3.274 \(\int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx\)
Optimal. Leaf size=88 \[ -\frac{i d^2 \text{PolyLog}\left (2,e^{4 i (a+b x)}\right )}{2 b^3}+\frac{2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac{2 i (c+d x)^2}{b} \]
[Out]
((-2*I)*(c + d*x)^2)/b - (2*(c + d*x)^2*Cot[2*a + 2*b*x])/b + (2*d*(c + d*x)*Log[1 - E^((4*I)*(a + b*x))])/b^2
- ((I/2)*d^2*PolyLog[2, E^((4*I)*(a + b*x))])/b^3
________________________________________________________________________________________
Rubi [A] time = 0.191444, antiderivative size = 88, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{4419, 4184, 3717, 2190, 2279, 2391} \[ -\frac{i d^2 \text{PolyLog}\left (2,e^{4 i (a+b x)}\right )}{2 b^3}+\frac{2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac{2 i (c+d x)^2}{b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^2*Csc[a + b*x]^2*Sec[a + b*x]^2,x]
[Out]
((-2*I)*(c + d*x)^2)/b - (2*(c + d*x)^2*Cot[2*a + 2*b*x])/b + (2*d*(c + d*x)*Log[1 - E^((4*I)*(a + b*x))])/b^2
- ((I/2)*d^2*PolyLog[2, E^((4*I)*(a + b*x))])/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 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 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 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 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2391
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rubi steps
\begin{align*} \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx &=4 \int (c+d x)^2 \csc ^2(2 a+2 b x) \, dx\\ &=-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac{(4 d) \int (c+d x) \cot (2 a+2 b x) \, dx}{b}\\ &=-\frac{2 i (c+d x)^2}{b}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac{(8 i d) \int \frac{e^{2 i (2 a+2 b x)} (c+d x)}{1-e^{2 i (2 a+2 b x)}} \, dx}{b}\\ &=-\frac{2 i (c+d x)^2}{b}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac{2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac{\left (2 d^2\right ) \int \log \left (1-e^{2 i (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 i (c+d x)^2}{b}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac{2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}+\frac{\left (i d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac{2 i (c+d x)^2}{b}-\frac{2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac{2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac{i d^2 \text{Li}_2\left (e^{4 i (a+b x)}\right )}{2 b^3}\\ \end{align*}
Mathematica [B] time = 1.75939, size = 277, normalized size = 3.15 \[ \frac{2 b^2 \csc (2 a) \sin (2 b x) (c+d x)^2 \csc (2 (a+b x))-\frac{i e^{4 i a} \left (-2 \left (1-e^{-4 i a}\right ) d^2 \text{PolyLog}\left (2,-e^{-i (a+b x)}\right )-2 \left (1-e^{-4 i a}\right ) d^2 \text{PolyLog}\left (2,e^{-i (a+b x)}\right )-\left (1-e^{-4 i a}\right ) d^2 \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+4 e^{-4 i a} b^2 (c+d x)^2+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1-e^{-i (a+b x)}\right )+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1+e^{-i (a+b x)}\right )+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1+e^{-2 i (a+b x)}\right )\right )}{-1+e^{4 i a}}}{b^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^2*Csc[a + b*x]^2*Sec[a + b*x]^2,x]
[Out]
(((-I)*E^((4*I)*a)*((4*b^2*(c + d*x)^2)/E^((4*I)*a) + (2*I)*b*d*(1 - E^((-4*I)*a))*(c + d*x)*Log[1 - E^((-I)*(
a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)*a))*(c + d*x)*Log[1 + E^((-I)*(a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)*a))
*(c + d*x)*Log[1 + E^((-2*I)*(a + b*x))] - 2*d^2*(1 - E^((-4*I)*a))*PolyLog[2, -E^((-I)*(a + b*x))] - 2*d^2*(1
- E^((-4*I)*a))*PolyLog[2, E^((-I)*(a + b*x))] - d^2*(1 - E^((-4*I)*a))*PolyLog[2, -E^((-2*I)*(a + b*x))]))/(
-1 + E^((4*I)*a)) + 2*b^2*(c + d*x)^2*Csc[2*a]*Csc[2*(a + b*x)]*Sin[2*b*x])/b^3
________________________________________________________________________________________
Maple [B] time = 0.332, size = 351, normalized size = 4. \begin{align*}{\frac{-4\,i{d}^{2}{x}^{2}}{b}}+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-8\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}-{\frac{8\,i{d}^{2}ax}{{b}^{2}}}-{\frac{4\,i{d}^{2}{a}^{2}}{{b}^{3}}}-{\frac{4\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) }}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-2\,{\frac{{d}^{2}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}+8\,{\frac{{d}^{2}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^2*csc(b*x+a)^2*sec(b*x+a)^2,x)
[Out]
-4*I*d^2/b*x^2+2*d/b^2*c*ln(exp(I*(b*x+a))-1)+2*d/b^2*c*ln(exp(I*(b*x+a))+1)+2*d/b^2*c*ln(exp(2*I*(b*x+a))+1)-
8*d/b^2*c*ln(exp(I*(b*x+a)))-2*I*d^2/b^3*polylog(2,-exp(I*(b*x+a)))+2*d^2/b^2*ln(exp(2*I*(b*x+a))+1)*x-I*d^2*p
olylog(2,-exp(2*I*(b*x+a)))/b^3+2*d^2/b^2*ln(1-exp(I*(b*x+a)))*x+2*d^2/b^3*ln(1-exp(I*(b*x+a)))*a-8*I*d^2/b^2*
a*x-4*I*d^2/b^3*a^2-4*I*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I*(b*x+a))+1)/(exp(2*I*(b*x+a))-1)-2*I*d^2*polylog(2,ex
p(I*(b*x+a)))/b^3+2*d^2/b^2*ln(exp(I*(b*x+a))+1)*x-2*d^2/b^3*a*ln(exp(I*(b*x+a))-1)+8*d^2/b^3*a*ln(exp(I*(b*x+
a)))
________________________________________________________________________________________
Maxima [B] time = 2.12905, size = 1049, normalized size = 11.92 \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)^2*sec(b*x+a)^2,x, algorithm="maxima")
[Out]
-(4*b^2*c^2 + (2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(4*b*x + 4*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(4*b*
x + 4*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(4*b
*x + 4*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(4*b*x + 4*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b*c*d*cos
(4*b*x + 4*a) + 2*I*b*c*d*sin(4*b*x + 4*a) - 2*b*c*d)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d^2*x*cos
(4*b*x + 4*a) + 2*I*b*d^2*x*sin(4*b*x + 4*a) - 2*b*d^2*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 4*(b^2*d^
2*x^2 + 2*b^2*c*d*x)*cos(4*b*x + 4*a) + (d^2*cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(2*I*b*
x + 2*I*a)) + 2*(d^2*cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(I*b*x + I*a)) + 2*(d^2*cos(4*b
*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*
d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(
2*b*x + 2*a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b
*x + 4*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*
b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x
+ a) + 1) - (-4*I*b^2*d^2*x^2 - 8*I*b^2*c*d*x)*sin(4*b*x + 4*a))/(-I*b^3*cos(4*b*x + 4*a) + b^3*sin(4*b*x + 4*
a) + I*b^3)
________________________________________________________________________________________
Fricas [B] time = 0.746377, size = 2573, normalized size = 29.24 \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)^2*sec(b*x+a)^2,x, algorithm="fricas")
[Out]
(b^2*d^2*x^2 + 2*b^2*c*d*x - I*d^2*cos(b*x + a)*dilog(cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) + I*d^2*cos(
b*x + a)*dilog(cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x
+ a))*sin(b*x + a) - I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) - I*d^2*cos(b*x + a
)*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)
)*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2*cos(b*x + a)*di
log(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + b^2*c^2 + (b*d^2*x + b*c*d)*cos(b*x + a)*log(cos(b*x + a) +
I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*
x + a) + (b*d^2*x + b*c*d)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*
cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*
x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1
)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*
x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*lo
g(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-1/2*cos(b*x +
a) - 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x +
a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) + (b*
d^2*x + a*d^2)*cos(b*x + a)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a
)*log(-cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(b*x + a)^
2)/(b^3*cos(b*x + a)*sin(b*x + a))
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**2*csc(b*x+a)**2*sec(b*x+a)**2,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^2*csc(b*x+a)^2*sec(b*x+a)^2,x, algorithm="giac")
[Out]
integrate((d*x + c)^2*csc(b*x + a)^2*sec(b*x + a)^2, x)