3.40 \(\int (c+d x)^3 \cot (a+b x) \csc (a+b x) \, dx\)

Optimal. Leaf size=146 \[ -\frac {6 d^3 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^4}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b} \]

[Out]

-6*d*(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b^2-(d*x+c)^3*csc(b*x+a)/b+6*I*d^2*(d*x+c)*polylog(2,-exp(I*(b*x+a)))/b
^3-6*I*d^2*(d*x+c)*polylog(2,exp(I*(b*x+a)))/b^3-6*d^3*polylog(3,-exp(I*(b*x+a)))/b^4+6*d^3*polylog(3,exp(I*(b
*x+a)))/b^4

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Rubi [A]  time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4410, 4183, 2531, 2282, 6589} \[ \frac {6 i d^2 (c+d x) \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac {6 d^3 \text {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \text {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^4}-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*d*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b^2 - ((c + d*x)^3*Csc[a + b*x])/b + ((6*I)*d^2*(c + d*x)*PolyLog[
2, -E^(I*(a + b*x))])/b^3 - ((6*I)*d^2*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^3 - (6*d^3*PolyLog[3, -E^(I*(a
+ b*x))])/b^4 + (6*d^3*PolyLog[3, E^(I*(a + b*x))])/b^4

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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4410

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp
[((c + d*x)^m*Csc[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Fr
eeQ[{a, b, c, d, n}, x] && EqQ[p, 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)^3 \cot (a+b x) \csc (a+b x) \, dx &=-\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {(3 d) \int (c+d x)^2 \csc (a+b x) \, dx}{b}\\ &=-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b}-\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}-\frac {\left (6 i d^3\right ) \int \text {Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^3}+\frac {\left (6 i d^3\right ) \int \text {Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}-\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}\\ &=-\frac {6 d (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \csc (a+b x)}{b}+\frac {6 i d^2 (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^3}-\frac {6 d^3 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^4}\\ \end {align*}

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Mathematica [B]  time = 1.17, size = 311, normalized size = 2.13 \[ -\frac {b^3 c^3 \csc (a+b x)+3 b^3 c^2 d x \csc (a+b x)+3 b^3 c d^2 x^2 \csc (a+b x)+b^3 d^3 x^3 \csc (a+b x)-3 b^2 c^2 d \log \left (1-e^{i (a+b x)}\right )+3 b^2 c^2 d \log \left (1+e^{i (a+b x)}\right )-6 b^2 c d^2 x \log \left (1-e^{i (a+b x)}\right )+6 b^2 c d^2 x \log \left (1+e^{i (a+b x)}\right )-3 b^2 d^3 x^2 \log \left (1-e^{i (a+b x)}\right )+3 b^2 d^3 x^2 \log \left (1+e^{i (a+b x)}\right )-6 i b d^2 (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )+6 i b d^2 (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )+6 d^3 \text {Li}_3\left (-e^{i (a+b x)}\right )-6 d^3 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*c^3*Csc[a + b*x] + 3*b^3*c^2*d*x*Csc[a + b*x] + 3*b^3*c*d^2*x^2*Csc[a + b*x] + b^3*d^3*x^3*Csc[a + b*x]
- 3*b^2*c^2*d*Log[1 - E^(I*(a + b*x))] - 6*b^2*c*d^2*x*Log[1 - E^(I*(a + b*x))] - 3*b^2*d^3*x^2*Log[1 - E^(I*
(a + b*x))] + 3*b^2*c^2*d*Log[1 + E^(I*(a + b*x))] + 6*b^2*c*d^2*x*Log[1 + E^(I*(a + b*x))] + 3*b^2*d^3*x^2*Lo
g[1 + E^(I*(a + b*x))] - (6*I)*b*d^2*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))] + (6*I)*b*d^2*(c + d*x)*PolyLog[2,
E^(I*(a + b*x))] + 6*d^3*PolyLog[3, -E^(I*(a + b*x))] - 6*d^3*PolyLog[3, E^(I*(a + b*x))])/b^4)

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fricas [C]  time = 0.55, size = 669, normalized size = 4.58 \[ -\frac {2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} - 6 \, d^{3} {\rm polylog}\left (3, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - 6 \, d^{3} {\rm polylog}\left (3, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 6 \, d^{3} {\rm polylog}\left (3, -\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 6 \, d^{3} {\rm polylog}\left (3, -\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (-6 i \, b d^{3} x - 6 i \, b c d^{2}\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - {\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 ) \sin \left (b x + a\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 ) \sin \left (b x + a\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right )}{2 \, b^{4} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.09, size = 433, normalized size = 2.97 \[ -\frac {6 i d^{3} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {6 d \,c^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 d^{3} a^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 d^{3} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{3} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {12 d^{2} c a \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i d^{2} c \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d^{2} c \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{b^{4}}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{4}}+\frac {6 i d^{3} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*I/b^3*d^2*c*polylog(2,exp(I*(b*x+a)))+6/b^2*d^2*c*ln(1-exp(I*(b*x+a)))*x+6/b^3*d^2*c*ln(1-exp(I*(b*x+a)))*a
-6/b^2*d^2*c*ln(exp(I*(b*x+a))+1)*x-6/b^3*d^2*c*ln(exp(I*(b*x+a))+1)*a-6/b^2*d*c^2*arctanh(exp(I*(b*x+a)))-6/b
^4*d^3*a^2*arctanh(exp(I*(b*x+a)))-6*d^3*polylog(3,-exp(I*(b*x+a)))/b^4+6*d^3*polylog(3,exp(I*(b*x+a)))/b^4+12
/b^3*d^2*c*a*arctanh(exp(I*(b*x+a)))+6*I/b^3*d^2*c*polylog(2,-exp(I*(b*x+a)))-6*I/b^3*d^3*polylog(2,exp(I*(b*x
+a)))*x-3/b^2*d^3*ln(exp(I*(b*x+a))+1)*x^2+3/b^4*d^3*ln(exp(I*(b*x+a))+1)*a^2-2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d
*x+c^3)*exp(I*(b*x+a))/b/(exp(2*I*(b*x+a))-1)+3/b^2*d^3*ln(1-exp(I*(b*x+a)))*x^2-3/b^4*d^3*ln(1-exp(I*(b*x+a))
)*a^2+6*I/b^3*d^3*polylog(2,-exp(I*(b*x+a)))*x

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maxima [B]  time = 0.56, size = 1770, normalized size = 12.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(a + b*x)*(c + d*x)^3)/sin(a + b*x)^2,x)

[Out]

int((cos(a + b*x)*(c + d*x)^3)/sin(a + b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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