3.1.85 \(\int x (d+e x^2) (a+b \csc ^{-1}(c x)) \, dx\) [85]

3.1.85.1 Optimal result
3.1.85.2 Mathematica [A] (verified)
3.1.85.3 Rubi [A] (verified)
3.1.85.4 Maple [A] (verified)
3.1.85.5 Fricas [A] (verification not implemented)
3.1.85.6 Sympy [A] (verification not implemented)
3.1.85.7 Maxima [A] (verification not implemented)
3.1.85.8 Giac [B] (verification not implemented)
3.1.85.9 Mupad [F(-1)]

3.1.85.1 Optimal result

Integrand size = 17, antiderivative size = 138 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {b \left (2 c^2 d+e\right ) x \sqrt {-1+c^2 x^2}}{4 c^3 \sqrt {c^2 x^2}}+\frac {b e x \left (-1+c^2 x^2\right )^{3/2}}{12 c^3 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}+\frac {b c d^2 x \arctan \left (\sqrt {-1+c^2 x^2}\right )}{4 e \sqrt {c^2 x^2}} \]

output
1/4*(e*x^2+d)^2*(a+b*arccsc(c*x))/e+1/12*b*e*x*(c^2*x^2-1)^(3/2)/c^3/(c^2* 
x^2)^(1/2)+1/4*b*c*d^2*x*arctan((c^2*x^2-1)^(1/2))/e/(c^2*x^2)^(1/2)+1/4*b 
*(2*c^2*d+e)*x*(c^2*x^2-1)^(1/2)/c^3/(c^2*x^2)^(1/2)
 
3.1.85.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {x \left (3 a c^3 x \left (2 d+e x^2\right )+b \sqrt {1-\frac {1}{c^2 x^2}} \left (2 e+c^2 \left (6 d+e x^2\right )\right )+3 b c^3 x \left (2 d+e x^2\right ) \csc ^{-1}(c x)\right )}{12 c^3} \]

input
Integrate[x*(d + e*x^2)*(a + b*ArcCsc[c*x]),x]
 
output
(x*(3*a*c^3*x*(2*d + e*x^2) + b*Sqrt[1 - 1/(c^2*x^2)]*(2*e + c^2*(6*d + e* 
x^2)) + 3*b*c^3*x*(2*d + e*x^2)*ArcCsc[c*x]))/(12*c^3)
 
3.1.85.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5760, 354, 99, 2009}

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.

\(\displaystyle \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx\)

\(\Big \downarrow \) 5760

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

\(\Big \downarrow \) 354

\(\displaystyle \frac {b c x \int \frac {\left (e x^2+d\right )^2}{x^2 \sqrt {c^2 x^2-1}}dx^2}{8 e \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {b c x \int \left (\frac {d^2}{x^2 \sqrt {c^2 x^2-1}}+\frac {e^2 \sqrt {c^2 x^2-1}}{c^2}+\frac {e \left (2 d c^2+e\right )}{c^2 \sqrt {c^2 x^2-1}}\right )dx^2}{8 e \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}+\frac {b c x \left (2 d^2 \arctan \left (\sqrt {c^2 x^2-1}\right )+\frac {2 e \sqrt {c^2 x^2-1} \left (2 c^2 d+e\right )}{c^4}+\frac {2 e^2 \left (c^2 x^2-1\right )^{3/2}}{3 c^4}\right )}{8 e \sqrt {c^2 x^2}}\)

input
Int[x*(d + e*x^2)*(a + b*ArcCsc[c*x]),x]
 
output
((d + e*x^2)^2*(a + b*ArcCsc[c*x]))/(4*e) + (b*c*x*((2*e*(2*c^2*d + e)*Sqr 
t[-1 + c^2*x^2])/c^4 + (2*e^2*(-1 + c^2*x^2)^(3/2))/(3*c^4) + 2*d^2*ArcTan 
[Sqrt[-1 + c^2*x^2]]))/(8*e*Sqrt[c^2*x^2])
 

3.1.85.3.1 Defintions of rubi rules used

rule 99
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 354
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5760
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcCsc[c*x])/(2*e*(p + 1))), x 
] + Simp[b*c*(x/(2*e*(p + 1)*Sqrt[c^2*x^2]))   Int[(d + e*x^2)^(p + 1)/(x*S 
qrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
 
3.1.85.4 Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.57

method result size
parts \(\frac {a \left (e \,x^{2}+d \right )^{2}}{4 e}+\frac {b \,\operatorname {arccsc}\left (c x \right ) e \,x^{4}}{4}+\frac {b \,\operatorname {arccsc}\left (c x \right ) x^{2} d}{2}+\frac {b \,d^{2} \operatorname {arccsc}\left (c x \right )}{4 e}+\frac {b \left (c^{2} x^{2}-1\right ) x e}{12 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {b \sqrt {c^{2} x^{2}-1}\, d^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{4 c e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {b \left (c^{2} x^{2}-1\right ) d}{2 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {b e \left (c^{2} x^{2}-1\right )}{6 c^{5} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}\) \(217\)
derivativedivides \(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \,c^{2} \operatorname {arccsc}\left (c x \right ) d^{2}}{4 e}+\frac {b \,\operatorname {arccsc}\left (c x \right ) d \,c^{2} x^{2}}{2}+\frac {b \,c^{2} e \,\operatorname {arccsc}\left (c x \right ) x^{4}}{4}-\frac {b c \sqrt {c^{2} x^{2}-1}\, d^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{4 e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {b \left (c^{2} x^{2}-1\right ) d}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {b e \left (c^{2} x^{2}-1\right ) x}{12 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}+\frac {b e \left (c^{2} x^{2}-1\right )}{6 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}}{c^{2}}\) \(238\)
default \(\frac {\frac {a \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}{4 c^{2} e}+\frac {b \,c^{2} \operatorname {arccsc}\left (c x \right ) d^{2}}{4 e}+\frac {b \,\operatorname {arccsc}\left (c x \right ) d \,c^{2} x^{2}}{2}+\frac {b \,c^{2} e \,\operatorname {arccsc}\left (c x \right ) x^{4}}{4}-\frac {b c \sqrt {c^{2} x^{2}-1}\, d^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{4 e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {b \left (c^{2} x^{2}-1\right ) d}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {b e \left (c^{2} x^{2}-1\right ) x}{12 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}+\frac {b e \left (c^{2} x^{2}-1\right )}{6 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}}{c^{2}}\) \(238\)

input
int(x*(e*x^2+d)*(a+b*arccsc(c*x)),x,method=_RETURNVERBOSE)
 
output
1/4*a*(e*x^2+d)^2/e+1/4*b*arccsc(c*x)*e*x^4+1/2*b*arccsc(c*x)*x^2*d+1/4*b* 
d^2*arccsc(c*x)/e+1/12*b/c^3*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)*x*e-1 
/4*b/c/e*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*d^2*arctan(1/(c^2 
*x^2-1)^(1/2))+1/2*b/c^3*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*d+1/6*b 
/c^5*e*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x
 
3.1.85.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {3 \, a c^{4} e x^{4} + 6 \, a c^{4} d x^{2} + 3 \, {\left (b c^{4} e x^{4} + 2 \, b c^{4} d x^{2}\right )} \operatorname {arccsc}\left (c x\right ) + {\left (b c^{2} e x^{2} + 6 \, b c^{2} d + 2 \, b e\right )} \sqrt {c^{2} x^{2} - 1}}{12 \, c^{4}} \]

input
integrate(x*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="fricas")
 
output
1/12*(3*a*c^4*e*x^4 + 6*a*c^4*d*x^2 + 3*(b*c^4*e*x^4 + 2*b*c^4*d*x^2)*arcc 
sc(c*x) + (b*c^2*e*x^2 + 6*b*c^2*d + 2*b*e)*sqrt(c^2*x^2 - 1))/c^4
 
3.1.85.6 Sympy [A] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.28 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {acsc}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {acsc}{\left (c x \right )}}{4} + \frac {b d \left (\begin {cases} \frac {\sqrt {c^{2} x^{2} - 1}}{c} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i \sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right )}{2 c} + \frac {b e \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} - 1}}{3 c} + \frac {2 \sqrt {c^{2} x^{2} - 1}}{3 c^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {2 i \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} & \text {otherwise} \end {cases}\right )}{4 c} \]

input
integrate(x*(e*x**2+d)*(a+b*acsc(c*x)),x)
 
output
a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*acsc(c*x)/2 + b*e*x**4*acsc(c*x)/4 + b* 
d*Piecewise((sqrt(c**2*x**2 - 1)/c, Abs(c**2*x**2) > 1), (I*sqrt(-c**2*x** 
2 + 1)/c, True))/(2*c) + b*e*Piecewise((x**2*sqrt(c**2*x**2 - 1)/(3*c) + 2 
*sqrt(c**2*x**2 - 1)/(3*c**3), Abs(c**2*x**2) > 1), (I*x**2*sqrt(-c**2*x** 
2 + 1)/(3*c) + 2*I*sqrt(-c**2*x**2 + 1)/(3*c**3), True))/(4*c)
 
3.1.85.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.71 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arccsc}\left (c x\right ) + \frac {x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c}\right )} b d + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arccsc}\left (c x\right ) + \frac {c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e \]

input
integrate(x*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="maxima")
 
output
1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/2*(x^2*arccsc(c*x) + x*sqrt(-1/(c^2*x^2) + 1 
)/c)*b*d + 1/12*(3*x^4*arccsc(c*x) + (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3 
*x*sqrt(-1/(c^2*x^2) + 1))/c^3)*b*e
 
3.1.85.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (118) = 236\).

Time = 0.33 (sec) , antiderivative size = 556, normalized size of antiderivative = 4.03 \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\frac {1}{192} \, {\left (\frac {3 \, b e x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4} \arcsin \left (\frac {1}{c x}\right )}{c} + \frac {3 \, a e x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4}}{c} + \frac {2 \, b e x^{3} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{3}}{c^{2}} + \frac {24 \, b d x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2} \arcsin \left (\frac {1}{c x}\right )}{c} + \frac {24 \, a d x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}}{c} + \frac {12 \, b e x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2} \arcsin \left (\frac {1}{c x}\right )}{c^{3}} + \frac {12 \, a e x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}}{c^{3}} + \frac {48 \, b d x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}}{c^{2}} + \frac {18 \, b e x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}}{c^{4}} + \frac {48 \, b d \arcsin \left (\frac {1}{c x}\right )}{c^{3}} + \frac {48 \, a d}{c^{3}} + \frac {18 \, b e \arcsin \left (\frac {1}{c x}\right )}{c^{5}} + \frac {18 \, a e}{c^{5}} - \frac {48 \, b d}{c^{4} x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}} - \frac {18 \, b e}{c^{6} x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}} + \frac {24 \, b d \arcsin \left (\frac {1}{c x}\right )}{c^{5} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} + \frac {24 \, a d}{c^{5} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} + \frac {12 \, b e \arcsin \left (\frac {1}{c x}\right )}{c^{7} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} + \frac {12 \, a e}{c^{7} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} - \frac {2 \, b e}{c^{8} x^{3} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{3}} + \frac {3 \, b e \arcsin \left (\frac {1}{c x}\right )}{c^{9} x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4}} + \frac {3 \, a e}{c^{9} x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4}}\right )} c \]

input
integrate(x*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="giac")
 
output
1/192*(3*b*e*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*arcsin(1/(c*x))/c + 3*a*e* 
x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4/c + 2*b*e*x^3*(sqrt(-1/(c^2*x^2) + 1) + 
 1)^3/c^2 + 24*b*d*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*arcsin(1/(c*x))/c + 
24*a*d*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2/c + 12*b*e*x^2*(sqrt(-1/(c^2*x^2 
) + 1) + 1)^2*arcsin(1/(c*x))/c^3 + 12*a*e*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1 
)^2/c^3 + 48*b*d*x*(sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 + 18*b*e*x*(sqrt(-1/(c 
^2*x^2) + 1) + 1)/c^4 + 48*b*d*arcsin(1/(c*x))/c^3 + 48*a*d/c^3 + 18*b*e*a 
rcsin(1/(c*x))/c^5 + 18*a*e/c^5 - 48*b*d/(c^4*x*(sqrt(-1/(c^2*x^2) + 1) + 
1)) - 18*b*e/(c^6*x*(sqrt(-1/(c^2*x^2) + 1) + 1)) + 24*b*d*arcsin(1/(c*x)) 
/(c^5*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) + 24*a*d/(c^5*x^2*(sqrt(-1/(c^2* 
x^2) + 1) + 1)^2) + 12*b*e*arcsin(1/(c*x))/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1 
) + 1)^2) + 12*a*e/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) - 2*b*e/(c^8*x 
^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3) + 3*b*e*arcsin(1/(c*x))/(c^9*x^4*(sqrt( 
-1/(c^2*x^2) + 1) + 1)^4) + 3*a*e/(c^9*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4) 
)*c
 
3.1.85.9 Mupad [F(-1)]

Timed out. \[ \int x \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx=\int x\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

input
int(x*(d + e*x^2)*(a + b*asin(1/(c*x))),x)
 
output
int(x*(d + e*x^2)*(a + b*asin(1/(c*x))), x)