3.84 \(\int x^3 (d+e x^2) (a+b \csc ^{-1}(c x)) \, dx\)

Optimal. Leaf size=153 \[ \frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {b x \left (c^2 x^2-1\right )^{3/2} \left (3 c^2 d+4 e\right )}{36 c^5 \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (3 c^2 d+2 e\right )}{12 c^5 \sqrt {c^2 x^2}}+\frac {b e x \left (c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {c^2 x^2}} \]

[Out]

1/4*d*x^4*(a+b*arccsc(c*x))+1/6*e*x^6*(a+b*arccsc(c*x))+1/36*b*(3*c^2*d+4*e)*x*(c^2*x^2-1)^(3/2)/c^5/(c^2*x^2)
^(1/2)+1/30*b*e*x*(c^2*x^2-1)^(5/2)/c^5/(c^2*x^2)^(1/2)+1/12*b*(3*c^2*d+2*e)*x*(c^2*x^2-1)^(1/2)/c^5/(c^2*x^2)
^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 5239, 12, 446, 77} \[ \frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {b x \left (c^2 x^2-1\right )^{3/2} \left (3 c^2 d+4 e\right )}{36 c^5 \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (3 c^2 d+2 e\right )}{12 c^5 \sqrt {c^2 x^2}}+\frac {b e x \left (c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x^2)*(a + b*ArcCsc[c*x]),x]

[Out]

(b*(3*c^2*d + 2*e)*x*Sqrt[-1 + c^2*x^2])/(12*c^5*Sqrt[c^2*x^2]) + (b*(3*c^2*d + 4*e)*x*(-1 + c^2*x^2)^(3/2))/(
36*c^5*Sqrt[c^2*x^2]) + (b*e*x*(-1 + c^2*x^2)^(5/2))/(30*c^5*Sqrt[c^2*x^2]) + (d*x^4*(a + b*ArcCsc[c*x]))/4 +
(e*x^6*(a + b*ArcCsc[c*x]))/6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5239

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int x^3 \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {(b c x) \int \frac {x^3 \left (3 d+2 e x^2\right )}{12 \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {(b c x) \int \frac {x^3 \left (3 d+2 e x^2\right )}{\sqrt {-1+c^2 x^2}} \, dx}{12 \sqrt {c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {(b c x) \operatorname {Subst}\left (\int \frac {x (3 d+2 e x)}{\sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{24 \sqrt {c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )+\frac {(b c x) \operatorname {Subst}\left (\int \left (\frac {3 c^2 d+2 e}{c^4 \sqrt {-1+c^2 x}}+\frac {\left (3 c^2 d+4 e\right ) \sqrt {-1+c^2 x}}{c^4}+\frac {2 e \left (-1+c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{24 \sqrt {c^2 x^2}}\\ &=\frac {b \left (3 c^2 d+2 e\right ) x \sqrt {-1+c^2 x^2}}{12 c^5 \sqrt {c^2 x^2}}+\frac {b \left (3 c^2 d+4 e\right ) x \left (-1+c^2 x^2\right )^{3/2}}{36 c^5 \sqrt {c^2 x^2}}+\frac {b e x \left (-1+c^2 x^2\right )^{5/2}}{30 c^5 \sqrt {c^2 x^2}}+\frac {1}{4} d x^4 \left (a+b \csc ^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \csc ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 97, normalized size = 0.63 \[ \frac {1}{180} x \left (15 a x^3 \left (3 d+2 e x^2\right )+\frac {b \sqrt {1-\frac {1}{c^2 x^2}} \left (3 c^4 \left (5 d x^2+2 e x^4\right )+c^2 \left (30 d+8 e x^2\right )+16 e\right )}{c^5}+15 b x^3 \csc ^{-1}(c x) \left (3 d+2 e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x^2)*(a + b*ArcCsc[c*x]),x]

[Out]

(x*(15*a*x^3*(3*d + 2*e*x^2) + (b*Sqrt[1 - 1/(c^2*x^2)]*(16*e + c^2*(30*d + 8*e*x^2) + 3*c^4*(5*d*x^2 + 2*e*x^
4)))/c^5 + 15*b*x^3*(3*d + 2*e*x^2)*ArcCsc[c*x]))/180

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fricas [A]  time = 0.60, size = 106, normalized size = 0.69 \[ \frac {30 \, a c^{6} e x^{6} + 45 \, a c^{6} d x^{4} + 15 \, {\left (2 \, b c^{6} e x^{6} + 3 \, b c^{6} d x^{4}\right )} \operatorname {arccsc}\left (c x\right ) + {\left (6 \, b c^{4} e x^{4} + 30 \, b c^{2} d + {\left (15 \, b c^{4} d + 8 \, b c^{2} e\right )} x^{2} + 16 \, b e\right )} \sqrt {c^{2} x^{2} - 1}}{180 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="fricas")

[Out]

1/180*(30*a*c^6*e*x^6 + 45*a*c^6*d*x^4 + 15*(2*b*c^6*e*x^6 + 3*b*c^6*d*x^4)*arccsc(c*x) + (6*b*c^4*e*x^4 + 30*
b*c^2*d + (15*b*c^4*d + 8*b*c^2*e)*x^2 + 16*b*e)*sqrt(c^2*x^2 - 1))/c^6

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giac [B]  time = 0.26, size = 920, normalized size = 6.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="giac")

[Out]

1/5760*(15*b*x^6*(sqrt(-1/(c^2*x^2) + 1) + 1)^6*arcsin(1/(c*x))*e/c + 15*a*x^6*(sqrt(-1/(c^2*x^2) + 1) + 1)^6*
e/c + 90*b*d*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*arcsin(1/(c*x))/c + 6*b*x^5*(sqrt(-1/(c^2*x^2) + 1) + 1)^5*e/c
^2 + 90*a*d*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4/c + 90*b*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*arcsin(1/(c*x))*e/c
^3 + 90*a*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*e/c^3 + 60*b*d*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3/c^2 + 360*b*d*x
^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*arcsin(1/(c*x))/c^3 + 50*b*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3*e/c^4 + 360*a*
d*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2/c^3 + 225*b*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*arcsin(1/(c*x))*e/c^5 + 22
5*a*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*e/c^5 + 540*b*d*x*(sqrt(-1/(c^2*x^2) + 1) + 1)/c^4 + 540*b*d*arcsin(1/(
c*x))/c^5 + 300*b*x*(sqrt(-1/(c^2*x^2) + 1) + 1)*e/c^6 + 540*a*d/c^5 + 300*b*arcsin(1/(c*x))*e/c^7 + 300*a*e/c
^7 - 540*b*d/(c^6*x*(sqrt(-1/(c^2*x^2) + 1) + 1)) + 360*b*d*arcsin(1/(c*x))/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1) +
1)^2) - 300*b*e/(c^8*x*(sqrt(-1/(c^2*x^2) + 1) + 1)) + 360*a*d/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) + 225
*b*arcsin(1/(c*x))*e/(c^9*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) + 225*a*e/(c^9*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^
2) - 60*b*d/(c^8*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3) + 90*b*d*arcsin(1/(c*x))/(c^9*x^4*(sqrt(-1/(c^2*x^2) + 1)
+ 1)^4) - 50*b*e/(c^10*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3) + 90*a*d/(c^9*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4)
+ 90*b*arcsin(1/(c*x))*e/(c^11*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4) + 90*a*e/(c^11*x^4*(sqrt(-1/(c^2*x^2) + 1)
+ 1)^4) - 6*b*e/(c^12*x^5*(sqrt(-1/(c^2*x^2) + 1) + 1)^5) + 15*b*arcsin(1/(c*x))*e/(c^13*x^6*(sqrt(-1/(c^2*x^2
) + 1) + 1)^6) + 15*a*e/(c^13*x^6*(sqrt(-1/(c^2*x^2) + 1) + 1)^6))*c

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maple [A]  time = 0.05, size = 134, normalized size = 0.88 \[ \frac {\frac {a \left (\frac {1}{6} e \,c^{6} x^{6}+\frac {1}{4} x^{4} c^{6} d \right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsc}\left (c x \right ) e \,c^{6} x^{6}}{6}+\frac {\mathrm {arccsc}\left (c x \right ) c^{6} x^{4} d}{4}+\frac {\left (c^{2} x^{2}-1\right ) \left (6 c^{4} e \,x^{4}+15 c^{4} d \,x^{2}+8 c^{2} e \,x^{2}+30 c^{2} d +16 e \right )}{180 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)*(a+b*arccsc(c*x)),x)

[Out]

1/c^4*(a/c^2*(1/6*e*c^6*x^6+1/4*x^4*c^6*d)+b/c^2*(1/6*arccsc(c*x)*e*c^6*x^6+1/4*arccsc(c*x)*c^6*x^4*d+1/180*(c
^2*x^2-1)*(6*c^4*e*x^4+15*c^4*d*x^2+8*c^2*e*x^2+30*c^2*d+16*e)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))

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maxima [A]  time = 0.33, size = 142, normalized size = 0.93 \[ \frac {1}{6} \, a e x^{6} + \frac {1}{4} \, a d x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arccsc}\left (c x\right ) + \frac {c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b d + \frac {1}{90} \, {\left (15 \, x^{6} \operatorname {arccsc}\left (c x\right ) + \frac {3 \, c^{4} x^{5} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} + 10 \, c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{5}}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)*(a+b*arccsc(c*x)),x, algorithm="maxima")

[Out]

1/6*a*e*x^6 + 1/4*a*d*x^4 + 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*d + 1/90*(15*x^6*arccsc(c*x) + (3*c^4*x^5*(-1/(c^2*x^2) + 1)^(5/2) + 10*c^2*x^3*(-1/(c^2*x^2) +
1)^(3/2) + 15*x*sqrt(-1/(c^2*x^2) + 1))/c^5)*b*e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.55, size = 272, normalized size = 1.78 \[ \frac {a d x^{4}}{4} + \frac {a e x^{6}}{6} + \frac {b d x^{4} \operatorname {acsc}{\left (c x \right )}}{4} + \frac {b e x^{6} \operatorname {acsc}{\left (c x \right )}}{6} + \frac {b d \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} - 1}}{3 c} + \frac {2 \sqrt {c^{2} x^{2} - 1}}{3 c^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {2 i \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} & \text {otherwise} \end {cases}\right )}{4 c} + \frac {b e \left (\begin {cases} \frac {x^{4} \sqrt {c^{2} x^{2} - 1}}{5 c} + \frac {4 x^{2} \sqrt {c^{2} x^{2} - 1}}{15 c^{3}} + \frac {8 \sqrt {c^{2} x^{2} - 1}}{15 c^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{4} \sqrt {- c^{2} x^{2} + 1}}{5 c} + \frac {4 i x^{2} \sqrt {- c^{2} x^{2} + 1}}{15 c^{3}} + \frac {8 i \sqrt {- c^{2} x^{2} + 1}}{15 c^{5}} & \text {otherwise} \end {cases}\right )}{6 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)*(a+b*acsc(c*x)),x)

[Out]

a*d*x**4/4 + a*e*x**6/6 + b*d*x**4*acsc(c*x)/4 + b*e*x**6*acsc(c*x)/6 + b*d*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) + b*e*Piecewise((x**4*sqrt(c**2*x**2 - 1)/(5*c) + 4*x**2*sqrt(c**2*x**2
- 1)/(15*c**3) + 8*sqrt(c**2*x**2 - 1)/(15*c**5), Abs(c**2*x**2) > 1), (I*x**4*sqrt(-c**2*x**2 + 1)/(5*c) + 4*
I*x**2*sqrt(-c**2*x**2 + 1)/(15*c**3) + 8*I*sqrt(-c**2*x**2 + 1)/(15*c**5), True))/(6*c)

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