3.286 \(\int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx\)
Optimal. Leaf size=235 \[ \frac{3 i x \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i x \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 i \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac{3 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{4 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{x \csc (a+b x)}{b^2}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b} \]
[Out]
((4*I)*x*ArcTan[E^(I*(a + b*x))])/b^2 - (3*x^2*ArcTanh[E^(I*(a + b*x))])/b - ArcTanh[Cos[a + b*x]]/b^3 - (x*Cs
c[a + b*x])/b^2 + ((3*I)*x*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 +
((2*I)*PolyLog[2, I*E^(I*(a + b*x))])/b^3 - ((3*I)*x*PolyLog[2, E^(I*(a + b*x))])/b^2 - (3*PolyLog[3, -E^(I*(a
+ b*x))])/b^3 + (3*PolyLog[3, E^(I*(a + b*x))])/b^3 + (3*x^2*Sec[a + b*x])/(2*b) - (x^2*Csc[a + b*x]^2*Sec[a
+ b*x])/(2*b)
________________________________________________________________________________________
Rubi [A] time = 0.53745, antiderivative size = 235, normalized size of antiderivative = 1.,
number of steps used = 29, number of rules used = 19, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.95, Rules used
= {2622, 288, 321, 207, 4420, 14, 6273, 12, 4183, 2531, 2282, 6589, 6742, 4181, 2279, 2391, 2621,
6271, 3770} \[ \frac{3 i x \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i x \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 i \text{PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac{3 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{4 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{x \csc (a+b x)}{b^2}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[x^2*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
[Out]
((4*I)*x*ArcTan[E^(I*(a + b*x))])/b^2 - (3*x^2*ArcTanh[E^(I*(a + b*x))])/b - ArcTanh[Cos[a + b*x]]/b^3 - (x*Cs
c[a + b*x])/b^2 + ((3*I)*x*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 +
((2*I)*PolyLog[2, I*E^(I*(a + b*x))])/b^3 - ((3*I)*x*PolyLog[2, E^(I*(a + b*x))])/b^2 - (3*PolyLog[3, -E^(I*(a
+ b*x))])/b^3 + (3*PolyLog[3, E^(I*(a + b*x))])/b^3 + (3*x^2*Sec[a + b*x])/(2*b) - (x^2*Csc[a + b*x]^2*Sec[a
+ b*x])/(2*b)
Rule 2622
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 4420
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Modul
e[{u = IntHide[Csc[a + b*x]^n*Sec[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)*u
, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]
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 6273
Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] && !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 4183
Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]
Rule 2531
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]
Rule 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 6589
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]
Rule 6742
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]
Rule 4181
Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Rule 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2391
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rule 2621
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rule 6271
Int[ArcTanh[u_], x_Symbol] :> Simp[x*ArcTanh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 - u^2), x], x] /; I
nverseFunctionFreeQ[u, x]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int x^2 \csc ^3(a+b x) \sec ^2(a+b x) \, dx &=-\frac{3 x^2 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}-2 \int x \left (-\frac{3 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 \sec (a+b x)}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b}\right ) \, dx\\ &=-\frac{3 x^2 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}-2 \int \left (-\frac{3 x \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{x \left (-3+\csc ^2(a+b x)\right ) \sec (a+b x)}{2 b}\right ) \, dx\\ &=-\frac{3 x^2 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{\int x \left (-3+\csc ^2(a+b x)\right ) \sec (a+b x) \, dx}{b}+\frac{3 \int x \tanh ^{-1}(\cos (a+b x)) \, dx}{b}\\ &=\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{\int \left (-3 x \sec (a+b x)+x \csc ^2(a+b x) \sec (a+b x)\right ) \, dx}{b}+\frac{3 \int b x^2 \csc (a+b x) \, dx}{2 b}\\ &=\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{3}{2} \int x^2 \csc (a+b x) \, dx+\frac{\int x \csc ^2(a+b x) \sec (a+b x) \, dx}{b}-\frac{3 \int x \sec (a+b x) \, dx}{b}\\ &=\frac{6 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{x \tanh ^{-1}(\sin (a+b x))}{b^2}-\frac{x \csc (a+b x)}{b^2}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{3 \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac{3 \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac{\int \left (\frac{\tanh ^{-1}(\sin (a+b x))}{b}-\frac{\csc (a+b x)}{b}\right ) \, dx}{b}-\frac{3 \int x \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac{3 \int x \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=\frac{6 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{x \tanh ^{-1}(\sin (a+b x))}{b^2}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}-\frac{(3 i) \int \text{Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{(3 i) \int \text{Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}-\frac{\int \tanh ^{-1}(\sin (a+b x)) \, dx}{b^2}+\frac{\int \csc (a+b x) \, dx}{b^2}\\ &=\frac{6 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{3 i \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac{\int b x \sec (a+b x) \, dx}{b^2}\\ &=\frac{6 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{3 i \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{\int x \sec (a+b x) \, dx}{b}\\ &=\frac{4 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{3 i \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}-\frac{\int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{4 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{3 i \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=\frac{4 i x \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 x^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b^3}-\frac{x \csc (a+b x)}{b^2}+\frac{3 i x \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{2 i \text{Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac{2 i \text{Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac{3 i x \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{3 \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{3 x^2 \sec (a+b x)}{2 b}-\frac{x^2 \csc ^2(a+b x) \sec (a+b x)}{2 b}\\ \end{align*}
Mathematica [B] time = 6.66329, size = 613, normalized size = 2.61 \[ \frac{6 i b x \text{PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))-6 i b x \text{PolyLog}(2,\cos (a+b x)+i \sin (a+b x))-6 \text{PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))+6 \text{PolyLog}(3,\cos (a+b x)+i \sin (a+b x))+3 b^2 x^2 \log (-i \sin (a+b x)-\cos (a+b x)+1)-3 b^2 x^2 \log (i \sin (a+b x)+\cos (a+b x)+1)+2 \log (-i \sin (a+b x)-\cos (a+b x)+1)-2 \log (i \sin (a+b x)+\cos (a+b x)+1)}{2 b^3}-\frac{2 \left (i \left (\text{PolyLog}\left (2,-e^{i \left (-a-b x+\frac{\pi }{2}\right )}\right )-\text{PolyLog}\left (2,e^{i \left (-a-b x+\frac{\pi }{2}\right )}\right )\right )+\left (-a-b x+\frac{\pi }{2}\right ) \left (\log \left (1-e^{i \left (-a-b x+\frac{\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-a-b x+\frac{\pi }{2}\right )}\right )\right )-\left (\frac{\pi }{2}-a\right ) \log \left (\tan \left (\frac{1}{2} \left (-a-b x+\frac{\pi }{2}\right )\right )\right )\right )}{b^3}+\frac{x \csc \left (\frac{a}{2}\right ) \sin \left (\frac{b x}{2}\right ) \csc \left (\frac{a}{2}+\frac{b x}{2}\right )}{2 b^2}-\frac{x \sec \left (\frac{a}{2}\right ) \sin \left (\frac{b x}{2}\right ) \sec \left (\frac{a}{2}+\frac{b x}{2}\right )}{2 b^2}+\frac{x \csc (a) \sec (a) (b x \sin (a)-\cos (a))}{b^2}-\frac{x^2 \csc ^2\left (\frac{a}{2}+\frac{b x}{2}\right )}{8 b}+\frac{x^2 \sec ^2\left (\frac{a}{2}+\frac{b x}{2}\right )}{8 b}+\frac{x^2 \sin \left (\frac{b x}{2}\right )}{b \left (\cos \left (\frac{a}{2}\right )-\sin \left (\frac{a}{2}\right )\right ) \left (\cos \left (\frac{a}{2}+\frac{b x}{2}\right )-\sin \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}-\frac{x^2 \sin \left (\frac{b x}{2}\right )}{b \left (\sin \left (\frac{a}{2}\right )+\cos \left (\frac{a}{2}\right )\right ) \left (\sin \left (\frac{a}{2}+\frac{b x}{2}\right )+\cos \left (\frac{a}{2}+\frac{b x}{2}\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*Csc[a + b*x]^3*Sec[a + b*x]^2,x]
[Out]
-(x^2*Csc[a/2 + (b*x)/2]^2)/(8*b) - (2*((-a + Pi/2 - b*x)*(Log[1 - E^(I*(-a + Pi/2 - b*x))] - Log[1 + E^(I*(-a
+ Pi/2 - b*x))]) - (-a + Pi/2)*Log[Tan[(-a + Pi/2 - b*x)/2]] + I*(PolyLog[2, -E^(I*(-a + Pi/2 - b*x))] - Poly
Log[2, E^(I*(-a + Pi/2 - b*x))])))/b^3 + (2*Log[1 - Cos[a + b*x] - I*Sin[a + b*x]] + 3*b^2*x^2*Log[1 - Cos[a +
b*x] - I*Sin[a + b*x]] - 2*Log[1 + Cos[a + b*x] + I*Sin[a + b*x]] - 3*b^2*x^2*Log[1 + Cos[a + b*x] + I*Sin[a
+ b*x]] + (6*I)*b*x*PolyLog[2, -Cos[a + b*x] - I*Sin[a + b*x]] - (6*I)*b*x*PolyLog[2, Cos[a + b*x] + I*Sin[a +
b*x]] - 6*PolyLog[3, -Cos[a + b*x] - I*Sin[a + b*x]] + 6*PolyLog[3, Cos[a + b*x] + I*Sin[a + b*x]])/(2*b^3) +
(x^2*Sec[a/2 + (b*x)/2]^2)/(8*b) + (x*Csc[a]*Sec[a]*(-Cos[a] + b*x*Sin[a]))/b^2 + (x*Csc[a/2]*Csc[a/2 + (b*x)
/2]*Sin[(b*x)/2])/(2*b^2) - (x*Sec[a/2]*Sec[a/2 + (b*x)/2]*Sin[(b*x)/2])/(2*b^2) + (x^2*Sin[(b*x)/2])/(b*(Cos[
a/2] - Sin[a/2])*(Cos[a/2 + (b*x)/2] - Sin[a/2 + (b*x)/2])) - (x^2*Sin[(b*x)/2])/(b*(Cos[a/2] + Sin[a/2])*(Cos
[a/2 + (b*x)/2] + Sin[a/2 + (b*x)/2]))
________________________________________________________________________________________
Maple [B] time = 0.402, size = 429, normalized size = 1.8 \begin{align*}{\frac{x \left ( 3\,bx{{\rm e}^{5\,i \left ( bx+a \right ) }}-2\,bx{{\rm e}^{3\,i \left ( bx+a \right ) }}-2\,i{{\rm e}^{5\,i \left ( bx+a \right ) }}+3\,bx{{\rm e}^{i \left ( bx+a \right ) }}+2\,i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}+{\frac{3\,{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{2\,{b}^{3}}}-{\frac{2\,i{\it dilog} \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{3\,{a}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}+{\frac{2\,i{\it dilog} \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{3\,ix{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+3\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}-2\,{\frac{\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}-{\frac{3\,ix{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{\ln \left ( 1+i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-2\,{\frac{\ln \left ( 1-i{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-{\frac{4\,ia\arctan \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{3\,\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{2\,b}}-{\frac{3\,\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{2\,b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}-{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{3}}}-3\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*csc(b*x+a)^3*sec(b*x+a)^2,x)
[Out]
x/b^2/(exp(2*I*(b*x+a))-1)^2/(exp(2*I*(b*x+a))+1)*(3*b*x*exp(5*I*(b*x+a))-2*b*x*exp(3*I*(b*x+a))-2*I*exp(5*I*(
b*x+a))+3*b*x*exp(I*(b*x+a))+2*I*exp(I*(b*x+a)))+3/2/b^3*a^2*ln(exp(I*(b*x+a))-1)-2*I/b^3*dilog(1+I*exp(I*(b*x
+a)))-3/2/b^3*a^2*ln(1-exp(I*(b*x+a)))+2*I/b^3*dilog(1-I*exp(I*(b*x+a)))+3*I*x*polylog(2,-exp(I*(b*x+a)))/b^2+
3*polylog(3,exp(I*(b*x+a)))/b^3+2/b^3*ln(1+I*exp(I*(b*x+a)))*a-2/b^3*ln(1-I*exp(I*(b*x+a)))*a-3*I*x*polylog(2,
exp(I*(b*x+a)))/b^2+2/b^2*ln(1+I*exp(I*(b*x+a)))*x-2/b^2*ln(1-I*exp(I*(b*x+a)))*x-4*I/b^3*a*arctan(exp(I*(b*x+
a)))+3/2/b*ln(1-exp(I*(b*x+a)))*x^2-3/2/b*ln(exp(I*(b*x+a))+1)*x^2+1/b^3*ln(exp(I*(b*x+a))-1)-1/b^3*ln(exp(I*(
b*x+a))+1)-3*polylog(3,-exp(I*(b*x+a)))/b^3
________________________________________________________________________________________
Maxima [B] time = 2.52132, size = 2996, normalized size = 12.75 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="maxima")
[Out]
1/4*(a^2*(2*(3*cos(b*x + a)^2 - 2)/(cos(b*x + a)^3 - cos(b*x + a)) - 3*log(cos(b*x + a) + 1) + 3*log(cos(b*x +
a) - 1)) + 4*((8*b*x*cos(6*b*x + 6*a) - 8*b*x*cos(4*b*x + 4*a) - 8*b*x*cos(2*b*x + 2*a) + 8*I*b*x*sin(6*b*x +
6*a) - 8*I*b*x*sin(4*b*x + 4*a) - 8*I*b*x*sin(2*b*x + 2*a) + 8*b*x)*arctan2(cos(b*x + a), sin(b*x + a) + 1) +
(8*b*x*cos(6*b*x + 6*a) - 8*b*x*cos(4*b*x + 4*a) - 8*b*x*cos(2*b*x + 2*a) + 8*I*b*x*sin(6*b*x + 6*a) - 8*I*b*
x*sin(4*b*x + 4*a) - 8*I*b*x*sin(2*b*x + 2*a) + 8*b*x)*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - (6*(b*x + a)
^2 - 12*(b*x + a)*a + 2*(3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*cos(6*b*x + 6*a) - 2*(3*(b*x + a)^2 - 6*(b*x + a)*
a + 2)*cos(4*b*x + 4*a) - 2*(3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*cos(2*b*x + 2*a) + (6*I*(b*x + a)^2 - 12*I*(b*
x + a)*a + 4*I)*sin(6*b*x + 6*a) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a - 4*I)*sin(4*b*x + 4*a) + (-6*I*(b*x +
a)^2 + 12*I*(b*x + a)*a - 4*I)*sin(2*b*x + 2*a) + 4)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (4*cos(6*b*x +
6*a) - 4*cos(4*b*x + 4*a) - 4*cos(2*b*x + 2*a) + 4*I*sin(6*b*x + 6*a) - 4*I*sin(4*b*x + 4*a) - 4*I*sin(2*b*x
+ 2*a) + 4)*arctan2(sin(b*x + a), cos(b*x + a) - 1) - (6*(b*x + a)^2 - 12*(b*x + a)*a + 6*((b*x + a)^2 - 2*(b*
x + a)*a)*cos(6*b*x + 6*a) - 6*((b*x + a)^2 - 2*(b*x + a)*a)*cos(4*b*x + 4*a) - 6*((b*x + a)^2 - 2*(b*x + a)*a
)*cos(2*b*x + 2*a) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a)*sin(6*b*x + 6*a) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a
)*a)*sin(4*b*x + 4*a) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), -cos(b*x
+ a) + 1) - (12*I*(b*x + a)^2 - 8*(b*x + a)*(3*I*a - 1) - 8*a)*cos(5*b*x + 5*a) - (-8*I*(b*x + a)^2 + 16*I*(b
*x + a)*a)*cos(3*b*x + 3*a) - (12*I*(b*x + a)^2 - 8*(b*x + a)*(3*I*a + 1) + 8*a)*cos(b*x + a) + (8*cos(6*b*x +
6*a) - 8*cos(4*b*x + 4*a) - 8*cos(2*b*x + 2*a) + 8*I*sin(6*b*x + 6*a) - 8*I*sin(4*b*x + 4*a) - 8*I*sin(2*b*x
+ 2*a) + 8)*dilog(I*e^(I*b*x + I*a)) - (8*cos(6*b*x + 6*a) - 8*cos(4*b*x + 4*a) - 8*cos(2*b*x + 2*a) + 8*I*sin
(6*b*x + 6*a) - 8*I*sin(4*b*x + 4*a) - 8*I*sin(2*b*x + 2*a) + 8)*dilog(-I*e^(I*b*x + I*a)) + (12*b*x*cos(6*b*x
+ 6*a) - 12*b*x*cos(4*b*x + 4*a) - 12*b*x*cos(2*b*x + 2*a) + 12*I*b*x*sin(6*b*x + 6*a) - 12*I*b*x*sin(4*b*x +
4*a) - 12*I*b*x*sin(2*b*x + 2*a) + 12*b*x)*dilog(-e^(I*b*x + I*a)) - (12*b*x*cos(6*b*x + 6*a) - 12*b*x*cos(4*
b*x + 4*a) - 12*b*x*cos(2*b*x + 2*a) + 12*I*b*x*sin(6*b*x + 6*a) - 12*I*b*x*sin(4*b*x + 4*a) - 12*I*b*x*sin(2*
b*x + 2*a) + 12*b*x)*dilog(e^(I*b*x + I*a)) - (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a + (-3*I*(b*x + a)^2 + 6*I*(b
*x + a)*a - 2*I)*cos(6*b*x + 6*a) + (3*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 2*I)*cos(4*b*x + 4*a) + (3*I*(b*x + a
)^2 - 6*I*(b*x + a)*a + 2*I)*cos(2*b*x + 2*a) + (3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*sin(6*b*x + 6*a) - (3*(b*x
+ a)^2 - 6*(b*x + a)*a + 2)*sin(4*b*x + 4*a) - (3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*sin(2*b*x + 2*a) - 2*I)*lo
g(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (3*I*(b*x + a)^2 - 6*I*(b*x + a)*a + (3*I*(b*x + a)^
2 - 6*I*(b*x + a)*a + 2*I)*cos(6*b*x + 6*a) + (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a - 2*I)*cos(4*b*x + 4*a) + (-
3*I*(b*x + a)^2 + 6*I*(b*x + a)*a - 2*I)*cos(2*b*x + 2*a) - (3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*sin(6*b*x + 6*
a) + (3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*sin(4*b*x + 4*a) + (3*(b*x + a)^2 - 6*(b*x + a)*a + 2)*sin(2*b*x + 2*
a) + 2*I)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (-4*I*b*x*cos(6*b*x + 6*a) + 4*I*b*x*cos
(4*b*x + 4*a) + 4*I*b*x*cos(2*b*x + 2*a) + 4*b*x*sin(6*b*x + 6*a) - 4*b*x*sin(4*b*x + 4*a) - 4*b*x*sin(2*b*x +
2*a) - 4*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - (4*I*b*x*cos(6*b*x + 6*a) - 4*I*b
*x*cos(4*b*x + 4*a) - 4*I*b*x*cos(2*b*x + 2*a) - 4*b*x*sin(6*b*x + 6*a) + 4*b*x*sin(4*b*x + 4*a) + 4*b*x*sin(2
*b*x + 2*a) + 4*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) - (-12*I*cos(6*b*x + 6*a) + 1
2*I*cos(4*b*x + 4*a) + 12*I*cos(2*b*x + 2*a) + 12*sin(6*b*x + 6*a) - 12*sin(4*b*x + 4*a) - 12*sin(2*b*x + 2*a)
- 12*I)*polylog(3, -e^(I*b*x + I*a)) - (12*I*cos(6*b*x + 6*a) - 12*I*cos(4*b*x + 4*a) - 12*I*cos(2*b*x + 2*a)
- 12*sin(6*b*x + 6*a) + 12*sin(4*b*x + 4*a) + 12*sin(2*b*x + 2*a) + 12*I)*polylog(3, e^(I*b*x + I*a)) + (12*(
b*x + a)^2 - (b*x + a)*(24*a + 8*I) + 8*I*a)*sin(5*b*x + 5*a) - 8*((b*x + a)^2 - 2*(b*x + a)*a)*sin(3*b*x + 3*
a) + (12*(b*x + a)^2 - (b*x + a)*(24*a - 8*I) - 8*I*a)*sin(b*x + a))/(-4*I*cos(6*b*x + 6*a) + 4*I*cos(4*b*x +
4*a) + 4*I*cos(2*b*x + 2*a) + 4*sin(6*b*x + 6*a) - 4*sin(4*b*x + 4*a) - 4*sin(2*b*x + 2*a) - 4*I))/b^3
________________________________________________________________________________________
Fricas [C] time = 0.770386, size = 3302, normalized size = 14.05 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="fricas")
[Out]
1/4*(6*b^2*x^2*cos(b*x + a)^2 - 4*b^2*x^2 + 4*b*x*cos(b*x + a)*sin(b*x + a) + (-6*I*b*x*cos(b*x + a)^3 + 6*I*b
*x*cos(b*x + a))*dilog(cos(b*x + a) + I*sin(b*x + a)) + (6*I*b*x*cos(b*x + a)^3 - 6*I*b*x*cos(b*x + a))*dilog(
cos(b*x + a) - I*sin(b*x + a)) + (4*I*cos(b*x + a)^3 - 4*I*cos(b*x + a))*dilog(I*cos(b*x + a) + sin(b*x + a))
+ (4*I*cos(b*x + a)^3 - 4*I*cos(b*x + a))*dilog(I*cos(b*x + a) - sin(b*x + a)) + (-4*I*cos(b*x + a)^3 + 4*I*co
s(b*x + a))*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-4*I*cos(b*x + a)^3 + 4*I*cos(b*x + a))*dilog(-I*cos(b*x
+ a) - sin(b*x + a)) + (-6*I*b*x*cos(b*x + a)^3 + 6*I*b*x*cos(b*x + a))*dilog(-cos(b*x + a) + I*sin(b*x + a))
+ (6*I*b*x*cos(b*x + a)^3 - 6*I*b*x*cos(b*x + a))*dilog(-cos(b*x + a) - I*sin(b*x + a)) - ((3*b^2*x^2 + 2)*cos
(b*x + a)^3 - (3*b^2*x^2 + 2)*cos(b*x + a))*log(cos(b*x + a) + I*sin(b*x + a) + 1) + 4*(a*cos(b*x + a)^3 - a*c
os(b*x + a))*log(cos(b*x + a) + I*sin(b*x + a) + I) - ((3*b^2*x^2 + 2)*cos(b*x + a)^3 - (3*b^2*x^2 + 2)*cos(b*
x + a))*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 4*(a*cos(b*x + a)^3 - a*cos(b*x + a))*log(cos(b*x + a) - I*si
n(b*x + a) + I) - 4*((b*x + a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(I*cos(b*x + a) + sin(b*x + a) + 1)
+ 4*((b*x + a)*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(I*cos(b*x + a) - sin(b*x + a) + 1) - 4*((b*x + a)
*cos(b*x + a)^3 - (b*x + a)*cos(b*x + a))*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 4*((b*x + a)*cos(b*x + a)^
3 - (b*x + a)*cos(b*x + a))*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + ((3*a^2 + 2)*cos(b*x + a)^3 - (3*a^2 + 2
)*cos(b*x + a))*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + ((3*a^2 + 2)*cos(b*x + a)^3 - (3*a^2 + 2)*
cos(b*x + a))*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) + 3*((b^2*x^2 - a^2)*cos(b*x + a)^3 - (b^2*x^2
- a^2)*cos(b*x + a))*log(-cos(b*x + a) + I*sin(b*x + a) + 1) + 4*(a*cos(b*x + a)^3 - a*cos(b*x + a))*log(-cos
(b*x + a) + I*sin(b*x + a) + I) + 3*((b^2*x^2 - a^2)*cos(b*x + a)^3 - (b^2*x^2 - a^2)*cos(b*x + a))*log(-cos(b
*x + a) - I*sin(b*x + a) + 1) - 4*(a*cos(b*x + a)^3 - a*cos(b*x + a))*log(-cos(b*x + a) - I*sin(b*x + a) + I)
+ 6*(cos(b*x + a)^3 - cos(b*x + a))*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 6*(cos(b*x + a)^3 - cos(b*x +
a))*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 6*(cos(b*x + a)^3 - cos(b*x + a))*polylog(3, -cos(b*x + a) + I
*sin(b*x + a)) - 6*(cos(b*x + a)^3 - cos(b*x + a))*polylog(3, -cos(b*x + a) - I*sin(b*x + a)))/(b^3*cos(b*x +
a)^3 - b^3*cos(b*x + a))
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*csc(b*x+a)**3*sec(b*x+a)**2,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*csc(b*x+a)^3*sec(b*x+a)^2,x, algorithm="giac")
[Out]
integrate(x^2*csc(b*x + a)^3*sec(b*x + a)^2, x)