3.52 \(\int \frac {(c+d x)^3}{a+b \cot (e+f x)} \, dx\)

Optimal. Leaf size=242 \[ -\frac {3 b d^2 (c+d x) \text {Li}_3\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^3 \left (a^2+b^2\right )}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}-\frac {3 i b d^3 \text {Li}_4\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{4 f^4 \left (a^2+b^2\right )}+\frac {(c+d x)^4}{4 d (a-i b)} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3731, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3 b d^2 (c+d x) \text {PolyLog}\left (3,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^3 \left (a^2+b^2\right )}+\frac {3 i b d (c+d x)^2 \text {PolyLog}\left (2,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}-\frac {3 i b d^3 \text {PolyLog}\left (4,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{4 f^4 \left (a^2+b^2\right )}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac {(c+d x)^4}{4 d (a-i b)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3/(a + b*Cot[e + f*x]),x]

[Out]

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

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 3731

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^
(m + 1)/(d*(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x])/((a +
I*b)^2 + (a^2 + b^2)*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Integer
Q[4*k] && NeQ[a^2 + b^2, 0] && 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 \frac {(c+d x)^3}{a+b \cot (e+f x)} \, dx &=\frac {(c+d x)^4}{4 (a-i b) d}+(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)^3}{(a-i b)^2+\left (-a^2-b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac {(c+d x)^4}{4 (a-i b) d}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {(3 b d) \int (c+d x)^2 \log \left (1+\frac {\left (-a^2-b^2\right ) e^{2 i (e+f x)}}{(a-i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f}\\ &=\frac {(c+d x)^4}{4 (a-i b) d}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2}-\frac {\left (3 i b d^2\right ) \int (c+d x) \text {Li}_2\left (-\frac {\left (-a^2-b^2\right ) e^{2 i (e+f x)}}{(a-i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f^2}\\ &=\frac {(c+d x)^4}{4 (a-i b) d}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^3}+\frac {\left (3 b d^3\right ) \int \text {Li}_3\left (-\frac {\left (-a^2-b^2\right ) e^{2 i (e+f x)}}{(a-i b)^2}\right ) \, dx}{2 \left (a^2+b^2\right ) f^3}\\ &=\frac {(c+d x)^4}{4 (a-i b) d}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^3}-\frac {\left (3 i b d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(a+i b) x}{a-i b}\right )}{x} \, dx,x,e^{2 i (e+f x)}\right )}{4 \left (a^2+b^2\right ) f^4}\\ &=\frac {(c+d x)^4}{4 (a-i b) d}-\frac {b (c+d x)^3 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^3}-\frac {3 i b d^3 \text {Li}_4\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{4 \left (a^2+b^2\right ) f^4}\\ \end {align*}

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Mathematica [A]  time = 2.14, size = 345, normalized size = 1.43 \[ \frac {x \sin (e) \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )}{4 (a \sin (e)+b \cos (e))}+\frac {b \left (\frac {3 d \left (b \left (1+e^{2 i e}\right )-i a \left (-1+e^{2 i e}\right )\right ) \left (2 f^2 (c+d x)^2 \text {Li}_2\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )+d \left (-2 i f (c+d x) \text {Li}_3\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )-d \text {Li}_4\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )\right )\right )}{f^4 \left (a^2+b^2\right )}-\frac {4 \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right ) (c+d x)^3 \log \left (1+\frac {(-a+i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f \left (a^2+b^2\right )}+\frac {2 i (c+d x)^4}{d (a+i b)}\right )}{4 \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(((2*I)*(c + d*x)^4)/((a + I*b)*d) - (4*(a*(-1 + E^((2*I)*e)) + I*b*(1 + E^((2*I)*e)))*(c + d*x)^3*Log[1 +
(-a + I*b)/((a + I*b)*E^((2*I)*(e + f*x)))])/((a^2 + b^2)*f) + (3*d*((-I)*a*(-1 + E^((2*I)*e)) + b*(1 + E^((2*
I)*e)))*(2*f^2*(c + d*x)^2*PolyLog[2, (a - I*b)/((a + I*b)*E^((2*I)*(e + f*x)))] + d*((-2*I)*f*(c + d*x)*PolyL
og[3, (a - I*b)/((a + I*b)*E^((2*I)*(e + f*x)))] - d*PolyLog[4, (a - I*b)/((a + I*b)*E^((2*I)*(e + f*x)))])))/
((a^2 + b^2)*f^4)))/(4*(a*(-1 + E^((2*I)*e)) + I*b*(1 + E^((2*I)*e)))) + (x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 +
d^3*x^3)*Sin[e])/(4*(b*Cos[e] + a*Sin[e]))

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fricas [C]  time = 0.59, size = 1029, normalized size = 4.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^3/(b*cot(f*x + e) + a), x)

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maple [B]  time = 2.56, size = 1386, normalized size = 5.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3/(a+b*cot(f*x+e)),x)

[Out]

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

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maxima [B]  time = 0.85, size = 987, normalized size = 4.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(18*c^2*d*e*(2*(f*x + e)*a/((a^2 + b^2)*f) - 2*b*log(a*tan(f*x + e) + b)/((a^2 + b^2)*f) + b*log(tan(f*x
+ e)^2 + 1)/((a^2 + b^2)*f)) - 6*(2*(f*x + e)*a/(a^2 + b^2) - 2*b*log(a*tan(f*x + e) + b)/(a^2 + b^2) + b*log
(tan(f*x + e)^2 + 1)/(a^2 + b^2))*c^3 - (3*(f*x + e)^4*(a + I*b)*d^3 - 12*I*b*d^3*polylog(4, (I*a - b)*e^(2*I*
f*x + 2*I*e)/(I*a + b)) - 12*((a + I*b)*d^3*e - (a + I*b)*c*d^2*f)*(f*x + e)^3 + 18*((a + I*b)*d^3*e^2 - 2*(a
+ I*b)*c*d^2*e*f + (a + I*b)*c^2*d*f^2)*(f*x + e)^2 - 12*((a + I*b)*d^3*e^3 - 3*(a + I*b)*c*d^2*e^2*f)*(f*x +
e) - (-12*I*b*d^3*e^3 + 36*I*b*c*d^2*e^2*f)*arctan2(b*cos(2*f*x + 2*e) + a*sin(2*f*x + 2*e) + b, a*cos(2*f*x +
2*e) - b*sin(2*f*x + 2*e) - a) - (16*I*(f*x + e)^3*b*d^3 + (-36*I*b*d^3*e + 36*I*b*c*d^2*f)*(f*x + e)^2 + (36
*I*b*d^3*e^2 - 72*I*b*c*d^2*e*f + 36*I*b*c^2*d*f^2)*(f*x + e))*arctan2(-(2*a*b*cos(2*f*x + 2*e) + (a^2 - b^2)*
sin(2*f*x + 2*e))/(a^2 + b^2), (2*a*b*sin(2*f*x + 2*e) + a^2 + b^2 - (a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)
) - (-24*I*(f*x + e)^2*b*d^3 - 18*I*b*d^3*e^2 + 36*I*b*c*d^2*e*f - 18*I*b*c^2*d*f^2 + (36*I*b*d^3*e - 36*I*b*c
*d^2*f)*(f*x + e))*dilog((I*a - b)*e^(2*I*f*x + 2*I*e)/(I*a + b)) + 6*(b*d^3*e^3 - 3*b*c*d^2*e^2*f)*log((a^2 +
b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 - 2*(a^2 - b^2)
*cos(2*f*x + 2*e)) - 2*(4*(f*x + e)^3*b*d^3 - 9*(b*d^3*e - b*c*d^2*f)*(f*x + e)^2 + 9*(b*d^3*e^2 - 2*b*c*d^2*e
*f + b*c^2*d*f^2)*(f*x + e))*log(((a^2 + b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*
f*x + 2*e)^2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)) - 6*(4*(f*x + e)*b*d^3 - 3*b*d^3*e + 3
*b*c*d^2*f)*polylog(3, (I*a - b)*e^(2*I*f*x + 2*I*e)/(I*a + b)))/((a^2 + b^2)*f^3))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/(a + b*cot(e + f*x)),x)

[Out]

int((c + d*x)^3/(a + b*cot(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3/(a+b*cot(f*x+e)),x)

[Out]

Integral((c + d*x)**3/(a + b*cot(e + f*x)), x)

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