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

Optimal. Leaf size=145 \[ -\frac {a^2 \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {2 i a \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {\log (a+b x)}{b^2}+\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)}{b^2}-\frac {4 a \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5259, 4427, 4190, 4183, 2279, 2391, 4184, 3475} \[ \frac {2 i a \text {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {2 i a \text {PolyLog}\left (2,e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {\log (a+b x)}{b^2}+\frac {(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)}{b^2}-\frac {4 a \csc ^{-1}(a+b x) \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCsc[a + b*x]^2,x]

[Out]

((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]*ArcCsc[a + b*x])/b^2 - (a^2*ArcCsc[a + b*x]^2)/(2*b^2) + (x^2*ArcCsc[a + b
*x]^2)/2 - (4*a*ArcCsc[a + b*x]*ArcTanh[E^(I*ArcCsc[a + b*x])])/b^2 + Log[a + b*x]/b^2 + ((2*I)*a*PolyLog[2, -
E^(I*ArcCsc[a + b*x])])/b^2 - ((2*I)*a*PolyLog[2, E^(I*ArcCsc[a + b*x])])/b^2

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 3475

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

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 4184

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

Rule 4190

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

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

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Mathematica [A]  time = 0.82, size = 213, normalized size = 1.47 \[ \frac {2 a \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \csc ^{-1}(a+b x)+2 b x \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \csc ^{-1}(a+b x)-a^2 \csc ^{-1}(a+b x)^2+b^2 x^2 \csc ^{-1}(a+b x)^2+4 i a \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )-4 i a \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )-2 \log \left (\frac {1}{a+b x}\right )+4 a \csc ^{-1}(a+b x) \log \left (1-e^{i \csc ^{-1}(a+b x)}\right )-4 a \csc ^{-1}(a+b x) \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*ArcCsc[a + b*x]^2,x]

[Out]

(2*a*Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*ArcCsc[a + b*x] + 2*b*x*Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x
^2)/(a + b*x)^2]*ArcCsc[a + b*x] - a^2*ArcCsc[a + b*x]^2 + b^2*x^2*ArcCsc[a + b*x]^2 + 4*a*ArcCsc[a + b*x]*Log
[1 - E^(I*ArcCsc[a + b*x])] - 4*a*ArcCsc[a + b*x]*Log[1 + E^(I*ArcCsc[a + b*x])] - 2*Log[(a + b*x)^(-1)] + (4*
I)*a*PolyLog[2, -E^(I*ArcCsc[a + b*x])] - (4*I)*a*PolyLog[2, E^(I*ArcCsc[a + b*x])])/(2*b^2)

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arccsc}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*arccsc(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccsc(b*x + a)^2, x)

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maple [A]  time = 0.34, size = 250, normalized size = 1.72 \[ -\frac {a^{2} \mathrm {arccsc}\left (b x +a \right )^{2}}{2 b^{2}}+\frac {2 a \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {2 a \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {2 i a \dilog \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {2 i a \dilog \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {x^{2} \mathrm {arccsc}\left (b x +a \right )^{2}}{2}+\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arccsc}\left (b x +a \right ) x}{b}+\frac {\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arccsc}\left (b x +a \right ) a}{b^{2}}-\frac {\ln \left (\frac {1}{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - \frac {1}{8} \, x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} + \int \frac {2 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} b x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )^{2} + {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + {\left (a^{2} - 1\right )} b x^{2} + 2 \, {\left (b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} - 1\right )} b x^{2} + {\left (a^{3} - a\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - 1/8*x^2*log(b^2*x^2 + 2*a*b*x + a^2)^2 + integrate
(1/2*(2*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*b*x^2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - 2*(b^3*x^4
+ 3*a*b^2*x^3 + (3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*log(b*x + a)^2 + (b^3*x^4 + 2*a*b^2*x^3 + (a^2 - 1)*b*x^2 +
2*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2 - 1)*b*x^2 + (a^3 - a)*x)*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 \[ \int x\,{\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asin(1/(a + b*x))^2,x)

[Out]

int(x*asin(1/(a + b*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acsc(a + b*x)**2, x)

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