3.205 \(\int \frac{(e+f x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=169 \[ -\frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}+\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{f \log (\sin (c+d x))}{a d^2}-\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x) \cot (c+d x)}{a d}+\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

[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) - ((e + f*x)*Cot[c
+ d*x])/(a*d) + (2*f*Log[Sin[c/2 + Pi/4 + (d*x)/2]])/(a*d^2) + (f*Log[Sin[c + d*x]])/(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 = 0.189351, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4535, 4184, 3475, 4183, 2279, 2391, 3318} \[ -\frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}+\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac{2 f \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )\right )}{a d^2}+\frac{f \log (\sin (c+d x))}{a d^2}-\frac{(e+f x) \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x) \cot (c+d x)}{a d}+\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Csc[c + d*x]^2)/(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) - ((e + f*x)*Cot[c
+ d*x])/(a*d) + (2*f*Log[Sin[c/2 + Pi/4 + (d*x)/2]])/(a*d^2) + (f*Log[Sin[c + d*x]])/(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 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]

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])

Rubi steps

\begin{align*} \int \frac{(e+f x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x) \csc ^2(c+d x) \, dx}{a}-\int \frac{(e+f x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac{(e+f x) \cot (c+d x)}{a d}-\frac{\int (e+f x) \csc (c+d x) \, dx}{a}+\frac{f \int \cot (c+d x) \, dx}{a d}+\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{(e+f x) \cot (c+d x)}{a d}+\frac{f \log (\sin (c+d x))}{a d^2}+\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{(e+f x) \cot (c+d x)}{a d}+\frac{f \log (\sin (c+d x))}{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{(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{(e+f x) \cot (c+d x)}{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{f \log (\sin (c+d x))}{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 = 1.70638, size = 396, normalized size = 2.34 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (-2 f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (i \left (\text{PolyLog}\left (2,-e^{i (c+d x)}\right )-\text{PolyLog}\left (2,e^{i (c+d x)}\right )\right )+(c+d x) \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )\right )\right )+4 d (e+f x) \sin \left (\frac{1}{2} (c+d x)\right )+d (e+f x) \sin \left (\frac{1}{2} (c+d x)\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right )+1\right )-d (e+f x) \cos \left (\frac{1}{2} (c+d x)\right ) \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right )-2 d e \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 f (c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 f \log (\sin (c+d x)) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 c f \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 a d^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-(d*(e + f*x)*Cos[(c + d*x)/2]*(1 + Cot[(c + d*x)/2])) + 4*d*(e + f*x)
*Sin[(c + d*x)/2] - 2*f*(c + d*x)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*f*Log[Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 2*f*Log[Sin[c + d*x]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) -
2*d*e*Log[Tan[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 2*c*f*Log[Tan[(c + d*x)/2]]*(Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]) - 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]) + d*(e + f*x)*Sin[(c
+ d*x)/2]*(1 + Tan[(c + d*x)/2])))/(2*a*d^2*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.185, size = 351, normalized size = 2.1 \begin{align*} -2\,{\frac{-2\,fx+i{{\rm e}^{i \left ( dx+c \right ) }}fx-2\,e+i{{\rm e}^{i \left ( dx+c \right ) }}e+fx{{\rm e}^{2\,i \left ( dx+c \right ) }}+e{{\rm e}^{2\,i \left ( dx+c \right ) }}}{ \left ({{\rm e}^{2\,i \left ( dx+c \right ) }}-1 \right ) \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \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}}}-{\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}}-4\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}-1 \right ) }{a{d}^{2}}}+{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ) }{a{d}^{2}}}+2\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \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{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+1 \right ) x}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

-2*(-2*f*x+I*exp(I*(d*x+c))*f*x-2*e+I*exp(I*(d*x+c))*e+f*x*exp(2*I*(d*x+c))+e*exp(2*I*(d*x+c)))/(exp(2*I*(d*x+
c))-1)/(exp(I*(d*x+c))+I)/d/a+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^2-1/d/a*e*ln(exp(I*(d*x+c))-1)+1/d/a*e*ln(exp(I*(d*x+c))+1)-4/d^2/a*f*ln(exp(I*(d*x+c)
))+1/d^2/a*f*ln(exp(I*(d*x+c))-1)+1/d^2/a*f*ln(exp(I*(d*x+c))+1)+2/d^2/a*f*ln(exp(I*(d*x+c))+I)-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 = 2.0077, size = 1727, normalized size = 10.22 \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)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-(8*d*f*x*cos(3*d*x + 3*c) + 8*I*d*f*x*sin(3*d*x + 3*c) + 8*I*d*e - (4*f*cos(3*d*x + 3*c) + 4*I*f*cos(2*d*x +
2*c) - 4*f*cos(d*x + c) + 4*I*f*sin(3*d*x + 3*c) - 4*f*sin(2*d*x + 2*c) - 4*I*f*sin(d*x + c) - 4*I*f)*arctan2(
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 + f)*cos(3*d*x + 3*c) + (2*I*d*
f*x + 2*I*d*e + 2*I*f)*cos(2*d*x + 2*c) - 2*(d*f*x + d*e + f)*cos(d*x + c) + (2*I*d*f*x + 2*I*d*e + 2*I*f)*sin
(3*d*x + 3*c) - 2*(d*f*x + d*e + f)*sin(2*d*x + 2*c) + (-2*I*d*f*x - 2*I*d*e - 2*I*f)*sin(d*x + c) - 2*I*f)*ar
ctan2(sin(d*x + c), cos(d*x + c) + 1) - (2*I*d*e - 2*(d*e - f)*cos(3*d*x + 3*c) + (-2*I*d*e + 2*I*f)*cos(2*d*x
+ 2*c) + 2*(d*e - f)*cos(d*x + c) + (-2*I*d*e + 2*I*f)*sin(3*d*x + 3*c) + 2*(d*e - f)*sin(2*d*x + 2*c) + (2*I
*d*e - 2*I*f)*sin(d*x + c) - 2*I*f)*arctan2(sin(d*x + c), cos(d*x + c) - 1) - (2*d*f*x*cos(3*d*x + 3*c) + 2*I*
d*f*x*cos(2*d*x + 2*c) - 2*d*f*x*cos(d*x + c) + 2*I*d*f*x*sin(3*d*x + 3*c) - 2*d*f*x*sin(2*d*x + 2*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) - (-4*I*d*f*x + 4*I*d*e)*cos(2*d*x + 2*
c) - 4*(d*f*x - d*e)*cos(d*x + c) + (2*f*cos(3*d*x + 3*c) + 2*I*f*cos(2*d*x + 2*c) - 2*f*cos(d*x + c) + 2*I*f*
sin(3*d*x + 3*c) - 2*f*sin(2*d*x + 2*c) - 2*I*f*sin(d*x + c) - 2*I*f)*dilog(-e^(I*d*x + I*c)) - (2*f*cos(3*d*x
+ 3*c) + 2*I*f*cos(2*d*x + 2*c) - 2*f*cos(d*x + c) + 2*I*f*sin(3*d*x + 3*c) - 2*f*sin(2*d*x + 2*c) - 2*I*f*si
n(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 - I*f)*cos(3*d*x + 3*c) - (d*f*x
+ d*e + f)*cos(2*d*x + 2*c) - (I*d*f*x + I*d*e + I*f)*cos(d*x + c) - (d*f*x + d*e + f)*sin(3*d*x + 3*c) - (I*
d*f*x + I*d*e + I*f)*sin(2*d*x + 2*c) + (d*f*x + d*e + f)*sin(d*x + c) + f)*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 - I*f)*cos(3*d*x + 3*c) - (d*f*x + d*e - f)*cos(2*d*
x + 2*c) + (-I*d*f*x - I*d*e + I*f)*cos(d*x + c) - (d*f*x + d*e - f)*sin(3*d*x + 3*c) + (-I*d*f*x - I*d*e + I*
f)*sin(2*d*x + 2*c) + (d*f*x + d*e - f)*sin(d*x + c) - f)*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*cos(d*x + c)
+ 1) - (-2*I*f*cos(3*d*x + 3*c) + 2*f*cos(2*d*x + 2*c) + 2*I*f*cos(d*x + c) + 2*f*sin(3*d*x + 3*c) + 2*I*f*si
n(2*d*x + 2*c) - 2*f*sin(d*x + c) - 2*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) - 4*(d*f*x - d*e)*sin(2*d*x + 2*c) - (4*I*d*f*x - 4*I*d*e)*sin(d*x + c))/(-2*I*a*d^2*cos
(3*d*x + 3*c) + 2*a*d^2*cos(2*d*x + 2*c) + 2*I*a*d^2*cos(d*x + c) + 2*a*d^2*sin(3*d*x + 3*c) + 2*I*a*d^2*sin(2
*d*x + 2*c) - 2*a*d^2*sin(d*x + c) - 2*a*d^2)

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Fricas [B]  time = 2.34288, size = 2284, normalized size = 13.51 \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)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*d*f*x - 4*(d*f*x + d*e)*cos(d*x + c)^2 + 2*d*e - 2*(d*f*x + d*e)*cos(d*x + c) + (-I*f*cos(d*x + c)^2 +
(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)^2 + (-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)^2 + (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)^2 + (-I*f*co
s(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*f*x + d*e + f)*cos(d
*x + c)^2 + d*e + (d*f*x + d*e + (d*f*x + d*e + f)*cos(d*x + c) + f)*sin(d*x + c) + f)*log(cos(d*x + c) + I*si
n(d*x + c) + 1) + (d*f*x - (d*f*x + d*e + f)*cos(d*x + c)^2 + d*e + (d*f*x + d*e + (d*f*x + d*e + f)*cos(d*x +
c) + f)*sin(d*x + c) + f)*log(cos(d*x + c) - I*sin(d*x + c) + 1) + ((d*e - (c + 1)*f)*cos(d*x + c)^2 - d*e +
(c + 1)*f - (d*e - (c + 1)*f + (d*e - (c + 1)*f)*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2*I*sin
(d*x + c) + 1/2) + ((d*e - (c + 1)*f)*cos(d*x + c)^2 - d*e + (c + 1)*f - (d*e - (c + 1)*f + (d*e - (c + 1)*f)*
cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - (d*f*x - (d*f*x + c*f)*cos(d*x
+ c)^2 + c*f + (d*f*x + c*f + (d*f*x + c*f)*cos(d*x + c))*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) +
1) - (d*f*x - (d*f*x + c*f)*cos(d*x + c)^2 + c*f + (d*f*x + c*f + (d*f*x + c*f)*cos(d*x + c))*sin(d*x + c))*lo
g(-cos(d*x + c) - I*sin(d*x + c) + 1) - 2*(f*cos(d*x + c)^2 - (f*cos(d*x + c) + f)*sin(d*x + c) - f)*log(sin(d
*x + c) + 1) - 2*(d*f*x + d*e + 2*(d*f*x + d*e)*cos(d*x + c))*sin(d*x + c))/(a*d^2*cos(d*x + c)^2 - a*d^2 - (a
*d^2*cos(d*x + c) + a*d^2)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e \csc ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f x \csc ^{2}{\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)**2/(a+a*sin(d*x+c)),x)

[Out]

(Integral(e*csc(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f*x*csc(c + d*x)**2/(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 )^{2}}{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)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*csc(d*x + c)^2/(a*sin(d*x + c) + a), x)