3.28 \(\int (c+d x)^3 \csc ^2(a+b x) \, dx\)

Optimal. Leaf size=113 \[ \frac {3 d^3 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {i (c+d x)^3}{b} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4184, 3717, 2190, 2531, 2282, 6589} \[ -\frac {3 i d^2 (c+d x) \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {i (c+d x)^3}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csc[a + b*x]^2,x]

[Out]

((-I)*(c + d*x)^3)/b - ((c + d*x)^3*Cot[a + b*x])/b + (3*d*(c + d*x)^2*Log[1 - E^((2*I)*(a + b*x))])/b^2 - ((3
*I)*d^2*(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b^3 + (3*d^3*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^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 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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && 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 \csc ^2(a+b x) \, dx &=-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {(3 d) \int (c+d x)^2 \cot (a+b x) \, dx}{b}\\ &=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {(6 i d) \int \frac {e^{2 i (a+b x)} (c+d x)^2}{1-e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {\left (6 d^2\right ) \int (c+d x) \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {\left (3 i d^3\right ) \int \text {Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}\\ &=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^3 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^4}\\ \end {align*}

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Mathematica [B]  time = 6.96, size = 478, normalized size = 4.23 \[ \frac {3 c^2 d \csc (a) (\sin (a) \log (\sin (a) \cos (b x)+\cos (a) \sin (b x))-b x \cos (a))}{b^2 \left (\sin ^2(a)+\cos ^2(a)\right )}-\frac {3 c d^2 \csc (a) \sec (a) \left (b^2 x^2 e^{i \tan ^{-1}(\tan (a))}+\frac {\tan (a) \left (i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\tan ^2(a)+1}}\right )}{b^3 \sqrt {\sec ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}-\frac {e^{i a} d^3 \csc (a) \left (2 e^{-2 i a} b^3 x^3+3 i \left (1-e^{-2 i a}\right ) b^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 i \left (1-e^{-2 i a}\right ) b^2 x^2 \log \left (1+e^{-i (a+b x)}\right )-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (-e^{-i (a+b x)}\right )-i \text {Li}_3\left (-e^{-i (a+b x)}\right )\right )-6 e^{-2 i a} \left (-1+e^{2 i a}\right ) \left (b x \text {Li}_2\left (e^{-i (a+b x)}\right )-i \text {Li}_3\left (e^{-i (a+b x)}\right )\right )\right )}{2 b^4}+\frac {\csc (a) \csc (a+b x) \left (c^3 \sin (b x)+3 c^2 d x \sin (b x)+3 c d^2 x^2 \sin (b x)+d^3 x^3 \sin (b x)\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Csc[a + b*x]^2,x]

[Out]

-1/2*(d^3*E^(I*a)*Csc[a]*((2*b^3*x^3)/E^((2*I)*a) + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2*Log[1 - E^((-I)*(a + b*x)
)] + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2*Log[1 + E^((-I)*(a + b*x))] - (6*(-1 + E^((2*I)*a))*(b*x*PolyLog[2, -E^(
(-I)*(a + b*x))] - I*PolyLog[3, -E^((-I)*(a + b*x))]))/E^((2*I)*a) - (6*(-1 + E^((2*I)*a))*(b*x*PolyLog[2, E^(
(-I)*(a + b*x))] - I*PolyLog[3, E^((-I)*(a + b*x))]))/E^((2*I)*a)))/b^4 + (3*c^2*d*Csc[a]*(-(b*x*Cos[a]) + Log
[Cos[b*x]*Sin[a] + Cos[a]*Sin[b*x]]*Sin[a]))/(b^2*(Cos[a]^2 + Sin[a]^2)) + (Csc[a]*Csc[a + b*x]*(c^3*Sin[b*x]
+ 3*c^2*d*x*Sin[b*x] + 3*c*d^2*x^2*Sin[b*x] + d^3*x^3*Sin[b*x]))/b - (3*c*d^2*Csc[a]*Sec[a]*(b^2*E^(I*ArcTan[T
an[a]])*x^2 + ((I*b*x*(-Pi + 2*ArcTan[Tan[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x + ArcTan[Tan[a]])*Log[1 -
E^((2*I)*(b*x + ArcTan[Tan[a]]))] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + I*Po
lyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))])*Tan[a])/Sqrt[1 + Tan[a]^2]))/(b^3*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]
^2)])

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fricas [C]  time = 0.90, size = 672, normalized size = 5.95 \[ \frac {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 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \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*csc(b*x+a)^2,x, algorithm="fricas")

[Out]

1/2*(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) - 2*(
b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos(b*x + a))/(b^4*sin(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.11, size = 541, normalized size = 4.79 \[ -\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\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 (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 {4 i d^{3} a^{3}}{b^{4}}-\frac {2 i d^{3} x^{3}}{b}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\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 {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 {12 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {6 i d^{3} a^{2} x}{b^{3}}-\frac {6 i d^{2} c \,x^{2}}{b}-\frac {6 i d^{2} c \,a^{2}}{b^{3}}-\frac {6 i d^{3} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 i d^{3} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{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 {12 i d^{2} c a x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.72, size = 1650, normalized size = 14.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*((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(2*b*x + 2*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*((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(2*b*x + 2*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*((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(2*b*x + 2*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/tan(b*x + a) + 6*a*c^2*d/(b*tan(b*x + a)) - 6*a^2*c*d^2/(b^2*tan(b*x + a)) + 2*a^3*d^3/(b^3*t
an(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(2*b*x + 2*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)) - (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))*co
s(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 + si
n(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(2*b*x + 2*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 \[ \int \frac {{\left (c+d\,x\right )}^3}{{\sin \left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/sin(a + b*x)^2,x)

[Out]

int((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 \[ \int \left (c + d x\right )^{3} \csc ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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