3.99 \(\int x^3 \text {sech}^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=57 \[ \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\tan ^{-1}\left (\sqrt {\frac {-a-b x^4+1}{a+b x^4+1}}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6715, 6313, 1961, 12, 203} \[ \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\tan ^{-1}\left (\sqrt {\frac {-a-b x^4+1}{a+b x^4+1}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSech[a + b*x^4],x]

[Out]

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

Rule 12

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

Rule 203

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

Rule 1961

Int[(u_)^(r_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[SimplifyIntegrand[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(1/n -
1)*(u /. x -> (-(a*e) + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r)/(b*e - d*x^q)^(1/n + 1), x], x], x, ((e*(a + b*x
^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && PolynomialQ[u, x] && FractionQ[p] && IntegerQ[1/
n] && IntegerQ[r]

Rule 6313

Int[ArcSech[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcSech[c + d*x])/d, x] + Int[Sqrt[(1 - c - d*x)/
(1 + c + d*x)]/(1 - c - d*x), x] /; FreeQ[{c, d}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 84, normalized size = 1.47 \[ \frac {\frac {\sqrt {1-\left (a+b x^4\right )^2} \sin ^{-1}\left (a+b x^4\right )}{\sqrt {-\frac {a+b x^4-1}{a+b x^4+1}} \left (a+b x^4+1\right )}+\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSech[a + b*x^4],x]

[Out]

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

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fricas [B]  time = 1.75, size = 283, normalized size = 4.96 \[ \frac {2 \, b x^{4} \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{b x^{4} + a}\right ) + a \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{x^{4}}\right ) - a \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - 1}{x^{4}}\right ) - 2 \, \arctan \left (\frac {{\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x^4+a),x, algorithm="fricas")

[Out]

1/8*(2*b*x^4*log(((b*x^4 + a)*sqrt(-(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) + 1)/(b*x^4 +
a)) + a*log(((b*x^4 + a)*sqrt(-(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) + 1)/x^4) - a*log
(((b*x^4 + a)*sqrt(-(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) - 1)/x^4) - 2*arctan((b^2*x^8
+ 2*a*b*x^4 + a^2)*sqrt(-(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)/(b^2*x^8 + 2*a*b*x^4 + a^2))/(b^2*x^8 + 2*a*b*x^4 +
a^2 - 1)))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arsech}\left (b x^{4} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x^4+a),x, algorithm="giac")

[Out]

integrate(x^3*arcsech(b*x^4 + a), x)

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maple [A]  time = 0.10, size = 62, normalized size = 1.09 \[ \frac {\mathrm {arcsech}\left (b \,x^{4}+a \right ) x^{4}}{4}+\frac {\mathrm {arcsech}\left (b \,x^{4}+a \right ) a}{4 b}-\frac {\arctan \left (\sqrt {\frac {1}{b \,x^{4}+a}-1}\, \sqrt {\frac {1}{b \,x^{4}+a}+1}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsech(b*x^4+a),x)

[Out]

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

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maxima [A]  time = 0.36, size = 38, normalized size = 0.67 \[ \frac {{\left (b x^{4} + a\right )} \operatorname {arsech}\left (b x^{4} + a\right ) - \arctan \left (\sqrt {\frac {1}{{\left (b x^{4} + a\right )}^{2}} - 1}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(b*x^4+a),x, algorithm="maxima")

[Out]

1/4*((b*x^4 + a)*arcsech(b*x^4 + a) - arctan(sqrt(1/(b*x^4 + a)^2 - 1)))/b

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mupad [B]  time = 2.99, size = 56, normalized size = 0.98 \[ \frac {\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{b\,x^4+a}-1}\,\sqrt {\frac {1}{b\,x^4+a}+1}}\right )}{4\,b}+\frac {\mathrm {acosh}\left (\frac {1}{b\,x^4+a}\right )\,\left (b\,x^4+a\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acosh(1/(a + b*x^4)),x)

[Out]

atan(1/((1/(a + b*x^4) - 1)^(1/2)*(1/(a + b*x^4) + 1)^(1/2)))/(4*b) + (acosh(1/(a + b*x^4))*(a + b*x^4))/(4*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {asech}{\left (a + b x^{4} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asech(b*x**4+a),x)

[Out]

Integral(x**3*asech(a + b*x**4), x)

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