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

Optimal. Leaf size=386 \[ \frac {i f \left (a^2-b^2\right ) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^2}+\frac {i f \left (a^2-b^2\right ) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b d^2}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a^2 b f}+\frac {i b f \text {Li}_2\left (e^{2 i (c+d x)}\right )}{2 a^2 d^2}-\frac {b (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {i b (e+f x)^2}{2 a^2 f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.78, antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 15, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.469, Rules used = {4543, 4408, 3296, 2638, 4410, 3770, 4404, 2635, 8, 3717, 2190, 2279, 2391, 4525, 4519} \[ \frac {i f \left (a^2-b^2\right ) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^2}+\frac {i f \left (a^2-b^2\right ) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d^2}+\frac {i b f \text {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{2 a^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d}-\frac {\left (a^2-b^2\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^2}{2 a^2 b f}-\frac {b (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {i b (e+f x)^2}{2 a^2 f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3770

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

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 4410

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp
[((c + d*x)^m*Csc[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Fr
eeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 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 (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \cos (c+d x) \cot ^2(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x) \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac {\int (e+f x) \cos (c+d x) \, dx}{a}+\frac {\int (e+f x) \cot (c+d x) \csc (c+d x) \, dx}{a}-\frac {b \int (e+f x) \cos ^2(c+d x) \cot (c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {(e+f x) \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2}\\ &=-\frac {(e+f x) \csc (c+d x)}{a d}-\frac {(e+f x) \sin (c+d x)}{a d}+\frac {\int (e+f x) \cos (c+d x) \, dx}{a}-\frac {b \int (e+f x) \cot (c+d x) \, dx}{a^2}-\left (1-\frac {b^2}{a^2}\right ) \int \frac {(e+f x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx+\frac {f \int \csc (c+d x) \, dx}{a d}+\frac {f \int \sin (c+d x) \, dx}{a d}\\ &=\frac {i b (e+f x)^2}{2 a^2 f}+\frac {i \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2}{2 b f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {f \cos (c+d x)}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d}+\frac {(2 i b) \int \frac {e^{2 i (c+d x)} (e+f x)}{1-e^{2 i (c+d x)}} \, dx}{a^2}-\left (1-\frac {b^2}{a^2}\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 (1-\frac {b^2}{a^2}\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}{a d}\\ &=\frac {i b (e+f x)^2}{2 a^2 f}+\frac {i \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2}{2 b f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {b (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {(b f) \int \log \left (1-e^{2 i (c+d x)}\right ) \, dx}{a^2 d}+\frac {\left (\left (1-\frac {b^2}{a^2}\right ) f\right ) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d}+\frac {\left (\left (1-\frac {b^2}{a^2}\right ) f\right ) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d}\\ &=\frac {i b (e+f x)^2}{2 a^2 f}+\frac {i \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2}{2 b f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {b (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}-\frac {(i b f) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a^2 d^2}-\frac {\left (i \left (1-\frac {b^2}{a^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 )}{b d^2}-\frac {\left (i \left (1-\frac {b^2}{a^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 )}{b d^2}\\ &=\frac {i b (e+f x)^2}{2 a^2 f}+\frac {i \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2}{2 b f}-\frac {f \tanh ^{-1}(\cos (c+d x))}{a d^2}-\frac {(e+f x) \csc (c+d x)}{a d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {b (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {i \left (1-\frac {b^2}{a^2}\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {i \left (1-\frac {b^2}{a^2}\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {i b f \text {Li}_2\left (e^{2 i (c+d x)}\right )}{2 a^2 d^2}\\ \end {align*}

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Mathematica [B]  time = 14.90, size = 2314, normalized size = 5.99 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-(d*e*Cos[(c + d*x)/2]) + c*f*Cos[(c + d*x)/2] - f*(c + d*x)*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(2*a*d^2) -
(b*e*Log[Sin[c + d*x]])/(a^2*d) + (b*c*f*Log[Sin[c + d*x]])/(a^2*d^2) + (f*Log[Tan[(c + d*x)/2]])/(a*d^2) - (
b*f*((c + d*x)*Log[1 - E^((2*I)*(c + d*x))] - (I/2)*((c + d*x)^2 + PolyLog[2, E^((2*I)*(c + d*x))])))/(a^2*d^2
) + (Sec[(c + d*x)/2]*(-(d*e*Sin[(c + d*x)/2]) + c*f*Sin[(c + d*x)/2] - f*(c + d*x)*Sin[(c + d*x)/2]))/(2*a*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[S
ec[(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]]*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])] - 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]))])*(-((e*Cos[c + d*x])/(a + b
*Sin[c + d*x])) + (b^2*e*Cos[c + d*x])/(a^2*(a + b*Sin[c + d*x])) + (c*f*Cos[c + d*x])/(d*(a + b*Sin[c + d*x])
) - (b^2*c*f*Cos[c + d*x])/(a^2*d*(a + b*Sin[c + d*x])) - (f*(c + d*x)*Cos[c + d*x])/(d*(a + b*Sin[c + d*x]))
+ (b^2*f*(c + d*x)*Cos[c + d*x])/(a^2*d*(a + b*Sin[c + d*x]))))/(d*(2*f*(c + d*x) - (4*I)*f*Log[(-2*I)/(-I + T
an[(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 - Sq
rt[-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*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])))

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fricas [B]  time = 0.71, size = 1427, normalized size = 3.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.58, size = 1732, normalized size = 4.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/d^2/b*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/b*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/a^2*b^3/d^2*f/(-a^2+b^2)*di
log((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))-I/a^2*b^3/d^2*f/(-a^2+b^2)*dilog((I*a+b*ex
p(I*(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))+1/d^2/b*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-
I*b)-2/d^2/b*f*c*ln(exp(I*(d*x+c)))-1/d/b*e*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+2/d/b*ln(exp(I*(d*
x+c)))*e-2*I*(f*x+e)*exp(I*(d*x+c))/d/a/(exp(2*I*(d*x+c))-1)-1/a^2/d*b*f*ln(exp(I*(d*x+c))+1)*x+1/a^2/d^2*b*f*
c*ln(exp(I*(d*x+c))-1)-I/a^2/d^2*b*f*dilog(exp(I*(d*x+c)))-1/a/d^2*f*ln(exp(I*(d*x+c))+1)+1/a/d^2*f*ln(exp(I*(
d*x+c))-1)+1/d/b*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^2*x+1/d^2/b
*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^2*c+1/d/b*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^2*x+1/d^2/b*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^2*c-2/d*b*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-2/d^2*b*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-2/d*b*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-2/d^2*b*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/a^2*b/d*e*ln(I*b*exp(2*I*(d*x
+c))-2*a*exp(I*(d*x+c))-I*b)+I/b/d^2*c^2*f-1/a^2/d*b*e*ln(exp(I*(d*x+c))+1)-1/a^2/d*b*e*ln(exp(I*(d*x+c))-1)+1
/2*I/b*f*x^2-I/b*e*x+2*I/b/d*c*f*x+2*I*b/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)))+2*I*b/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)))
+I/a^2*b/d^2*f*dilog(exp(I*(d*x+c))+1)-1/a^2*b/d^2*f*c*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+1/a^2*b
^3/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)))*c+1/a^2*b^3/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)))*c+1/a^2*b^3/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)))*x+1/a^2*b^3/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)))*x

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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)*cot(d*x+c)^2/(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)*cot(c + d*x)^2*(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 \[ \int \frac {\left (e + f x\right ) \cos {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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