3.222 \(\int (c+d x)^3 \sin ^2(a+b x) \tan (a+b x) \, dx\)

Optimal. Leaf size=251 \[ -\frac {3 i d^3 \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}-\frac {3 d^2 (c+d x) \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 b}-\frac {3 d^3 x}{8 b^3}+\frac {(c+d x)^3}{4 b}+\frac {i (c+d x)^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4407, 4404, 3311, 32, 2635, 8, 3719, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3 d^2 (c+d x) \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \text {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}-\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 b}-\frac {3 d^3 x}{8 b^3}+\frac {(c+d x)^3}{4 b}+\frac {i (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d^3*x)/(8*b^3) + (c + d*x)^3/(4*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])/(8*
b^4) - (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/(4*b^2) + (3*d^2*(c + d*x)*Sin[a + b*x]^2)/(4*b^3) - ((c +
d*x)^3*Sin[a + b*x]^2)/(2*b)

Rule 8

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

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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int (c+d x)^3 \sin ^2(a+b x) \tan (a+b x) \, dx &=-\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 {(c+d x)^3 \sin ^2(a+b x)}{2 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 {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 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 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 {(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 {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)}{8 b^4}-\frac {3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 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}{8 b^3}+\frac {(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 {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)}{8 b^4}-\frac {3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 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}{8 b^3}+\frac {(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 {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)}{8 b^4}-\frac {3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 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}{8 b^3}+\frac {(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 {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)}{8 b^4}-\frac {3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}+\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^3 \sin ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [B]  time = 7.12, size = 1720, normalized size = 6.85 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/4*I)*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]
*Cos[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
[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]])*Lo
g[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 +
Sin[a]^2)]) + Sec[a]*(Cos[2*a + 2*b*x]/(64*b^4) - ((I/64)*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] - 1
2*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] + (32*I)*b^4*c^3
*x*Cos[a + 2*b*x] + (48*I)*b^4*c^2*d*x^2*Cos[a + 2*b*x] + (32*I)*b^4*c*d^2*x^3*Cos[a + 2*b*x] + (8*I)*b^4*d^3*
x^4*Cos[a + 2*b*x] - (32*I)*b^4*c^3*x*Cos[3*a + 2*b*x] - (48*I)*b^4*c^2*d*x^2*Cos[3*a + 2*b*x] - (32*I)*b^4*c*
d^2*x^3*Cos[3*a + 2*b*x] - (8*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*C
os[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*Co
s[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] - 32*b^4*c^3*x*Sin[a + 2*b*x] - 48*b^4*c^2*d*x^2*S
in[a + 2*b*x] - 32*b^4*c*d^2*x^3*Sin[a + 2*b*x] - 8*b^4*d^3*x^4*Sin[a + 2*b*x] + 32*b^4*c^3*x*Sin[3*a + 2*b*x]
+ 48*b^4*c^2*d*x^2*Sin[3*a + 2*b*x] + 32*b^4*c*d^2*x^3*Sin[3*a + 2*b*x] + 8*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*Si
n[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])

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fricas [C]  time = 0.61, size = 1134, normalized size = 4.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 - 24*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) + 24*I*d^3*polylog(
4, I*cos(b*x + a) - sin(b*x + a)) + 24*I*d^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) - 24*I*d^3*polylog(4,
-I*cos(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 - (-12*I*b^2*d^3*x^2 - 24*I*b^2*c*d^2*x - 12*I*b^2*c^2*d)*dilog(I*cos(b*x + a) + si
n(b*x + a)) - (12*I*b^2*d^3*x^2 + 24*I*b^2*c*d^2*x + 12*I*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - (1
2*I*b^2*d^3*x^2 + 24*I*b^2*c*d^2*x + 12*I*b^2*c^2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - (-12*I*b^2*d^3*x^
2 - 24*I*b^2*c*d^2*x - 12*I*b^2*c^2*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + 4*(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) + 4*(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) + 4*(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) + 4*(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) + 4*(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) + 4*(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*co
s(b*x + a) - sin(b*x + a) + 1) + 4*(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*s
in(b*x + a) + I) + 4*(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) + 24*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) + sin(b*x + a)) + 24*(b*d^3*x + b*c*d^2)*polylog(3, I*co
s(b*x + a) - sin(b*x + a)) + 24*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) + 24*(b*d^3*x +
b*c*d^2)*polylog(3, -I*cos(b*x + a) - sin(b*x + a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.22, size = 641, normalized size = 2.55 \[ -\frac {3 d^{3} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{2 b^{3}}-\frac {d^{3} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {6 i c \,d^{2} a^{2} x}{b^{2}}+\frac {6 i c^{2} d a x}{b}-\frac {2 d^{3} a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 i a^{4} d^{3}}{2 b^{4}}+i c \,d^{2} x^{3}+\frac {3 i c^{2} d \,x^{2}}{2}-\frac {3 c \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}-\frac {3 i d^{3} \polylog \left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {c^{3} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {3 i c \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}-i c^{3} x +\frac {i d^{3} x^{4}}{4}+\frac {2 c^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {6 c \,d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 i c^{2} d \,a^{2}}{b^{2}}+\frac {2 i a^{3} d^{3} x}{b^{3}}-\frac {4 i c \,d^{2} a^{3}}{b^{3}}-\frac {6 c^{2} d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 i d^{3} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{2 b^{2}}+\frac {3 i c^{2} d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}-\frac {3 c \,d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {\left (4 d^{3} x^{3} b^{3}+12 b^{3} c \,d^{2} x^{2}+6 i b^{2} d^{3} x^{2}+12 b^{3} c^{2} d x +12 i b^{2} c \,d^{2} x +4 b^{3} c^{3}+6 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b -3 i d^{3}\right ) {\mathrm e}^{2 i \left (b x +a \right )}}{32 b^{4}}+\frac {\left (4 d^{3} x^{3} b^{3}+12 b^{3} c \,d^{2} x^{2}-6 i b^{2} d^{3} x^{2}+12 b^{3} c^{2} d x -12 i b^{2} c \,d^{2} x +4 b^{3} c^{3}-6 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b +3 i d^{3}\right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{32 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*d^3*x^4+I*c*d^2*x^3-I*c^3*x+3*I/b^2*a^2*c^2*d-4*I/b^3*a^3*c*d^2+2*I/b^3*d^3*a^3*x-1/b*c^3*ln(1+exp(2*I*(
b*x+a)))-2/b^4*d^3*a^3*ln(exp(I*(b*x+a)))+1/32*(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))+1/32*(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))-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-3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4+2/b*c^3*ln(exp(I*(b*x+a)))+3*I/b^2*c*d^2*polylog(2,-exp(2*I*
(b*x+a)))*x+3/2*I*c^2*d*x^2+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)))-1/b*d^3*ln(1+e
xp(2*I*(b*x+a)))*x^3+3/2*I/b^2*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2-6*I/b^2*a^2*c*d^2*x+6*I/b*a*c^2*d*x+3/2*I/
b^2*c^2*d*polylog(2,-exp(2*I*(b*x+a)))+3/2*I/b^4*d^3*a^4-3/b*c^2*d*ln(1+exp(2*I*(b*x+a)))*x-3/b*c*d^2*ln(1+exp
(2*I*(b*x+a)))*x^2

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maxima [B]  time = 0.56, size = 685, normalized size = 2.73 \[ -\frac {24 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} c^{3} - \frac {72 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a c^{2} d}{b} + \frac {72 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a^{2} c d^{2}}{b^{2}} - \frac {24 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a^{3} d^{3}}{b^{3}} + \frac {-12 i \, {\left (b x + a\right )}^{4} d^{3} + {\left (-48 i \, b c d^{2} + 48 i \, a d^{3}\right )} {\left (b x + a\right )}^{3} + 48 i \, d^{3} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + {\left (-72 i \, b^{2} c^{2} d + 144 i \, a b c d^{2} - 72 i \, a^{2} d^{3}\right )} {\left (b x + a\right )}^{2} + {\left (64 i \, {\left (b x + a\right )}^{3} d^{3} + {\left (144 i \, b c d^{2} - 144 i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + {\left (144 i \, b^{2} c^{2} d - 288 i \, a b c d^{2} + 144 i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (2 \, {\left (b x + a\right )}^{3} d^{3} - 3 \, b c d^{2} + 3 \, a d^{3} + 6 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (2 \, b^{2} c^{2} d - 4 \, a b c d^{2} + {\left (2 \, a^{2} - 1\right )} d^{3}\right )} {\left (b x + a\right )}\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (-72 i \, b^{2} c^{2} d + 144 i \, a b c d^{2} - 96 i \, {\left (b x + a\right )}^{2} d^{3} - 72 i \, a^{2} d^{3} + {\left (-144 i \, b c d^{2} + 144 i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 8 \, {\left (4 \, {\left (b x + a\right )}^{3} d^{3} + 9 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 9 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} {\left (b x + a\right )}\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 ) + 24 \, {\left (3 \, b c d^{2} + 4 \, {\left (b x + a\right )} d^{3} - 3 \, a d^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 9 \, {\left (2 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, {\left (b x + a\right )}^{2} d^{3} + {\left (2 \, a^{2} - 1\right )} d^{3} + 4 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} \sin \left (2 \, b x + 2 \, a\right )}{b^{3}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(24*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*c^3 - 72*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*a*c^2
*d/b + 72*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*a^2*c*d^2/b^2 - 24*(sin(b*x + a)^2 + log(sin(b*x + a)^2 -
1))*a^3*d^3/b^3 + (-12*I*(b*x + a)^4*d^3 + (-48*I*b*c*d^2 + 48*I*a*d^3)*(b*x + a)^3 + 48*I*d^3*polylog(4, -e^
(2*I*b*x + 2*I*a)) + (-72*I*b^2*c^2*d + 144*I*a*b*c*d^2 - 72*I*a^2*d^3)*(b*x + a)^2 + (64*I*(b*x + a)^3*d^3 +
(144*I*b*c*d^2 - 144*I*a*d^3)*(b*x + a)^2 + (144*I*b^2*c^2*d - 288*I*a*b*c*d^2 + 144*I*a^2*d^3)*(b*x + a))*arc
tan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(2*(b*x + a)^3*d^3 - 3*b*c*d^2 + 3*a*d^3 + 6*(b*c*d^2 - a*d^3
)*(b*x + a)^2 + 3*(2*b^2*c^2*d - 4*a*b*c*d^2 + (2*a^2 - 1)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (-72*I*b^2*c^2*d
+ 144*I*a*b*c*d^2 - 96*I*(b*x + a)^2*d^3 - 72*I*a^2*d^3 + (-144*I*b*c*d^2 + 144*I*a*d^3)*(b*x + a))*dilog(-e^
(2*I*b*x + 2*I*a)) + 8*(4*(b*x + a)^3*d^3 + 9*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + a^2
*d^3)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 24*(3*b*c*d^2 + 4*(b*
x + a)*d^3 - 3*a*d^3)*polylog(3, -e^(2*I*b*x + 2*I*a)) + 9*(2*b^2*c^2*d - 4*a*b*c*d^2 + 2*(b*x + a)^2*d^3 + (2
*a^2 - 1)*d^3 + 4*(b*c*d^2 - a*d^3)*(b*x + a))*sin(2*b*x + 2*a))/b^3)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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