3.166 \(\int (c+d x)^2 \cos ^2(a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=181 \[ \frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}-\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {c d x}{2 b}+\frac {d^2 x^2}{4 b}-\frac {i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4408, 4404, 3310, 3717, 2190, 2531, 2282, 6589} \[ -\frac {i d (c+d x) \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac {d^2 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}+\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {c d x}{2 b}+\frac {d^2 x^2}{4 b}-\frac {i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cos[a + b*x]^2*Cot[a + b*x],x]

[Out]

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

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 3310

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

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 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 4408

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[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]

Rubi steps

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

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Mathematica [B]  time = 2.90, size = 564, normalized size = 3.12 \[ \frac {48 b^3 c d x^2 \cot (a)-48 b^3 c d x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt {\sec ^2(a)}+48 b^2 c^2 \log (\sin (a+b x))-6 b^2 c^2 \csc (a) \sin (a+2 b x)+6 b^2 c^2 \csc (a) \sin (3 a+2 b x)-96 i b^2 c d x \tan ^{-1}(\tan (a))+96 b^2 c d x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-12 b^2 c d x \csc (a) \sin (a+2 b x)+12 b^2 c d x \csc (a) \sin (3 a+2 b x)+48 b^2 d^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+48 b^2 d^2 x^2 \log \left (1+e^{-i (a+b x)}\right )-6 b^2 d^2 x^2 \csc (a) \sin (a+2 b x)+6 b^2 d^2 x^2 \csc (a) \sin (3 a+2 b x)-48 i b c d \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+96 b c d \tan ^{-1}(\tan (a)) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-6 b c d \csc (a) \cos (a+2 b x)+6 b c d \csc (a) \cos (3 a+2 b x)-96 b c d \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+96 i b d^2 x \text {Li}_2\left (-e^{-i (a+b x)}\right )+96 i b d^2 x \text {Li}_2\left (e^{-i (a+b x)}\right )+96 d^2 \text {Li}_3\left (-e^{-i (a+b x)}\right )+96 d^2 \text {Li}_3\left (e^{-i (a+b x)}\right )-6 b d^2 x \csc (a) \cos (a+2 b x)+6 b d^2 x \csc (a) \cos (3 a+2 b x)+3 d^2 \csc (a) \sin (a+2 b x)-3 d^2 \csc (a) \sin (3 a+2 b x)+16 i b^3 d^2 x^3+48 i \pi b^2 c d x+48 \pi b c d \log \left (1+e^{-2 i b x}\right )-48 \pi b c d \log (\cos (b x))}{48 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Cos[a + b*x]^2*Cot[a + b*x],x]

[Out]

((48*I)*b^2*c*d*Pi*x + (16*I)*b^3*d^2*x^3 - (96*I)*b^2*c*d*x*ArcTan[Tan[a]] + 48*b^3*c*d*x^2*Cot[a] - 6*b*c*d*
Cos[a + 2*b*x]*Csc[a] - 6*b*d^2*x*Cos[a + 2*b*x]*Csc[a] + 6*b*c*d*Cos[3*a + 2*b*x]*Csc[a] + 6*b*d^2*x*Cos[3*a
+ 2*b*x]*Csc[a] + 48*b*c*d*Pi*Log[1 + E^((-2*I)*b*x)] + 48*b^2*d^2*x^2*Log[1 - E^((-I)*(a + b*x))] + 48*b^2*d^
2*x^2*Log[1 + E^((-I)*(a + b*x))] + 96*b^2*c*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 96*b*c*d*ArcTan[T
an[a]]*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] - 48*b*c*d*Pi*Log[Cos[b*x]] + 48*b^2*c^2*Log[Sin[a + b*x]] -
96*b*c*d*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + (96*I)*b*d^2*x*PolyLog[2, -E^((-I)*(a + b*x))] + (96*
I)*b*d^2*x*PolyLog[2, E^((-I)*(a + b*x))] - (48*I)*b*c*d*PolyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 96*d^2
*PolyLog[3, -E^((-I)*(a + b*x))] + 96*d^2*PolyLog[3, E^((-I)*(a + b*x))] - 48*b^3*c*d*E^(I*ArcTan[Tan[a]])*x^2
*Cot[a]*Sqrt[Sec[a]^2] - 6*b^2*c^2*Csc[a]*Sin[a + 2*b*x] + 3*d^2*Csc[a]*Sin[a + 2*b*x] - 12*b^2*c*d*x*Csc[a]*S
in[a + 2*b*x] - 6*b^2*d^2*x^2*Csc[a]*Sin[a + 2*b*x] + 6*b^2*c^2*Csc[a]*Sin[3*a + 2*b*x] - 3*d^2*Csc[a]*Sin[3*a
+ 2*b*x] + 12*b^2*c*d*x*Csc[a]*Sin[3*a + 2*b*x] + 6*b^2*d^2*x^2*Csc[a]*Sin[3*a + 2*b*x])/(48*b^3)

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fricas [C]  time = 0.56, size = 594, normalized size = 3.28 \[ -\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 4 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 4 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 4 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - {\left (-4 i \, b d^{2} x - 4 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - {\left (4 i \, b d^{2} x + 4 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (4 i \, b d^{2} x + 4 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - {\left (-4 i \, b d^{2} x - 4 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)^2*cot(b*x+a),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)^2*cot(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^2*cos(b*x + a)^2*cot(b*x + a), x)

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maple [B]  time = 0.49, size = 535, normalized size = 2.96 \[ \frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}-\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {4 i d^{2} a^{3}}{3 b^{3}}-i c d \,x^{2}-\frac {4 i c d a x}{b}+i c^{2} x -\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {i d^{2} x^{3}}{3}+\frac {2 d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d \left (d x +c \right ) \sin \left (2 b x +2 a \right )}{4 b^{2}}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {4 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 i d^{2} a^{2} x}{b^{2}}-\frac {2 i c d \,a^{2}}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {\left (2 d^{2} x^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}-d^{2}\right ) \cos \left (2 b x +2 a \right )}{8 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*cos(b*x+a)^2*cot(b*x+a),x)

[Out]

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

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maxima [B]  time = 0.46, size = 522, normalized size = 2.88 \[ -\frac {12 \, {\left (\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )\right )} c^{2} - \frac {24 \, {\left (\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )\right )} a c d}{b} + \frac {12 \, {\left (\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )\right )} a^{2} d^{2}}{b^{2}} - \frac {-8 i \, {\left (b x + a\right )}^{3} d^{2} + {\left (-24 i \, b c d + 24 i \, a d^{2}\right )} {\left (b x + a\right )}^{2} + 48 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) + 48 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) + {\left (24 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (48 i \, b c d - 48 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (-24 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (-48 i \, b c d + 48 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} d^{2} + 4 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (-48 i \, b c d - 48 i \, {\left (b x + a\right )} d^{2} + 48 i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (-48 i \, b c d - 48 i \, {\left (b x + a\right )} d^{2} + 48 i \, a d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 12 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + 12 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - 6 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{b^{2}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)^2*cot(b*x+a),x, algorithm="maxima")

[Out]

-1/24*(12*(sin(b*x + a)^2 - log(sin(b*x + a)^2))*c^2 - 24*(sin(b*x + a)^2 - log(sin(b*x + a)^2))*a*c*d/b + 12*
(sin(b*x + a)^2 - log(sin(b*x + a)^2))*a^2*d^2/b^2 - (-8*I*(b*x + a)^3*d^2 + (-24*I*b*c*d + 24*I*a*d^2)*(b*x +
a)^2 + 48*d^2*polylog(3, -e^(I*b*x + I*a)) + 48*d^2*polylog(3, e^(I*b*x + I*a)) + (24*I*(b*x + a)^2*d^2 + (48
*I*b*c*d - 48*I*a*d^2)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (-24*I*(b*x + a)^2*d^2 + (-48*I*b*
c*d + 48*I*a*d^2)*(b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 3*(2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^
2)*(b*x + a) - d^2)*cos(2*b*x + 2*a) + (-48*I*b*c*d - 48*I*(b*x + a)*d^2 + 48*I*a*d^2)*dilog(-e^(I*b*x + I*a))
+ (-48*I*b*c*d - 48*I*(b*x + a)*d^2 + 48*I*a*d^2)*dilog(e^(I*b*x + I*a)) + 12*((b*x + a)^2*d^2 + 2*(b*c*d - a
*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + 12*((b*x + a)^2*d^2 + 2*(b*c*d -
a*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - 6*(b*c*d + (b*x + a)*d^2 - a*d^2
)*sin(2*b*x + 2*a))/b^2)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (a+b\,x\right )}^2\,\mathrm {cot}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*cot(a + b*x)*(c + d*x)^2,x)

[Out]

int(cos(a + b*x)^2*cot(a + b*x)*(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cos(b*x+a)**2*cot(b*x+a),x)

[Out]

Integral((c + d*x)**2*cos(a + b*x)**2*cot(a + b*x), x)

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