3.331 \(\int \frac {(e+f x) \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=379 \[ -\frac {i f \left (a^2-b^2\right ) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f \left (a^2-b^2\right ) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}-\frac {i f \text {Li}_2\left (e^{2 i (c+d x)}\right )}{2 a d^2}+\frac {(e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {i (e+f x)^2}{2 a f}-\frac {f \cos (c+d x)}{b d^2}-\frac {(e+f x) \sin (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.63, antiderivative size = 379, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 13, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {4543, 4408, 4404, 2635, 8, 3717, 2190, 2279, 2391, 4525, 3296, 2638, 4519} \[ -\frac {i f \left (a^2-b^2\right ) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f \left (a^2-b^2\right ) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d^2}-\frac {i f \text {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{2 a d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}+\frac {(e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {i (e+f x)^2}{2 a f}-\frac {f \cos (c+d x)}{b d^2}-\frac {(e+f x) \sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Cos[c + d*x]^2*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

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

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4404

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

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 4519

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

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 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]

Rubi steps

\begin {align*} \int \frac {(e+f x) \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \cos ^2(c+d x) \cot (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac {\int (e+f x) \cot (c+d x) \, dx}{a}-\frac {\int (e+f x) \cos (c+d x) \, dx}{b}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {(e+f x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx\\ &=-\frac {i (e+f x)^2}{2 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}-\frac {(e+f x) \sin (c+d x)}{b d}-\frac {(2 i) \int \frac {e^{2 i (c+d x)} (e+f x)}{1-e^{2 i (c+d x)}} \, dx}{a}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {e^{i (c+d x)} (e+f x)}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {e^{i (c+d x)} (e+f x)}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx+\frac {f \int \sin (c+d x) \, dx}{b d}\\ &=-\frac {i (e+f x)^2}{2 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}-\frac {f \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {(e+f x) \sin (c+d x)}{b d}-\frac {f \int \log \left (1-e^{2 i (c+d x)}\right ) \, dx}{a d}-\frac {\left (\left (a^2-b^2\right ) f\right ) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{a b^2 d}-\frac {\left (\left (a^2-b^2\right ) f\right ) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{a b^2 d}\\ &=-\frac {i (e+f x)^2}{2 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}-\frac {f \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {(e+f x) \sin (c+d x)}{b d}+\frac {(i f) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a d^2}+\frac {\left (i \left (a^2-b^2\right ) f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^2 d^2}+\frac {\left (i \left (a^2-b^2\right ) f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^2 d^2}\\ &=-\frac {i (e+f x)^2}{2 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a b^2 f}-\frac {f \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f \text {Li}_2\left (e^{2 i (c+d x)}\right )}{2 a d^2}-\frac {(e+f x) \sin (c+d x)}{b d}\\ \end {align*}

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Mathematica [B]  time = 14.94, size = 2209, normalized size = 5.83 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((f*Cos[c + d*x])/(b*d^2)) + (e*Log[Sin[c + d*x]])/(a*d) - (c*f*Log[Sin[c + d*x]])/(a*d^2) + (f*((c + d*x)*Lo
g[1 - E^((2*I)*(c + d*x))] - (I/2)*((c + d*x)^2 + PolyLog[2, E^((2*I)*(c + d*x))])))/(a*d^2) - ((d*e - c*f + f
*(c + d*x))*Sin[c + d*x])/(b*d^2) + ((f*(c + d*x)^2 + (2*I)*d*e*Log[Sec[(c + d*x)/2]^2] - (2*I)*c*f*Log[Sec[(c
+ d*x)/2]^2] - (2*I)*d*e*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] + (2*I)*c*f*Log[Sec[(c + d*x)/2]^2*(a +
b*Sin[c + d*x])] - (4*I)*f*(c + d*x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Lo
g[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])] + 2*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]))] + 2*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])] - 2*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])] + 4*f*PolyLog[2, -Cos[
c + d*x] + I*Sin[c + d*x]] + 2*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))] - 2*f
*PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))] + 2*f*PolyLog[2, (a*(I + Tan[(c + d*x
)/2]))/(I*a - b + Sqrt[-a^2 + b^2])] - 2*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]
))])*((a*e*Cos[c + d*x])/(b*(a + b*Sin[c + d*x])) - (b*e*Cos[c + d*x])/(a*(a + b*Sin[c + d*x])) - (a*c*f*Cos[c
+ d*x])/(b*d*(a + b*Sin[c + d*x])) + (b*c*f*Cos[c + d*x])/(a*d*(a + b*Sin[c + d*x])) + (a*f*(c + d*x)*Cos[c +
d*x])/(b*d*(a + b*Sin[c + d*x])) - (b*f*(c + d*x)*Cos[c + d*x])/(a*d*(a + b*Sin[c + d*x]))))/(d*(2*f*(c + d*x
) - (4*I)*f*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - (4*f*Log[1 + Cos[c + d*x] - I*Sin[c + d*x]]*(I*Cos[c + d*x]
+ Sin[c + d*x]))/(-Cos[c + d*x] + I*Sin[c + d*x]) + (I*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)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d
*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[(b + Sqrt[-a^2
+ b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) + (
I*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)/(1 + I*Tan[(c + d
*x)/2]) - (I*f*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^
2)/(1 + I*Tan[(c + d*x)/2]) - (I*f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2]
)]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) + (2*I)*d*e*Tan[(c + d*x)/2] - (2*I)*c*f*Tan[(c + d*x)/2] + ((
2*I)*f*(c + d*x)*Sec[(c + d*x)/2]^2)/(-I + Tan[(c + d*x)/2]) - (f*Log[1 - (a*(I + Tan[(c + d*x)/2]))/(I*a - b
+ Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(I + Tan[(c + d*x)/2]) + (I*a*f*Log[1 - (a + I*a*Tan[(c + d*x)/2])/(a
+ I*(-b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(a + I*a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2
]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c +
d*x)/2]^2)/(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)
/(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[
-a^2 + b^2] + a*Tan[(c + d*x)/2]) - ((2*I)*d*e*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c
+ d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x]) + ((2*I)*c*f*Cos[(c + d*x)/2]^2*(b*Co
s[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x]
)))

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fricas [B]  time = 0.73, size = 1301, normalized size = 3.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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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)*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 2.12, size = 1694, normalized size = 4.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d^2*f*c*ln(exp(I*(d*x+c))-1)+1/a/d*ln(exp(I*(d*x+c))+1)*f*x+1/a/d*e*ln(exp(I*(d*x+c))+1)+1/a/d*e*ln(exp(I
*(d*x+c))-1)+I/d^2*f*dilog(exp(I*(d*x+c)))/a+2/d^2*f/(-a^2+b^2)*ln((-I*a-b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(-
I*a+(-a^2+b^2)^(1/2)))*a*c+2/d*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))
*a*x+2/d^2*f/(-a^2+b^2)*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*a*c+2/d*f/(-a^2+b^2
)*ln((-I*a-b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))*a*x+2/d^2/b^2*a*f*c*ln(exp(I*(d*x+c)))-
1/d^2/b^2*a*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-I/d^2/b^2*a*f*c^2-2*I/d^2*f/(-a^2+b^2)*dilog((
-I*a-b*exp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))*a-2*I/d^2*f/(-a^2+b^2)*dilog((I*a+b*exp(I*(d*
x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*a+1/2*I*(d*f*x+I*f+d*e)/b/d^2*exp(I*(d*x+c))-1/d*e/a*ln(I*b*ex
p(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+1/d^2*f*c/a*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+1/d/b^2*a*e
*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-2/d/b^2*a*e*ln(exp(I*(d*x+c)))-I/d^2*f/a*dilog(exp(I*(d*x+c))
+1)-1/2*I*a/b^2*f*x^2+I*a/b^2*e*x-1/2*I*(d*f*x-I*f+d*e)/b/d^2*exp(-I*(d*x+c))-2*I/d/b^2*a*f*c*x-1/d*b^2*f/a/(-
a^2+b^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*b^2*f/a/(-a^2+b^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*b^2*f/a/(-a^2+b^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*b^2*f/a/(-a^2+b^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^2/b^2*a^3*f/(-a^2+b^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/b^2*a^3*f/(-a^2+b^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/b^2*a^3*f/(-a^2+b^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/b^2*
a^3*f/(-a^2+b^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/b^2*a^3*f/(-a^2+b^
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/b^2*a^3*f/(-a^2+b^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*b^2*f/a/(-a^2+b^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*b^2*f/a/(-a^2+b^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 = 0.00, 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)*cos(d*x+c)^2*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 details)Is 4*b^2-4*a^2 positive or negative?

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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((cos(c + d*x)^2*cot(c + d*x)*(e + f*x))/(a + b*sin(c + d*x)),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right ) \cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)**2*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*cos(c + d*x)**2*cot(c + d*x)/(a + b*sin(c + d*x)), x)

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