3.335 \(\int \frac {(e+f x) \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=524 \[ -\frac {f \left (a^2-b^2\right ) \sin (c+d x)}{a b^2 d^2}+\frac {\left (a^2-b^2\right ) (e+f x) \cos (c+d x)}{a b^2 d}+\frac {f \left (a^2-b^2\right )^{3/2} \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {f \left (a^2-b^2\right )^{3/2} \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^3 d}+\frac {e x \left (a^2-b^2\right )}{b^3}+\frac {f x^2 \left (a^2-b^2\right )}{2 b^3}+\frac {i f \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {f \sin (c+d x)}{a d^2}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \cos ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {e x}{2 b}-\frac {f x^2}{4 b} \]

[Out]

-1/2*e*x/b+(a^2-b^2)*e*x/b^3-1/4*f*x^2/b+1/2*(a^2-b^2)*f*x^2/b^3-2*(f*x+e)*arctanh(exp(I*(d*x+c)))/a/d+(f*x+e)
*cos(d*x+c)/a/d+(a^2-b^2)*(f*x+e)*cos(d*x+c)/a/b^2/d-1/4*f*cos(d*x+c)^2/b/d^2+I*(a^2-b^2)^(3/2)*(f*x+e)*ln(1-I
*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^3/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*(a^2-b^2)^(3/2)*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^3/d+(a^2-b^2)^(3/2)*f*pol
ylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^3/d^2-(a^2-b^2)^(3/2)*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^
2-b^2)^(1/2)))/a/b^3/d^2-f*sin(d*x+c)/a/d^2-(a^2-b^2)*f*sin(d*x+c)/a/b^2/d^2-1/2*(f*x+e)*cos(d*x+c)*sin(d*x+c)
/b/d

________________________________________________________________________________________

Rubi [A]  time = 0.90, antiderivative size = 524, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 14, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4543, 4408, 4405, 2633, 3296, 2637, 4183, 2279, 2391, 4525, 3310, 3323, 2264, 2190} \[ \frac {f \left (a^2-b^2\right )^{3/2} \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {f \left (a^2-b^2\right )^{3/2} \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^3 d^2}+\frac {i f \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {i f \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {f \left (a^2-b^2\right ) \sin (c+d x)}{a b^2 d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^3 d}+\frac {\left (a^2-b^2\right ) (e+f x) \cos (c+d x)}{a b^2 d}+\frac {e x \left (a^2-b^2\right )}{b^3}+\frac {f x^2 \left (a^2-b^2\right )}{2 b^3}-\frac {f \sin (c+d x)}{a d^2}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \cos ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {e x}{2 b}-\frac {f x^2}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

-(e*x)/(2*b) + ((a^2 - b^2)*e*x)/b^3 - (f*x^2)/(4*b) + ((a^2 - b^2)*f*x^2)/(2*b^3) - (2*(e + f*x)*ArcTanh[E^(I
*(c + d*x))])/(a*d) + ((e + f*x)*Cos[c + d*x])/(a*d) + ((a^2 - b^2)*(e + f*x)*Cos[c + d*x])/(a*b^2*d) - (f*Cos
[c + d*x]^2)/(4*b*d^2) + (I*(a^2 - b^2)^(3/2)*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/
(a*b^3*d) - (I*(a^2 - b^2)^(3/2)*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*b^3*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) + ((a^2 - b^2)^(3/2)*f*P
olyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*b^3*d^2) - ((a^2 - b^2)^(3/2)*f*PolyLog[2, (I*b*E^(
I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*b^3*d^2) - (f*Sin[c + d*x])/(a*d^2) - ((a^2 - b^2)*f*Sin[c + d*x])/(a
*b^2*d^2) - ((e + f*x)*Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 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 4405

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[((c +
d*x)^m*Cos[a + b*x]^(n + 1))/(b*(n + 1)), x] + Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Cos[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 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 ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \cos ^3(c+d x) \cot (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x) \cos ^4(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac {\int (e+f x) \cos (c+d x) \cot (c+d x) \, dx}{a}-\frac {\int (e+f x) \cos ^2(c+d x) \, dx}{b}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {(e+f x) \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\\ &=-\frac {f \cos ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int (e+f x) \csc (c+d x) \, dx}{a}-\frac {\int (e+f x) \sin (c+d x) \, dx}{a}+\left (\frac {1}{a}-\frac {a}{b^2}\right ) \int (e+f x) \sin (c+d x) \, dx-\frac {\int (e+f x) \, dx}{2 b}+\frac {\left (a^2-b^2\right ) \int (e+f x) \, dx}{b^3}-\frac {\left (a^2-b^2\right )^2 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{a b^3}\\ &=-\frac {e x}{2 b}+\frac {\left (a^2-b^2\right ) e x}{b^3}-\frac {f x^2}{4 b}+\frac {\left (a^2-b^2\right ) f x^2}{2 b^3}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x) \cos (c+d x)}{d}-\frac {f \cos ^2(c+d x)}{4 b d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (2 \left (a^2-b^2\right )^2\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}{a b^3}-\frac {f \int \cos (c+d x) \, dx}{a d}-\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 {\left (\left (\frac {1}{a}-\frac {a}{b^2}\right ) f\right ) \int \cos (c+d x) \, dx}{d}\\ &=-\frac {e x}{2 b}+\frac {\left (a^2-b^2\right ) e x}{b^3}-\frac {f x^2}{4 b}+\frac {\left (a^2-b^2\right ) f x^2}{2 b^3}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x) \cos (c+d x)}{d}-\frac {f \cos ^2(c+d x)}{4 b d^2}-\frac {f \sin (c+d x)}{a d^2}+\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (2 i \left (a^2-b^2\right )^{3/2}\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 b^2}-\frac {\left (2 i \left (a^2-b^2\right )^{3/2}\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 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 {e x}{2 b}+\frac {\left (a^2-b^2\right ) e x}{b^3}-\frac {f x^2}{4 b}+\frac {\left (a^2-b^2\right ) f x^2}{2 b^3}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x) \cos (c+d x)}{d}-\frac {f \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 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 {f \sin (c+d x)}{a d^2}+\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (i \left (a^2-b^2\right )^{3/2} 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^3 d}+\frac {\left (i \left (a^2-b^2\right )^{3/2} 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^3 d}\\ &=-\frac {e x}{2 b}+\frac {\left (a^2-b^2\right ) e x}{b^3}-\frac {f x^2}{4 b}+\frac {\left (a^2-b^2\right ) f x^2}{2 b^3}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x) \cos (c+d x)}{d}-\frac {f \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 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 {f \sin (c+d x)}{a d^2}+\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (\left (a^2-b^2\right )^{3/2} 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^3 d^2}+\frac {\left (\left (a^2-b^2\right )^{3/2} 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^3 d^2}\\ &=-\frac {e x}{2 b}+\frac {\left (a^2-b^2\right ) e x}{b^3}-\frac {f x^2}{4 b}+\frac {\left (a^2-b^2\right ) f x^2}{2 b^3}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x) \cos (c+d x)}{d}-\frac {f \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 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 (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {\left (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {f \sin (c+d x)}{a d^2}+\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x) \sin (c+d x)}{2 b d}\\ \end {align*}

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Mathematica [A]  time = 11.76, size = 934, normalized size = 1.78 \[ -\frac {(d e+d 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 {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 )\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 {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 )\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 {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 )\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 {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 )\right )}{\sqrt {b^2-a^2}}\right ) \left (a^2-b^2\right )^2}{a b^3 d^2 \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 )}-\frac {\left (3 b^2-2 a^2\right ) (c+d x) (2 d e-2 c f+f (c+d x))}{4 b^3 d^2}+\frac {a (d e-c f+f (c+d x)) \cos (c+d x)}{b^2 d^2}-\frac {f \cos (2 (c+d x))}{8 b d^2}+\frac {e \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}-\frac {c f \log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d^2}+\frac {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 {Li}_2\left (-e^{i (c+d x)}\right )-\text {Li}_2\left (e^{i (c+d x)}\right )\right )\right )}{a d^2}-\frac {a f \sin (c+d x)}{b^2 d^2}-\frac {(d e-c f+f (c+d x)) \sin (2 (c+d x))}{4 b d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

-1/4*((-2*a^2 + 3*b^2)*(c + d*x)*(2*d*e - 2*c*f + f*(c + d*x)))/(b^3*d^2) + (a*(d*e - c*f + f*(c + d*x))*Cos[c
+ d*x])/(b^2*d^2) - (f*Cos[2*(c + d*x)])/(8*b*d^2) + (e*Log[Tan[(c + d*x)/2]])/(a*d) - (c*f*Log[Tan[(c + d*x)
/2]])/(a*d^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*d^2) - ((a^2 - b^2)^2*(d*e + d*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*Ta
n[(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])]))/Sq
rt[-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]))]))/Sqrt[-a^2 + b^2]
))/(a*b^3*d^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*f*Sin[c +
d*x])/(b^2*d^2) - ((d*e - c*f + f*(c + d*x))*Sin[2*(c + d*x)])/(4*b*d^2)

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fricas [B]  time = 0.80, size = 1619, normalized size = 3.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*((2*a^3 - 3*a*b^2)*d^2*f*x^2 - a*b^2*f*cos(d*x + c)^2 + 2*(2*a^3 - 3*a*b^2)*d^2*e*x - 2*I*b^3*f*dilog(cos(
d*x + c) + I*sin(d*x + c)) + 2*I*b^3*f*dilog(cos(d*x + c) - I*sin(d*x + c)) - 2*I*b^3*f*dilog(-cos(d*x + c) +
I*sin(d*x + c)) + 2*I*b^3*f*dilog(-cos(d*x + c) - I*sin(d*x + c)) - 2*I*(a^2*b - b^3)*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 - b^3)*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 - b^3)*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 - b^3)*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*((a^2*b -
b^3)*d*e - (a^2*b - b^3)*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*((a^2*b - b^3)*d*e - (a^2*b - b^3)*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*((a^2*b - b^3)*d*e - (a^2*b - b^3)*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*((a^2*b
- b^3)*d*e - (a^2*b - b^3)*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*((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*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 - b^3)*d*f*x + (a^2*b - b^3)*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 - b^3)*d*f*x + (a^2*b -
b^3)*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 - b^3)*d*f*x + (a^2*b - b^3)*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) + 4*(a^2*b*d*f*x + a^2*b*d*e)*cos(d*x + c) - 2*(b^3*d*f*x + b^3*d*e)*log(cos(d*x + c) + I*sin(d*
x + c) + 1) - 2*(b^3*d*f*x + b^3*d*e)*log(cos(d*x + c) - I*sin(d*x + c) + 1) + 2*(b^3*d*e - b^3*c*f)*log(-1/2*
cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2) + 2*(b^3*d*e - b^3*c*f)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) +
1/2) + 2*(b^3*d*f*x + b^3*c*f)*log(-cos(d*x + c) + I*sin(d*x + c) + 1) + 2*(b^3*d*f*x + b^3*c*f)*log(-cos(d*x
+ c) - I*sin(d*x + c) + 1) - 2*(2*a^2*b*f + (a*b^2*d*f*x + a*b^2*d*e)*cos(d*x + c))*sin(d*x + c))/(a*b^3*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)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 2.14, size = 1850, normalized size = 3.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

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))+1/2*a*(d*f*x+I*f+d*e)
/b^2/d^2*exp(I*(d*x+c))+1/2*a*(d*f*x-I*f+d*e)/b^2/d^2*exp(-I*(d*x+c))-2/d/b*a*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-2/d^2/b*a*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-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+2*
I*a^3/b^3/d^2*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))+a^3/b^3/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+a^3/b^3/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*a^3/b^3/d*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))-I*a^3/b^3/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)))-1/a/d*e*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-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))-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/a*dilog(exp(I*(d*x+c))+1)+I/d^2*f*dilog(exp(I*(d*x+c)))/a-4*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))+I/b^3/d^2*a^3*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/b^3/d^2*a^3*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/b/d*a*f/(-a^2+b^2)^(1/2)*ln((-I*a-b*e
xp(I*(d*x+c))+(-a^2+b^2)^(1/2))/(-I*a+(-a^2+b^2)^(1/2)))*x+2/b/d^2*a*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-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^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^3/d*a^3*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+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)))+4*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))-3/4*f*x^2/b+2*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)))-2*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)))+1/2*a^2*f*x^2/b^3-1/4*(
f*x+e)/d/b*sin(2*d*x+2*c)+a^2*e*x/b^3-3/2*e*x/b-1/8*f/d^2/b*cos(2*d*x+2*c)

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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)^3*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)^3*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 \[ \int \frac {\left (e + f x\right ) \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)**3*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*cos(c + d*x)**3*cot(c + d*x)/(a + b*sin(c + d*x)), x)

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