3.180 \(\int (c+d x)^2 \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=168 \[ -\frac {d^2 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \log (\sin (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \cot (a+b x)}{b^2}-\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \cot ^2(a+b x)}{2 b}-\frac {c d x}{b}-\frac {d^2 x^2}{2 b}+\frac {i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3720, 3475, 3717, 2190, 2531, 2282, 6589} \[ \frac {i d (c+d x) \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \cot (a+b x)}{b^2}+\frac {d^2 \log (\sin (a+b x))}{b^3}-\frac {(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \cot ^2(a+b x)}{2 b}-\frac {c d x}{b}-\frac {d^2 x^2}{2 b}+\frac {i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cot[a + b*x]^3,x]

[Out]

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

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 3720

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

Rubi steps

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

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Mathematica [B]  time = 6.68, size = 540, normalized size = 3.21 \[ \frac {d^2 \csc (a) (\sin (a) \log (\sin (a) \cos (b x)+\cos (a) \sin (b x))-b x \cos (a))}{b^3 \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac {\csc (a) \csc (a+b x) \left (c d \sin (b x)+d^2 x \sin (b x)\right )}{b^2}+\frac {c d \csc (a) \sec (a) \left (b^2 x^2 e^{i \tan ^{-1}(\tan (a))}+\frac {\tan (a) \left (i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\tan ^2(a)+1}}\right )}{b^2 \sqrt {\sec ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac {e^{i a} d^2 \csc (a) \left (2 e^{-2 i a} b^3 x^3+3 i \left (1-e^{-2 i a}\right ) b^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 i \left (1-e^{-2 i a}\right ) b^2 x^2 \log \left (1+e^{-i (a+b x)}\right )-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (-e^{-i (a+b x)}\right )-i \text {Li}_3\left (-e^{-i (a+b x)}\right )\right )-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (e^{-i (a+b x)}\right )-i \text {Li}_3\left (e^{-i (a+b x)}\right )\right )\right )}{6 b^3}-\frac {c^2 \csc (a) (\sin (a) \log (\sin (a) \cos (b x)+\cos (a) \sin (b x))-b x \cos (a))}{b \left (\sin ^2(a)+\cos ^2(a)\right )}-\frac {(c+d x)^2 \csc ^2(a+b x)}{2 b}-\frac {1}{3} x \cot (a) \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [C]  time = 0.47, size = 655, normalized size = 3.90 \[ \frac {4 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c d x + 4 \, b^{2} c^{2} + {\left (-2 i \, b d^{2} x - 2 i \, b c d + {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\left (2 i \, b d^{2} x + 2 i \, b c d + {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} - 1\right )} d^{2} - {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} - 1\right )} d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} - 1\right )} d^{2} - {\left (b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} - 1\right )} d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, 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} - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, 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} - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 4 \, {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, {\left (b^{3} \cos \left (2 \, b x + 2 \, a\right ) - b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*(4*b^2*d^2*x^2 + 8*b^2*c*d*x + 4*b^2*c^2 + (-2*I*b*d^2*x - 2*I*b*c*d + (2*I*b*d^2*x + 2*I*b*c*d)*cos(2*b*x
+ 2*a))*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + (2*I*b*d^2*x + 2*I*b*c*d + (-2*I*b*d^2*x - 2*I*b*c*d)*
cos(2*b*x + 2*a))*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) + 2*(b^2*c^2 - 2*a*b*c*d + (a^2 - 1)*d^2 - (b^2
*c^2 - 2*a*b*c*d + (a^2 - 1)*d^2)*cos(2*b*x + 2*a))*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2)
+ 2*(b^2*c^2 - 2*a*b*c*d + (a^2 - 1)*d^2 - (b^2*c^2 - 2*a*b*c*d + (a^2 - 1)*d^2)*cos(2*b*x + 2*a))*log(-1/2*co
s(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*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 - (b^2*d^2
*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cos(2*b*x + 2*a))*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) +
2*(b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2 - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cos(2*b
*x + 2*a))*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) - (d^2*cos(2*b*x + 2*a) - d^2)*polylog(3, cos(2*b*x
+ 2*a) + I*sin(2*b*x + 2*a)) - (d^2*cos(2*b*x + 2*a) - d^2)*polylog(3, cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a))
+ 4*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a))/(b^3*cos(2*b*x + 2*a) - b^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \cot \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*cot(b*x + a)^3, x)

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maple [B]  time = 0.11, size = 635, normalized size = 3.78 \[ -\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}+\frac {2 b \,d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}+4 b c d x \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b \,c^{2} {\mathrm e}^{2 i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{2 i \left (b x +a \right )}+2 i d^{2} x +2 i d c}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-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 {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 {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*cot(b*x+a)^3,x)

[Out]

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

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maxima [B]  time = 0.71, size = 1966, normalized size = 11.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(a + b*x)^3*(c + d*x)^2,x)

[Out]

int(cot(a + b*x)^3*(c + d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \cot ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cot(b*x+a)**3,x)

[Out]

Integral((c + d*x)**2*cot(a + b*x)**3, x)

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