3.199 \(\int \frac{(e+f x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=134 \[ \frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

[Out]

(-2*(e + f*x)*ArcTanh[E^(I*(c + d*x))])/(a*d) + ((e + f*x)*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) - (2*f*Log[Sin[c/2
+ Pi/4 + (d*x)/2]])/(a*d^2) + (I*f*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - (I*f*PolyLog[2, E^(I*(c + d*x))])/
(a*d^2)

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Rubi [A]  time = 0.156537, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4535, 4183, 2279, 2391, 3318, 4184, 3475} \[ \frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(-2*(e + f*x)*ArcTanh[E^(I*(c + d*x))])/(a*d) + ((e + f*x)*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) - (2*f*Log[Sin[c/2
+ Pi/4 + (d*x)/2]])/(a*d^2) + (I*f*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - (I*f*PolyLog[2, E^(I*(c + d*x))])/
(a*d^2)

Rule 4535

Int[(Csc[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/a, Int[(e + f*x)^m*Csc[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csc[c + d*x]^(n - 1))/(a +
b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 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 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]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

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 3475

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

Rubi steps

\begin{align*} \int \frac{(e+f x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x) \csc (c+d x) \, dx}{a}-\int \frac{e+f x}{a+a \sin (c+d x)} \, dx\\ &=-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{\int (e+f x) \csc ^2\left (\frac{1}{2} \left (c+\frac{\pi }{2}\right )+\frac{d x}{2}\right ) \, dx}{2 a}-\frac{f \int \log \left (1-e^{i (c+d x)}\right ) \, dx}{a d}+\frac{f \int \log \left (1+e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(i f) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^2}-\frac{(i f) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^2}-\frac{f \int \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )\right )}{a d^2}+\frac{i f \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{i f \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}\\ \end{align*}

Mathematica [B]  time = 1.06422, size = 300, normalized size = 2.24 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (i \left (\text{PolyLog}\left (2,-e^{i (c+d x)}\right )-\text{PolyLog}\left (2,e^{i (c+d x)}\right )\right )+(c+d x) \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )\right )\right )-2 d (e+f x) \sin \left (\frac{1}{2} (c+d x)\right )+d e \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+f (c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-c f \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a d^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.144, size = 245, normalized size = 1.8 \begin{align*} 2\,{\frac{fx+e}{da \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }}-2\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }{a{d}^{2}}}+{\frac{e\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}-1 \right ) }{da}}-{\frac{e\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ) }{da}}-{\frac{fc\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}-1 \right ) }{a{d}^{2}}}-{\frac{if{\it polylog} \left ( 2,{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+{\frac{if{\it polylog} \left ( 2,-{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+2\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( dx+c \right ) }} \right ) fx}{da}}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( dx+c \right ) }} \right ) cf}{a{d}^{2}}}-{\frac{\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ) fx}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

2*(f*x+e)/d/a/(exp(I*(d*x+c))+I)-2/d^2/a*f*ln(exp(I*(d*x+c))+I)+1/d/a*e*ln(exp(I*(d*x+c))-1)-1/d/a*e*ln(exp(I*
(d*x+c))+1)-1/d^2/a*f*c*ln(exp(I*(d*x+c))-1)-I*f*polylog(2,exp(I*(d*x+c)))/a/d^2+I*f*polylog(2,-exp(I*(d*x+c))
)/a/d^2+2/d^2/a*f*ln(exp(I*(d*x+c)))+1/d/a*ln(1-exp(I*(d*x+c)))*f*x+1/d^2/a*ln(1-exp(I*(d*x+c)))*c*f-1/d/a*ln(
exp(I*(d*x+c))+1)*f*x

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Maxima [B]  time = 1.45336, size = 698, normalized size = 5.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

(4*d*f*x*cos(d*x + c) + 4*I*d*f*x*sin(d*x + c) - 4*I*d*e - (4*f*cos(d*x + c) + 4*I*f*sin(d*x + c) + 4*I*f)*arc
tan2(cos(c) + sin(d*x), cos(d*x) + sin(c)) - (2*I*d*f*x + 2*I*d*e + 2*(d*f*x + d*e)*cos(d*x + c) + (2*I*d*f*x
+ 2*I*d*e)*sin(d*x + c))*arctan2(sin(d*x + c), cos(d*x + c) + 1) + (2*d*e*cos(d*x + c) + 2*I*d*e*sin(d*x + c)
+ 2*I*d*e)*arctan2(sin(d*x + c), cos(d*x + c) - 1) - (2*d*f*x*cos(d*x + c) + 2*I*d*f*x*sin(d*x + c) + 2*I*d*f*
x)*arctan2(sin(d*x + c), -cos(d*x + c) + 1) + (2*f*cos(d*x + c) + 2*I*f*sin(d*x + c) + 2*I*f)*dilog(-e^(I*d*x
+ I*c)) - (2*f*cos(d*x + c) + 2*I*f*sin(d*x + c) + 2*I*f)*dilog(e^(I*d*x + I*c)) - (d*f*x + d*e + (-I*d*f*x -
I*d*e)*cos(d*x + c) + (d*f*x + d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1) +
(d*f*x + d*e - (I*d*f*x + I*d*e)*cos(d*x + c) + (d*f*x + d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^
2 - 2*cos(d*x + c) + 1) + 2*(I*f*cos(d*x + c) - f*sin(d*x + c) - f)*log(cos(d*x)^2 + cos(c)^2 + 2*cos(c)*sin(d
*x) + sin(d*x)^2 + 2*cos(d*x)*sin(c) + sin(c)^2))/(-2*I*a*d^2*cos(d*x + c) + 2*a*d^2*sin(d*x + c) + 2*a*d^2)

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Fricas [B]  time = 2.10089, size = 1654, normalized size = 12.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f x \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

(Integral(e*csc(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f*x*csc(c + d*x)/(sin(c + d*x) + 1), x))/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \csc \left (d x + c\right )}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*csc(d*x + c)/(a*sin(d*x + c) + a), x)