3.230 \(\int (c+d x)^2 \csc (a+b x) \sec (a+b x) \, dx\)
Optimal. Leaf size=127 \[ \frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d (c+d x) \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{d^2 \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]
[Out]
(-2*(c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/b + (I*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*
(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 - (d^2*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^3) + (d^2*PolyLog
[3, E^((2*I)*(a + b*x))])/(2*b^3)
________________________________________________________________________________________
Rubi [A] time = 0.116645, antiderivative size = 127, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{4419, 4183, 2531, 2282, 6589} \[ \frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d (c+d x) \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{d^2 \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^2*Csc[a + b*x]*Sec[a + b*x],x]
[Out]
(-2*(c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/b + (I*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^2 - (I*d*
(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 - (d^2*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^3) + (d^2*PolyLog
[3, E^((2*I)*(a + b*x))])/(2*b^3)
Rule 4419
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]
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 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 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 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) \sec (a+b x) \, dx &=2 \int (c+d x)^2 \csc (2 a+2 b x) \, dx\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac{(2 d) \int (c+d x) \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac{(2 d) \int (c+d x) \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac{i d (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d (c+d x) \text{Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac{\left (i d^2\right ) \int \text{Li}_2\left (-e^{i (2 a+2 b x)}\right ) \, dx}{b^2}+\frac{\left (i d^2\right ) \int \text{Li}_2\left (e^{i (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac{i d (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d (c+d x) \text{Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}+\frac{i d (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{i d (c+d x) \text{Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{d^2 \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}\\ \end{align*}
Mathematica [A] time = 0.628039, size = 213, normalized size = 1.68 \[ \frac{2 i b d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )-2 i b d (c+d x) \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )-d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )+d^2 \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )-4 b^2 c^2 \tanh ^{-1}\left (e^{2 i (a+b x)}\right )+4 b^2 c d x \log \left (1-e^{2 i (a+b x)}\right )-4 b^2 c d x \log \left (1+e^{2 i (a+b x)}\right )+2 b^2 d^2 x^2 \log \left (1-e^{2 i (a+b x)}\right )-2 b^2 d^2 x^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^2*Csc[a + b*x]*Sec[a + b*x],x]
[Out]
(-4*b^2*c^2*ArcTanh[E^((2*I)*(a + b*x))] + 4*b^2*c*d*x*Log[1 - E^((2*I)*(a + b*x))] + 2*b^2*d^2*x^2*Log[1 - E^
((2*I)*(a + b*x))] - 4*b^2*c*d*x*Log[1 + E^((2*I)*(a + b*x))] - 2*b^2*d^2*x^2*Log[1 + E^((2*I)*(a + b*x))] + (
2*I)*b*d*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))] - (2*I)*b*d*(c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))] - d^
2*PolyLog[3, -E^((2*I)*(a + b*x))] + d^2*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3)
________________________________________________________________________________________
Maple [B] time = 0.297, size = 469, normalized size = 3.7 \begin{align*} -{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}-{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){a}^{2}}{{b}^{3}}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}+{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}-{\frac{{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}-{\frac{2\,idc{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{2\,idc{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+{\frac{idc{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,i{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ){d}^{2}x}{{b}^{2}}}-2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}+{\frac{{a}^{2}{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}-2\,{\frac{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^2*csc(b*x+a)*sec(b*x+a),x)
[Out]
-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3-1/b*d^2*ln(exp(2*I*(b*x+a))+1)*x^2-1/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^
2+1/b*d^2*ln(exp(I*(b*x+a))+1)*x^2+1/b*d^2*ln(1-exp(I*(b*x+a)))*x^2-1/b*c^2*ln(exp(2*I*(b*x+a))+1)-2*I/b^2*c*d
*polylog(2,-exp(I*(b*x+a)))-2*I/b^2*c*d*polylog(2,exp(I*(b*x+a)))+I/b^2*d^2*polylog(2,-exp(2*I*(b*x+a)))*x-2*I
/b^2*d^2*polylog(2,-exp(I*(b*x+a)))*x+I/b^2*c*d*polylog(2,-exp(2*I*(b*x+a)))+2/b*c*d*ln(exp(I*(b*x+a))+1)*x+1/
b*c^2*ln(exp(I*(b*x+a))+1)+1/b*c^2*ln(exp(I*(b*x+a))-1)+2*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+2*d^2*polylog(3,e
xp(I*(b*x+a)))/b^3-2*I/b^2*d^2*polylog(2,exp(I*(b*x+a)))*x-2/b*c*d*ln(exp(2*I*(b*x+a))+1)*x+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+1/b^3*d^2*a^2*ln(exp(I*(b*x+a))-1)-2/b^2*c*d*a*ln(exp(I*(b*x+a))
-1)
________________________________________________________________________________________
Maxima [B] time = 1.97525, size = 797, normalized size = 6.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)*sec(b*x+a),x, algorithm="maxima")
[Out]
-1/2*(c^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 2*a*c*d*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a
)^2))/b + a^2*d^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 + (d^2*polylog(3, -e^(2*I*b*x + 2*I*a))
- 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(2*b*x + 2*a), cos(2*b*x + 2*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) + (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*c*d - 2*I*(b*x + a)*d^2 + 2*I*a*d^2)*dilog(-e^
(2*I*b*x + 2*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)) + (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(2*b*
x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*lo
g(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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Fricas [C] time = 0.737504, size = 2850, normalized size = 22.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)*sec(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, I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 2*d^2*polylog(3
, -I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, -I*cos(b*x + a) - 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)*dilo
g(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(I*cos(b*x + a) + sin(b*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(I*cos(b*x + a) - sin(b
*x + a)) + (2*I*b*d^2*x + 2*I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-2*I*b*d^2*x - 2*I*b*c*d)*dilog(
-I*cos(b*x + a) - 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)) + (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(cos(b*x + a) + I*sin(b*x + a) + I) + (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(cos(b*
x + a) - I*sin(b*x + a) + I) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos(b*x + a) + 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(I*cos(b*x + a) - 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(-I*cos(b*x + a) + 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(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-1/2*co
s(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*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (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*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) - I*sin
(b*x + a) + I))/b^3
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \csc{\left (a + b x \right )} \sec{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**2*csc(b*x+a)*sec(b*x+a),x)
[Out]
Integral((c + d*x)**2*csc(a + b*x)*sec(a + b*x), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \csc \left (b x + a\right ) \sec \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)*sec(b*x+a),x, algorithm="giac")
[Out]
integrate((d*x + c)^2*csc(b*x + a)*sec(b*x + a), x)