3.327 \(\int \frac{(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=351 \[ -\frac{f \sqrt{a^2-b^2} \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{f \sqrt{a^2-b^2} \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^2}+\frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{i \sqrt{a^2-b^2} (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{e x}{b}-\frac{f x^2}{2 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.660472, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 11, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.367, Rules used = {4543, 4408, 3296, 2637, 4183, 2279, 2391, 4525, 3323, 2264, 2190} \[ -\frac{f \sqrt{a^2-b^2} \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d^2}+\frac{f \sqrt{a^2-b^2} \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d^2}+\frac{i f \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{i \sqrt{a^2-b^2} (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a b d}+\frac{i \sqrt{a^2-b^2} (e+f x) \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a b d}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{e x}{b}-\frac{f x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4543

Int[(Cos[(c_.) + (d_.)*(x_)]^(p_.)*Cot[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin
[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[(e + f*x)^m*Cos[c + d*x]^p*Cot[c + d*x]^n, x], x] - Dist[b/a
, Int[((e + f*x)^m*Cos[c + d*x]^(p + 1)*Cot[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] && IGtQ[p, 0]

Rule 4408

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*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 4525

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

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 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 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]

Rubi steps

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

Mathematica [B]  time = 6.79811, size = 812, normalized size = 2.31 \[ \frac{\frac{(c+d x) (c f-d (2 e+f x))}{b}+\frac{2 d e \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a}-\frac{2 c f \log \left (\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a}+\frac{2 f \left ((c+d x) \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )\right )+i \left (\text{PolyLog}\left (2,-e^{i (c+d x)}\right )-\text{PolyLog}\left (2,e^{i (c+d x)}\right )\right )\right )}{a}+\frac{2 \left (a^2-b^2\right ) d (e+f x) \left (\frac{2 (d e-c f) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{-i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\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 )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{i a+b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\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 )\right )}{\sqrt{b^2-a^2}}+\frac{i f \left (\log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right ) \log \left (\frac{-b-a \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt{b^2-a^2}}{i a-b+\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{a \left (\tan \left (\frac{1}{2} (c+d x)\right )+i\right )}{i a-b+\sqrt{b^2-a^2}}\right )\right )}{\sqrt{b^2-a^2}}-\frac{i f \left (\log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \log \left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )-\sqrt{b^2-a^2}}{i a+b-\sqrt{b^2-a^2}}\right )+\text{PolyLog}\left (2,\frac{i \tan \left (\frac{1}{2} (c+d x)\right ) a+a}{a+i \left (\sqrt{b^2-a^2}-b\right )}\right )\right )}{\sqrt{b^2-a^2}}\right )}{a b \left (d e-c f+i f \log \left (1-i \tan \left (\frac{1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((c + d*x)*(c*f - d*(2*e + f*x)))/b + (2*d*e*Log[Tan[(c + d*x)/2]])/a - (2*c*f*Log[Tan[(c + d*x)/2]])/a + (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))])))/a + (2*(a^2 - b^2)*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])] + PolyLo
g[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]))/Sqrt[-a^2 + b^2]))/(a*b*(d*e - c*f + I*f*Lo
g[1 - I*Tan[(c + d*x)/2]] - I*f*Log[1 + I*Tan[(c + d*x)/2]])))/(2*d^2)

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Maple [B]  time = 0.306, size = 1207, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.31358, size = 3267, normalized size = 9.31 \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)*cos(d*x+c)*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*a*d^2*f*x^2 + 4*a*d^2*e*x - 2*I*b*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*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*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*f*sqrt(-(a^2 - b^2)/b^2)*di
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 + 1) + 2*I*b*f*dilog(cos(d*x + c) + I*sin(d*x + c)) - 2*I*b*f*dilog(cos(d*x + c) - I*sin(d*x + c))
+ 2*I*b*f*dilog(-cos(d*x + c) + I*sin(d*x + c)) - 2*I*b*f*dilog(-cos(d*x + c) - I*sin(d*x + c)) - 2*(b*d*e - b
*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*d*e - b*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*d*e - b*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*d*e - b*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*d*f*x + b*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*d*f*x +
b*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*d*f*x + b*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*d*f*x +
b*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*d*f*x + b*d*e)*log(cos(d*x + c) + I*sin(d*x + c) + 1) + 2*(b*
d*f*x + b*d*e)*log(cos(d*x + c) - I*sin(d*x + c) + 1) - 2*(b*d*e - b*c*f)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*
x + c) + 1/2) - 2*(b*d*e - b*c*f)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - 2*(b*d*f*x + b*c*f)*log(
-cos(d*x + c) + I*sin(d*x + c) + 1) - 2*(b*d*f*x + b*c*f)*log(-cos(d*x + c) - I*sin(d*x + c) + 1))/(a*b*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out