3.30 \(\int \csc ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=86 \[ -\frac {2 i \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}+\frac {4 \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b} \]

[Out]

(b*x+a)*arccsc(b*x+a)^2/b+4*arccsc(b*x+a)*arctanh(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/b-2*I*polylog(2,-I/(b*x+a)-
(1-1/(b*x+a)^2)^(1/2))/b+2*I*polylog(2,I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/b

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Rubi [A]  time = 0.06, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5253, 5217, 3758, 4183, 2279, 2391} \[ -\frac {2 i \text {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {PolyLog}\left (2,e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}+\frac {4 \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[a + b*x]^2,x]

[Out]

((a + b*x)*ArcCsc[a + b*x]^2)/b + (4*ArcCsc[a + b*x]*ArcTanh[E^(I*ArcCsc[a + b*x])])/b - ((2*I)*PolyLog[2, -E^
(I*ArcCsc[a + b*x])])/b + ((2*I)*PolyLog[2, E^(I*ArcCsc[a + b*x])])/b

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3758

Int[Cot[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csc[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(x^
(m - n + 1)*Csc[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csc[a + b*x^n]^p, x], x] /
; FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 5217

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Dist[c^(-1), Subst[Int[(a + b*x)^n*Csc[x]*Cot[x], x
], x, ArcCsc[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

Rule 5253

Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCsc[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \csc ^{-1}(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \csc ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int x^2 \cot (x) \csc (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}-\frac {2 \operatorname {Subst}\left (\int x \csc (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}+\frac {4 \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {2 \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}-\frac {2 \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}+\frac {4 \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^2}{b}+\frac {4 \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {2 i \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {2 i \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 99, normalized size = 1.15 \[ \frac {-2 i \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )+2 i \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )+\csc ^{-1}(a+b x) \left ((a+b x) \csc ^{-1}(a+b x)-2 \log \left (1-e^{i \csc ^{-1}(a+b x)}\right )+2 \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCsc[a + b*x]^2,x]

[Out]

(ArcCsc[a + b*x]*((a + b*x)*ArcCsc[a + b*x] - 2*Log[1 - E^(I*ArcCsc[a + b*x])] + 2*Log[1 + E^(I*ArcCsc[a + b*x
])]) - (2*I)*PolyLog[2, -E^(I*ArcCsc[a + b*x])] + (2*I)*PolyLog[2, E^(I*ArcCsc[a + b*x])])/b

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arccsc}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(arccsc(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arccsc}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(arccsc(b*x + a)^2, x)

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maple [A]  time = 0.24, size = 167, normalized size = 1.94 \[ x \mathrm {arccsc}\left (b x +a \right )^{2}+\frac {\mathrm {arccsc}\left (b x +a \right )^{2} a}{b}-\frac {2 \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}+\frac {2 \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}-\frac {2 i \dilog \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}+\frac {2 i \dilog \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsc(b*x+a)^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{4} \, x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} + \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )^{2} + {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} - 1\right )} b x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsc(b*x+a)^2,x, algorithm="maxima")

[Out]

x*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/4*x*log(b^2*x^2 + 2*a*b*x + a^2)^2 + integrate((2*sqrt
(b*x + a + 1)*sqrt(b*x + a - 1)*b*x*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - (b^3*x^3 + 3*a*b^2*x^2 +
a^3 + (3*a^2 - 1)*b*x - a)*log(b*x + a)^2 + (b^3*x^3 + 2*a*b^2*x^2 + (a^2 - 1)*b*x + (b^3*x^3 + 3*a*b^2*x^2 +
a^3 + (3*a^2 - 1)*b*x - a)*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2
- 1)*b*x - a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(1/(a + b*x))^2,x)

[Out]

int(asin(1/(a + b*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {acsc}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsc(b*x+a)**2,x)

[Out]

Integral(acsc(a + b*x)**2, x)

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