3.1.100 \(\int x^{-1+n} \text {sech}^{-1}(a+b x^n) \, dx\) [100]

Optimal. Leaf size=58 \[ \frac {\left (a+b x^n\right ) \text {sech}^{-1}\left (a+b x^n\right )}{b n}-\frac {2 \text {ArcTan}\left (\sqrt {\frac {1-a-b x^n}{1+a+b x^n}}\right )}{b n} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6847, 6448, 1983, 12, 209} \begin {gather*} \frac {\left (a+b x^n\right ) \text {sech}^{-1}\left (a+b x^n\right )}{b n}-\frac {2 \text {ArcTan}\left (\sqrt {\frac {-a-b x^n+1}{a+b x^n+1}}\right )}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + n)*ArcSech[a + b*x^n],x]

[Out]

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

Rule 12

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

Rule 209

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

Rule 1983

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)/(b*e - d*x^q)^(1/n + 1))*(u /. x -> ((-a)*e + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r, 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 6448

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 6847

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

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Mathematica [A]
time = 0.15, size = 106, normalized size = 1.83 \begin {gather*} \frac {\left (a+b x^n\right ) \text {sech}^{-1}\left (a+b x^n\right )+\frac {2 \sqrt {-\frac {-1+a+b x^n}{1+a+b x^n}} \sqrt {1-\left (a+b x^n\right )^2} \text {ArcTan}\left (\frac {\sqrt {1-a-b x^n}}{\sqrt {1+a+b x^n}}\right )}{-1+a+b x^n}}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + n)*ArcSech[a + b*x^n],x]

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int x^{-1+n} \mathrm {arcsech}\left (a +b \,x^{n}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+n)*arcsech(a+b*x^n),x)

[Out]

int(x^(-1+n)*arcsech(a+b*x^n),x)

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Maxima [A]
time = 0.25, size = 40, normalized size = 0.69 \begin {gather*} \frac {{\left (b x^{n} + a\right )} \operatorname {arsech}\left (b x^{n} + a\right ) - \arctan \left (\sqrt {\frac {1}{{\left (b x^{n} + a\right )}^{2}} - 1}\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+n)*arcsech(a+b*x^n),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (54) = 108\).
time = 0.42, size = 385, normalized size = 6.64 \begin {gather*} \frac {2 \, {\left (b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right )\right )} \log \left (\frac {\sqrt {-\frac {2 \, a b + {\left (a^{2} + b^{2} - 1\right )} \cosh \left (n \log \left (x\right )\right ) - {\left (a^{2} - b^{2} - 1\right )} \sinh \left (n \log \left (x\right )\right )}{\cosh \left (n \log \left (x\right )\right ) - \sinh \left (n \log \left (x\right )\right )}} + 1}{b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a}\right ) + a \log \left (\frac {\sqrt {-\frac {2 \, a b + {\left (a^{2} + b^{2} - 1\right )} \cosh \left (n \log \left (x\right )\right ) - {\left (a^{2} - b^{2} - 1\right )} \sinh \left (n \log \left (x\right )\right )}{\cosh \left (n \log \left (x\right )\right ) - \sinh \left (n \log \left (x\right )\right )}} + 1}{\cosh \left (n \log \left (x\right )\right ) + \sinh \left (n \log \left (x\right )\right )}\right ) - a \log \left (\frac {\sqrt {-\frac {2 \, a b + {\left (a^{2} + b^{2} - 1\right )} \cosh \left (n \log \left (x\right )\right ) - {\left (a^{2} - b^{2} - 1\right )} \sinh \left (n \log \left (x\right )\right )}{\cosh \left (n \log \left (x\right )\right ) - \sinh \left (n \log \left (x\right )\right )}} - 1}{\cosh \left (n \log \left (x\right )\right ) + \sinh \left (n \log \left (x\right )\right )}\right ) - 2 \, \arctan \left (\frac {{\left (b \cosh \left (n \log \left (x\right )\right ) + b \sinh \left (n \log \left (x\right )\right ) + a\right )} \sqrt {-\frac {2 \, a b + {\left (a^{2} + b^{2} - 1\right )} \cosh \left (n \log \left (x\right )\right ) - {\left (a^{2} - b^{2} - 1\right )} \sinh \left (n \log \left (x\right )\right )}{\cosh \left (n \log \left (x\right )\right ) - \sinh \left (n \log \left (x\right )\right )}}}{b^{2} \cosh \left (n \log \left (x\right )\right )^{2} + b^{2} \sinh \left (n \log \left (x\right )\right )^{2} + 2 \, a b \cosh \left (n \log \left (x\right )\right ) + a^{2} + 2 \, {\left (b^{2} \cosh \left (n \log \left (x\right )\right ) + a b\right )} \sinh \left (n \log \left (x\right )\right ) - 1}\right )}{2 \, b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+n)*arcsech(a+b*x^n),x, algorithm="fricas")

[Out]

1/2*(2*(b*cosh(n*log(x)) + b*sinh(n*log(x)))*log((sqrt(-(2*a*b + (a^2 + b^2 - 1)*cosh(n*log(x)) - (a^2 - b^2 -
1)*sinh(n*log(x)))/(cosh(n*log(x)) - sinh(n*log(x)))) + 1)/(b*cosh(n*log(x)) + b*sinh(n*log(x)) + a)) + a*log
((sqrt(-(2*a*b + (a^2 + b^2 - 1)*cosh(n*log(x)) - (a^2 - b^2 - 1)*sinh(n*log(x)))/(cosh(n*log(x)) - sinh(n*log
(x)))) + 1)/(cosh(n*log(x)) + sinh(n*log(x)))) - a*log((sqrt(-(2*a*b + (a^2 + b^2 - 1)*cosh(n*log(x)) - (a^2 -
b^2 - 1)*sinh(n*log(x)))/(cosh(n*log(x)) - sinh(n*log(x)))) - 1)/(cosh(n*log(x)) + sinh(n*log(x)))) - 2*arcta
n((b*cosh(n*log(x)) + b*sinh(n*log(x)) + a)*sqrt(-(2*a*b + (a^2 + b^2 - 1)*cosh(n*log(x)) - (a^2 - b^2 - 1)*si
nh(n*log(x)))/(cosh(n*log(x)) - sinh(n*log(x))))/(b^2*cosh(n*log(x))^2 + b^2*sinh(n*log(x))^2 + 2*a*b*cosh(n*l
og(x)) + a^2 + 2*(b^2*cosh(n*log(x)) + a*b)*sinh(n*log(x)) - 1)))/(b*n)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+n)*asech(a+b*x**n),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+n)*arcsech(a+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(n - 1)*arcsech(b*x^n + a), x)

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Mupad [B]
time = 2.35, size = 54, normalized size = 0.93 \begin {gather*} \frac {\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{a+b\,x^n}-1}\,\sqrt {\frac {1}{a+b\,x^n}+1}}\right )+\mathrm {acosh}\left (\frac {1}{a+b\,x^n}\right )\,\left (a+b\,x^n\right )}{b\,n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n - 1)*acosh(1/(a + b*x^n)),x)

[Out]

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

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