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

Optimal. Leaf size=48 \[ \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6715, 5251, 372, 266, 63, 206} \[ \frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b} \]

Antiderivative was successfully verified.

[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 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]

Rule 6715

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*} \int x^3 \csc ^{-1}\left (a+b x^4\right ) \, dx &=\frac {1}{4} \operatorname {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} \operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}\right )}{4 b}\\ \end {align*}

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Mathematica [B]  time = 0.25, size = 127, normalized size = 2.65 \[ \frac {\sqrt {\left (a+b x^4\right )^2-1} \left (\log \left (\frac {a+b x^4}{\sqrt {\left (a+b x^4\right )^2-1}}+1\right )-\log \left (1-\frac {a+b x^4}{\sqrt {\left (a+b x^4\right )^2-1}}\right )\right )}{8 b \left (a+b x^4\right ) \sqrt {1-\frac {1}{\left (a+b x^4\right )^2}}}+\frac {\left (a+b x^4\right ) \csc ^{-1}\left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[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)])

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fricas [B]  time = 0.74, size = 88, normalized size = 1.83 \[ \frac {b x^{4} \operatorname {arccsc}\left (b x^{4} + a\right ) - 2 \, a \arctan \left (-b x^{4} - a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right ) - \log \left (-b x^{4} - a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.25, size = 91, normalized size = 1.90 \[ \frac {1}{8} \, b {\left (\frac {2 \, {\left (b x^{4} + a\right )} \arcsin \left (-\frac {1}{{\left (b x^{4} + a\right )} {\left (\frac {a}{b x^{4} + a} - 1\right )} - a}\right )}{b^{2}} + \frac {\log \left (\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right )}{b^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.11, size = 65, normalized size = 1.35 \[ \frac {\mathrm {arccsc}\left (b \,x^{4}+a \right ) x^{4}}{4}+\frac {\mathrm {arccsc}\left (b \,x^{4}+a \right ) a}{4 b}+\frac {\ln \left (b \,x^{4}+a +\left (b \,x^{4}+a \right ) \sqrt {1-\frac {1}{\left (b \,x^{4}+a \right )^{2}}}\right )}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 63, normalized size = 1.31 \[ \frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arccsc}\left (b x^{4} + a\right ) + \log \left (\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.03, size = 44, normalized size = 0.92 \[ \frac {\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{{\left (b\,x^4+a\right )}^2}}}\right )}{4\,b}+\frac {\mathrm {asin}\left (\frac {1}{b\,x^4+a}\right )\,\left (b\,x^4+a\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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