3.37 \(\int (c+d x)^3 (a+b \cot (e+f x)) \, dx\)
Optimal. Leaf size=147 \[ \frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,e^{2 i (e+f x)}\right )}{2 f^3}-\frac{3 i b d (c+d x)^2 \text{PolyLog}\left (2,e^{2 i (e+f x)}\right )}{2 f^2}+\frac{3 i b d^3 \text{PolyLog}\left (4,e^{2 i (e+f x)}\right )}{4 f^4}+\frac{a (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b (c+d x)^4}{4 d} \]
[Out]
(a*(c + d*x)^4)/(4*d) - ((I/4)*b*(c + d*x)^4)/d + (b*(c + d*x)^3*Log[1 - E^((2*I)*(e + f*x))])/f - (((3*I)/2)*
b*d*(c + d*x)^2*PolyLog[2, E^((2*I)*(e + f*x))])/f^2 + (3*b*d^2*(c + d*x)*PolyLog[3, E^((2*I)*(e + f*x))])/(2*
f^3) + (((3*I)/4)*b*d^3*PolyLog[4, E^((2*I)*(e + f*x))])/f^4
________________________________________________________________________________________
Rubi [A] time = 0.255047, antiderivative size = 147, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used =
{3722, 3717, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,e^{2 i (e+f x)}\right )}{2 f^3}-\frac{3 i b d (c+d x)^2 \text{PolyLog}\left (2,e^{2 i (e+f x)}\right )}{2 f^2}+\frac{3 i b d^3 \text{PolyLog}\left (4,e^{2 i (e+f x)}\right )}{4 f^4}+\frac{a (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b (c+d x)^4}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^3*(a + b*Cot[e + f*x]),x]
[Out]
(a*(c + d*x)^4)/(4*d) - ((I/4)*b*(c + d*x)^4)/d + (b*(c + d*x)^3*Log[1 - E^((2*I)*(e + f*x))])/f - (((3*I)/2)*
b*d*(c + d*x)^2*PolyLog[2, E^((2*I)*(e + f*x))])/f^2 + (3*b*d^2*(c + d*x)*PolyLog[3, E^((2*I)*(e + f*x))])/(2*
f^3) + (((3*I)/4)*b*d^3*PolyLog[4, E^((2*I)*(e + f*x))])/f^4
Rule 3722
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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 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 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 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]
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 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)^3 (a+b \cot (e+f x)) \, dx &=\int \left (a (c+d x)^3+b (c+d x)^3 \cot (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^4}{4 d}+b \int (c+d x)^3 \cot (e+f x) \, dx\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}-(2 i b) \int \frac{e^{2 i (e+f x)} (c+d x)^3}{1-e^{2 i (e+f x)}} \, dx\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{(3 b d) \int (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (e^{2 i (e+f x)}\right )}{2 f^2}+\frac{\left (3 i b d^2\right ) \int (c+d x) \text{Li}_2\left (e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (e^{2 i (e+f x)}\right )}{2 f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (e^{2 i (e+f x)}\right )}{2 f^3}-\frac{\left (3 b d^3\right ) \int \text{Li}_3\left (e^{2 i (e+f x)}\right ) \, dx}{2 f^3}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (e^{2 i (e+f x)}\right )}{2 f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (e^{2 i (e+f x)}\right )}{2 f^3}+\frac{\left (3 i b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{4 f^4}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{i b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{3 i b d (c+d x)^2 \text{Li}_2\left (e^{2 i (e+f x)}\right )}{2 f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (e^{2 i (e+f x)}\right )}{2 f^3}+\frac{3 i b d^3 \text{Li}_4\left (e^{2 i (e+f x)}\right )}{4 f^4}\\ \end{align*}
Mathematica [B] time = 7.44631, size = 730, normalized size = 4.97 \[ -\frac{3 b c^2 d \csc (e) \sec (e) \left (\frac{\tan (e) \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )+i f x \left (2 \tan ^{-1}(\tan (e))-\pi \right )-2 \left (\tan ^{-1}(\tan (e))+f x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )+2 \tan ^{-1}(\tan (e)) \log \left (\sin \left (\tan ^{-1}(\tan (e))+f x\right )\right )-\pi \log \left (1+e^{-2 i f x}\right )+\pi \log (\cos (f x))\right )}{\sqrt{\tan ^2(e)+1}}+f^2 x^2 e^{i \tan ^{-1}(\tan (e))}\right )}{2 f^2 \sqrt{\sec ^2(e) \left (\sin ^2(e)+\cos ^2(e)\right )}}-\frac{b c d^2 e^{i e} \csc (e) \left (-6 e^{-2 i e} \left (-1+e^{2 i e}\right ) \left (f x \text{PolyLog}\left (2,-e^{-i (e+f x)}\right )-i \text{PolyLog}\left (3,-e^{-i (e+f x)}\right )\right )-6 e^{-2 i e} \left (-1+e^{2 i e}\right ) \left (f x \text{PolyLog}\left (2,e^{-i (e+f x)}\right )-i \text{PolyLog}\left (3,e^{-i (e+f x)}\right )\right )+2 e^{-2 i e} f^3 x^3+3 i \left (1-e^{-2 i e}\right ) f^2 x^2 \log \left (1-e^{-i (e+f x)}\right )+3 i \left (1-e^{-2 i e}\right ) f^2 x^2 \log \left (1+e^{-i (e+f x)}\right )\right )}{2 f^3}-\frac{b d^3 e^{i e} \csc (e) \left (-6 e^{-2 i e} \left (-1+e^{2 i e}\right ) \left (f^2 x^2 \text{PolyLog}\left (2,-e^{-i (e+f x)}\right )-2 i f x \text{PolyLog}\left (3,-e^{-i (e+f x)}\right )-2 \text{PolyLog}\left (4,-e^{-i (e+f x)}\right )\right )-6 e^{-2 i e} \left (-1+e^{2 i e}\right ) \left (f^2 x^2 \text{PolyLog}\left (2,e^{-i (e+f x)}\right )-2 i f x \text{PolyLog}\left (3,e^{-i (e+f x)}\right )-2 \text{PolyLog}\left (4,e^{-i (e+f x)}\right )\right )+e^{-2 i e} f^4 x^4+2 i \left (1-e^{-2 i e}\right ) f^3 x^3 \log \left (1-e^{-i (e+f x)}\right )+2 i \left (1-e^{-2 i e}\right ) f^3 x^3 \log \left (1+e^{-i (e+f x)}\right )\right )}{4 f^4}+\frac{1}{4} x \csc (e) \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) (a \sin (e)+b \cos (e))+\frac{b c^3 \csc (e) (\sin (e) \log (\sin (e) \cos (f x)+\cos (e) \sin (f x))-f x \cos (e))}{f \left (\sin ^2(e)+\cos ^2(e)\right )} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*x)^3*(a + b*Cot[e + f*x]),x]
[Out]
-(b*c*d^2*E^(I*e)*Csc[e]*((2*f^3*x^3)/E^((2*I)*e) + (3*I)*(1 - E^((-2*I)*e))*f^2*x^2*Log[1 - E^((-I)*(e + f*x)
)] + (3*I)*(1 - E^((-2*I)*e))*f^2*x^2*Log[1 + E^((-I)*(e + f*x))] - (6*(-1 + E^((2*I)*e))*(f*x*PolyLog[2, -E^(
(-I)*(e + f*x))] - I*PolyLog[3, -E^((-I)*(e + f*x))]))/E^((2*I)*e) - (6*(-1 + E^((2*I)*e))*(f*x*PolyLog[2, E^(
(-I)*(e + f*x))] - I*PolyLog[3, E^((-I)*(e + f*x))]))/E^((2*I)*e)))/(2*f^3) - (b*d^3*E^(I*e)*Csc[e]*((f^4*x^4)
/E^((2*I)*e) + (2*I)*(1 - E^((-2*I)*e))*f^3*x^3*Log[1 - E^((-I)*(e + f*x))] + (2*I)*(1 - E^((-2*I)*e))*f^3*x^3
*Log[1 + E^((-I)*(e + f*x))] - (6*(-1 + E^((2*I)*e))*(f^2*x^2*PolyLog[2, -E^((-I)*(e + f*x))] - (2*I)*f*x*Poly
Log[3, -E^((-I)*(e + f*x))] - 2*PolyLog[4, -E^((-I)*(e + f*x))]))/E^((2*I)*e) - (6*(-1 + E^((2*I)*e))*(f^2*x^2
*PolyLog[2, E^((-I)*(e + f*x))] - (2*I)*f*x*PolyLog[3, E^((-I)*(e + f*x))] - 2*PolyLog[4, E^((-I)*(e + f*x))])
)/E^((2*I)*e)))/(4*f^4) + (x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Csc[e]*(b*Cos[e] + a*Sin[e]))/4 + (b*
c^3*Csc[e]*(-(f*x*Cos[e]) + Log[Cos[f*x]*Sin[e] + Cos[e]*Sin[f*x]]*Sin[e]))/(f*(Cos[e]^2 + Sin[e]^2)) - (3*b*c
^2*d*Csc[e]*Sec[e]*(E^(I*ArcTan[Tan[e]])*f^2*x^2 + ((I*f*x*(-Pi + 2*ArcTan[Tan[e]]) - Pi*Log[1 + E^((-2*I)*f*x
)] - 2*(f*x + ArcTan[Tan[e]])*Log[1 - E^((2*I)*(f*x + ArcTan[Tan[e]]))] + Pi*Log[Cos[f*x]] + 2*ArcTan[Tan[e]]*
Log[Sin[f*x + ArcTan[Tan[e]]]] + I*PolyLog[2, E^((2*I)*(f*x + ArcTan[Tan[e]]))])*Tan[e])/Sqrt[1 + Tan[e]^2]))/
(2*f^2*Sqrt[Sec[e]^2*(Cos[e]^2 + Sin[e]^2)])
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Maple [B] time = 0.396, size = 857, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^3*(a+b*cot(f*x+e)),x)
[Out]
I*b*c^3*x+a*c^3*x+a*c*d^2*x^3+1/4*a*d^3*x^4-I*b*c*d^2*x^3-3/2*I*b*c^2*d*x^2+6*b/f^3*d^3*polylog(3,-exp(I*(f*x+
e)))*x+6*b/f^3*d^3*polylog(3,exp(I*(f*x+e)))*x-1/4*I*b*d^3*x^4+b/f*d^3*ln(exp(I*(f*x+e))+1)*x^3+3/2*a*c^2*d*x^
2+3*b/f^3*c*d^2*e^2*ln(exp(I*(f*x+e))-1)-3*b/f^2*c^2*d*e*ln(exp(I*(f*x+e))-1)+3*b/f*c^2*d*ln(1-exp(I*(f*x+e)))
*x+3*b/f^2*c^2*d*ln(1-exp(I*(f*x+e)))*e+3*b/f*c^2*d*ln(exp(I*(f*x+e))+1)*x+b/f*c^3*ln(exp(I*(f*x+e))+1)+b/f*c^
3*ln(exp(I*(f*x+e))-1)-2*b/f*c^3*ln(exp(I*(f*x+e)))-3*I*b/f^2*d^3*polylog(2,exp(I*(f*x+e)))*x^2+6*b/f^3*c*d^2*
polylog(3,exp(I*(f*x+e)))+2*b/f^4*d^3*e^3*ln(exp(I*(f*x+e)))-b/f^4*d^3*e^3*ln(exp(I*(f*x+e))-1)+6*b/f^3*c*d^2*
polylog(3,-exp(I*(f*x+e)))-3/2*I*b/f^4*d^3*e^4+6*I*b/f^4*d^3*polylog(4,exp(I*(f*x+e)))+6*I*b/f^4*d^3*polylog(4
,-exp(I*(f*x+e)))+3*b/f*c*d^2*ln(1-exp(I*(f*x+e)))*x^2-3*b/f^3*c*d^2*e^2*ln(1-exp(I*(f*x+e)))+3*b/f*c*d^2*ln(e
xp(I*(f*x+e))+1)*x^2+6*b/f^2*c^2*d*e*ln(exp(I*(f*x+e)))-6*b/f^3*c*d^2*e^2*ln(exp(I*(f*x+e)))+4*I*b/f^3*c*d^2*e
^3-3*I*b/f^2*c^2*d*e^2-2*I*b/f^3*d^3*e^3*x-3*I*b/f^2*c^2*d*polylog(2,-exp(I*(f*x+e)))-3*I*b/f^2*d^3*polylog(2,
-exp(I*(f*x+e)))*x^2-3*I*b/f^2*c^2*d*polylog(2,exp(I*(f*x+e)))-6*I*b/f*c^2*d*e*x-6*I*b/f^2*c*d^2*polylog(2,exp
(I*(f*x+e)))*x-6*I*b/f^2*c*d^2*polylog(2,-exp(I*(f*x+e)))*x+6*I*b/f^2*c*d^2*e^2*x+b/f*d^3*ln(1-exp(I*(f*x+e)))
*x^3+b/f^4*d^3*ln(1-exp(I*(f*x+e)))*e^3
________________________________________________________________________________________
Maxima [B] time = 2.54188, size = 1304, normalized size = 8.87 \begin{align*} \text{result too large to display} \end{align*}
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]
1/4*(4*(f*x + e)*a*c^3 + (f*x + e)^4*a*d^3/f^3 - 4*(f*x + e)^3*a*d^3*e/f^3 + 6*(f*x + e)^2*a*d^3*e^2/f^3 - 4*(
f*x + e)*a*d^3*e^3/f^3 + 4*(f*x + e)^3*a*c*d^2/f^2 - 12*(f*x + e)^2*a*c*d^2*e/f^2 + 12*(f*x + e)*a*c*d^2*e^2/f
^2 + 6*(f*x + e)^2*a*c^2*d/f - 12*(f*x + e)*a*c^2*d*e/f + 4*b*c^3*log(sin(f*x + e)) - 4*b*d^3*e^3*log(sin(f*x
+ e))/f^3 + 12*b*c*d^2*e^2*log(sin(f*x + e))/f^2 - 12*b*c^2*d*e*log(sin(f*x + e))/f + (-I*(f*x + e)^4*b*d^3 +
24*I*b*d^3*polylog(4, -e^(I*f*x + I*e)) + 24*I*b*d^3*polylog(4, e^(I*f*x + I*e)) + (4*I*b*d^3*e - 4*I*b*c*d^2*
f)*(f*x + e)^3 + (-6*I*b*d^3*e^2 + 12*I*b*c*d^2*e*f - 6*I*b*c^2*d*f^2)*(f*x + e)^2 + (4*I*(f*x + e)^3*b*d^3 +
(-12*I*b*d^3*e + 12*I*b*c*d^2*f)*(f*x + e)^2 + (12*I*b*d^3*e^2 - 24*I*b*c*d^2*e*f + 12*I*b*c^2*d*f^2)*(f*x + e
))*arctan2(sin(f*x + e), cos(f*x + e) + 1) + (-4*I*(f*x + e)^3*b*d^3 + (12*I*b*d^3*e - 12*I*b*c*d^2*f)*(f*x +
e)^2 + (-12*I*b*d^3*e^2 + 24*I*b*c*d^2*e*f - 12*I*b*c^2*d*f^2)*(f*x + e))*arctan2(sin(f*x + e), -cos(f*x + e)
+ 1) + (-12*I*(f*x + e)^2*b*d^3 - 12*I*b*d^3*e^2 + 24*I*b*c*d^2*e*f - 12*I*b*c^2*d*f^2 + (24*I*b*d^3*e - 24*I*
b*c*d^2*f)*(f*x + e))*dilog(-e^(I*f*x + I*e)) + (-12*I*(f*x + e)^2*b*d^3 - 12*I*b*d^3*e^2 + 24*I*b*c*d^2*e*f -
12*I*b*c^2*d*f^2 + (24*I*b*d^3*e - 24*I*b*c*d^2*f)*(f*x + e))*dilog(e^(I*f*x + I*e)) + 2*((f*x + e)^3*b*d^3 -
3*(b*d^3*e - b*c*d^2*f)*(f*x + e)^2 + 3*(b*d^3*e^2 - 2*b*c*d^2*e*f + b*c^2*d*f^2)*(f*x + e))*log(cos(f*x + e)
^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) + 2*((f*x + e)^3*b*d^3 - 3*(b*d^3*e - b*c*d^2*f)*(f*x + e)^2 + 3*(b*
d^3*e^2 - 2*b*c*d^2*e*f + b*c^2*d*f^2)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*cos(f*x + e) + 1) +
24*((f*x + e)*b*d^3 - b*d^3*e + b*c*d^2*f)*polylog(3, -e^(I*f*x + I*e)) + 24*((f*x + e)*b*d^3 - b*d^3*e + b*c*
d^2*f)*polylog(3, e^(I*f*x + I*e)))/f^3)/f
________________________________________________________________________________________
Fricas [C] time = 1.92029, size = 1524, normalized size = 10.37 \begin{align*} \text{result too large to display} \end{align*}
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, cos(2*f*x
+ 2*e) + I*sin(2*f*x + 2*e)) - 3*I*b*d^3*polylog(4, cos(2*f*x + 2*e) - I*sin(2*f*x + 2*e)) + (-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(cos(2*f*x + 2*e) + I*sin(2*f*x + 2*e)) + (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(cos(2*f*x + 2*e) - I*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*cos(2*f*x + 2*e) + 1/2*I*sin(2*f*x + 2*e) + 1/2) - 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*cos(2*f*x + 2*e) - 1/2*I*sin(2*f*x + 2*e) + 1/2) +
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(-
cos(2*f*x + 2*e) + I*sin(2*f*x + 2*e) + 1) + 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(-cos(2*f*x + 2*e) - I*sin(2*f*x + 2*e) + 1) + 6*(b*d^3*f*x + b*c*d^
2*f)*polylog(3, cos(2*f*x + 2*e) + I*sin(2*f*x + 2*e)) + 6*(b*d^3*f*x + b*c*d^2*f)*polylog(3, cos(2*f*x + 2*e)
- I*sin(2*f*x + 2*e)))/f^4
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \cot{\left (e + f x \right )}\right ) \left (c + d x\right )^{3}\, dx \end{align*}
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((a + b*cot(e + f*x))*(c + d*x)**3, x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3}{\left (b \cot \left (f x + e\right ) + a\right )}\,{d x} \end{align*}
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)