\(\int x^3 \csc ^{-1}(a+b x^4) \, dx\) [38]

Optimal result

Integrand size = 12, antiderivative size = 48 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6847, 5359, 379, 272, 65, 212} \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b}+\frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b} \]

[In]

Int[x^3*ArcCsc[a + b*x^4],x]

[Out]

((a + b*x^4)*ArcCsc[a + b*x^4])/(4*b) + ArcTanh[Sqrt[1 - (a + b*x^4)^(-2)]]/(4*b)

Rule 65

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 212

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

Rule 272

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 379

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 5359

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]

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*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \csc ^{-1}(a+b x) \, dx,x,x^4\right ) \\ & = \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}} \, dx,x,x^4\right ) \\ & = \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {1}{x^2}} x} \, dx,x,a+b x^4\right )}{4 b} \\ & = \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{\left (a+b x^4\right )^2}\right )}{8 b} \\ & = \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b} \\ & = \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(127\) vs. \(2(48)=96\).

Time = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.65 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\sqrt {-1+\left (a+b x^4\right )^2} \left (-\log \left (1-\frac {a+b x^4}{\sqrt {-1+\left (a+b x^4\right )^2}}\right )+\log \left (1+\frac {a+b x^4}{\sqrt {-1+\left (a+b x^4\right )^2}}\right )\right )}{8 b \left (a+b x^4\right ) \sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}} \]

[In]

Integrate[x^3*ArcCsc[a + b*x^4],x]

[Out]

((a + b*x^4)*ArcCsc[a + b*x^4])/(4*b) + (Sqrt[-1 + (a + b*x^4)^2]*(-Log[1 - (a + b*x^4)/Sqrt[-1 + (a + b*x^4)^
2]] + Log[1 + (a + b*x^4)/Sqrt[-1 + (a + b*x^4)^2]]))/(8*b*(a + b*x^4)*Sqrt[1 - (a + b*x^4)^(-2)])

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {\operatorname {arccsc}\left (b \,x^{4}+a \right ) \left (b \,x^{4}+a \right )+\ln \left (b \,x^{4}+a +\left (b \,x^{4}+a \right ) \sqrt {1-\frac {1}{\left (b \,x^{4}+a \right )^{2}}}\right )}{4 b}\)	\(54\)
default	\(\frac {\operatorname {arccsc}\left (b \,x^{4}+a \right ) \left (b \,x^{4}+a \right )+\ln \left (b \,x^{4}+a +\left (b \,x^{4}+a \right ) \sqrt {1-\frac {1}{\left (b \,x^{4}+a \right )^{2}}}\right )}{4 b}\)	\(54\)

[In]

int(x^3*arccsc(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

1/4/b*(arccsc(b*x^4+a)*(b*x^4+a)+ln(b*x^4+a+(b*x^4+a)*(1-1/(b*x^4+a)^2)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (42) = 84\).

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.83 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {b x^{4} \operatorname {arccsc}\left (b x^{4} + a\right ) - 2 \, a \arctan \left (-b x^{4} - a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right ) - \log \left (-b x^{4} - a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )}{4 \, b} \]

[In]

integrate(x^3*arccsc(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*(b*x^4*arccsc(b*x^4 + a) - 2*a*arctan(-b*x^4 - a + sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)) - log(-b*x^4 - a +
sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 - 1)))/b

Sympy [F(-1)]

Timed out. \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\text {Timed out} \]

[In]

integrate(x**3*acsc(b*x**4+a),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.31 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arccsc}\left (b x^{4} + a\right ) + \log \left (\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right )}{8 \, b} \]

[In]

integrate(x^3*arccsc(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*(2*(b*x^4 + a)*arccsc(b*x^4 + a) + log(sqrt(-1/(b*x^4 + a)^2 + 1) + 1) - log(-sqrt(-1/(b*x^4 + a)^2 + 1) +
1))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (42) = 84\).

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.90 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {1}{8} \, b {\left (\frac {2 \, {\left (b x^{4} + a\right )} \arcsin \left (-\frac {1}{{\left (b x^{4} + a\right )} {\left (\frac {a}{b x^{4} + a} - 1\right )} - a}\right )}{b^{2}} + \frac {\log \left (\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right )}{b^{2}}\right )} \]

[In]

integrate(x^3*arccsc(b*x^4+a),x, algorithm="giac")

[Out]

1/8*b*(2*(b*x^4 + a)*arcsin(-1/((b*x^4 + a)*(a/(b*x^4 + a) - 1) - a))/b^2 + (log(sqrt(-1/(b*x^4 + a)^2 + 1) +
1) - log(-sqrt(-1/(b*x^4 + a)^2 + 1) + 1))/b^2)

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx=\frac {\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{{\left (b\,x^4+a\right )}^2}}}\right )}{4\,b}+\frac {\mathrm {asin}\left (\frac {1}{b\,x^4+a}\right )\,\left (b\,x^4+a\right )}{4\,b} \]

[In]

int(x^3*asin(1/(a + b*x^4)),x)

[Out]

atanh(1/(1 - 1/(a + b*x^4)^2)^(1/2))/(4*b) + (asin(1/(a + b*x^4))*(a + b*x^4))/(4*b)