3.21 \(\int \csc ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=36 \[ \frac {(a+b x) \csc ^{-1}(a+b x)}{b}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b} \]

[Out]

(b*x+a)*arccsc(b*x+a)/b+arctanh((1-1/(b*x+a)^2)^(1/2))/b

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5251, 372, 266, 63, 206} \[ \frac {(a+b x) \csc ^{-1}(a+b x)}{b}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[a + b*x],x]

[Out]

((a + b*x)*ArcCsc[a + b*x])/b + ArcTanh[Sqrt[1 - (a + b*x)^(-2)]]/b

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 5251

Int[ArcCsc[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcCsc[c + d*x])/d, x] + Int[1/((c + d*x)*Sqrt[1 -
1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [B]  time = 0.12, size = 114, normalized size = 3.17 \[ \frac {(a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (\tanh ^{-1}\left (\frac {a+b x}{\sqrt {a^2+2 a b x+b^2 x^2-1}}\right )-a \tan ^{-1}\left (\sqrt {(a+b x)^2-1}\right )\right )}{b \sqrt {a^2+2 a b x+b^2 x^2-1}}+x \csc ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCsc[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.55, size = 75, normalized size = 2.08 \[ \frac {b x \operatorname {arccsc}\left (b x + a\right ) - 2 \, a \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.16, size = 81, normalized size = 2.25 \[ \frac {1}{2} \, b {\left (\frac {2 \, {\left (b x + a\right )} \arcsin \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{2}} + \frac {\log \left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} + 1\right )}{b^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a),x, algorithm="giac")

[Out]

1/2*b*(2*(b*x + a)*arcsin(-1/((b*x + a)*(a/(b*x + a) - 1) - a))/b^2 + (log(sqrt(-1/(b*x + a)^2 + 1) + 1) - log
(-sqrt(-1/(b*x + a)^2 + 1) + 1))/b^2)

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maple [A]  time = 0.05, size = 50, normalized size = 1.39 \[ x \,\mathrm {arccsc}\left (b x +a \right )+\frac {\mathrm {arccsc}\left (b x +a \right ) a}{b}+\frac {\ln \left (b x +a +\left (b x +a \right ) \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsc(b*x+a),x)

[Out]

x*arccsc(b*x+a)+1/b*arccsc(b*x+a)*a+1/b*ln(b*x+a+(b*x+a)*(1-1/(b*x+a)^2)^(1/2))

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maxima [A]  time = 0.33, size = 55, normalized size = 1.53 \[ \frac {2 \, {\left (b x + a\right )} \operatorname {arccsc}\left (b x + a\right ) + \log \left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a),x, algorithm="maxima")

[Out]

1/2*(2*(b*x + a)*arccsc(b*x + a) + log(sqrt(-1/(b*x + a)^2 + 1) + 1) - log(-sqrt(-1/(b*x + a)^2 + 1) + 1))/b

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mupad [B]  time = 0.88, size = 33, normalized size = 0.92 \[ \frac {\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{{\left (a+b\,x\right )}^2}}}\right )+\mathrm {asin}\left (\frac {1}{a+b\,x}\right )\,\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(1/(a + b*x)),x)

[Out]

(atanh(1/(1 - 1/(a + b*x)^2)^(1/2)) + asin(1/(a + b*x))*(a + b*x))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsc(b*x+a),x)

[Out]

Integral(acsc(a + b*x), x)

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