\(\int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx\) [268]

Optimal result

Integrand size = 22, antiderivative size = 219 \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b} \]

[Out]

4*I*d*(d*x+c)*arctan(exp(I*(b*x+a)))/b^2-2*(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b+2*I*d*(d*x+c)*polylog(2,-exp(I*
(b*x+a)))/b^2-2*I*d^2*polylog(2,-I*exp(I*(b*x+a)))/b^3+2*I*d^2*polylog(2,I*exp(I*(b*x+a)))/b^3-2*I*d*(d*x+c)*p
olylog(2,exp(I*(b*x+a)))/b^2-2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2*polylog(3,exp(I*(b*x+a)))/b^3+(d*x+c)^
2*sec(b*x+a)/b

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.682, Rules used = {2702, 327, 213, 4505, 6873, 12, 6874, 6408, 4268, 2611, 2320, 6724, 4266, 2317, 2438} \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b} \]

[In]

Int[(c + d*x)^2*Csc[a + b*x]*Sec[a + b*x]^2,x]

[Out]

((4*I)*d*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 - (2*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b + ((2*I)*d*(c + d
*x)*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 + ((2*I)*d^2*PolyLog[
2, I*E^(I*(a + b*x))])/b^3 - ((2*I)*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 - (2*d^2*PolyLog[3, -E^(I*(a
+ b*x))])/b^3 + (2*d^2*PolyLog[3, E^(I*(a + b*x))])/b^3 + ((c + d*x)^2*Sec[a + b*x])/b

Rule 12

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

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 2317

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 2320

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 2438

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 2611

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 2702

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 4266

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 4268

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 4505

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 6408

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 6724

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 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = -\frac {(c+d x)^2 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}-(2 d) \int (c+d x) \left (-\frac {\text {arctanh}(\cos (a+b x))}{b}+\frac {\sec (a+b x)}{b}\right ) \, dx \\ & = -\frac {(c+d x)^2 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}-(2 d) \int \frac {(c+d x) (-\text {arctanh}(\cos (a+b x))+\sec (a+b x))}{b} \, dx \\ & = -\frac {(c+d x)^2 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {(2 d) \int (c+d x) (-\text {arctanh}(\cos (a+b x))+\sec (a+b x)) \, dx}{b} \\ & = -\frac {(c+d x)^2 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {(2 d) \int (-((c+d x) \text {arctanh}(\cos (a+b x)))+(c+d x) \sec (a+b x)) \, dx}{b} \\ & = -\frac {(c+d x)^2 \text {arctanh}(\cos (a+b x))}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}+\frac {(2 d) \int (c+d x) \text {arctanh}(\cos (a+b x)) \, dx}{b}-\frac {(2 d) \int (c+d x) \sec (a+b x) \, dx}{b} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}+\frac {\int b (c+d x)^2 \csc (a+b x) \, dx}{b}+\frac {\left (2 d^2\right ) \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac {\left (2 d^2\right ) \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^2} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac {\left (2 i d^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\int (c+d x)^2 \csc (a+b x) \, dx \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {(2 d) \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac {(2 d) \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {\left (2 i d^2\right ) \int \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (2 i d^2\right ) \int \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right ) \, dx}{b^2} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3} \\ & = \frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.45 \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\frac {-4 b c d \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )-4 d^2 \arctan (\cot (a)) \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )+b^2 (c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-b^2 (c+d x)^2 \log \left (1+e^{i (a+b x)}\right )+\frac {2 d^2 \csc (a) \left ((b x-\arctan (\cot (a))) \left (\log \left (1-e^{i (b x-\arctan (\cot (a)))}\right )-\log \left (1+e^{i (b x-\arctan (\cot (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x-\arctan (\cot (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {\csc ^2(a)}}+2 i d \left (b (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )\right )+2 d \left (-i b (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )+d \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )\right )+b^2 (c+d x)^2 \sec (a+b x)}{b^3} \]

[In]

Integrate[(c + d*x)^2*Csc[a + b*x]*Sec[a + b*x]^2,x]

[Out]

(-4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] - 4*d^2*ArcTan[Cot[a]]*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] +
b^2*(c + d*x)^2*Log[1 - E^(I*(a + b*x))] - b^2*(c + d*x)^2*Log[1 + E^(I*(a + b*x))] + (2*d^2*Csc[a]*((b*x - A
rcTan[Cot[a]])*(Log[1 - E^(I*(b*x - ArcTan[Cot[a]]))] - Log[1 + E^(I*(b*x - ArcTan[Cot[a]]))]) + I*PolyLog[2,
-E^(I*(b*x - ArcTan[Cot[a]]))] - I*PolyLog[2, E^(I*(b*x - ArcTan[Cot[a]]))]))/Sqrt[Csc[a]^2] + (2*I)*d*(b*(c +
d*x)*PolyLog[2, -E^(I*(a + b*x))] + I*d*PolyLog[3, -E^(I*(a + b*x))]) + 2*d*((-I)*b*(c + d*x)*PolyLog[2, E^(I
*(a + b*x))] + d*PolyLog[3, E^(I*(a + b*x))]) + b^2*(c + d*x)^2*Sec[a + b*x])/b^3

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (196 ) = 392\).

Time = 1.65 (sec) , antiderivative size = 568, normalized size of antiderivative = 2.59

method	result	size
risch	\(-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}-\frac {2 i d c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b}+\frac {2 d c \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{2}}+\frac {2 d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b}-\frac {2 d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {c^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x^{2}}{b}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a^{2}}{b^{3}}-\frac {2 d c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{b}+\frac {2 d^{2} \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {2 \,{\mathrm e}^{i \left (x b +a \right )} \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}+\frac {2 d^{2} \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{3}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{3}}+\frac {2 i d^{2} \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {4 i d^{2} a \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {2 i c d \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{2}}+\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {4 i d c \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}\)	\(568\)

[In]

int((d*x+c)^2*csc(b*x+a)*sec(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^3*d^2*a^2*ln(exp(I*(b*x+a))-1)+2*I/b^2*d^2*polylog(2,-exp(I*(b*x+a)))*x+2*I/b^2*c*d*polylog(2,-exp(I*(b*x+
a)))-2*I/b^2*d^2*polylog(2,exp(I*(b*x+a)))*x+4*I/b^2*d*c*arctan(exp(I*(b*x+a)))-4*I/b^3*d^2*a*arctan(exp(I*(b*
x+a)))-1/b*c^2*ln(exp(I*(b*x+a))+1)+1/b*c^2*ln(exp(I*(b*x+a))-1)-1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+1/b*d^2*ln
(1-exp(I*(b*x+a)))*x^2-1/b*d^2*ln(exp(I*(b*x+a))+1)*x^2+2*exp(I*(b*x+a))*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I*(b*x
+a))+1)+2*I/b^3*d^2*dilog(1-I*exp(I*(b*x+a)))-2/b*d*c*ln(exp(I*(b*x+a))+1)*x+2/b^2*d*c*ln(1-exp(I*(b*x+a)))*a-
2/b^2*c*d*a*ln(exp(I*(b*x+a))-1)+2/b*d*c*ln(1-exp(I*(b*x+a)))*x-2*I/b^2*d*c*polylog(2,exp(I*(b*x+a)))+2/b^2*d^
2*ln(1+I*exp(I*(b*x+a)))*x+2/b^3*d^2*ln(1+I*exp(I*(b*x+a)))*a-2/b^2*d^2*ln(1-I*exp(I*(b*x+a)))*x-2/b^3*d^2*ln(
1-I*exp(I*(b*x+a)))*a-2*I/b^3*d^2*dilog(1+I*exp(I*(b*x+a)))-2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2*polylog
(3,exp(I*(b*x+a)))/b^3

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1035 vs. \(2 (185) = 370\).

Time = 0.32 (sec) , antiderivative size = 1035, normalized size of antiderivative = 4.73 \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^2*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="fricas")

[Out]

1/2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 + 2*I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a)) + 2*I
*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 2*I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x
+ a)) - 2*I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + 2*d^2*cos(b*x + a)*polylog(3, cos(b*x + a
) + I*sin(b*x + a)) + 2*d^2*cos(b*x + a)*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 2*d^2*cos(b*x + a)*polylo
g(3, -cos(b*x + a) + I*sin(b*x + a)) - 2*d^2*cos(b*x + a)*polylog(3, -cos(b*x + a) - I*sin(b*x + a)) - 2*(I*b*
d^2*x + I*b*c*d)*cos(b*x + a)*dilog(cos(b*x + a) + I*sin(b*x + a)) - 2*(-I*b*d^2*x - I*b*c*d)*cos(b*x + a)*dil
og(cos(b*x + a) - I*sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d)*cos(b*x + a)*dilog(-cos(b*x + a) + I*sin(b*x + a))
- 2*(-I*b*d^2*x - I*b*c*d)*cos(b*x + a)*dilog(-cos(b*x + a) - I*sin(b*x + a)) - (b^2*d^2*x^2 + 2*b^2*c*d*x +
b^2*c^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - 2*(b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a)
+ I*sin(b*x + a) + I) - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a)
+ 1) + 2*(b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + I) - 2*(b*d^2*x + a*d^2)*cos(b*x + a
)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 2*(b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a)
+ 1) - 2*(b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 2*(b*d^2*x + a*d^2)*cos(b*x
+ a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cos(b*x + a)*log(-1/2*cos(b*x +
a) + 1/2*I*sin(b*x + a) + 1/2) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cos(b*x + a)*log(-1/2*cos(b*x + a) - 1/2*I*s
in(b*x + a) + 1/2) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(
b*x + a) + 1) - 2*(b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b^2*d^2*x^2 + 2*b^2*
c*d*x + 2*a*b*c*d - a^2*d^2)*cos(b*x + a)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) + 2*(b*c*d - a*d^2)*cos(b*x
+ a)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/(b^3*cos(b*x + a))

Sympy [F]

\[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\int \left (c + d x\right )^{2} \csc {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate((d*x+c)**2*csc(b*x+a)*sec(b*x+a)**2,x)

[Out]

Integral((c + d*x)**2*csc(a + b*x)*sec(a + b*x)**2, x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1590 vs. \(2 (185) = 370\).

Time = 0.47 (sec) , antiderivative size = 1590, normalized size of antiderivative = 7.26 \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^2*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="maxima")

[Out]

1/2*(c^2*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1)) - 2*a*c*d*(2/cos(b*x + a) - log(cos(
b*x + a) + 1) + log(cos(b*x + a) - 1))/b + a^2*d^2*(2/cos(b*x + a) - log(cos(b*x + a) + 1) + log(cos(b*x + a)
- 1))/b^2 + 2*(4*(b*c*d + (b*x + a)*d^2 - a*d^2 + (b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (I*b*c*d
+ I*(b*x + a)*d^2 - I*a*d^2)*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), sin(b*x + a) + 1) + 4*(b*c*d + (b*x + a)*
d^2 - a*d^2 + (b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (I*b*c*d + I*(b*x + a)*d^2 - I*a*d^2)*sin(2*b
*x + 2*a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + ((b*x
+ a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - (-I*(b*x + a)^2*d^2 + 2*(-I*b*c*d + I*a*d^2)*(b*
x + a))*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*
x + a) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - (-I*(b*x + a)^2*d^2 + 2*(-I*b*c*d
+ I*a*d^2)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 4*(-I*(b*x + a)^2*d^2 + 2*(
-I*b*c*d + I*a*d^2)*(b*x + a))*cos(b*x + a) + 4*(d^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) + d^2)*dilog(I*
e^(I*b*x + I*a)) - 4*(d^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) + d^2)*dilog(-I*e^(I*b*x + I*a)) + 4*(b*c*
d + (b*x + a)*d^2 - a*d^2 + (b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (I*b*c*d + I*(b*x + a)*d^2 - I*
a*d^2)*sin(2*b*x + 2*a))*dilog(-e^(I*b*x + I*a)) - 4*(b*c*d + (b*x + a)*d^2 - a*d^2 + (b*c*d + (b*x + a)*d^2 -
a*d^2)*cos(2*b*x + 2*a) - (-I*b*c*d - I*(b*x + a)*d^2 + I*a*d^2)*sin(2*b*x + 2*a))*dilog(e^(I*b*x + I*a)) - (
-I*(b*x + a)^2*d^2 - 2*(I*b*c*d - I*a*d^2)*(b*x + a) + (-I*(b*x + a)^2*d^2 - 2*(I*b*c*d - I*a*d^2)*(b*x + a))*
cos(2*b*x + 2*a) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(
b*x + a)^2 + 2*cos(b*x + a) + 1) - (I*(b*x + a)^2*d^2 - 2*(-I*b*c*d + I*a*d^2)*(b*x + a) + (I*(b*x + a)^2*d^2
- 2*(-I*b*c*d + I*a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*sin(2*b
*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 2*(I*b*c*d + I*(b*x + a)*d^2 - I*a*d^2
+ (I*b*c*d + I*(b*x + a)*d^2 - I*a*d^2)*cos(2*b*x + 2*a) - (b*c*d + (b*x + a)*d^2 - a*d^2)*sin(2*b*x + 2*a))*l
og(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + 2*(-I*b*c*d - I*(b*x + a)*d^2 + I*a*d^2 + (-I*b*c*d
- I*(b*x + a)*d^2 + I*a*d^2)*cos(2*b*x + 2*a) + (b*c*d + (b*x + a)*d^2 - a*d^2)*sin(2*b*x + 2*a))*log(cos(b*x
+ a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1) + 4*(I*d^2*cos(2*b*x + 2*a) - d^2*sin(2*b*x + 2*a) + I*d^2)*pol
ylog(3, -e^(I*b*x + I*a)) + 4*(-I*d^2*cos(2*b*x + 2*a) + d^2*sin(2*b*x + 2*a) - I*d^2)*polylog(3, e^(I*b*x + I
*a)) + 4*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*sin(b*x + a))/(-2*I*b^2*cos(2*b*x + 2*a) + 2*b^2*sin(
2*b*x + 2*a) - 2*I*b^2))/b

Giac [F]

\[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*x+c)^2*csc(b*x+a)*sec(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*csc(b*x + a)*sec(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \csc (a+b x) \sec ^2(a+b x) \, dx=\text {Hanged} \]

[In]

int((c + d*x)^2/(cos(a + b*x)^2*sin(a + b*x)),x)

[Out]

\text{Hanged}