\(\int \frac {x^3 (a+b \csc ^{-1}(c x))}{\sqrt {1-c^4 x^4}} \, dx\) [176]

Optimal result

Integrand size = 26, antiderivative size = 126 \[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=-\frac {b x \sqrt {1-c^4 x^4}}{2 c^3 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}+\frac {b x \arctan \left (\frac {\sqrt {1-c^4 x^4}}{\sqrt {-1+c^2 x^2}}\right )}{2 c^3 \sqrt {c^2 x^2}} \] Output:

-1/2*b*x*(-c^4*x^4+1)^(1/2)/c^3/(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)-1/2*(-c^ 
4*x^4+1)^(1/2)*(a+b*arccsc(c*x))/c^4+1/2*b*x*arctan((-c^4*x^4+1)^(1/2)/(c^ 
2*x^2-1)^(1/2))/c^3/(c^2*x^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {\left (a-b c \sqrt {1-\frac {1}{c^2 x^2}} x-a c^2 x^2\right ) \sqrt {1-c^4 x^4}-b \left (-1+c^2 x^2\right ) \sqrt {1-c^4 x^4} \csc ^{-1}(c x)+\left (b-b c^2 x^2\right ) \arctan \left (\frac {c \sqrt {1-\frac {1}{c^2 x^2}} x}{\sqrt {1-c^4 x^4}}\right )}{2 c^4 \left (-1+c^2 x^2\right )} \] Input:

Integrate[(x^3*(a + b*ArcCsc[c*x]))/Sqrt[1 - c^4*x^4],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5770, 27, 1896, 1388, 243, 60, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx\)

\(\Big \downarrow \) 5770

\(\displaystyle \frac {b \int -\frac {\sqrt {1-c^4 x^4}}{2 c^4 \sqrt {1-\frac {1}{c^2 x^2}} x^2}dx}{c}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {\sqrt {1-c^4 x^4}}{\sqrt {1-\frac {1}{c^2 x^2}} x^2}dx}{2 c^5}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 1896

\(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {1-c^4 x^4}}{x \sqrt {1-c^2 x^2}}dx}{2 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 1388

\(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {c^2 x^2+1}}{x}dx}{2 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {c^2 x^2+1}}{x^2}dx^2}{4 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 60

\(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \left (\int \frac {1}{x^2 \sqrt {c^2 x^2+1}}dx^2+2 \sqrt {c^2 x^2+1}\right )}{4 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \left (\frac {2 \int \frac {1}{\frac {x^4}{c^2}-\frac {1}{c^2}}d\sqrt {c^2 x^2+1}}{c^2}+2 \sqrt {c^2 x^2+1}\right )}{4 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\sqrt {1-c^4 x^4} \left (a+b \csc ^{-1}(c x)\right )}{2 c^4}-\frac {b \sqrt {1-c^2 x^2} \left (2 \sqrt {c^2 x^2+1}-2 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )\right )}{4 c^5 x \sqrt {1-\frac {1}{c^2 x^2}}}\)

Input:

Int[(x^3*(a + b*ArcCsc[c*x]))/Sqrt[1 - c^4*x^4],x]
 

Output:

-1/2*(Sqrt[1 - c^4*x^4]*(a + b*ArcCsc[c*x]))/c^4 - (b*Sqrt[1 - c^2*x^2]*(2 
*Sqrt[1 + c^2*x^2] - 2*ArcTanh[Sqrt[1 + c^2*x^2]]))/(4*c^5*Sqrt[1 - 1/(c^2 
*x^2)]*x)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 1896

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^( 
p_.), x_Symbol] :> Simp[(e^IntPart[q]*((d + e*x^mn)^FracPart[q]/(1 + d*(1/( 
x^mn*e)))^FracPart[q]))/x^(mn*FracPart[q])   Int[x^(m + mn*q)*(1 + d*(1/(x^ 
mn*e)))^q*(a + c*x^n2)^p, x], x] /; FreeQ[{a, c, d, e, m, mn, p, q}, x] && 
EqQ[n2, -2*mn] &&  !IntegerQ[p] &&  !IntegerQ[q] && PosQ[n2]
 

rule 5770

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide 
[u, x]}, Simp[(a + b*ArcCsc[c*x])   v, x] + Simp[b/c   Int[SimplifyIntegran 
d[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]] 
/; FreeQ[{a, b, c}, x]
 

Maple [F]

\[\int \frac {x^{3} \left (a +b \,\operatorname {arccsc}\left (c x \right )\right )}{\sqrt {-x^{4} c^{4}+1}}d x\]

Input:

int(x^3*(a+b*arccsc(c*x))/(-c^4*x^4+1)^(1/2),x)
 

Output:

int(x^3*(a+b*arccsc(c*x))/(-c^4*x^4+1)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=-\frac {\sqrt {-c^{4} x^{4} + 1} \sqrt {c^{2} x^{2} - 1} b - {\left (b c^{2} x^{2} - b\right )} \arctan \left (\frac {\sqrt {-c^{4} x^{4} + 1} \sqrt {c^{2} x^{2} - 1}}{c^{4} x^{4} - 1}\right ) + \sqrt {-c^{4} x^{4} + 1} {\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - b\right )} \operatorname {arccsc}\left (c x\right ) - a\right )}}{2 \, {\left (c^{6} x^{2} - c^{4}\right )}} \] Input:

integrate(x^3*(a+b*arccsc(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-1/2*(sqrt(-c^4*x^4 + 1)*sqrt(c^2*x^2 - 1)*b - (b*c^2*x^2 - b)*arctan(sqrt 
(-c^4*x^4 + 1)*sqrt(c^2*x^2 - 1)/(c^4*x^4 - 1)) + sqrt(-c^4*x^4 + 1)*(a*c^ 
2*x^2 + (b*c^2*x^2 - b)*arccsc(c*x) - a))/(c^6*x^2 - c^4)
 

Sympy [F]

\[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right ) \left (c^{2} x^{2} + 1\right )}}\, dx \] Input:

integrate(x**3*(a+b*acsc(c*x))/(-c**4*x**4+1)**(1/2),x)
 

Output:

Integral(x**3*(a + b*acsc(c*x))/sqrt(-(c*x - 1)*(c*x + 1)*(c**2*x**2 + 1)) 
, x)
 

Maxima [F]

\[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int { \frac {{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{3}}{\sqrt {-c^{4} x^{4} + 1}} \,d x } \] Input:

integrate(x^3*(a+b*arccsc(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int { \frac {{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{3}}{\sqrt {-c^{4} x^{4} + 1}} \,d x } \] Input:

integrate(x^3*(a+b*arccsc(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="giac")
 

Output:

integrate((b*arccsc(c*x) + a)*x^3/sqrt(-c^4*x^4 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {1-c^4\,x^4}} \,d x \] Input:

int((x^3*(a + b*asin(1/(c*x))))/(1 - c^4*x^4)^(1/2),x)
 

Output:

int((x^3*(a + b*asin(1/(c*x))))/(1 - c^4*x^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {-\sqrt {-c^{4} x^{4}+1}\, a -2 \left (\int \frac {\sqrt {-c^{4} x^{4}+1}\, \mathit {acsc} \left (c x \right ) x^{3}}{c^{4} x^{4}-1}d x \right ) b \,c^{4}}{2 c^{4}} \] Input:

int(x^3*(a+b*acsc(c*x))/(-c^4*x^4+1)^(1/2),x)
 

Output:

( - sqrt( - c**4*x**4 + 1)*a - 2*int((sqrt( - c**4*x**4 + 1)*acsc(c*x)*x** 
3)/(c**4*x**4 - 1),x)*b*c**4)/(2*c**4)