3.383 \(\int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx\)

Optimal. Leaf size=242 \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b^2}-\frac{3 d^3 \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}+\frac{3 d^3 x}{2 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d} \]

[Out]

(3*d^3*x)/(2*b^3) - (c + d*x)^3/b - ((I/4)*(c + d*x)^4)/d + ((c + d*x)^3*Log[1 + E^((2*I)*(a + b*x))])/b - (((
3*I)/2)*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))
])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3*d^3*Cos[a + b*x]*Sin[a + b*x])/(2*b^4)
+ (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/b^2 - (3*d^2*(c + d*x)*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^3*Sin[
a + b*x]^2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.445514, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {4431, 4404, 3311, 32, 2635, 8, 4407, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b^2}-\frac{3 d^3 \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}+\frac{3 d^3 x}{2 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^3*x)/(2*b^3) - (c + d*x)^3/b - ((I/4)*(c + d*x)^4)/d + ((c + d*x)^3*Log[1 + E^((2*I)*(a + b*x))])/b - (((
3*I)/2)*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))
])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3*d^3*Cos[a + b*x]*Sin[a + b*x])/(2*b^4)
+ (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/b^2 - (3*d^2*(c + d*x)*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^3*Sin[
a + b*x]^2)/b

Rule 4431

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3311

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4407

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Int[
(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3719

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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]

Rubi steps

\begin{align*} \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^3 \cos (a+b x) \sin (a+b x)-(c+d x)^3 \sin ^2(a+b x) \tan (a+b x)\right ) \, dx\\ &=3 \int (c+d x)^3 \cos (a+b x) \sin (a+b x) \, dx-\int (c+d x)^3 \sin ^2(a+b x) \tan (a+b x) \, dx\\ &=\frac{3 (c+d x)^3 \sin ^2(a+b x)}{2 b}-\frac{(9 d) \int (c+d x)^2 \sin ^2(a+b x) \, dx}{2 b}+\int (c+d x)^3 \cos (a+b x) \sin (a+b x) \, dx-\int (c+d x)^3 \tan (a+b x) \, dx\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{9 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{9 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}+2 i \int \frac{e^{2 i (a+b x)} (c+d x)^3}{1+e^{2 i (a+b x)}} \, dx-\frac{(3 d) \int (c+d x)^2 \sin ^2(a+b x) \, dx}{2 b}-\frac{(9 d) \int (c+d x)^2 \, dx}{4 b}+\frac{\left (9 d^3\right ) \int \sin ^2(a+b x) \, dx}{4 b^3}\\ &=-\frac{3 (c+d x)^3}{4 b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{9 d^3 \cos (a+b x) \sin (a+b x)}{8 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}-\frac{(3 d) \int (c+d x)^2 \, dx}{4 b}-\frac{(3 d) \int (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}+\frac{\left (3 d^3\right ) \int \sin ^2(a+b x) \, dx}{4 b^3}+\frac{\left (9 d^3\right ) \int 1 \, dx}{8 b^3}\\ &=\frac{9 d^3 x}{8 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}+\frac{\left (3 i d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (3 d^3\right ) \int 1 \, dx}{8 b^3}\\ &=\frac{3 d^3 x}{2 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}-\frac{\left (3 d^3\right ) \int \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{3 d^3 x}{2 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}+\frac{\left (3 i d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=\frac{3 d^3 x}{2 b^3}-\frac{(c+d x)^3}{b}-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i d^3 \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac{2 (c+d x)^3 \sin ^2(a+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 6.60769, size = 1719, normalized size = 7.1 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/4)*c*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)) + (I/8)*d^3*E^(I*a)*((2*x^4)/E^((2*I)*a) - ((4*I)*(1 + E^((-2*I)*a))*x^3*Log[1 + E^((-2*I)*(a + b*x
))])/b + (3*(1 + E^((2*I)*a))*(2*b^2*x^2*PolyLog[2, -E^((-2*I)*(a + b*x))] - (2*I)*b*x*PolyLog[3, -E^((-2*I)*(
a + b*x))] - PolyLog[4, -E^((-2*I)*(a + b*x))]))/(b^4*E^((2*I)*a)))*Sec[a] + (c^3*Sec[a]*(Cos[a]*Log[Cos[a]*Co
s[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) + (3*c^2*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Co
t[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 - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Si
n[a]^2)]) + Sec[a]*(Cos[2*a + 2*b*x]/(16*b^4) - ((I/16)*Sin[2*a + 2*b*x])/b^4)*(-8*b^3*c^3*Cos[a] + (12*I)*b^2
*c^2*d*Cos[a] + 12*b*c*d^2*Cos[a] - (6*I)*d^3*Cos[a] - 24*b^3*c^2*d*x*Cos[a] + (24*I)*b^2*c*d^2*x*Cos[a] + 12*
b*d^3*x*Cos[a] - 24*b^3*c*d^2*x^2*Cos[a] + (12*I)*b^2*d^3*x^2*Cos[a] - 8*b^3*d^3*x^3*Cos[a] - (8*I)*b^4*c^3*x*
Cos[a + 2*b*x] - (12*I)*b^4*c^2*d*x^2*Cos[a + 2*b*x] - (8*I)*b^4*c*d^2*x^3*Cos[a + 2*b*x] - (2*I)*b^4*d^3*x^4*
Cos[a + 2*b*x] + (8*I)*b^4*c^3*x*Cos[3*a + 2*b*x] + (12*I)*b^4*c^2*d*x^2*Cos[3*a + 2*b*x] + (8*I)*b^4*c*d^2*x^
3*Cos[3*a + 2*b*x] + (2*I)*b^4*d^3*x^4*Cos[3*a + 2*b*x] - 4*b^3*c^3*Cos[3*a + 4*b*x] - (6*I)*b^2*c^2*d*Cos[3*a
+ 4*b*x] + 6*b*c*d^2*Cos[3*a + 4*b*x] + (3*I)*d^3*Cos[3*a + 4*b*x] - 12*b^3*c^2*d*x*Cos[3*a + 4*b*x] - (12*I)
*b^2*c*d^2*x*Cos[3*a + 4*b*x] + 6*b*d^3*x*Cos[3*a + 4*b*x] - 12*b^3*c*d^2*x^2*Cos[3*a + 4*b*x] - (6*I)*b^2*d^3
*x^2*Cos[3*a + 4*b*x] - 4*b^3*d^3*x^3*Cos[3*a + 4*b*x] - 4*b^3*c^3*Cos[5*a + 4*b*x] - (6*I)*b^2*c^2*d*Cos[5*a
+ 4*b*x] + 6*b*c*d^2*Cos[5*a + 4*b*x] + (3*I)*d^3*Cos[5*a + 4*b*x] - 12*b^3*c^2*d*x*Cos[5*a + 4*b*x] - (12*I)*
b^2*c*d^2*x*Cos[5*a + 4*b*x] + 6*b*d^3*x*Cos[5*a + 4*b*x] - 12*b^3*c*d^2*x^2*Cos[5*a + 4*b*x] - (6*I)*b^2*d^3*
x^2*Cos[5*a + 4*b*x] - 4*b^3*d^3*x^3*Cos[5*a + 4*b*x] + 8*b^4*c^3*x*Sin[a + 2*b*x] + 12*b^4*c^2*d*x^2*Sin[a +
2*b*x] + 8*b^4*c*d^2*x^3*Sin[a + 2*b*x] + 2*b^4*d^3*x^4*Sin[a + 2*b*x] - 8*b^4*c^3*x*Sin[3*a + 2*b*x] - 12*b^4
*c^2*d*x^2*Sin[3*a + 2*b*x] - 8*b^4*c*d^2*x^3*Sin[3*a + 2*b*x] - 2*b^4*d^3*x^4*Sin[3*a + 2*b*x] - (4*I)*b^3*c^
3*Sin[3*a + 4*b*x] + 6*b^2*c^2*d*Sin[3*a + 4*b*x] + (6*I)*b*c*d^2*Sin[3*a + 4*b*x] - 3*d^3*Sin[3*a + 4*b*x] -
(12*I)*b^3*c^2*d*x*Sin[3*a + 4*b*x] + 12*b^2*c*d^2*x*Sin[3*a + 4*b*x] + (6*I)*b*d^3*x*Sin[3*a + 4*b*x] - (12*I
)*b^3*c*d^2*x^2*Sin[3*a + 4*b*x] + 6*b^2*d^3*x^2*Sin[3*a + 4*b*x] - (4*I)*b^3*d^3*x^3*Sin[3*a + 4*b*x] - (4*I)
*b^3*c^3*Sin[5*a + 4*b*x] + 6*b^2*c^2*d*Sin[5*a + 4*b*x] + (6*I)*b*c*d^2*Sin[5*a + 4*b*x] - 3*d^3*Sin[5*a + 4*
b*x] - (12*I)*b^3*c^2*d*x*Sin[5*a + 4*b*x] + 12*b^2*c*d^2*x*Sin[5*a + 4*b*x] + (6*I)*b*d^3*x*Sin[5*a + 4*b*x]
- (12*I)*b^3*c*d^2*x^2*Sin[5*a + 4*b*x] + 6*b^2*d^3*x^2*Sin[5*a + 4*b*x] - (4*I)*b^3*d^3*x^3*Sin[5*a + 4*b*x])

________________________________________________________________________________________

Maple [B]  time = 0.214, size = 639, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4+1/b*c^3*ln(exp(2*I*(b*x+a))+1)+3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b*
x+a)))+3/2/b^3*d^3*polylog(3,-exp(2*I*(b*x+a)))*x-3/2*I/b^4*d^3*a^4-3/2*I/b^2*c^2*d*polylog(2,-exp(2*I*(b*x+a)
))-3/2*I/b^2*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2+4*I/b^3*a^3*c*d^2-3*I/b^2*a^2*c^2*d+3/b*c^2*d*ln(exp(2*I*(b*
x+a))+1)*x+3/b*c*d^2*ln(exp(2*I*(b*x+a))+1)*x^2+1/b*d^3*ln(exp(2*I*(b*x+a))+1)*x^3-1/8*(4*d^3*x^3*b^3+6*I*b^2*
d^3*x^2+12*b^3*c*d^2*x^2+12*I*b^2*c*d^2*x+12*b^3*c^2*d*x+6*I*b^2*c^2*d+4*b^3*c^3-6*b*d^3*x-3*I*d^3-6*c*d^2*b)/
b^4*exp(2*I*(b*x+a))-2*I/b^3*d^3*a^3*x-6/b^3*c*d^2*a^2*ln(exp(I*(b*x+a)))+6/b^2*c^2*d*a*ln(exp(I*(b*x+a)))+I*c
^3*x-1/8*(4*d^3*x^3*b^3-6*I*b^2*d^3*x^2+12*b^3*c*d^2*x^2-12*I*b^2*c*d^2*x+12*b^3*c^2*d*x-6*I*b^2*c^2*d+4*b^3*c
^3-6*b*d^3*x+3*I*d^3-6*c*d^2*b)/b^4*exp(-2*I*(b*x+a))-I*c*d^2*x^3-3/2*I*c^2*d*x^2+2/b^4*d^3*a^3*ln(exp(I*(b*x+
a)))-3*I/b^2*polylog(2,-exp(2*I*(b*x+a)))*c*d^2*x-6*I/b*a*c^2*d*x+6*I/b^2*a^2*c*d^2*x-2/b*c^3*ln(exp(I*(b*x+a)
))-1/4*I*d^3*x^4

________________________________________________________________________________________

Maxima [B]  time = 1.50017, size = 597, normalized size = 2.47 \begin{align*} -\frac{c^{3}{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )\right )}}{2 \, b} + \frac{-3 i \, b^{4} d^{3} x^{4} - 12 i \, b^{4} c d^{2} x^{3} - 18 i \, b^{4} c^{2} d x^{2} + 12 i \, d^{3}{\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) +{\left (16 i \, b^{3} d^{3} x^{3} + 36 i \, b^{3} c d^{2} x^{2} + 36 i \, b^{3} c^{2} d x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \,{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} - 3 \, b c d^{2} + 3 \,{\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (-24 i \, b^{2} d^{3} x^{2} - 36 i \, b^{2} c d^{2} x - 18 i \, b^{2} c^{2} d\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \,{\left (4 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 9 \, b^{3} c^{2} d x\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \,{\left (4 \, b d^{3} x + 3 \, b c d^{2}\right )}{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 9 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{12 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^3*(2*cos(2*b*x + 2*a) - log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*
b*x)*sin(2*a) + sin(2*a)^2))/b + 1/12*(-3*I*b^4*d^3*x^4 - 12*I*b^4*c*d^2*x^3 - 18*I*b^4*c^2*d*x^2 + 12*I*d^3*p
olylog(4, -e^(2*I*b*x + 2*I*a)) + (16*I*b^3*d^3*x^3 + 36*I*b^3*c*d^2*x^2 + 36*I*b^3*c^2*d*x)*arctan2(sin(2*b*x
+ 2*a), cos(2*b*x + 2*a) + 1) - 6*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 - 3*b*c*d^2 + 3*(2*b^3*c^2*d - b*d^3)*x)*c
os(2*b*x + 2*a) + (-24*I*b^2*d^3*x^2 - 36*I*b^2*c*d^2*x - 18*I*b^2*c^2*d)*dilog(-e^(2*I*b*x + 2*I*a)) + 2*(4*b
^3*d^3*x^3 + 9*b^3*c*d^2*x^2 + 9*b^3*c^2*d*x)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a)
+ 1) + 6*(4*b*d^3*x + 3*b*c*d^2)*polylog(3, -e^(2*I*b*x + 2*I*a)) + 9*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*
c^2*d - d^3)*sin(2*b*x + 2*a))/b^4

________________________________________________________________________________________

Fricas [C]  time = 0.865917, size = 2703, normalized size = 11.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 - 6*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4,
I*cos(b*x + a) - sin(b*x + a)) + 6*I*d^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*polylog(4, -I*co
s(b*x + a) - sin(b*x + a)) - 2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 2*b^3*c^3 - 3*b*c*d^2 + 3*(2*b^3*c^2*d - b*d
^3)*x)*cos(b*x + a)^2 + 3*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d - d^3)*cos(b*x + a)*sin(b*x + a) + 3*(2
*b^3*c^2*d - b*d^3)*x + (3*I*b^2*d^3*x^2 + 6*I*b^2*c*d^2*x + 3*I*b^2*c^2*d)*dilog(I*cos(b*x + a) + sin(b*x + a
)) + (-3*I*b^2*d^3*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) + (-3*I*b^2*d^3
*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (3*I*b^2*d^3*x^2 + 6*I*b^2*c*d
^2*x + 3*I*b^2*c^2*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d
^3)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x +
a) - I*sin(b*x + a) + I) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a
^3*d^3)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*
d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c
^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b
^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) - sin(b*x + a) + 1
) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b^3*c^3 - 3
*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) - I*sin(b*x + a) + I) + 6*(b*d^3*x + b*c*d^2)*polylo
g(3, I*cos(b*x + a) + sin(b*x + a)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 6*(b*d
^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a)
- sin(b*x + a)))/b^4

________________________________________________________________________________________

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((d*x+c)**3*sec(b*x+a)*sin(3*b*x+3*a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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