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]

((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)

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Rubi [A]  time = 0.115012, antiderivative size = 57, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.202167, 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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Maple [A]  time = 0.289, size = 62, normalized size = 1.1 \begin{align*}{\frac{{\rm arcsech} \left (b{x}^{4}+a\right ){x}^{4}}{4}}+{\frac{{\rm arcsech} \left (b{x}^{4}+a\right )a}{4\,b}}-{\frac{1}{4\,b}\arctan \left ( \sqrt{ \left ( b{x}^{4}+a \right ) ^{-1}-1}\sqrt{ \left ( b{x}^{4}+a \right ) ^{-1}+1} \right ) } \end{align*}

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 = 0.993601, size = 51, normalized size = 0.89 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 2.47328, size = 613, normalized size = 10.75 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsech}\left (b x^{4} + a\right )\,{d x} \end{align*}

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)