3.312 \(\int (c+d x)^2 \csc (a+b x) \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=201 \[ -\frac {d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac {c d x}{b}+\frac {d^2 x^2}{2 b} \]

[Out]

c*d*x/b+1/2*d^2*x^2/b-2*(d*x+c)^2*arctanh(exp(2*I*(b*x+a)))/b-d^2*ln(cos(b*x+a))/b^3+I*d*(d*x+c)*polylog(2,-ex
p(2*I*(b*x+a)))/b^2-I*d*(d*x+c)*polylog(2,exp(2*I*(b*x+a)))/b^2-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/2*d
^2*polylog(3,exp(2*I*(b*x+a)))/b^3-d*(d*x+c)*tan(b*x+a)/b^2+1/2*(d*x+c)^2*tan(b*x+a)^2/b

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Rubi [A]  time = 0.41, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {2620, 14, 4420, 6741, 12, 6742, 2551, 4183, 2531, 2282, 6589, 3720, 3475} \[ \frac {i d (c+d x) \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \tan (a+b x)}{b^2}-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac {c d x}{b}+\frac {d^2 x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*d*x)/b + (d^2*x^2)/(2*b) - (2*(c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/b - (d^2*Log[Cos[a + b*x]])/b^3 + (
I*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 - (d
^2*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^3) + (d^2*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3) - (d*(c + d*x)*Ta
n[a + b*x])/b^2 + ((c + d*x)^2*Tan[a + b*x]^2)/(2*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 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 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 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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

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 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 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 6741

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

Rule 6742

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

Rubi steps

\begin {align*} \int (c+d x)^2 \csc (a+b x) \sec ^3(a+b x) \, dx &=\frac {(c+d x)^2 \log (\tan (a+b x))}{b}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-(2 d) \int (c+d x) \left (\frac {\log (\tan (a+b x))}{b}+\frac {\tan ^2(a+b x)}{2 b}\right ) \, dx\\ &=\frac {(c+d x)^2 \log (\tan (a+b x))}{b}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-(2 d) \int \frac {(c+d x) \left (2 \log (\tan (a+b x))+\tan ^2(a+b x)\right )}{2 b} \, dx\\ &=\frac {(c+d x)^2 \log (\tan (a+b x))}{b}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \left (2 \log (\tan (a+b x))+\tan ^2(a+b x)\right ) \, dx}{b}\\ &=\frac {(c+d x)^2 \log (\tan (a+b x))}{b}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {d \int \left (2 (c+d x) \log (\tan (a+b x))+(c+d x) \tan ^2(a+b x)\right ) \, dx}{b}\\ &=\frac {(c+d x)^2 \log (\tan (a+b x))}{b}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {d \int (c+d x) \tan ^2(a+b x) \, dx}{b}-\frac {(2 d) \int (c+d x) \log (\tan (a+b x)) \, dx}{b}\\ &=-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}+\frac {\int 2 b (c+d x)^2 \csc (2 a+2 b x) \, dx}{b}+\frac {d \int (c+d x) \, dx}{b}+\frac {d^2 \int \tan (a+b x) \, dx}{b^2}\\ &=\frac {c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {d^2 \log (\cos (a+b x))}{b^3}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}+2 \int (c+d x)^2 \csc (2 a+2 b x) \, dx\\ &=\frac {c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d^2 \log (\cos (a+b x))}{b^3}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {(2 d) \int (c+d x) \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac {(2 d) \int (c+d x) \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=\frac {c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {\left (i d^2\right ) \int \text {Li}_2\left (-e^{i (2 a+2 b x)}\right ) \, dx}{b^2}+\frac {\left (i d^2\right ) \int \text {Li}_2\left (e^{i (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=\frac {c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}\\ &=\frac {c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d^2 \log (\cos (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \tan (a+b x)}{b^2}+\frac {(c+d x)^2 \tan ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [B]  time = 6.76, size = 875, normalized size = 4.35 \[ -\frac {\sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a)) c^2}{b \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac {\csc (a) (\log (\cos (b x) \sin (a)+\cos (a) \sin (b x)) \sin (a)-b x \cos (a)) c^2}{b \left (\cos ^2(a)+\sin ^2(a)\right )}-\frac {d \csc (a) \sec (a) \left (b^2 e^{i \tan ^{-1}(\tan (a))} x^2+\frac {\left (i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-\pi \log \left (1+e^{-2 i b x}\right )-2 \left (b x+\tan ^{-1}(\tan (a))\right ) \log \left (1-e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+\pi \log (\cos (b x))+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (b x+\tan ^{-1}(\tan (a))\right )\right )+i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )\right ) \tan (a)}{\sqrt {\tan ^2(a)+1}}\right ) c}{b^2 \sqrt {\sec ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}-\frac {d \csc (a) \left (b^2 e^{-i \tan ^{-1}(\cot (a))} x^2-\frac {\cot (a) \left (i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-\pi \log \left (1+e^{-2 i b x}\right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+\pi \log (\cos (b x))-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )+i \text {Li}_2\left (e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )\right )}{\sqrt {\cot ^2(a)+1}}\right ) \sec (a) c}{b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}+\frac {(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac {d^2 e^{i a} \csc (a) \left (2 b^3 e^{-2 i a} x^3+3 i b^2 \left (1-e^{-2 i a}\right ) \log \left (1-e^{-i (a+b x)}\right ) x^2+3 i b^2 \left (1-e^{-2 i a}\right ) \log \left (1+e^{-i (a+b x)}\right ) x^2-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (-e^{-i (a+b x)}\right )-i \text {Li}_3\left (-e^{-i (a+b x)}\right )\right )-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (e^{-i (a+b x)}\right )-i \text {Li}_3\left (e^{-i (a+b x)}\right )\right )\right )}{6 b^3}+\frac {1}{3} x \left (3 c^2+3 d x c+d^2 x^2\right ) \csc (a) \sec (a)-\frac {i d^2 e^{-i a} \left (2 b^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right ) x^2+6 b \left (1+e^{2 i a}\right ) \text {Li}_2\left (-e^{-2 i (a+b x)}\right ) x-3 i \left (1+e^{2 i a}\right ) \text {Li}_3\left (-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{12 b^3}+\frac {\sec (a) \sec (a+b x) \left (-x \sin (b x) d^2-c \sin (b x) d\right )}{b^2}-\frac {d^2 \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a))}{b^3 \left (\cos ^2(a)+\sin ^2(a)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(d^2*E^(I*a)*Csc[a]*((2*b^3*x^3)/E^((2*I)*a) + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2*Log[1 - E^((-I)*(a + b*x)
)] + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2*Log[1 + E^((-I)*(a + b*x))] - (6*(-1 + E^((2*I)*a))*(b*x*PolyLog[2, -E^(
(-I)*(a + b*x))] - I*PolyLog[3, -E^((-I)*(a + b*x))]))/E^((2*I)*a) - (6*(-1 + E^((2*I)*a))*(b*x*PolyLog[2, E^(
(-I)*(a + b*x))] - I*PolyLog[3, E^((-I)*(a + b*x))]))/E^((2*I)*a)))/b^3 + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Csc[a
]*Sec[a])/3 - ((I/12)*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + 6*b*(1
+ E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))]
)*Sec[a])/(b^3*E^(I*a)) + ((c + d*x)^2*Sec[a + b*x]^2)/(2*b) - (c^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a
]*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) - (d^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*
x]] + b*x*Sin[a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) + (c^2*Csc[a]*(-(b*x*Cos[a]) + Log[Cos[b*x]*Sin[a] + Cos[a]*Sin
[b*x]]*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) - (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi
- 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot
[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - Ar
cTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + (Sec[a]*Sec[a + b*x
]*(-(c*d*Sin[b*x]) - d^2*x*Sin[b*x]))/b^2 - (c*d*Csc[a]*Sec[a]*(b^2*E^(I*ArcTan[Tan[a]])*x^2 + ((I*b*x*(-Pi +
2*ArcTan[Tan[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a
]]))] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + I*PolyLog[2, E^((2*I)*(b*x + ArcT
an[Tan[a]]))])*Tan[a])/Sqrt[1 + Tan[a]^2]))/(b^2*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]^2)])

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fricas [C]  time = 0.65, size = 1396, normalized size = 6.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*d^2*x^2 + 2*b^2*c*d*x + 2*d^2*cos(b*x + a)^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 2*d^2*cos(b*
x + a)^2*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 2*d^2*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) + sin(b*x
+ a)) - 2*d^2*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 2*d^2*cos(b*x + a)^2*polylog(3, -I*co
s(b*x + a) + sin(b*x + a)) - 2*d^2*cos(b*x + a)^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) + 2*d^2*cos(b*x +
a)^2*polylog(3, -cos(b*x + a) + I*sin(b*x + a)) + 2*d^2*cos(b*x + a)^2*polylog(3, -cos(b*x + a) - I*sin(b*x +
a)) + b^2*c^2 + (-2*I*b*d^2*x - 2*I*b*c*d)*cos(b*x + a)^2*dilog(cos(b*x + a) + I*sin(b*x + a)) + (2*I*b*d^2*x
+ 2*I*b*c*d)*cos(b*x + a)^2*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*b*c*d)*cos(b*x + a)^2*
dilog(I*cos(b*x + a) + sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x
+ a)) + (2*I*b*d^2*x + 2*I*b*c*d)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*
b*c*d)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*cos(b*x + a)^2*dilog(-
cos(b*x + a) + I*sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*b*c*d)*cos(b*x + a)^2*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)^2*log(cos(b*x + a) + I*sin(b*x + a) + 1) - (b^2*c^2 -
2*a*b*c*d + (a^2 + 1)*d^2)*cos(b*x + a)^2*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)^2*log(cos(b*x + a) - I*sin(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + (a^2 + 1)*d^2)*cos(b
*x + a)^2*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)^2*log(I*cos(b*x + a) + sin(b*x + a) + 1) - (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)^
2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (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)^2*lo
g(-I*cos(b*x + a) + sin(b*x + a) + 1) - (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)^2*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)^2*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)^2*log(-1/2*cos(b*x + a) - 1/2*I*sin(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)^2*log(-cos(b*x + a) + I*sin(b*x
+ a) + 1) - (b^2*c^2 - 2*a*b*c*d + (a^2 + 1)*d^2)*cos(b*x + a)^2*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)^2*log(-cos(b*x + a) - I*sin(b*x + a) + 1) - (b^2*c
^2 - 2*a*b*c*d + (a^2 + 1)*d^2)*cos(b*x + a)^2*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 2*(b*d^2*x + b*c*d)*c
os(b*x + a)*sin(b*x + a))/(b^3*cos(b*x + a)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \csc \left (b x + a\right ) \sec \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2*csc(b*x + a)*sec(b*x + a)^3, x)

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maple [B]  time = 0.13, size = 614, normalized size = 3.05 \[ \frac {2 b \,d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}+4 b c d x \,{\mathrm e}^{2 i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b \,c^{2} {\mathrm e}^{2 i \left (b x +a \right )}-2 i d^{2} x -2 i c d}{b^{2} \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )^{2}}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}-\frac {c^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {2 d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {i d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {i c d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*csc(b*x+a)*sec(b*x+a)^3,x)

[Out]

-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/b^3*d^2*a^2*ln(exp(I*(b*x+a))-1)+1/b*d^2*ln(1-exp(I*(b*x+a)))*x^2-
1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+1/b*d^2*ln(exp(I*(b*x+a))+1)*x^2+2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2
*polylog(3,exp(I*(b*x+a)))/b^3-1/b*c^2*ln(1+exp(2*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*d^2*ln(1+exp(2*I*(b*x+a)))*x^2+2*(b*d^2*x^2*exp(2*I*(b*x+a))-I*d^2*x*exp(2*I*(b*x+a))+2*b*c*d*x*
exp(2*I*(b*x+a))-I*c*d*exp(2*I*(b*x+a))+b*c^2*exp(2*I*(b*x+a))-I*d^2*x-I*c*d)/b^2/(1+exp(2*I*(b*x+a)))^2+I/b^2
*d^2*polylog(2,-exp(2*I*(b*x+a)))*x+I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))-2/b*c*d*ln(1+exp(2*I*(b*x+a)))*x+2/
b*c*d*ln(1-exp(I*(b*x+a)))*x+2/b^2*c*d*ln(1-exp(I*(b*x+a)))*a+2/b*c*d*ln(exp(I*(b*x+a))+1)*x-2/b^2*c*d*a*ln(ex
p(I*(b*x+a))-1)-2*I/b^2*d^2*polylog(2,-exp(I*(b*x+a)))*x-2*I/b^2*d^2*polylog(2,exp(I*(b*x+a)))*x-2*I/b^2*c*d*p
olylog(2,-exp(I*(b*x+a)))-2*I/b^2*c*d*polylog(2,exp(I*(b*x+a)))-1/b^3*d^2*ln(1+exp(2*I*(b*x+a)))+2/b^3*d^2*ln(
exp(I*(b*x+a)))

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maxima [B]  time = 0.81, size = 2442, normalized size = 12.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c^2*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 2*a*c*d*(1/(sin(b*x + a)^
2 - 1) + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b + a^2*d^2*(1/(sin(b*x + a)^2 - 1) + log(sin(b*x + a)
^2 - 1) - log(sin(b*x + a)^2))/b^2 - 2*(4*(b*x + a)*d^2*cos(4*b*x + 4*a) + 4*I*(b*x + a)*d^2*sin(4*b*x + 4*a)
- 4*b*c*d + 4*a*d^2 - (2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*(b*x + a) + 2*d^2 + 2*((b*x + a)^2*d^2 + 2*(b*c*d
- a*d^2)*(b*x + a) + d^2)*cos(4*b*x + 4*a) + 4*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + d^2)*cos(2*b*
x + 2*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I*a*d^2)*(b*x + a) + 2*I*d^2)*sin(4*b*x + 4*a) + (4*I*(b*x +
a)^2*d^2 + (8*I*b*c*d - 8*I*a*d^2)*(b*x + a) + 4*I*d^2)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x
+ 2*a) + 1) + (2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*(b*x + a) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x +
a))*cos(4*b*x + 4*a) + 4*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - (-2*I*(b*x + a)^2
*d^2 + (-4*I*b*c*d + 4*I*a*d^2)*(b*x + a))*sin(4*b*x + 4*a) - (-4*I*(b*x + a)^2*d^2 + (-8*I*b*c*d + 8*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 + 4*(b*c*d - a*d^2)
*(b*x + a) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(4*b*x + 4*a) + 4*((b*x + a)^2*d^2 + 2*(b*c*
d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I*a*d^2)*(b*x + a))*sin(4*b*x +
4*a) + (4*I*(b*x + a)^2*d^2 + (8*I*b*c*d - 8*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 + 4*b*c*d - 4*a*d^2 + (8*I*b*c*d - 4*(2*I*a + 1)*d^2)*(b*x + a))*cos(2*b
*x + 2*a) + (2*b*c*d + 2*(b*x + a)*d^2 - 2*a*d^2 + 2*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) + 4*(b*c
*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (-2*I*b*c*d - 2*I*(b*x + a)*d^2 + 2*I*a*d^2)*sin(4*b*x + 4*a) -
(-4*I*b*c*d - 4*I*(b*x + a)*d^2 + 4*I*a*d^2)*sin(2*b*x + 2*a))*dilog(-e^(2*I*b*x + 2*I*a)) - (4*b*c*d + 4*(b*
x + a)*d^2 - 4*a*d^2 + 4*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) + 8*(b*c*d + (b*x + a)*d^2 - a*d^2)*
cos(2*b*x + 2*a) + (4*I*b*c*d + 4*I*(b*x + a)*d^2 - 4*I*a*d^2)*sin(4*b*x + 4*a) + (8*I*b*c*d + 8*I*(b*x + a)*d
^2 - 8*I*a*d^2)*sin(2*b*x + 2*a))*dilog(-e^(I*b*x + I*a)) - (4*b*c*d + 4*(b*x + a)*d^2 - 4*a*d^2 + 4*(b*c*d +
(b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) + 8*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (4*I*b*c*d + 4*
I*(b*x + a)*d^2 - 4*I*a*d^2)*sin(4*b*x + 4*a) + (8*I*b*c*d + 8*I*(b*x + a)*d^2 - 8*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 + 2*I*a*d^2)*(b*x + a) - I*d^2 + (-I*(b*x + a)^2*d^
2 + (-2*I*b*c*d + 2*I*a*d^2)*(b*x + a) - I*d^2)*cos(4*b*x + 4*a) + (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d + 4*I*a
*d^2)*(b*x + a) - 2*I*d^2)*cos(2*b*x + 2*a) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + d^2)*sin(4*b*x
+ 4*a) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + d^2)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + si
n(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (I*(b*x + a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)*(b*x + a) + (I*(b*x
+ a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)*(b*x + a))*cos(4*b*x + 4*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*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(4*b*x + 4*a) - 2*((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 - 2*I*a*d^2)*(b*x + a) + (I*(b*x + a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)
*(b*x + a))*cos(4*b*x + 4*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*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(4*b*x + 4*a) - 2*((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*d^2*cos(4*b*x + 4*a) -
2*I*d^2*cos(2*b*x + 2*a) + d^2*sin(4*b*x + 4*a) + 2*d^2*sin(2*b*x + 2*a) - I*d^2)*polylog(3, -e^(2*I*b*x + 2*I
*a)) - (4*I*d^2*cos(4*b*x + 4*a) + 8*I*d^2*cos(2*b*x + 2*a) - 4*d^2*sin(4*b*x + 4*a) - 8*d^2*sin(2*b*x + 2*a)
+ 4*I*d^2)*polylog(3, -e^(I*b*x + I*a)) - (4*I*d^2*cos(4*b*x + 4*a) + 8*I*d^2*cos(2*b*x + 2*a) - 4*d^2*sin(4*b
*x + 4*a) - 8*d^2*sin(2*b*x + 2*a) + 4*I*d^2)*polylog(3, e^(I*b*x + I*a)) + (4*(b*x + a)^2*d^2 - 4*I*b*c*d + 4
*I*a*d^2 + (8*b*c*d - (8*a - 4*I)*d^2)*(b*x + a))*sin(2*b*x + 2*a))/(-2*I*b^2*cos(4*b*x + 4*a) - 4*I*b^2*cos(2
*b*x + 2*a) + 2*b^2*sin(4*b*x + 4*a) + 4*b^2*sin(2*b*x + 2*a) - 2*I*b^2))/b

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/(cos(a + b*x)^3*sin(a + b*x)),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \csc {\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*csc(b*x+a)*sec(b*x+a)**3,x)

[Out]

Integral((c + d*x)**2*csc(a + b*x)*sec(a + b*x)**3, x)

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