\(\int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx\) [234]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 325 \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {i f \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {b f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {b f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2016\) vs. \(2(325)=650\).

Time = 16.59 (sec) , antiderivative size = 2016, normalized size of antiderivative = 6.20 \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

(e*Log[Tan[(c + d*x)/2]])/(a*d) - (c*f*Log[Tan[(c + d*x)/2]])/(a*d^2) + (f 
*((c + d*x)*(Log[1 - E^(I*(c + d*x))] - Log[1 + E^(I*(c + d*x))]) + I*(Pol 
yLog[2, -E^(I*(c + d*x))] - PolyLog[2, E^(I*(c + d*x))])))/(a*d^2) + (((2* 
(d*e - c*f)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b 
^2] - (I*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[( 
c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b^2] + (I*f*Log[1 
- I*Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I 
*a - b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (I*f*Log[1 - I*Tan[(c + d 
*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt 
[-a^2 + b^2])])/Sqrt[-a^2 + b^2] + (I*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b 
 + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])])/S 
qrt[-a^2 + b^2] - (I*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + 
 Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyLog[2, (a*(1 + I*Tan[(c 
+ d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*f*PolyL 
og[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])])/Sqrt[-a^2 
+ b^2] - (I*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 
 + b^2]))])/Sqrt[-a^2 + b^2])*(-((b*e)/(a*(a + b*Sin[c + d*x]))) + (b*c*f) 
/(a*d*(a + b*Sin[c + d*x])) - (b*f*(c + d*x))/(a*d*(a + b*Sin[c + d*x])))) 
/(d*((f*Log[1 - (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]) 
)]*Sec[(c + d*x)/2]^2)/(2*Sqrt[-a^2 + b^2]*(1 - I*Tan[(c + d*x)/2])) + ...
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {5046, 3042, 3804, 2694, 27, 2620, 2715, 2838, 4671, 2715, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5046

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {b \int \frac {e+f x}{a+b \sin (c+d x)}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {b \int \frac {e+f x}{a+b \sin (c+d x)}dx}{a}\)

\(\Big \downarrow \) 3804

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \int \frac {e^{i (c+d x)} (e+f x)}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{a}\)

\(\Big \downarrow \) 2694

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {f \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \left (\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{b d^2}+\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\int (e+f x) \csc (c+d x)dx}{a}-\frac {2 b \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {-\frac {f \int \log \left (1-e^{i (c+d x)}\right )dx}{d}+\frac {f \int \log \left (1+e^{i (c+d x)}\right )dx}{d}-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{d}}{a}-\frac {2 b \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {i f \int e^{-i (c+d x)} \log \left (1-e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {i f \int e^{-i (c+d x)} \log \left (1+e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{d}}{a}-\frac {2 b \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{d^2}}{a}-\frac {2 b \left (\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {i f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}\)

Input:

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

Output:

((-2*(e + f*x)*ArcTanh[E^(I*(c + d*x))])/d + (I*f*PolyLog[2, -E^(I*(c + d* 
x))])/d^2 - (I*f*PolyLog[2, E^(I*(c + d*x))])/d^2)/a - (2*b*(((-1/2*I)*b*( 
((e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - ( 
I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d^2)))/Sqr 
t[a^2 - b^2] + ((I/2)*b*(((e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqr 
t[a^2 - b^2])])/(b*d) - (I*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^ 
2 - b^2])])/(b*d^2)))/Sqrt[a^2 - b^2]))/a
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2694
Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838
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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3804
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 4671
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] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 5046
Int[(Csc[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_. 
)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(e + f*x)^m*Csc[c + 
 d*x]^n, x], x] - Simp[b/a   Int[(e + f*x)^m*(Csc[c + d*x]^(n - 1)/(a + b*S 
in[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ 
[n, 0]
 
Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 650 vs. \(2 (287 ) = 574\).

Time = 1.12 (sec) , antiderivative size = 651, normalized size of antiderivative = 2.00

method result size
risch \(\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {f b \ln \left (\frac {-i a -{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d a \sqrt {-a^{2}+b^{2}}}-\frac {2 i e b \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d a \sqrt {-a^{2}+b^{2}}}+\frac {i f \operatorname {dilog}\left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a}-\frac {c f \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d^{2} a}-\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}+\frac {i f \operatorname {dilog}\left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d^{2} a}+\frac {f b \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d a \sqrt {-a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {f b \ln \left (\frac {-i a -{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) x}{d a}+\frac {i f b \operatorname {dilog}\left (\frac {-i a -{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{-i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {i f b \operatorname {dilog}\left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}}+\frac {2 i c f b \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}}\) \(651\)

Input:

int((f*x+e)*csc(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/a*e*ln(exp(I*(d*x+c))-1)-1/d*f*b/a/(-a^2+b^2)^(1/2)*ln((-I*a-exp(I*(d* 
x+c))*b+(-a^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))*x-2*I/d*e*b/a/(-a^2+b^2 
)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+I/d^2*f/a* 
dilog(exp(I*(d*x+c)))-1/d^2/a*c*f*ln(exp(I*(d*x+c))-1)-1/d/a*e*ln(exp(I*(d 
*x+c))+1)+I/d^2*f/a*dilog(exp(I*(d*x+c))+1)+1/d*f*b/a/(-a^2+b^2)^(1/2)*ln( 
(I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x+1/d^2*f* 
b/a/(-a^2+b^2)^(1/2)*ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2 
+b^2)^(1/2)))*c-1/d^2*f*b/a/(-a^2+b^2)^(1/2)*ln((-I*a-exp(I*(d*x+c))*b+(-a 
^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))*c-1/d/a*f*ln(exp(I*(d*x+c))+1)*x+I 
/d^2*f*b/a/(-a^2+b^2)^(1/2)*dilog((-I*a-exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2)) 
/(-I*a+(-a^2+b^2)^(1/2)))-I/d^2*f*b/a/(-a^2+b^2)^(1/2)*dilog((I*a+exp(I*(d 
*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))+2*I/d^2*c*f*b/a/(-a^2+b 
^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1428 vs. \(2 (275) = 550\).

Time = 0.32 (sec) , antiderivative size = 1428, normalized size of antiderivative = 4.39 \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (e + f x\right ) \csc {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F(-1)]

Timed out. \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b e +\left (\int \frac {\csc \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{3} d f -\left (\int \frac {\csc \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a \,b^{2} d f +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} e -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2} e}{a d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*b 
*e + int((csc(c + d*x)*x)/(sin(c + d*x)*b + a),x)*a**3*d*f - int((csc(c + 
d*x)*x)/(sin(c + d*x)*b + a),x)*a*b**2*d*f + log(tan((c + d*x)/2))*a**2*e 
- log(tan((c + d*x)/2))*b**2*e)/(a*d*(a**2 - b**2))