3.53 \(\int \frac {(c+d x)^2}{a+b \cot (e+f x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3731, 2190, 2531, 2282, 6589} \[ \frac {i b d (c+d x) \text {PolyLog}\left (2,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f^2 \left (a^2+b^2\right )}-\frac {b d^2 \text {PolyLog}\left (3,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^3 \left (a^2+b^2\right )}-\frac {b (c+d x)^2 \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac {(c+d x)^3}{3 d (a-i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.55, size = 289, normalized size = 1.60 \[ \frac {x \sin (e) \left (3 c^2+3 c d x+d^2 x^2\right )}{3 (a \sin (e)+b \cos (e))}+\frac {b \left (\frac {3 d \left (b \left (1+e^{2 i e}\right )-i a \left (-1+e^{2 i e}\right )\right ) \left (2 f (c+d x) \text {Li}_2\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )-i d \text {Li}_3\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )\right )}{f^3 \left (a^2+b^2\right )}-\frac {6 \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right ) (c+d x)^2 \log \left (1+\frac {(-a+i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f \left (a^2+b^2\right )}+\frac {4 i (c+d x)^3}{d (a+i b)}\right )}{6 \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.75, size = 730, normalized size = 4.03 \[ \frac {4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, b d^{2} {\rm polylog}\left (3, \frac {{\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 3 \, b d^{2} {\rm polylog}\left (3, \frac {{\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) - 6 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\frac {1}{2} \, a^{2} + i \, a b - \frac {1}{2} \, b^{2} - \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) - 6 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\frac {1}{2} \, a^{2} + i \, a b + \frac {1}{2} \, b^{2} + \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{12 \, {\left (a^{2} + b^{2}\right )} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*a*d^2*f^3*x^3 + 12*a*c*d*f^3*x^2 + 12*a*c^2*f^3*x - 3*b*d^2*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)) - 3*b*d^2*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*I*b*d^2*f*x + 6*I*b*c*d*f)*dil
og(-(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^2*f*x - 6*I*b*c*d*f)*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*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*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)) - 6*(b*d^2*e^2 - 2*b*c
*d*e*f + b*c^2*f^2)*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)*si
n(2*f*x + 2*e)) - 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*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^2*f^2*x^2 + 2*b*c*d
*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*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)))/((a^2 + b^2)*f^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2/(b*cot(f*x + e) + a), x)

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maple [B]  time = 2.27, size = 881, normalized size = 4.87 \[ \frac {d^{2} x^{3}}{3 i b +3 a}+\frac {c d \,x^{2}}{i b +a}+\frac {c^{2} x}{i b +a}-\frac {i b \,c^{2} \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i {\mathrm e}^{2 i \left (f x +e \right )} b -a +i b \right )}{f \left (-i a +b \right ) \left (i b -a \right )}+\frac {i b \,d^{2} \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) x^{2}}{f \left (-i a +b \right ) \left (-i b +a \right )}+\frac {i b \,d^{2} \polylog \left (3, \frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right )}{2 f^{3} \left (-i a +b \right ) \left (-i b +a \right )}+\frac {2 i b c d e \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i {\mathrm e}^{2 i \left (f x +e \right )} b -a +i b \right )}{f^{2} \left (-i a +b \right ) \left (i b -a \right )}+\frac {2 b \,d^{2} x^{3}}{3 \left (-i a +b \right ) \left (-i b +a \right )}-\frac {2 b \,d^{2} e^{2} x}{f^{2} \left (-i a +b \right ) \left (-i b +a \right )}-\frac {4 b \,d^{2} e^{3}}{3 f^{3} \left (-i a +b \right ) \left (-i b +a \right )}-\frac {4 i b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (-i a +b \right ) \left (i b -a \right )}+\frac {2 i b c d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) x}{f \left (-i a +b \right ) \left (-i b +a \right )}+\frac {2 i b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3} \left (-i a +b \right ) \left (i b -a \right )}+\frac {2 i b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (-i a +b \right ) \left (i b -a \right )}-\frac {i b \,d^{2} e^{2} \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i {\mathrm e}^{2 i \left (f x +e \right )} b -a +i b \right )}{f^{3} \left (-i a +b \right ) \left (i b -a \right )}+\frac {b \,d^{2} \polylog \left (2, \frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) x}{f^{2} \left (-i a +b \right ) \left (-i b +a \right )}-\frac {i b \,d^{2} \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) e^{2}}{f^{3} \left (-i a +b \right ) \left (-i b +a \right )}+\frac {2 i b c d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) e}{f^{2} \left (-i a +b \right ) \left (-i b +a \right )}+\frac {2 b c d \,x^{2}}{\left (-i a +b \right ) \left (-i b +a \right )}+\frac {4 b c d e x}{f \left (-i a +b \right ) \left (-i b +a \right )}+\frac {2 b c d \,e^{2}}{f^{2} \left (-i a +b \right ) \left (-i b +a \right )}+\frac {b c d \polylog \left (2, \frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right )}{f^{2} \left (-i a +b \right ) \left (-i b +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2/(a+b*cot(f*x+e)),x)

[Out]

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

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maxima [B]  time = 0.82, size = 719, normalized size = 3.97 \[ -\frac {6 \, c d e {\left (\frac {2 \, {\left (f x + e\right )} a}{{\left (a^{2} + b^{2}\right )} f} - \frac {2 \, b \log \left (a \tan \left (f x + e\right ) + b\right )}{{\left (a^{2} + b^{2}\right )} f} + \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{{\left (a^{2} + b^{2}\right )} f}\right )} - 3 \, {\left (\frac {2 \, {\left (f x + e\right )} a}{a^{2} + b^{2}} - \frac {2 \, b \log \left (a \tan \left (f x + e\right ) + b\right )}{a^{2} + b^{2}} + \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{2} - \frac {2 \, {\left (f x + e\right )}^{3} {\left (a + i \, b\right )} d^{2} + 6 \, {\left (f x + e\right )} {\left (a + i \, b\right )} d^{2} e^{2} - 6 i \, b d^{2} e^{2} \arctan \left (b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) - b \sin \left (2 \, f x + 2 \, e\right ) - a\right ) - 3 \, b d^{2} e^{2} \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) - 3 \, b d^{2} {\rm Li}_{3}(\frac {{\left (i \, a - b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{i \, a + b}) - 6 \, {\left ({\left (a + i \, b\right )} d^{2} e - {\left (a + i \, b\right )} c d f\right )} {\left (f x + e\right )}^{2} - {\left (6 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (-12 i \, b d^{2} e + 12 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (-\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - {\left (-6 i \, {\left (f x + e\right )} b d^{2} + 6 i \, b d^{2} e - 6 i \, b c d f\right )} {\rm Li}_2\left (\frac {{\left (i \, a - b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{i \, a + b}\right ) - 3 \, {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{{\left (a^{2} + b^{2}\right )} f^{2}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*c*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)) - 3*(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^2 - (2*(f*x + e)^3*(a + I*b)*d^2 + 6*(f*x + e)*(a + I*b)*d^2*e^2 - 6*I*b*d^2*e
^2*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) - 3*b*d^2
*e^2*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)) - 3*b*d^2*polylog(3, (I*a - b)*e^(2*I*f*x + 2*I*e)/(I*a + b)) - 6*((a + I*b)
*d^2*e - (a + I*b)*c*d*f)*(f*x + e)^2 - (6*I*(f*x + e)^2*b*d^2 + (-12*I*b*d^2*e + 12*I*b*c*d*f)*(f*x + e))*arc
tan2(-(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)) - (-6*I*(f*x + e)*b*d^2 + 6*I*b*d^2*e - 6*I*b*c*d*f)*dilog((I*a
- b)*e^(2*I*f*x + 2*I*e)/(I*a + b)) - 3*((f*x + e)^2*b*d^2 - 2*(b*d^2*e - b*c*d*f)*(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)))/((a^2 + b^2)*f^2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/(a + b*cot(e + f*x)),x)

[Out]

int((c + d*x)^2/(a + b*cot(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2/(a+b*cot(f*x+e)),x)

[Out]

Integral((c + d*x)**2/(a + b*cot(e + f*x)), x)

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