3.1263 \(\int \frac{\cos (c+d x) \cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=303 \[ -\frac{10 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^6 d}+\frac{\left (-20 a^2 b^2+3 a^4+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}-\frac{5 \left (-12 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))} \]

[Out]

(-10*b*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^6*d) - (5*(3*a^4 - 12*a^2*b^2 +
8*b^4)*ArcTanh[Cos[c + d*x]])/(8*a^6*d) + ((3*a^4 - 20*a^2*b^2 + 15*b^4)*Cot[c + d*x])/(3*a^5*b*d) + (5*(5*a^2
- 4*b^2)*Cot[c + d*x]*Csc[c + d*x])/(8*a^4*d) - Cot[c + d*x]/(b*d*(a + b*Sin[c + d*x])) - ((6*a^2 - 5*b^2)*Co
t[c + d*x]*Csc[c + d*x])/(3*a^3*d*(a + b*Sin[c + d*x])) + (5*b*Cot[c + d*x]*Csc[c + d*x]^2)/(12*a^2*d*(a + b*S
in[c + d*x])) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.21099, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2896, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac{10 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^6 d}+\frac{\left (-20 a^2 b^2+3 a^4+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}-\frac{5 \left (-12 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^5)/(a + b*Sin[c + d*x])^2,x]

[Out]

(-10*b*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^6*d) - (5*(3*a^4 - 12*a^2*b^2 +
8*b^4)*ArcTanh[Cos[c + d*x]])/(8*a^6*d) + ((3*a^4 - 20*a^2*b^2 + 15*b^4)*Cot[c + d*x])/(3*a^5*b*d) + (5*(5*a^2
- 4*b^2)*Cot[c + d*x]*Csc[c + d*x])/(8*a^4*d) - Cot[c + d*x]/(b*d*(a + b*Sin[c + d*x])) - ((6*a^2 - 5*b^2)*Co
t[c + d*x]*Csc[c + d*x])/(3*a^3*d*(a + b*Sin[c + d*x])) + (5*b*Cot[c + d*x]*Csc[c + d*x]^2)/(12*a^2*d*(a + b*S
in[c + d*x])) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d*(a + b*Sin[c + d*x]))

Rule 2896

Int[cos[(e_.) + (f_.)*(x_)]^6*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b^2*d^2*(n + 1)*(n + 2)*(m + n + 5)*(m + n + 6)), Int[(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*
x])^m*Simp[a^4*(n + 1)*(n + 2)*(n + 3)*(n + 5) - a^2*b^2*(n + 2)*(2*n + 1)*(m + n + 5)*(m + n + 6) + b^4*(m +
n + 2)*(m + n + 3)*(m + n + 5)*(m + n + 6) + a*b*m*(a^2*(n + 1)*(n + 2) - b^2*(m + n + 5)*(m + n + 6))*Sin[e +
f*x] - (a^4*(n + 1)*(n + 2)*(4 + n)*(n + 5) + b^4*(m + n + 2)*(m + n + 4)*(m + n + 5)*(m + n + 6) - a^2*b^2*(
n + 1)*(n + 2)*(m + n + 5)*(2*n + 2*m + 13))*Sin[e + f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(d*
Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*d^2*f*(n + 1)*(n + 2)), x] - Simp[(a*(n + 5)*Cos[e +
f*x]*(d*Sin[e + f*x])^(n + 3)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d^3*f*(m + n + 5)*(m + n + 6)), x] + Simp[(Co
s[e + f*x]*(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^(m + 1))/(b*d^4*f*(m + n + 6)), x]) /; FreeQ[{a, b, d
, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && NeQ[n, -1] && NeQ[n, -2] && NeQ[m + n + 5, 0]
&& NeQ[m + n + 6, 0] &&  !IGtQ[m, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^3(c+d x) \left (-2 b^2 \left (27 a^2-20 b^2\right )-4 a b \left (6 a^2-b^2\right ) \sin (c+d x)+6 b^2 \left (4 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{24 a^2 b^2}\\ &=-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^3(c+d x) \left (-30 b^2 \left (5 a^4-9 a^2 b^2+4 b^4\right )-2 a b \left (12 a^4-17 a^2 b^2+5 b^4\right ) \sin (c+d x)+16 b^2 \left (6 a^4-11 a^2 b^2+5 b^4\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{24 a^3 b^2 \left (a^2-b^2\right )}\\ &=\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^2(c+d x) \left (-16 b \left (3 a^6-23 a^4 b^2+35 a^2 b^4-15 b^6\right )+2 a b^2 \left (21 a^4-41 a^2 b^2+20 b^4\right ) \sin (c+d x)-30 b^3 \left (5 a^4-9 a^2 b^2+4 b^4\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{48 a^4 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (30 b^2 \left (3 a^6-15 a^4 b^2+20 a^2 b^4-8 b^6\right )-30 a b^3 \left (5 a^4-9 a^2 b^2+4 b^4\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{48 a^5 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}-\frac{\left (5 b \left (a^2-b^2\right )^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^6}+\frac{\left (5 \left (3 a^4-12 a^2 b^2+8 b^4\right )\right ) \int \csc (c+d x) \, dx}{8 a^6}\\ &=-\frac{5 \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac{\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}-\frac{\left (10 b \left (a^2-b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^6 d}\\ &=-\frac{5 \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac{\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}+\frac{\left (20 b \left (a^2-b^2\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^6 d}\\ &=-\frac{10 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^6 d}-\frac{5 \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac{\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{3 a^5 b d}+\frac{5 \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}-\frac{\cot (c+d x)}{b d (a+b \sin (c+d x))}-\frac{\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^3 d (a+b \sin (c+d x))}+\frac{5 b \cot (c+d x) \csc ^2(c+d x)}{12 a^2 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.20382, size = 487, normalized size = 1.61 \[ \frac{3 \left (3 a^2-4 b^2\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 a^4 d}-\frac{3 \left (3 a^2-4 b^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 a^4 d}+\frac{5 \left (-12 a^2 b^2+3 a^4+8 b^4\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^6 d}-\frac{5 \left (-12 a^2 b^2+3 a^4+8 b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^6 d}+\frac{-2 a^2 b^2 \cos (c+d x)+a^4 \cos (c+d x)+b^4 \cos (c+d x)}{a^5 d (a+b \sin (c+d x))}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \left (6 b^3 \cos \left (\frac{1}{2} (c+d x)\right )-7 a^2 b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 d}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (7 a^2 b \sin \left (\frac{1}{2} (c+d x)\right )-6 b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 d}-\frac{10 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^6 d}+\frac{b \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{12 a^3 d}-\frac{b \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{12 a^3 d}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 a^2 d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 a^2 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^5)/(a + b*Sin[c + d*x])^2,x]

[Out]

(-10*b*(a^2 - b^2)^(3/2)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])
/(a^6*d) + ((-7*a^2*b*Cos[(c + d*x)/2] + 6*b^3*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(3*a^5*d) + (3*(3*a^2 - 4*b
^2)*Csc[(c + d*x)/2]^2)/(32*a^4*d) + (b*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(12*a^3*d) - Csc[(c + d*x)/2]^4/(
64*a^2*d) - (5*(3*a^4 - 12*a^2*b^2 + 8*b^4)*Log[Cos[(c + d*x)/2]])/(8*a^6*d) + (5*(3*a^4 - 12*a^2*b^2 + 8*b^4)
*Log[Sin[(c + d*x)/2]])/(8*a^6*d) - (3*(3*a^2 - 4*b^2)*Sec[(c + d*x)/2]^2)/(32*a^4*d) + Sec[(c + d*x)/2]^4/(64
*a^2*d) + (Sec[(c + d*x)/2]*(7*a^2*b*Sin[(c + d*x)/2] - 6*b^3*Sin[(c + d*x)/2]))/(3*a^5*d) + (a^4*Cos[c + d*x]
- 2*a^2*b^2*Cos[c + d*x] + b^4*Cos[c + d*x])/(a^5*d*(a + b*Sin[c + d*x])) - (b*Sec[(c + d*x)/2]^2*Tan[(c + d*
x)/2])/(12*a^3*d)

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Maple [B]  time = 0.187, size = 718, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x)

[Out]

1/64/d/a^2*tan(1/2*d*x+1/2*c)^4-1/12/d/a^3*tan(1/2*d*x+1/2*c)^3*b-1/4/d/a^2*tan(1/2*d*x+1/2*c)^2+3/8/d/a^4*b^2
*tan(1/2*d*x+1/2*c)^2+9/4/d/a^3*tan(1/2*d*x+1/2*c)*b-2/d/a^5*b^3*tan(1/2*d*x+1/2*c)+2/d/a^2*b/(tan(1/2*d*x+1/2
*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)-4/d/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
*tan(1/2*d*x+1/2*c)*b^3+2/d*b^5/a^6/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d/a
/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-4/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b
^2+2/d*b^4/a^5/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-10/d*b/a^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta
n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+20/d*b^3/a^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a
^2-b^2)^(1/2))-10/d*b^5/a^6/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/64/d/a^
2/tan(1/2*d*x+1/2*c)^4+1/4/d/a^2/tan(1/2*d*x+1/2*c)^2-3/8/d/a^4/tan(1/2*d*x+1/2*c)^2*b^2+15/8/d/a^2*ln(tan(1/2
*d*x+1/2*c))-15/2/d/a^4*ln(tan(1/2*d*x+1/2*c))*b^2+5/d/a^6*ln(tan(1/2*d*x+1/2*c))*b^4+1/12/d/a^3*b/tan(1/2*d*x
+1/2*c)^3-9/4/d*b/a^3/tan(1/2*d*x+1/2*c)+2/d*b^3/a^5/tan(1/2*d*x+1/2*c)

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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(cos(d*x+c)^6*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.28851, size = 3613, normalized size = 11.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/48*(16*(3*a^5 - 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5 - 10*(15*a^5 - 68*a^3*b^2 + 48*a*b^4)*cos(d*x + c)^3
- 120*((a^3*b - a*b^3)*cos(d*x + c)^4 + a^3*b - a*b^3 - 2*(a^3*b - a*b^3)*cos(d*x + c)^2 + ((a^2*b^2 - b^4)*co
s(d*x + c)^4 + a^2*b^2 - b^4 - 2*(a^2*b^2 - b^4)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 -
b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(
-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 30*(3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*
x + c) - 15*(3*a^5 - 12*a^3*b^2 + 8*a*b^4 + (3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c)^4 - 2*(3*a^5 - 12*a^3*
b^2 + 8*a*b^4)*cos(d*x + c)^2 + (3*a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4
- 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 15*(3*a^5 - 12*
a^3*b^2 + 8*a*b^4 + (3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c)^4 - 2*(3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x +
c)^2 + (3*a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^
3 + 8*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 10*((17*a^4*b - 12*a^2*b^3)*cos(d*x +
c)^3 - 3*(5*a^4*b - 4*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^7*d*cos(d*x + c)^4 - 2*a^7*d*cos(d*x + c)^2 + a^
7*d + (a^6*b*d*cos(d*x + c)^4 - 2*a^6*b*d*cos(d*x + c)^2 + a^6*b*d)*sin(d*x + c)), 1/48*(16*(3*a^5 - 20*a^3*b^
2 + 15*a*b^4)*cos(d*x + c)^5 - 10*(15*a^5 - 68*a^3*b^2 + 48*a*b^4)*cos(d*x + c)^3 + 240*((a^3*b - a*b^3)*cos(d
*x + c)^4 + a^3*b - a*b^3 - 2*(a^3*b - a*b^3)*cos(d*x + c)^2 + ((a^2*b^2 - b^4)*cos(d*x + c)^4 + a^2*b^2 - b^4
- 2*(a^2*b^2 - b^4)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^
2)*cos(d*x + c))) + 30*(3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c) - 15*(3*a^5 - 12*a^3*b^2 + 8*a*b^4 + (3*a^5
- 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c)^4 - 2*(3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c)^2 + (3*a^4*b - 12*a^2*
b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)
*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 15*(3*a^5 - 12*a^3*b^2 + 8*a*b^4 + (3*a^5 - 12*a^3*b^2 + 8*a*b^4)
*cos(d*x + c)^4 - 2*(3*a^5 - 12*a^3*b^2 + 8*a*b^4)*cos(d*x + c)^2 + (3*a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b -
12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*
cos(d*x + c) + 1/2) + 10*((17*a^4*b - 12*a^2*b^3)*cos(d*x + c)^3 - 3*(5*a^4*b - 4*a^2*b^3)*cos(d*x + c))*sin(d
*x + c))/(a^7*d*cos(d*x + c)^4 - 2*a^7*d*cos(d*x + c)^2 + a^7*d + (a^6*b*d*cos(d*x + c)^4 - 2*a^6*b*d*cos(d*x
+ c)^2 + a^6*b*d)*sin(d*x + c))]

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Sympy [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(cos(d*x+c)**6*csc(d*x+c)**5/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29556, size = 641, normalized size = 2.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/192*(120*(3*a^4 - 12*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c)))/a^6 - 1920*(a^4*b - 2*a^2*b^3 + b^5)*(p
i*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2
)*a^6) + 384*(a^4*b*tan(1/2*d*x + 1/2*c) - 2*a^2*b^3*tan(1/2*d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c) + a^5 - 2
*a^3*b^2 + a*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^6) + (3*a^6*tan(1/2*d*x + 1/2*c
)^4 - 16*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 48*a^6*tan(1/2*d*x + 1/2*c)^2 + 72*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 + 43
2*a^5*b*tan(1/2*d*x + 1/2*c) - 384*a^3*b^3*tan(1/2*d*x + 1/2*c))/a^8 - (750*a^4*tan(1/2*d*x + 1/2*c)^4 - 3000*
a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 2000*b^4*tan(1/2*d*x + 1/2*c)^4 + 432*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 384*a*b^
3*tan(1/2*d*x + 1/2*c)^3 - 48*a^4*tan(1/2*d*x + 1/2*c)^2 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 16*a^3*b*tan(1/
2*d*x + 1/2*c) + 3*a^4)/(a^6*tan(1/2*d*x + 1/2*c)^4))/d