3.32 \(\int (c+d x)^4 \cot (a+b x) \, dx\)

Optimal. Leaf size=151 \[ -\frac {3 d^4 \text {Li}_5\left (e^{2 i (a+b x)}\right )}{2 b^5}+\frac {3 i d^3 (c+d x) \text {Li}_4\left (e^{2 i (a+b x)}\right )}{b^4}+\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{b^3}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i (c+d x)^5}{5 d} \]

[Out]

-1/5*I*(d*x+c)^5/d+(d*x+c)^4*ln(1-exp(2*I*(b*x+a)))/b-2*I*d*(d*x+c)^3*polylog(2,exp(2*I*(b*x+a)))/b^2+3*d^2*(d
*x+c)^2*polylog(3,exp(2*I*(b*x+a)))/b^3+3*I*d^3*(d*x+c)*polylog(4,exp(2*I*(b*x+a)))/b^4-3/2*d^4*polylog(5,exp(
2*I*(b*x+a)))/b^5

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3717, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 d^2 (c+d x)^2 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 (c+d x) \text {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{b^4}-\frac {2 i d (c+d x)^3 \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {3 d^4 \text {PolyLog}\left (5,e^{2 i (a+b x)}\right )}{2 b^5}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i (c+d x)^5}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Cot[a + b*x],x]

[Out]

((-I/5)*(c + d*x)^5)/d + ((c + d*x)^4*Log[1 - E^((2*I)*(a + b*x))])/b - ((2*I)*d*(c + d*x)^3*PolyLog[2, E^((2*
I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)^2*PolyLog[3, E^((2*I)*(a + b*x))])/b^3 + ((3*I)*d^3*(c + d*x)*PolyLog[4
, E^((2*I)*(a + b*x))])/b^4 - (3*d^4*PolyLog[5, E^((2*I)*(a + b*x))])/(2*b^5)

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 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)^4 \cot (a+b x) \, dx &=-\frac {i (c+d x)^5}{5 d}-2 i \int \frac {e^{2 i (a+b x)} (c+d x)^4}{1-e^{2 i (a+b x)}} \, dx\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(4 d) \int (c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {\left (6 i d^2\right ) \int (c+d x)^2 \text {Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{b^3}-\frac {\left (6 d^3\right ) \int (c+d x) \text {Li}_3\left (e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 (c+d x) \text {Li}_4\left (e^{2 i (a+b x)}\right )}{b^4}-\frac {\left (3 i d^4\right ) \int \text {Li}_4\left (e^{2 i (a+b x)}\right ) \, dx}{b^4}\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 (c+d x) \text {Li}_4\left (e^{2 i (a+b x)}\right )}{b^4}-\frac {\left (3 d^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^5}\\ &=-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^2 (c+d x)^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 (c+d x) \text {Li}_4\left (e^{2 i (a+b x)}\right )}{b^4}-\frac {3 d^4 \text {Li}_5\left (e^{2 i (a+b x)}\right )}{2 b^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.11, size = 799, normalized size = 5.29 \[ \frac {1}{5} i d^4 x^5+i c d^3 x^4+\frac {d^4 \log \left (1-e^{-i (a+b x)}\right ) x^4}{b}+\frac {d^4 \log \left (1+e^{-i (a+b x)}\right ) x^4}{b}+2 i c^2 d^2 x^3+\frac {4 c d^3 \log \left (1-e^{-i (a+b x)}\right ) x^3}{b}+\frac {4 c d^3 \log \left (1+e^{-i (a+b x)}\right ) x^3}{b}+2 c^3 d \cot (a) x^2+\frac {6 c^2 d^2 \log \left (1-e^{-i (a+b x)}\right ) x^2}{b}+\frac {6 c^2 d^2 \log \left (1+e^{-i (a+b x)}\right ) x^2}{b}+\frac {12 d^4 \text {Li}_3\left (-e^{-i (a+b x)}\right ) x^2}{b^3}+\frac {12 d^4 \text {Li}_3\left (e^{-i (a+b x)}\right ) x^2}{b^3}-2 c^3 d e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt {\sec ^2(a)} x^2-\frac {4 i c^3 d \tan ^{-1}(\tan (a)) x}{b}+\frac {4 c^3 d \log \left (1-e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right ) x}{b}+\frac {4 i d^2 \left (3 c^2+3 d x c+d^2 x^2\right ) \text {Li}_2\left (-e^{-i (a+b x)}\right ) x}{b^2}+\frac {4 i d^2 \left (3 c^2+3 d x c+d^2 x^2\right ) \text {Li}_2\left (e^{-i (a+b x)}\right ) x}{b^2}+\frac {24 c d^3 \text {Li}_3\left (-e^{-i (a+b x)}\right ) x}{b^3}+\frac {24 c d^3 \text {Li}_3\left (e^{-i (a+b x)}\right ) x}{b^3}-\frac {24 i d^4 \text {Li}_4\left (-e^{-i (a+b x)}\right ) x}{b^4}-\frac {24 i d^4 \text {Li}_4\left (e^{-i (a+b x)}\right ) x}{b^4}+\frac {2 i c^3 d \pi x}{b}+\frac {2 c^3 d \pi \log \left (1+e^{-2 i b x}\right )}{b^2}+\frac {4 c^3 d \tan ^{-1}(\tan (a)) \log \left (1-e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )}{b^2}-\frac {2 c^3 d \pi \log (\cos (b x))}{b^2}+\frac {c^4 \log (\sin (a+b x))}{b}-\frac {4 c^3 d \tan ^{-1}(\tan (a)) \log \left (\sin \left (b x+\tan ^{-1}(\tan (a))\right )\right )}{b^2}-\frac {2 i c^3 d \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )}{b^2}+\frac {12 c^2 d^2 \text {Li}_3\left (-e^{-i (a+b x)}\right )}{b^3}+\frac {12 c^2 d^2 \text {Li}_3\left (e^{-i (a+b x)}\right )}{b^3}-\frac {24 i c d^3 \text {Li}_4\left (-e^{-i (a+b x)}\right )}{b^4}-\frac {24 i c d^3 \text {Li}_4\left (e^{-i (a+b x)}\right )}{b^4}-\frac {24 d^4 \text {Li}_5\left (-e^{-i (a+b x)}\right )}{b^5}-\frac {24 d^4 \text {Li}_5\left (e^{-i (a+b x)}\right )}{b^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^4*Cot[a + b*x],x]

[Out]

((2*I)*c^3*d*Pi*x)/b + (2*I)*c^2*d^2*x^3 + I*c*d^3*x^4 + (I/5)*d^4*x^5 - ((4*I)*c^3*d*x*ArcTan[Tan[a]])/b + 2*
c^3*d*x^2*Cot[a] + (2*c^3*d*Pi*Log[1 + E^((-2*I)*b*x)])/b^2 + (6*c^2*d^2*x^2*Log[1 - E^((-I)*(a + b*x))])/b +
(4*c*d^3*x^3*Log[1 - E^((-I)*(a + b*x))])/b + (d^4*x^4*Log[1 - E^((-I)*(a + b*x))])/b + (6*c^2*d^2*x^2*Log[1 +
E^((-I)*(a + b*x))])/b + (4*c*d^3*x^3*Log[1 + E^((-I)*(a + b*x))])/b + (d^4*x^4*Log[1 + E^((-I)*(a + b*x))])/
b + (4*c^3*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))])/b + (4*c^3*d*ArcTan[Tan[a]]*Log[1 - E^((2*I)*(b*x +
ArcTan[Tan[a]]))])/b^2 - (2*c^3*d*Pi*Log[Cos[b*x]])/b^2 + (c^4*Log[Sin[a + b*x]])/b - (4*c^3*d*ArcTan[Tan[a]]*
Log[Sin[b*x + ArcTan[Tan[a]]]])/b^2 + ((4*I)*d^2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*PolyLog[2, -E^((-I)*(a + b*x))]
)/b^2 + ((4*I)*d^2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*PolyLog[2, E^((-I)*(a + b*x))])/b^2 - ((2*I)*c^3*d*PolyLog[2,
E^((2*I)*(b*x + ArcTan[Tan[a]]))])/b^2 + (12*c^2*d^2*PolyLog[3, -E^((-I)*(a + b*x))])/b^3 + (24*c*d^3*x*PolyL
og[3, -E^((-I)*(a + b*x))])/b^3 + (12*d^4*x^2*PolyLog[3, -E^((-I)*(a + b*x))])/b^3 + (12*c^2*d^2*PolyLog[3, E^
((-I)*(a + b*x))])/b^3 + (24*c*d^3*x*PolyLog[3, E^((-I)*(a + b*x))])/b^3 + (12*d^4*x^2*PolyLog[3, E^((-I)*(a +
b*x))])/b^3 - ((24*I)*c*d^3*PolyLog[4, -E^((-I)*(a + b*x))])/b^4 - ((24*I)*d^4*x*PolyLog[4, -E^((-I)*(a + b*x
))])/b^4 - ((24*I)*c*d^3*PolyLog[4, E^((-I)*(a + b*x))])/b^4 - ((24*I)*d^4*x*PolyLog[4, E^((-I)*(a + b*x))])/b
^4 - (24*d^4*PolyLog[5, -E^((-I)*(a + b*x))])/b^5 - (24*d^4*PolyLog[5, E^((-I)*(a + b*x))])/b^5 - 2*c^3*d*E^(I
*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2]

________________________________________________________________________________________

fricas [C]  time = 0.60, size = 1204, normalized size = 7.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*csc(b*x+a),x, algorithm="fricas")

[Out]

-1/2*(24*d^4*polylog(5, cos(b*x + a) + I*sin(b*x + a)) + 24*d^4*polylog(5, cos(b*x + a) - I*sin(b*x + a)) + 24
*d^4*polylog(5, -cos(b*x + a) + I*sin(b*x + a)) + 24*d^4*polylog(5, -cos(b*x + a) - I*sin(b*x + a)) - (-4*I*b^
3*d^4*x^3 - 12*I*b^3*c*d^3*x^2 - 12*I*b^3*c^2*d^2*x - 4*I*b^3*c^3*d)*dilog(cos(b*x + a) + I*sin(b*x + a)) - (4
*I*b^3*d^4*x^3 + 12*I*b^3*c*d^3*x^2 + 12*I*b^3*c^2*d^2*x + 4*I*b^3*c^3*d)*dilog(cos(b*x + a) - I*sin(b*x + a))
- (4*I*b^3*d^4*x^3 + 12*I*b^3*c*d^3*x^2 + 12*I*b^3*c^2*d^2*x + 4*I*b^3*c^3*d)*dilog(-cos(b*x + a) + I*sin(b*x
+ a)) - (-4*I*b^3*d^4*x^3 - 12*I*b^3*c*d^3*x^2 - 12*I*b^3*c^2*d^2*x - 4*I*b^3*c^3*d)*dilog(-cos(b*x + a) - I*
sin(b*x + a)) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + b^4*c^4)*log(cos(b*x + a)
+ I*sin(b*x + a) + 1) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + b^4*c^4)*log(cos
(b*x + a) - I*sin(b*x + a) + 1) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(
-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 +
a^4*d^4)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x
^2 + 4*b^4*c^3*d*x + 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(-cos(b*x + a) + I*sin(b*
x + a) + 1) - (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 4*a*b^3*c^3*d - 6*a^2*b^2*c
^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) - (24*I*b*d^4*x + 24*I*b*c*d^3)*poly
log(4, cos(b*x + a) + I*sin(b*x + a)) - (-24*I*b*d^4*x - 24*I*b*c*d^3)*polylog(4, cos(b*x + a) - I*sin(b*x + a
)) - (-24*I*b*d^4*x - 24*I*b*c*d^3)*polylog(4, -cos(b*x + a) + I*sin(b*x + a)) - (24*I*b*d^4*x + 24*I*b*c*d^3)
*polylog(4, -cos(b*x + a) - I*sin(b*x + a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, cos(b*
x + a) + I*sin(b*x + a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, cos(b*x + a) - I*sin(b*x
+ a)) - 12*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, -cos(b*x + a) + I*sin(b*x + a)) - 12*(b^2*d^
4*x^2 + 2*b^2*c*d^3*x + b^2*c^2*d^2)*polylog(3, -cos(b*x + a) - I*sin(b*x + a)))/b^5

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{4} \cos \left (b x + a\right ) \csc \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*csc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^4*cos(b*x + a)*csc(b*x + a), x)

________________________________________________________________________________________

maple [B]  time = 0.17, size = 1150, normalized size = 7.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4*cos(b*x+a)*csc(b*x+a),x)

[Out]

1/b^5*d^4*a^4*ln(exp(I*(b*x+a))-1)-2/b^5*d^4*a^4*ln(exp(I*(b*x+a)))+12/b^3*c^2*d^2*polylog(3,-exp(I*(b*x+a)))+
12/b^3*c^2*d^2*polylog(3,exp(I*(b*x+a)))-1/b^5*d^4*a^4*ln(1-exp(I*(b*x+a)))+12/b^3*d^4*polylog(3,exp(I*(b*x+a)
))*x^2+12/b^3*d^4*polylog(3,-exp(I*(b*x+a)))*x^2+8/5*I/b^5*d^4*a^5-I*c*d^3*x^4-2*I*c^2*d^2*x^3-2*I*c^3*d*x^2-2
4*d^4*polylog(5,-exp(I*(b*x+a)))/b^5-24*d^4*polylog(5,exp(I*(b*x+a)))/b^5+I*c^4*x-2/b*c^4*ln(exp(I*(b*x+a)))+1
/b*c^4*ln(exp(I*(b*x+a))+1)+1/b*c^4*ln(exp(I*(b*x+a))-1)-1/5*I*d^4*x^5+4/b*c^3*d*ln(exp(I*(b*x+a))+1)*x+4/b*c^
3*d*ln(1-exp(I*(b*x+a)))*x+4/b^2*c^3*d*ln(1-exp(I*(b*x+a)))*a+6/b*c^2*d^2*ln(exp(I*(b*x+a))+1)*x^2+24/b^3*c*d^
3*polylog(3,-exp(I*(b*x+a)))*x-6/b^3*c^2*d^2*a^2*ln(1-exp(I*(b*x+a)))+6/b*c^2*d^2*ln(1-exp(I*(b*x+a)))*x^2+24/
b^3*c*d^3*polylog(3,exp(I*(b*x+a)))*x+24*I/b^4*c*d^3*polylog(4,-exp(I*(b*x+a)))+24*I/b^4*c*d^3*polylog(4,exp(I
*(b*x+a)))+2*I/b^4*d^4*a^4*x-4*I/b^2*c^3*d*a^2+8*I/b^3*c^2*d^2*a^3-6*I/b^4*c*d^3*a^4-4*I/b^2*d^4*polylog(2,exp
(I*(b*x+a)))*x^3+24*I/b^4*d^4*polylog(4,exp(I*(b*x+a)))*x-4*I/b^2*d^4*polylog(2,-exp(I*(b*x+a)))*x^3+24*I/b^4*
d^4*polylog(4,-exp(I*(b*x+a)))*x-4*I/b^2*c^3*d*polylog(2,-exp(I*(b*x+a)))-4*I/b^2*c^3*d*polylog(2,exp(I*(b*x+a
)))+8/b^2*c^3*d*a*ln(exp(I*(b*x+a)))-4/b^4*c*d^3*a^3*ln(exp(I*(b*x+a))-1)+8/b^4*c*d^3*a^3*ln(exp(I*(b*x+a)))+6
/b^3*c^2*d^2*a^2*ln(exp(I*(b*x+a))-1)-12/b^3*c^2*d^2*a^2*ln(exp(I*(b*x+a)))-4/b^2*c^3*d*a*ln(exp(I*(b*x+a))-1)
+1/b*d^4*ln(1-exp(I*(b*x+a)))*x^4+1/b*d^4*ln(exp(I*(b*x+a))+1)*x^4-8*I/b^3*c*d^3*a^3*x+12*I/b^2*c^2*d^2*a^2*x-
8*I/b*c^3*d*a*x-12*I/b^2*c*d^3*polylog(2,-exp(I*(b*x+a)))*x^2-12*I/b^2*c^2*d^2*polylog(2,-exp(I*(b*x+a)))*x-12
*I/b^2*c^2*d^2*polylog(2,exp(I*(b*x+a)))*x-12*I/b^2*c*d^3*polylog(2,exp(I*(b*x+a)))*x^2+4/b*c*d^3*ln(exp(I*(b*
x+a))+1)*x^3+4/b*c*d^3*ln(1-exp(I*(b*x+a)))*x^3+4/b^4*c*d^3*ln(1-exp(I*(b*x+a)))*a^3

________________________________________________________________________________________

maxima [B]  time = 0.62, size = 1262, normalized size = 8.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cos(b*x+a)*csc(b*x+a),x, algorithm="maxima")

[Out]

1/10*(10*c^4*log(sin(b*x + a)) - 40*a*c^3*d*log(sin(b*x + a))/b + 60*a^2*c^2*d^2*log(sin(b*x + a))/b^2 - 40*a^
3*c*d^3*log(sin(b*x + a))/b^3 + 10*a^4*d^4*log(sin(b*x + a))/b^4 + (-2*I*(b*x + a)^5*d^4 + (-10*I*b*c*d^3 + 10
*I*a*d^4)*(b*x + a)^4 - 240*d^4*polylog(5, -e^(I*b*x + I*a)) - 240*d^4*polylog(5, e^(I*b*x + I*a)) + (-20*I*b^
2*c^2*d^2 + 40*I*a*b*c*d^3 - 20*I*a^2*d^4)*(b*x + a)^3 + (-20*I*b^3*c^3*d + 60*I*a*b^2*c^2*d^2 - 60*I*a^2*b*c*
d^3 + 20*I*a^3*d^4)*(b*x + a)^2 + (10*I*(b*x + a)^4*d^4 + (40*I*b*c*d^3 - 40*I*a*d^4)*(b*x + a)^3 + (60*I*b^2*
c^2*d^2 - 120*I*a*b*c*d^3 + 60*I*a^2*d^4)*(b*x + a)^2 + (40*I*b^3*c^3*d - 120*I*a*b^2*c^2*d^2 + 120*I*a^2*b*c*
d^3 - 40*I*a^3*d^4)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (-10*I*(b*x + a)^4*d^4 + (-40*I*b*c*d
^3 + 40*I*a*d^4)*(b*x + a)^3 + (-60*I*b^2*c^2*d^2 + 120*I*a*b*c*d^3 - 60*I*a^2*d^4)*(b*x + a)^2 + (-40*I*b^3*c
^3*d + 120*I*a*b^2*c^2*d^2 - 120*I*a^2*b*c*d^3 + 40*I*a^3*d^4)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a)
+ 1) + (-40*I*b^3*c^3*d + 120*I*a*b^2*c^2*d^2 - 120*I*a^2*b*c*d^3 - 40*I*(b*x + a)^3*d^4 + 40*I*a^3*d^4 + (-12
0*I*b*c*d^3 + 120*I*a*d^4)*(b*x + a)^2 + (-120*I*b^2*c^2*d^2 + 240*I*a*b*c*d^3 - 120*I*a^2*d^4)*(b*x + a))*dil
og(-e^(I*b*x + I*a)) + (-40*I*b^3*c^3*d + 120*I*a*b^2*c^2*d^2 - 120*I*a^2*b*c*d^3 - 40*I*(b*x + a)^3*d^4 + 40*
I*a^3*d^4 + (-120*I*b*c*d^3 + 120*I*a*d^4)*(b*x + a)^2 + (-120*I*b^2*c^2*d^2 + 240*I*a*b*c*d^3 - 120*I*a^2*d^4
)*(b*x + a))*dilog(e^(I*b*x + I*a)) + 5*((b*x + a)^4*d^4 + 4*(b*c*d^3 - a*d^4)*(b*x + a)^3 + 6*(b^2*c^2*d^2 -
2*a*b*c*d^3 + a^2*d^4)*(b*x + a)^2 + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*(b*x + a))*log(
cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + 5*((b*x + a)^4*d^4 + 4*(b*c*d^3 - a*d^4)*(b*x + a)^3 +
6*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a)^2 + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^
4)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + (240*I*b*c*d^3 + 240*I*(b*x + a)*d^4
- 240*I*a*d^4)*polylog(4, -e^(I*b*x + I*a)) + (240*I*b*c*d^3 + 240*I*(b*x + a)*d^4 - 240*I*a*d^4)*polylog(4,
e^(I*b*x + I*a)) + 120*(b^2*c^2*d^2 - 2*a*b*c*d^3 + (b*x + a)^2*d^4 + a^2*d^4 + 2*(b*c*d^3 - a*d^4)*(b*x + a))
*polylog(3, -e^(I*b*x + I*a)) + 120*(b^2*c^2*d^2 - 2*a*b*c*d^3 + (b*x + a)^2*d^4 + a^2*d^4 + 2*(b*c*d^3 - a*d^
4)*(b*x + a))*polylog(3, e^(I*b*x + I*a)))/b^4)/b

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^4}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(a + b*x)*(c + d*x)^4)/sin(a + b*x),x)

[Out]

int((cos(a + b*x)*(c + d*x)^4)/sin(a + b*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{4} \cos {\left (a + b x \right )} \csc {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*cos(b*x+a)*csc(b*x+a),x)

[Out]

Integral((c + d*x)**4*cos(a + b*x)*csc(a + b*x), x)

________________________________________________________________________________________