\(\int \frac {(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [327]

Optimal result

Integrand size = 30, antiderivative size = 351 \[ \int \frac {(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {e x}{b}-\frac {f x^2}{2 b}-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {i \sqrt {a^2-b^2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d}+\frac {i \sqrt {a^2-b^2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d}+\frac {i f \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {\sqrt {a^2-b^2} f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^2}+\frac {\sqrt {a^2-b^2} f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {4639, 4493, 3377, 2717, 4268, 2317, 2438, 4621, 3404, 2296, 2221} \[ \int \frac {(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {f \sqrt {a^2-b^2} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^2}+\frac {f \sqrt {a^2-b^2} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^2}-\frac {i \sqrt {a^2-b^2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d}+\frac {i \sqrt {a^2-b^2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b d}-\frac {2 (e+f x) \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}+\frac {i f \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {e x}{b}-\frac {f x^2}{2 b} \]

[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 2221

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]
- 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 2296

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 2317

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 2438

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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 3404

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 4268

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 4493

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 4621

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 4639

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]

Rubi steps \begin{align*} \text {integral}& = \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) \text {arctanh}\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) \text {arctanh}\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) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^2}-\frac {(i f) \text {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) \text {arctanh}\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 \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 {\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) \text {arctanh}\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 \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 {\left (\sqrt {a^2-b^2} f\right ) \text {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 ) \text {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) \text {arctanh}\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 \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 {\sqrt {a^2-b^2} f \operatorname {PolyLog}\left (2,\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 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^2} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(876\) vs. \(2(351)=702\).

Time = 6.19 (sec) , antiderivative size = 876, normalized size of antiderivative = 2.50 \[ \int \frac {(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\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 (\operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )-\operatorname {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) \arctan \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 \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-b+\sqrt {-a^2+b^2}-a \tan \left (\frac {1}{2} (c+d x)\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b-\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b-\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{-i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a-i \left (b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^2}}+\frac {i f \operatorname {PolyLog}\left (2,\frac {a \left (i+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{i a-b+\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {i f \operatorname {PolyLog}\left (2,\frac {a+i a \tan \left (\frac {1}{2} (c+d x)\right )}{a+i \left (-b+\sqrt {-a^2+b^2}\right )}\right )}{\sqrt {-a^2+b^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 (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}}{2 d^2} \]

[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])])/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*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*P
olyLog[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*
(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]))/(a*b*(d*e - c*f + I*f*Log[1 - I*Tan[(c + d*x)/2]]
- I*f*Log[1 + I*Tan[(c + d*x)/2]])))/(2*d^2)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1188 vs. \(2 (311 ) = 622\).

Time = 0.61 (sec) , antiderivative size = 1189, normalized size of antiderivative = 3.39

method	result	size
risch	\(\text {Expression too large to display}\)	\(1189\)

[In]

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

[Out]

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

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. 1273 vs. \(2 (304) = 608\).

Time = 0.50 (sec) , antiderivative size = 1273, normalized size of antiderivative = 3.63 \[ \int \frac {(e+f x) \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[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)

Maxima [F(-2)]

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

[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 >> 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 de

Giac [F(-1)]

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

[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

Mupad [F(-1)]

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

[In]

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

[Out]

\text{Hanged}