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

Optimal. Leaf size=114 \[ -\frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4408, 4404, 2635, 8, 3717, 2190, 2279, 2391} \[ -\frac {i d \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \sin (a+b x) \cos (a+b x)}{4 b^2}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {(c+d x) \sin ^2(a+b x)}{2 b}+\frac {d x}{4 b}-\frac {i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2635

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 131, normalized size = 1.15 \[ \frac {d \left ((a+b x) \log \left (1-e^{2 i (a+b x)}\right )-\frac {1}{2} i \left ((a+b x)^2+\text {Li}_2\left (e^{2 i (a+b x)}\right )\right )\right )}{b^2}-\frac {d \sin (2 (a+b x))}{8 b^2}-\frac {a d \log (\sin (a+b x))}{b^2}-\frac {c \sin ^2(a+b x)}{2 b}+\frac {c \log (\sin (a+b x))}{b}+\frac {d x \cos (2 (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Cos[2*(a + b*x)])/(4*b) + (c*Log[Sin[a + b*x]])/b - (a*d*Log[Sin[a + b*x]])/b^2 + (d*((a + b*x)*Log[1 - E
^((2*I)*(a + b*x))] - (I/2)*((a + b*x)^2 + PolyLog[2, E^((2*I)*(a + b*x))])))/b^2 - (c*Sin[a + b*x]^2)/(2*b) -
(d*Sin[2*(a + b*x)])/(8*b^2)

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fricas [B]  time = 0.59, size = 292, normalized size = 2.56 \[ -\frac {b d x - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*d*x - 2*(b*d*x + b*c)*cos(b*x + a)^2 + d*cos(b*x + a)*sin(b*x + a) + 2*I*d*dilog(cos(b*x + a) + I*sin(
b*x + a)) - 2*I*d*dilog(cos(b*x + a) - I*sin(b*x + a)) - 2*I*d*dilog(-cos(b*x + a) + I*sin(b*x + a)) + 2*I*d*d
ilog(-cos(b*x + a) - I*sin(b*x + a)) - 2*(b*d*x + b*c)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - 2*(b*d*x + b*c
)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 2*(b*c - a*d)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 2
*(b*c - a*d)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - 2*(b*d*x + a*d)*log(-cos(b*x + a) + I*sin(b*x
+ a) + 1) - 2*(b*d*x + a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + 1))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.41, size = 249, normalized size = 2.18 \[ i c x -\frac {i d \,x^{2}}{2}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {i d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {i d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {i d \,a^{2}}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 i d a x}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {\left (d x +c \right ) \cos \left (2 b x +2 a \right )}{4 b}-\frac {d \sin \left (2 b x +2 a \right )}{8 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.79, size = 222, normalized size = 1.95 \[ \frac {-4 i \, b^{2} d x^{2} - 8 i \, b^{2} c x - 8 i \, b d x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 8 i \, b c \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + {\left (8 i \, b d x + 8 i \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + 2 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - 8 i \, d {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 8 i \, d {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - d \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-4*I*b^2*d*x^2 - 8*I*b^2*c*x - 8*I*b*d*x*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 8*I*b*c*arctan2(sin(b
*x + a), cos(b*x + a) - 1) + (8*I*b*d*x + 8*I*b*c)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + 2*(b*d*x + b*c)*c
os(2*b*x + 2*a) - 8*I*d*dilog(-e^(I*b*x + I*a)) - 8*I*d*dilog(e^(I*b*x + I*a)) + 4*(b*d*x + b*c)*log(cos(b*x +
a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + 4*(b*d*x + b*c)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x
+ a) + 1) - d*sin(2*b*x + 2*a))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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