+\frac {3 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{40 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcSech[c*x]),x]

[Out]

(-3*b*x*Sqrt[1 - c*x])/(40*c^4*Sqrt[(1 + c*x)^(-1)]) - (b*x^3*Sqrt[1 - c*x])/(20*c^2*Sqrt[(1 + c*x)^(-1)]) + (
x^5*(a + b*ArcSech[c*x]))/5 + (3*b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(40*c^5)

Rule 12

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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^4 \left (a+b \text {sech}^{-1}(c x)\right ) \, dx &=\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{5} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^4}{\sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{5} x^5 \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 {3 x^2}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{20 c^2}\\ &=-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {\left (3 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^2}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{20 c^2}\\ &=-\frac {3 b x \sqrt {1-c x}}{40 c^4 \sqrt {\frac {1}{1+c x}}}-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {\left (3 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{40 c^4}\\ &=-\frac {3 b x \sqrt {1-c x}}{40 c^4 \sqrt {\frac {1}{1+c x}}}-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {\left (3 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{40 c^4}\\ &=-\frac {3 b x \sqrt {1-c x}}{40 c^4 \sqrt {\frac {1}{1+c x}}}-\frac {b x^3 \sqrt {1-c x}}{20 c^2 \sqrt {\frac {1}{1+c x}}}+\frac {1}{5} x^5 \left (a+b \text {sech}^{-1}(c x)\right )+\frac {3 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{40 c^5}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 123, normalized size = 1.12 \[ \frac {a x^5}{5}+\frac {3 i b \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{40 c^5}+b \sqrt {\frac {1-c x}{c x+1}} \left (-\frac {3 x}{40 c^4}-\frac {3 x^2}{40 c^3}-\frac {x^3}{20 c^2}-\frac {x^4}{20 c}\right )+\frac {1}{5} b x^5 \text {sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcSech[c*x]),x]

[Out]

(a*x^5)/5 + b*Sqrt[(1 - c*x)/(1 + c*x)]*((-3*x)/(40*c^4) - (3*x^2)/(40*c^3) - x^3/(20*c^2) - x^4/(20*c)) + (b*
x^5*ArcSech[c*x])/5 + (((3*I)/40)*b*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)])/c^5

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fricas [B]  time = 0.55, size = 174, normalized size = 1.58 \[ \frac {8 \, a c^{5} x^{5} - 8 \, b c^{5} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) - 6 \, b \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) + 8 \, {\left (b c^{5} x^{5} - b c^{5}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (2 \, b c^{4} x^{4} + 3 \, b c^{2} x^{2}\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{40 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="fricas")

[Out]

1/40*(8*a*c^5*x^5 - 8*b*c^5*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) - 6*b*arctan((c*x*sqrt(-(c^2*x^2 -
1)/(c^2*x^2)) - 1)/(c*x)) + 8*(b*c^5*x^5 - b*c^5)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (2*b*
c^4*x^4 + 3*b*c^2*x^2)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/c^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)*x^4, x)

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maple [A]  time = 0.06, size = 118, normalized size = 1.07 \[ \frac {\frac {c^{5} x^{5} a}{5}+b \left (\frac {c^{5} x^{5} \mathrm {arcsech}\left (c x \right )}{5}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (-2 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}-3 c x \sqrt {-c^{2} x^{2}+1}+3 \arcsin \left (c x \right )\right )}{40 \sqrt {-c^{2} x^{2}+1}}\right )}{c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsech(c*x)),x)

[Out]

1/c^5*(1/5*c^5*x^5*a+b*(1/5*c^5*x^5*arcsech(c*x)+1/40*(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(-2*c^3*x^3
*(-c^2*x^2+1)^(1/2)-3*c*x*(-c^2*x^2+1)^(1/2)+3*arcsin(c*x))/(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.41, size = 106, normalized size = 0.96 \[ \frac {1}{5} \, a x^{5} + \frac {1}{40} \, {\left (8 \, x^{5} \operatorname {arsech}\left (c x\right ) - \frac {\frac {3 \, {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 5 \, \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{4}} + \frac {3 \, \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}{c^{4}}}{c}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsech(c*x)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/40*(8*x^5*arcsech(c*x) - ((3*(1/(c^2*x^2) - 1)^(3/2) + 5*sqrt(1/(c^2*x^2) - 1))/(c^4*(1/(c^2*x^2
) - 1)^2 + 2*c^4*(1/(c^2*x^2) - 1) + c^4) + 3*arctan(sqrt(1/(c^2*x^2) - 1))/c^4)/c)*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*acosh(1/(c*x))),x)

[Out]

int(x^4*(a + b*acosh(1/(c*x))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \left (a + b \operatorname {asech}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asech(c*x)),x)

[Out]

Integral(x**4*(a + b*asech(c*x)), x)

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