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

Optimal. Leaf size=79 \[ -\frac {a^2 \csc ^{-1}(a+b x)}{2 b^2}+\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{2 b^2}-\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5259, 4427, 3773, 3770, 3767, 8} \[ -\frac {a^2 \csc ^{-1}(a+b x)}{2 b^2}+\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{2 b^2}-\frac {a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCsc[a + b*x],x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

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

Rule 3773

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

Rule 4427

Int[Cot[(c_.) + (d_.)*(x_)]*Csc[(c_.) + (d_.)*(x_)]*(Csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.)*((e_.) + (f_.
)*(x_))^(m_.), x_Symbol] :> -Simp[((e + f*x)^m*(a + b*Csc[c + d*x])^(n + 1))/(b*d*(n + 1)), x] + Dist[(f*m)/(b
*d*(n + 1)), Int[(e + f*x)^(m - 1)*(a + b*Csc[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
IGtQ[m, 0] && NeQ[n, -1]

Rule 5259

Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1))
^(-1), Subst[Int[(a + b*x)^p*Csc[x]*Cot[x]*(d*e - c*f + f*Csc[x])^m, x], x, ArcCsc[c + d*x]], x] /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 110, normalized size = 1.39 \[ \frac {(a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}-2 a \log \left ((a+b x) \left (\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )+a^2 \left (-\sin ^{-1}\left (\frac {1}{a+b x}\right )\right )+b^2 x^2 \csc ^{-1}(a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCsc[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.64, size = 102, normalized size = 1.29 \[ \frac {b^{2} x^{2} \operatorname {arccsc}\left (b x + a\right ) + 2 \, a^{2} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 2 \, a \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 134, normalized size = 1.70 \[ -\frac {1}{4} \, b {\left (\frac {2 \, {\left (b x + a\right )}^{2} {\left (\frac {2 \, a}{b x + a} - 1\right )} \arcsin \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{3}} - \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 4 \, a \log \left (-{\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} {\left | b x + a \right |}\right ) - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*(2*(b*x + a)^2*(2*a/(b*x + a) - 1)*arcsin(-1/((b*x + a)*(a/(b*x + a) - 1) - a))/b^3 - ((b*x + a)*(sqrt(
-1/(b*x + a)^2 + 1) - 1) + 4*a*log(-(sqrt(-1/(b*x + a)^2 + 1) - 1)*abs(b*x + a)) - 1/((b*x + a)*(sqrt(-1/(b*x
+ a)^2 + 1) - 1)))/b^3)

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maple [A]  time = 0.06, size = 127, normalized size = 1.61 \[ \frac {x^{2} \mathrm {arccsc}\left (b x +a \right )}{2}-\frac {a^{2} \mathrm {arccsc}\left (b x +a \right )}{2 b^{2}}-\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {-1+\left (b x +a \right )^{2}}{2 b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + \int \frac {{\left (b^{2} x^{3} + a b x^{2}\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + 1\right ) + \frac {1}{2} \, \log \left (b x + a - 1\right )\right )}}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\log \left (b x + a + 1\right ) + \log \left (b x + a - 1\right )\right )} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + integrate(1/2*(b^2*x^3 + a*b*x^2)*e^(1/2*log(b*x + a
+ 1) + 1/2*log(b*x + a - 1))/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x + a^2 - 1)*e^(log(b*x + a + 1) + l
og(b*x + a - 1)) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asin(1/(a + b*x)),x)

[Out]

int(x*asin(1/(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acsc(a + b*x), x)

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