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

Optimal. Leaf size=325 \[ \frac {b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}-\frac {b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}+\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d \sqrt {a^2-b^2}}+\frac {i f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.62, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4535, 4183, 2279, 2391, 3323, 2264, 2190} \[ \frac {b f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d^2 \sqrt {a^2-b^2}}-\frac {b f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d^2 \sqrt {a^2-b^2}}+\frac {i f \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {i b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a d \sqrt {a^2-b^2}}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2264

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]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(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 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 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 6.37, size = 764, normalized size = 2.35 \[ \frac {-\frac {b d (e+f x) \left (\frac {2 (d e-c f) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {i f \left (\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}-i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{a-i \left (b+\sqrt {b^2-a^2}\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}+\frac {i f \left (\text {Li}_2\left (\frac {a \left (\tan \left (\frac {1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt {b^2-a^2}}\right )+\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {b^2-a^2}-a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {b^2-a^2}+i a-b}\right )\right )}{\sqrt {b^2-a^2}}-\frac {i f \left (\text {Li}_2\left (\frac {i \tan \left (\frac {1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt {b^2-a^2}-b\right )}\right )+\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-\sqrt {b^2-a^2}+a \tan \left (\frac {1}{2} (c+d x)\right )+b}{-\sqrt {b^2-a^2}+i a+b}\right )\right )}{\sqrt {b^2-a^2}}\right )}{i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )-c f+d e}+d e \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+f \left (i \left (\text {Li}_2\left (-e^{i (c+d x)}\right )-\text {Li}_2\left (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 )-c f \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*e*Log[Tan[(c + d*x)/2]] - c*f*Log[Tan[(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))])) - (b*d*(e + f*x)*((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])] + 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*(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])] + 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*(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])] + PolyLog[2, (a*(I + 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])] + PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]))/S
qrt[-a^2 + b^2]))/(d*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]]))/(a*d^2)

________________________________________________________________________________________

fricas [B]  time = 0.77, size = 1444, normalized size = 4.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*I*b^2*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) -
I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 2*I*b^2*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*c
os(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) -
2*I*b^2*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*
sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*b^2*f*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d
*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*I*
(a^2 - b^2)*f*dilog(cos(d*x + c) + I*sin(d*x + c)) - 2*I*(a^2 - b^2)*f*dilog(cos(d*x + c) - I*sin(d*x + c)) +
2*I*(a^2 - b^2)*f*dilog(-cos(d*x + c) + I*sin(d*x + c)) - 2*I*(a^2 - b^2)*f*dilog(-cos(d*x + c) - I*sin(d*x +
c)) + 2*(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) + 2*(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) - 2*(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) - 2*(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) + 2*(b^2*d*f*x + b^2*c*f)*sqrt(-(a^2
- b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2
- b^2)/b^2) + 2*b)/b) - 2*(b^2*d*f*x + b^2*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(
d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(b^2*d*f*x + b^2*c*f)*sq
rt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*s
qrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^2*d*f*x + b^2*c*f)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c)
+ 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 - b^2)*d
*f*x + (a^2 - b^2)*d*e)*log(cos(d*x + c) + I*sin(d*x + c) + 1) + 2*((a^2 - b^2)*d*f*x + (a^2 - b^2)*d*e)*log(c
os(d*x + c) - I*sin(d*x + c) + 1) - 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*
x + c) + 1/2) - 2*((a^2 - b^2)*d*e - (a^2 - b^2)*c*f)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - 2*((
a^2 - b^2)*d*f*x + (a^2 - b^2)*c*f)*log(-cos(d*x + c) + I*sin(d*x + c) + 1) - 2*((a^2 - b^2)*d*f*x + (a^2 - b^
2)*c*f)*log(-cos(d*x + c) - I*sin(d*x + c) + 1))/((a^3 - a*b^2)*d^2)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.25, size = 660, normalized size = 2.03 \[ -\frac {2 i e b \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d a \sqrt {-a^{2}+b^{2}}}-\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a d}-\frac {f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a \,d^{2}}+\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a d}-\frac {f b \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{d a \sqrt {-a^{2}+b^{2}}}-\frac {f b \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {-a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{d a \sqrt {-a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {-a^{2}+b^{2}}}+\frac {i f b \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {i f b \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) f x}{a d}+\frac {i f \dilog \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d^{2} a}+\frac {i f \dilog \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a}+\frac {2 i f c b \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right ) \csc {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________