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

Optimal. Leaf size=197 \[ -\frac {3 i d^3 \text {Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 i d^3 \text {Li}_4\left (e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 d^2 (c+d x) \text {Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d^2 (c+d x) \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i d (c+d x)^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d (c+d x)^2 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4419, 4183, 2531, 6609, 2282, 6589} \[ -\frac {3 d^2 (c+d x) \text {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d^2 (c+d x) \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d (c+d x)^2 \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \text {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 i d^3 \text {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

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

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Mathematica [A]  time = 0.95, size = 350, normalized size = 1.78 \[ \frac {-8 b^3 c^3 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )+12 b^3 c^2 d x \log \left (1-e^{2 i (a+b x)}\right )-12 b^3 c^2 d x \log \left (1+e^{2 i (a+b x)}\right )+12 b^3 c d^2 x^2 \log \left (1-e^{2 i (a+b x)}\right )-12 b^3 c d^2 x^2 \log \left (1+e^{2 i (a+b x)}\right )+4 b^3 d^3 x^3 \log \left (1-e^{2 i (a+b x)}\right )-4 b^3 d^3 x^3 \log \left (1+e^{2 i (a+b x)}\right )+6 i b^2 d (c+d x)^2 \text {Li}_2\left (-e^{2 i (a+b x)}\right )-6 i b^2 d (c+d x)^2 \text {Li}_2\left (e^{2 i (a+b x)}\right )-6 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (a+b x)}\right )+6 b c d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )+6 b d^3 x \text {Li}_3\left (e^{2 i (a+b x)}\right )-3 i d^3 \text {Li}_4\left (-e^{2 i (a+b x)}\right )+3 i d^3 \text {Li}_4\left (e^{2 i (a+b x)}\right )}{4 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^3*c^3*ArcTanh[E^((2*I)*(a + b*x))] + 12*b^3*c^2*d*x*Log[1 - E^((2*I)*(a + b*x))] + 12*b^3*c*d^2*x^2*Log[
1 - E^((2*I)*(a + b*x))] + 4*b^3*d^3*x^3*Log[1 - E^((2*I)*(a + b*x))] - 12*b^3*c^2*d*x*Log[1 + E^((2*I)*(a + b
*x))] - 12*b^3*c*d^2*x^2*Log[1 + E^((2*I)*(a + b*x))] - 4*b^3*d^3*x^3*Log[1 + E^((2*I)*(a + b*x))] + (6*I)*b^2
*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))] - (6*I)*b^2*d*(c + d*x)^2*PolyLog[2, E^((2*I)*(a + b*x))] - 6*
b*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))] + 6*b*c*d^2*PolyLog[3, E^((2*I)*(a + b*x))] + 6*b*d^3*x*PolyL
og[3, E^((2*I)*(a + b*x))] - (3*I)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))] + (3*I)*d^3*PolyLog[4, E^((2*I)*(a + b
*x))])/(4*b^4)

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fricas [C]  time = 0.60, size = 1778, normalized size = 9.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*I*d^3*polylog(4, cos(b*x + a) + I*sin(b*x + a)) - 6*I*d^3*polylog(4, cos(b*x + a) - I*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*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*pol
ylog(4, -cos(b*x + a) + I*sin(b*x + a)) + 6*I*d^3*polylog(4, -cos(b*x + a) - I*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(cos(b*x + a) + I*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(cos(b*x + a) - I*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))
+ (-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(-cos(b*x + a) + I*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(-cos(b*x + a) - I*sin(b*x + a)) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x +
b^3*c^3)*log(cos(b*x + a) + I*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*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*log(cos(b*x + a) -
I*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*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(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + (b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) + (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(-cos(b*x + a) + I*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*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(-cos(b*x + a) - I*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) + 6*(b*d^3*x + b*c*d^
2)*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, cos(b*x + a) - I*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)) - 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, -cos(b*x + a) + I*sin(b*x +
a)) + 6*(b*d^3*x + b*c*d^2)*polylog(3, -cos(b*x + a) - I*sin(b*x + a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 816, normalized size = 4.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*I/b^2*c*d^2*polylog(2,exp(I*(b*x+a)))*x-6*I/b^2*c*d^2*polylog(2,-exp(I*(b*x+a)))*x+6*I*d^3*polylog(4,exp(I*
(b*x+a)))/b^4-1/b*c^3*ln(1+exp(2*I*(b*x+a)))-1/b^4*d^3*a^3*ln(exp(I*(b*x+a))-1)+6/b^3*c*d^2*polylog(3,-exp(I*(
b*x+a)))+6/b^3*c*d^2*polylog(3,exp(I*(b*x+a)))+6/b^3*d^3*polylog(3,exp(I*(b*x+a)))*x+6/b^3*d^3*polylog(3,-exp(
I*(b*x+a)))*x+6*I/b^4*d^3*polylog(4,-exp(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*po
lylog(3,-exp(2*I*(b*x+a)))*x-3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4+1/b*c^3*ln(exp(I*(b*x+a))-1)+1/b*c^3*l
n(exp(I*(b*x+a))+1)+3*I/b^2*c*d^2*polylog(2,-exp(2*I*(b*x+a)))*x+3/b^3*c*d^2*a^2*ln(exp(I*(b*x+a))-1)-3*I/b^2*
c^2*d*polylog(2,exp(I*(b*x+a)))-3*I/b^2*c^2*d*polylog(2,-exp(I*(b*x+a)))-3*I/b^2*d^3*polylog(2,exp(I*(b*x+a)))
*x^2-3*I/b^2*d^3*polylog(2,-exp(I*(b*x+a)))*x^2+3/b*c^2*d*ln(exp(I*(b*x+a))+1)*x+3/b*c^2*d*ln(1-exp(I*(b*x+a))
)*x+3/b^2*c^2*d*ln(1-exp(I*(b*x+a)))*a-3/b^3*c*d^2*a^2*ln(1-exp(I*(b*x+a)))+3/b*c*d^2*ln(1-exp(I*(b*x+a)))*x^2
+3/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2-3/b^2*c^2*d*a*ln(exp(I*(b*x+a))-1)+1/b*d^3*ln(1-exp(I*(b*x+a)))*x^3+1/b^4*
d^3*ln(1-exp(I*(b*x+a)))*a^3+1/b*d^3*ln(exp(I*(b*x+a))+1)*x^3-1/b*d^3*ln(1+exp(2*I*(b*x+a)))*x^3+3/2*I/b^2*d^3
*polylog(2,-exp(2*I*(b*x+a)))*x^2+3/2*I/b^2*c^2*d*polylog(2,-exp(2*I*(b*x+a)))-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 = 0.61, size = 1063, normalized size = 5.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*c^3*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 9*a*c^2*d*(log(sin(b*x + a)^2 - 1) - log(sin(b*x
+ a)^2))/b + 9*a^2*c*d^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 - 3*a^3*d^3*(log(sin(b*x + a)^2
- 1) - log(sin(b*x + a)^2))/b^3 + (6*I*d^3*polylog(4, -e^(2*I*b*x + 2*I*a)) - 36*I*d^3*polylog(4, -e^(I*b*x +
I*a)) - 36*I*d^3*polylog(4, e^(I*b*x + I*a)) + (8*I*(b*x + a)^3*d^3 + (18*I*b*c*d^2 - 18*I*a*d^3)*(b*x + a)^2
+ (18*I*b^2*c^2*d - 36*I*a*b*c*d^2 + 18*I*a^2*d^3)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1)
+ (-6*I*(b*x + a)^3*d^3 + (-18*I*b*c*d^2 + 18*I*a*d^3)*(b*x + a)^2 + (-18*I*b^2*c^2*d + 36*I*a*b*c*d^2 - 18*I*
a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*I*(b*x + a)^3*d^3 + (18*I*b*c*d^2 - 18*I*a*d^
3)*(b*x + a)^2 + (18*I*b^2*c^2*d - 36*I*a*b*c*d^2 + 18*I*a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x +
a) + 1) + (-9*I*b^2*c^2*d + 18*I*a*b*c*d^2 - 12*I*(b*x + a)^2*d^3 - 9*I*a^2*d^3 + (-18*I*b*c*d^2 + 18*I*a*d^3)
*(b*x + a))*dilog(-e^(2*I*b*x + 2*I*a)) + (18*I*b^2*c^2*d - 36*I*a*b*c*d^2 + 18*I*(b*x + a)^2*d^3 + 18*I*a^2*d
^3 + (36*I*b*c*d^2 - 36*I*a*d^3)*(b*x + a))*dilog(-e^(I*b*x + I*a)) + (18*I*b^2*c^2*d - 36*I*a*b*c*d^2 + 18*I*
(b*x + a)^2*d^3 + 18*I*a^2*d^3 + (36*I*b*c*d^2 - 36*I*a*d^3)*(b*x + a))*dilog(e^(I*b*x + I*a)) + (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) - 3*((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*
(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - 3*(
(b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b
*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 3*(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)) - 36*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*polylog(3, -e^(I*b*x + I*a)) - 36*(b*c*d^2 + (b*x + a)*
d^3 - a*d^3)*polylog(3, e^(I*b*x + I*a)))/b^3)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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