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

Optimal. Leaf size=123 \[ -\frac {2 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac {2 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4183, 2531, 2282, 6589} \[ \frac {2 i d (c+d x) \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \text {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csc[a + b*x],x]

[Out]

(-2*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b + ((2*I)*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*d*
(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 - (2*d^2*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (2*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 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 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 (a+b x) \, dx &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(2 d) \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac {(2 d) \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {\left (2 i d^2\right ) \int \text {Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (2 i d^2\right ) \int \text {Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac {2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 148, normalized size = 1.20 \[ \frac {\frac {2 i d \left (b (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )+i d \text {Li}_3\left (-e^{i (a+b x)}\right )\right )}{b^2}+\frac {2 d \left (d \text {Li}_3\left (e^{i (a+b x)}\right )-i b (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )\right )}{b^2}+(c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-(c+d x)^2 \log \left (1+e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Csc[a + b*x],x]

[Out]

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

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fricas [C]  time = 0.78, size = 500, normalized size = 4.07 \[ \frac {2 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (-2 i \, b d^{2} x - 2 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 361, normalized size = 2.93 \[ -\frac {2 c^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {2 d^{2} a^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {4 c d a \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{b^{3}}+\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.68, size = 392, normalized size = 3.19 \[ -\frac {2 \, c^{2} \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right ) - \frac {4 \, a c d \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right )}{b} + \frac {2 \, a^{2} d^{2} \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right )}{b^{2}} + \frac {4 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) + {\left (2 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (4 i \, b c d - 4 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (2 i \, {\left (b x + a\right )}^{2} d^{2} + {\left (4 i \, b c d - 4 i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + {\left (-4 i \, b c d - 4 i \, {\left (b x + a\right )} d^{2} + 4 i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (4 i \, b c d + 4 i \, {\left (b x + a\right )} d^{2} - 4 i \, a d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right )}{b^{2}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/sin(a + b*x),x)

[Out]

int((c + d*x)^2/sin(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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