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

Optimal. Leaf size=180 \[ -\frac {d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}+\frac {i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b} \]

[Out]

-(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b-d^2*arctanh(cos(b*x+a))/b^3-d*(d*x+c)*csc(b*x+a)/b^2-1/2*(d*x+c)^2*cot(b*
x+a)*csc(b*x+a)/b+I*d*(d*x+c)*polylog(2,-exp(I*(b*x+a)))/b^2-I*d*(d*x+c)*polylog(2,exp(I*(b*x+a)))/b^2-d^2*pol
ylog(3,-exp(I*(b*x+a)))/b^3+d^2*polylog(3,exp(I*(b*x+a)))/b^3

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4186, 3770, 4183, 2531, 2282, 6589} \[ \frac {i d (c+d x) \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \text {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csc[a + b*x]^3,x]

[Out]

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

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

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)^2 \csc ^3(a+b x) \, dx &=-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^2 \csc (a+b x) \, dx+\frac {d^2 \int \csc (a+b x) \, dx}{b^2}\\ &=-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b}-\frac {d \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac {d \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {\left (i d^2\right ) \int \text {Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (i d^2\right ) \int \text {Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac {(c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d^2 \tanh ^{-1}(\cos (a+b x))}{b^3}-\frac {d (c+d x) \csc (a+b x)}{b^2}-\frac {(c+d x)^2 \cot (a+b x) \csc (a+b x)}{2 b}+\frac {i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 7.64, size = 471, normalized size = 2.62 \[ \frac {\csc \left (\frac {a}{2}\right ) \csc \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (c d \sin \left (\frac {b x}{2}\right )+d^2 x \sin \left (\frac {b x}{2}\right )\right )}{2 b^2}+\frac {\sec \left (\frac {a}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (d^2 (-x) \sin \left (\frac {b x}{2}\right )-c d \sin \left (\frac {b x}{2}\right )\right )}{2 b^2}-\frac {d \csc (a) (c+d x)}{b^2}+\frac {b^2 c^2 \log \left (1-e^{i (a+b x)}\right )-b^2 c^2 \log \left (1+e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1-e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1+e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1-e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1+e^{i (a+b x)}\right )+2 i b d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )-2 i b d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )-2 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )+2 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )+2 d^2 \log \left (1-e^{i (a+b x)}\right )-2 d^2 \log \left (1+e^{i (a+b x)}\right )}{2 b^3}+\frac {\left (-c^2-2 c d x-d^2 x^2\right ) \csc ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {\left (c^2+2 c d x+d^2 x^2\right ) \sec ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(c + d*x)*Csc[a])/b^2) + ((-c^2 - 2*c*d*x - d^2*x^2)*Csc[a/2 + (b*x)/2]^2)/(8*b) + (b^2*c^2*Log[1 - E^(I*
(a + b*x))] + 2*d^2*Log[1 - E^(I*(a + b*x))] + 2*b^2*c*d*x*Log[1 - E^(I*(a + b*x))] + b^2*d^2*x^2*Log[1 - E^(I
*(a + b*x))] - b^2*c^2*Log[1 + E^(I*(a + b*x))] - 2*d^2*Log[1 + E^(I*(a + b*x))] - 2*b^2*c*d*x*Log[1 + E^(I*(a
+ b*x))] - b^2*d^2*x^2*Log[1 + E^(I*(a + b*x))] + (2*I)*b*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))] - (2*I)*b*
d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))] - 2*d^2*PolyLog[3, -E^(I*(a + b*x))] + 2*d^2*PolyLog[3, E^(I*(a + b*x)
)])/(2*b^3) + ((c^2 + 2*c*d*x + d^2*x^2)*Sec[a/2 + (b*x)/2]^2)/(8*b) + (Sec[a/2]*Sec[a/2 + (b*x)/2]*(-(c*d*Sin
[(b*x)/2]) - d^2*x*Sin[(b*x)/2]))/(2*b^2) + (Csc[a/2]*Csc[a/2 + (b*x)/2]*(c*d*Sin[(b*x)/2] + d^2*x*Sin[(b*x)/2
]))/(2*b^2)

________________________________________________________________________________________

fricas [C]  time = 0.87, size = 968, normalized size = 5.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.12, size = 548, normalized size = 3.04 \[ \frac {d^{2} x^{2} b \,{\mathrm e}^{3 i \left (b x +a \right )}+2 c d x b \,{\mathrm e}^{3 i \left (b x +a \right )}+b \,c^{2} {\mathrm e}^{3 i \left (b x +a \right )}+d^{2} x^{2} b \,{\mathrm e}^{i \left (b x +a \right )}+2 c d x b \,{\mathrm e}^{i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{3 i \left (b x +a \right )}+b \,c^{2} {\mathrm e}^{i \left (b x +a \right )}-2 i d c \,{\mathrm e}^{3 i \left (b x +a \right )}+2 i d^{2} x \,{\mathrm e}^{i \left (b x +a \right )}+2 i d c \,{\mathrm e}^{i \left (b x +a \right )}}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{2 b^{3}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{2 b^{3}}-\frac {c^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {d^{2} a^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{2 b}-\frac {i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{2 b}-\frac {i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}+\frac {c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 c d a \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2/(exp(2*I*(b*x+a))-1)^2*(d^2*x^2*b*exp(3*I*(b*x+a))+2*c*d*x*b*exp(3*I*(b*x+a))+b*c^2*exp(3*I*(b*x+a))+d^2
*x^2*b*exp(I*(b*x+a))+2*c*d*x*b*exp(I*(b*x+a))-2*I*d^2*x*exp(3*I*(b*x+a))+b*c^2*exp(I*(b*x+a))-2*I*d*c*exp(3*I
*(b*x+a))+2*I*d^2*x*exp(I*(b*x+a))+2*I*d*c*exp(I*(b*x+a)))+1/2/b^3*d^2*ln(exp(I*(b*x+a))+1)*a^2-1/2/b^3*d^2*ln
(1-exp(I*(b*x+a)))*a^2-1/b*c^2*arctanh(exp(I*(b*x+a)))-1/b^3*d^2*a^2*arctanh(exp(I*(b*x+a)))-d^2*polylog(3,-ex
p(I*(b*x+a)))/b^3+d^2*polylog(3,exp(I*(b*x+a)))/b^3-2/b^3*d^2*arctanh(exp(I*(b*x+a)))-1/2/b*d^2*ln(exp(I*(b*x+
a))+1)*x^2-I/b^2*c*d*polylog(2,exp(I*(b*x+a)))+1/2/b*d^2*ln(1-exp(I*(b*x+a)))*x^2+I/b^2*c*d*polylog(2,-exp(I*(
b*x+a)))-1/b*c*d*ln(exp(I*(b*x+a))+1)*x-1/b^2*c*d*ln(exp(I*(b*x+a))+1)*a+1/b*c*d*ln(1-exp(I*(b*x+a)))*x+1/b^2*
c*d*ln(1-exp(I*(b*x+a)))*a+2/b^2*c*d*a*arctanh(exp(I*(b*x+a)))-I/b^2*polylog(2,exp(I*(b*x+a)))*d^2*x+I/b^2*pol
ylog(2,-exp(I*(b*x+a)))*d^2*x

________________________________________________________________________________________

maxima [B]  time = 1.88, size = 1934, normalized size = 10.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c^2*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1)) - 2*a*c*d*(2*co
s(b*x + a)/(cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1))/b + a^2*d^2*(2*cos(b*x + a)/(
cos(b*x + a)^2 - 1) - log(cos(b*x + a) + 1) + log(cos(b*x + a) - 1))/b^2 - 4*((2*(b*x + a)^2*d^2 + 4*(b*c*d -
a*d^2)*(b*x + a) + 4*d^2 + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(4*b*x + 4*a) - 4*((b*
x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*cos(2*b*x + 2*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I
*a*d^2)*(b*x + a) + 4*I*d^2)*sin(4*b*x + 4*a) + (-4*I*(b*x + a)^2*d^2 + (-8*I*b*c*d + 8*I*a*d^2)*(b*x + a) - 8
*I*d^2)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (4*d^2*cos(4*b*x + 4*a) - 8*d^2*cos(2*b*x
+ 2*a) + 4*I*d^2*sin(4*b*x + 4*a) - 8*I*d^2*sin(2*b*x + 2*a) + 4*d^2)*arctan2(sin(b*x + a), cos(b*x + a) - 1)
+ (2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*(b*x + a) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(4*b
*x + 4*a) - 4*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b
*c*d - 4*I*a*d^2)*(b*x + a))*sin(4*b*x + 4*a) + (-4*I*(b*x + a)^2*d^2 + (-8*I*b*c*d + 8*I*a*d^2)*(b*x + a))*si
n(2*b*x + 2*a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (4*I*(b*x + a)^2*d^2 + 8*b*c*d - 8*a*d^2 + (8*I*b*c
*d - 8*(I*a - 1)*d^2)*(b*x + a))*cos(3*b*x + 3*a) + (4*I*(b*x + a)^2*d^2 - 8*b*c*d + 8*a*d^2 - 8*(-I*b*c*d + (
I*a + 1)*d^2)*(b*x + a))*cos(b*x + a) - (4*b*c*d + 4*(b*x + a)*d^2 - 4*a*d^2 + 4*(b*c*d + (b*x + a)*d^2 - a*d^
2)*cos(4*b*x + 4*a) - 8*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (-4*I*b*c*d - 4*I*(b*x + a)*d^2 + 4
*I*a*d^2)*sin(4*b*x + 4*a) - (8*I*b*c*d + 8*I*(b*x + a)*d^2 - 8*I*a*d^2)*sin(2*b*x + 2*a))*dilog(-e^(I*b*x + I
*a)) + (4*b*c*d + 4*(b*x + a)*d^2 - 4*a*d^2 + 4*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) - 8*(b*c*d +
(b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (4*I*b*c*d + 4*I*(b*x + a)*d^2 - 4*I*a*d^2)*sin(4*b*x + 4*a) + (-8*I
*b*c*d - 8*I*(b*x + a)*d^2 + 8*I*a*d^2)*sin(2*b*x + 2*a))*dilog(e^(I*b*x + I*a)) + (-I*(b*x + a)^2*d^2 + (-2*I
*b*c*d + 2*I*a*d^2)*(b*x + a) - 2*I*d^2 + (-I*(b*x + a)^2*d^2 + (-2*I*b*c*d + 2*I*a*d^2)*(b*x + a) - 2*I*d^2)*
cos(4*b*x + 4*a) + (2*I*(b*x + a)^2*d^2 + (4*I*b*c*d - 4*I*a*d^2)*(b*x + a) + 4*I*d^2)*cos(2*b*x + 2*a) + ((b*
x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*sin(4*b*x + 4*a) - 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*
(b*x + a) + 2*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (I*(b*x + a)^
2*d^2 + (2*I*b*c*d - 2*I*a*d^2)*(b*x + a) + 2*I*d^2 + (I*(b*x + a)^2*d^2 + (2*I*b*c*d - 2*I*a*d^2)*(b*x + a) +
2*I*d^2)*cos(4*b*x + 4*a) + (-2*I*(b*x + a)^2*d^2 + (-4*I*b*c*d + 4*I*a*d^2)*(b*x + a) - 4*I*d^2)*cos(2*b*x +
2*a) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) + 2*d^2)*sin(4*b*x + 4*a) + 2*((b*x + a)^2*d^2 + 2*(b*c
*d - a*d^2)*(b*x + a) + 2*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + (
-4*I*d^2*cos(4*b*x + 4*a) + 8*I*d^2*cos(2*b*x + 2*a) + 4*d^2*sin(4*b*x + 4*a) - 8*d^2*sin(2*b*x + 2*a) - 4*I*d
^2)*polylog(3, -e^(I*b*x + I*a)) + (4*I*d^2*cos(4*b*x + 4*a) - 8*I*d^2*cos(2*b*x + 2*a) - 4*d^2*sin(4*b*x + 4*
a) + 8*d^2*sin(2*b*x + 2*a) + 4*I*d^2)*polylog(3, e^(I*b*x + I*a)) - (4*(b*x + a)^2*d^2 - 8*I*b*c*d + 8*I*a*d^
2 + (8*b*c*d - (8*a + 8*I)*d^2)*(b*x + a))*sin(3*b*x + 3*a) - (4*(b*x + a)^2*d^2 + 8*I*b*c*d - 8*I*a*d^2 + (8*
b*c*d - (8*a - 8*I)*d^2)*(b*x + a))*sin(b*x + a))/(-4*I*b^2*cos(4*b*x + 4*a) + 8*I*b^2*cos(2*b*x + 2*a) + 4*b^
2*sin(4*b*x + 4*a) - 8*b^2*sin(2*b*x + 2*a) - 4*I*b^2))/b

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/sin(a + b*x)^3,x)

[Out]

\text{Hanged}

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________