3.231 \(\int (c+d x) \csc (a+b x) \sec (a+b x) \, dx\)

Optimal. Leaf size=71 \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{2 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0555391, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4419, 4183, 2279, 2391} \[ \frac{i d \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{i d \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{2 (c+d x) \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Csc[a + b*x]*Sec[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.11855, size = 141, normalized size = 1.99 \[ \frac{d \left (i \left (\text{PolyLog}\left (2,-e^{i (2 a+2 b x)}\right )-\text{PolyLog}\left (2,e^{i (2 a+2 b x)}\right )\right )+(2 a+2 b x) \left (\log \left (1-e^{i (2 a+2 b x)}\right )-\log \left (1+e^{i (2 a+2 b x)}\right )\right )-2 a \log \left (\tan \left (\frac{1}{2} (2 a+2 b x)\right )\right )\right )}{2 b^2}+\frac{c \log (\sin (a+b x))}{b}-\frac{c \log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Csc[a + b*x]*Sec[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.234, size = 208, normalized size = 2.9 \begin{align*} -{\frac{c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}+{\frac{c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}+{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{id{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-{\frac{id{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{ad\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)*csc(b*x+a)*sec(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.94968, size = 360, normalized size = 5.07 \begin{align*} -\frac{2 i \, b d x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 i \, b c \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) +{\left (2 i \, b d x + 2 i \, b c\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-2 i \, b d x - 2 i \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - i \, d{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 i \, d{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 2 i \, d{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) +{\left (b d x + b c\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) -{\left (b d x + b c\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 (b d x + b c\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 )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.630837, size = 1544, normalized size = 21.75 \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)*csc(b*x+a)*sec(b*x+a),x, algorithm="fricas")

[Out]

1/2*(-I*d*dilog(cos(b*x + a) + I*sin(b*x + a)) + I*d*dilog(cos(b*x + a) - I*sin(b*x + a)) - I*d*dilog(I*cos(b*
x + a) + sin(b*x + a)) + I*d*dilog(I*cos(b*x + a) - sin(b*x + a)) + I*d*dilog(-I*cos(b*x + a) + sin(b*x + a))
- I*d*dilog(-I*cos(b*x + a) - sin(b*x + a)) + I*d*dilog(-cos(b*x + a) + I*sin(b*x + a)) - I*d*dilog(-cos(b*x +
a) - I*sin(b*x + a)) + (b*d*x + b*c)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - (b*c - a*d)*log(cos(b*x + a) +
I*sin(b*x + a) + I) + (b*d*x + b*c)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - (b*c - a*d)*log(cos(b*x + a) - I*
sin(b*x + a) + I) - (b*d*x + a*d)*log(I*cos(b*x + a) + sin(b*x + a) + 1) - (b*d*x + a*d)*log(I*cos(b*x + a) -
sin(b*x + a) + 1) - (b*d*x + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b*d*x + a*d)*log(-I*cos(b*x + a)
- sin(b*x + a) + 1) + (b*c - a*d)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + (b*c - a*d)*log(-1/2*cos
(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) + (b*d*x + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - (b*c - a*d)*lo
g(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*d*x + a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) - (b*c - a*d)*lo
g(-cos(b*x + a) - I*sin(b*x + a) + I))/b^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \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)*csc(b*x+a)*sec(b*x+a),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \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)*csc(b*x+a)*sec(b*x+a),x, algorithm="giac")

[Out]

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