3.1271 \(\int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=395 \[ -\frac{6 \left (a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{\left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{6 \left (a^2 b^4+a^6-2 b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 d \sqrt{a^2-b^2}}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{x}{b^3} \]

[Out]

-(x/b^3) - (6*Sqrt[a^2 - b^2]*(a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*d) + (Sqr
t[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*d) + (6*(a^6 + a^2*b^4 -
2*b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*b^3*Sqrt[a^2 - b^2]*d) - ArcTanh[Cos[c + d*x]]/
(2*a^3*d) + (3*(a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a^5*d) + (3*b*Cot[c + d*x])/(a^4*d) - (Cot[c + d*x]*Csc[c
+ d*x])/(2*a^3*d) + ((a^2 - b^2)^2*Cos[c + d*x])/(2*a^3*b^2*d*(a + b*Sin[c + d*x])^2) + (3*(a^2 - b^2)*Cos[c
+ d*x])/(2*a^2*b^2*d*(a + b*Sin[c + d*x])) - (3*(a^4 - b^4)*Cos[c + d*x])/(a^4*b^2*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.523451, antiderivative size = 395, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 3768, 2664, 2754, 12, 2660, 618, 204} \[ -\frac{6 \left (a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{\left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{6 \left (a^2 b^4+a^6-2 b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 d \sqrt{a^2-b^2}}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/b^3) - (6*Sqrt[a^2 - b^2]*(a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*d) + (Sqr
t[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*d) + (6*(a^6 + a^2*b^4 -
2*b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*b^3*Sqrt[a^2 - b^2]*d) - ArcTanh[Cos[c + d*x]]/
(2*a^3*d) + (3*(a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a^5*d) + (3*b*Cot[c + d*x])/(a^4*d) - (Cot[c + d*x]*Csc[c
+ d*x])/(2*a^3*d) + ((a^2 - b^2)^2*Cos[c + d*x])/(2*a^3*b^2*d*(a + b*Sin[c + d*x])^2) + (3*(a^2 - b^2)*Cos[c
+ d*x])/(2*a^2*b^2*d*(a + b*Sin[c + d*x])) - (3*(a^4 - b^4)*Cos[c + d*x])/(a^4*b^2*d*(a + b*Sin[c + d*x]))

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

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

Rule 3767

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

Rule 8

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

Rule 3768

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

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 ^3(c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (-\frac{1}{b^3}-\frac{3 \left (a^2-2 b^2\right ) \csc (c+d x)}{a^5}-\frac{3 b \csc ^2(c+d x)}{a^4}+\frac{\csc ^3(c+d x)}{a^3}+\frac{\left (a^2-b^2\right )^3}{a^3 b^3 (a+b \sin (c+d x))^3}-\frac{3 \left (a^2-b^2\right )^2 \left (a^2+b^2\right )}{a^4 b^3 (a+b \sin (c+d x))^2}+\frac{3 \left (a^6+a^2 b^4-2 b^6\right )}{a^5 b^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{x}{b^3}+\frac{\int \csc ^3(c+d x) \, dx}{a^3}-\frac{(3 b) \int \csc ^2(c+d x) \, dx}{a^4}-\frac{\left (3 \left (a^2-2 b^2\right )\right ) \int \csc (c+d x) \, dx}{a^5}+\frac{\left (a^2-b^2\right )^3 \int \frac{1}{(a+b \sin (c+d x))^3} \, dx}{a^3 b^3}-\frac{\left (3 \left (a^2-b^2\right )^2 \left (a^2+b^2\right )\right ) \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^4 b^3}+\frac{\left (3 \left (a^6+a^2 b^4-2 b^6\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^5 b^3}\\ &=-\frac{x}{b^3}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}+\frac{\int \csc (c+d x) \, dx}{2 a^3}-\frac{\left (a^2-b^2\right )^2 \int \frac{-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 a^3 b^3}+\frac{\left (3 \left (a^2-b^2\right )^2 \left (a^2+b^2\right )\right ) \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^4 b^3 \left (-a^2+b^2\right )}+\frac{(3 b) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^4 d}+\frac{\left (6 \left (a^6+a^2 b^4-2 b^6\right )\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^5 b^3 d}\\ &=-\frac{x}{b^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}-\left (3 \left (\frac{a}{b^3}-\frac{b}{a^3}\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx+\frac{\left (a^2-b^2\right ) \int \frac{2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 a^3 b^3}-\frac{\left (12 \left (a^6+a^2 b^4-2 b^6\right )\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^5 b^3 d}\\ &=-\frac{x}{b^3}+\frac{6 \left (a^6+a^2 b^4-2 b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}+\frac{\left (\left (a^2-b^2\right ) \left (2 a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 a^3 b^3}-\frac{\left (6 \left (\frac{a}{b^3}-\frac{b}{a^3}\right )\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 )}{d}\\ &=-\frac{x}{b^3}+\frac{6 \left (a^6+a^2 b^4-2 b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}+\frac{\left (12 \left (\frac{a}{b^3}-\frac{b}{a^3}\right )\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 )}{d}+\frac{\left (\left (a^2-b^2\right ) \left (2 a^2+b^2\right )\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^3 b^3 d}\\ &=-\frac{x}{b^3}-\frac{6 \left (\frac{a}{b^3}-\frac{b}{a^3}\right ) \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} d}+\frac{6 \left (a^6+a^2 b^4-2 b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}-\frac{\left (2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right )\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^3 b^3 d}\\ &=-\frac{x}{b^3}-\frac{6 \left (\frac{a}{b^3}-\frac{b}{a^3}\right ) \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} d}+\frac{\sqrt{a^2-b^2} \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{6 \left (a^6+a^2 b^4-2 b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{3 b \cot (c+d x)}{a^4 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac{3 \left (a^4-b^4\right ) \cos (c+d x)}{a^4 b^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.25991, size = 384, normalized size = 0.97 \[ \frac{\left (12 b^2-5 a^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac{\left (5 a^2-12 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}-\frac{3 \left (a^2 b^2 \cos (c+d x)+a^4 \cos (c+d x)-2 b^4 \cos (c+d x)\right )}{2 a^4 b^2 d (a+b \sin (c+d x))}+\frac{-2 a^2 b^2 \cos (c+d x)+a^4 \cos (c+d x)+b^4 \cos (c+d x)}{2 a^3 b^2 d (a+b \sin (c+d x))^2}+\frac{\left (-a^4 b^2+11 a^2 b^4+2 a^6-12 b^6\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^5 b^3 d \sqrt{a^2-b^2}}-\frac{3 b \tan \left (\frac{1}{2} (c+d x)\right )}{2 a^4 d}+\frac{3 b \cot \left (\frac{1}{2} (c+d x)\right )}{2 a^4 d}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^3 d}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^3 d}-\frac{c+d x}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)/(b^3*d)) + ((2*a^6 - a^4*b^2 + 11*a^2*b^4 - 12*b^6)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] +
a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^5*b^3*Sqrt[a^2 - b^2]*d) + (3*b*Cot[(c + d*x)/2])/(2*a^4*d) - Csc[(
c + d*x)/2]^2/(8*a^3*d) + ((5*a^2 - 12*b^2)*Log[Cos[(c + d*x)/2]])/(2*a^5*d) + ((-5*a^2 + 12*b^2)*Log[Sin[(c +
d*x)/2]])/(2*a^5*d) + Sec[(c + d*x)/2]^2/(8*a^3*d) + (a^4*Cos[c + d*x] - 2*a^2*b^2*Cos[c + d*x] + b^4*Cos[c +
d*x])/(2*a^3*b^2*d*(a + b*Sin[c + d*x])^2) - (3*(a^4*Cos[c + d*x] + a^2*b^2*Cos[c + d*x] - 2*b^4*Cos[c + d*x]
))/(2*a^4*b^2*d*(a + b*Sin[c + d*x])) - (3*b*Tan[(c + d*x)/2])/(2*a^4*d)

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Maple [B]  time = 0.221, size = 943, 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)^3/(a+b*sin(d*x+c))^3,x)

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^4*tan(1/2*d*x+1/2*c)*b-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))-1/d/b/(tan(1/
2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3-7/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)^2*tan(1/2*d*x+1/2*c)^3+8/d/a^4*b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2
*d*x+1/2*c)^3-2/d*a/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-9/d/(tan(1/2*
d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^2-3/d/a^3*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/
2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2+14/d/a^5*b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(
1/2*d*x+1/2*c)^2-7/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)-13/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)^2*tan(1/2*d*x+1/2*c)+20/d/a^4*b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1
/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)-2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a-5/d/a/(t
an(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+7/d/a^3*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a
)^2+2/d*a/b^3/(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/d/a/b/(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))+11/d/a^3*b/(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))-12/d/a^5*b^3/(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))-1/8/d/a^3/tan(1/2*d*x+1/2*c)^2-5/2/d/a^3*ln(tan(1/2*d*x+1/2*c))+6/d/a^5*ln(tan(1/2*d*x+1/2*c))*b^2+3/2
/d*b/a^4/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)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.21889, size = 3688, normalized size = 9.34 \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)^3/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*a^5*b^2*d*x*cos(d*x + c)^4 - 4*(a^7 + 2*a^5*b^2)*d*x*cos(d*x + c)^2 - 2*(2*a^6*b + 5*a^4*b^3 - 18*a^2
*b^5)*cos(d*x + c)^3 + 4*(a^7 + a^5*b^2)*d*x - (2*a^6 + 3*a^4*b^2 + 13*a^2*b^4 + 12*b^6 + (2*a^4*b^2 + a^2*b^4
+ 12*b^6)*cos(d*x + c)^4 - (2*a^6 + 5*a^4*b^2 + 14*a^2*b^4 + 24*b^6)*cos(d*x + c)^2 + 2*(2*a^5*b + a^3*b^3 +
12*a*b^5 - (2*a^5*b + a^3*b^3 + 12*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*c
os(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)) + 4*(a^6*b + 3*a^4*b^3 - 9*a^2*b^5)*cos(d*x + c)
- (5*a^4*b^3 - 7*a^2*b^5 - 12*b^7 + (5*a^2*b^5 - 12*b^7)*cos(d*x + c)^4 - (5*a^4*b^3 - 2*a^2*b^5 - 24*b^7)*cos
(d*x + c)^2 + 2*(5*a^3*b^4 - 12*a*b^6 - (5*a^3*b^4 - 12*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x +
c) + 1/2) + (5*a^4*b^3 - 7*a^2*b^5 - 12*b^7 + (5*a^2*b^5 - 12*b^7)*cos(d*x + c)^4 - (5*a^4*b^3 - 2*a^2*b^5 -
24*b^7)*cos(d*x + c)^2 + 2*(5*a^3*b^4 - 12*a*b^6 - (5*a^3*b^4 - 12*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(-1
/2*cos(d*x + c) + 1/2) - 2*(4*a^6*b*d*x*cos(d*x + c)^2 - 4*a^6*b*d*x + 3*(a^5*b^2 + a^3*b^4 - 4*a*b^6)*cos(d*x
+ c)^3 - (3*a^5*b^2 - a^3*b^4 - 12*a*b^6)*cos(d*x + c))*sin(d*x + c))/(a^5*b^5*d*cos(d*x + c)^4 - (a^7*b^3 +
2*a^5*b^5)*d*cos(d*x + c)^2 + (a^7*b^3 + a^5*b^5)*d - 2*(a^6*b^4*d*cos(d*x + c)^2 - a^6*b^4*d)*sin(d*x + c)),
-1/4*(4*a^5*b^2*d*x*cos(d*x + c)^4 - 4*(a^7 + 2*a^5*b^2)*d*x*cos(d*x + c)^2 - 2*(2*a^6*b + 5*a^4*b^3 - 18*a^2*
b^5)*cos(d*x + c)^3 + 4*(a^7 + a^5*b^2)*d*x + 2*(2*a^6 + 3*a^4*b^2 + 13*a^2*b^4 + 12*b^6 + (2*a^4*b^2 + a^2*b^
4 + 12*b^6)*cos(d*x + c)^4 - (2*a^6 + 5*a^4*b^2 + 14*a^2*b^4 + 24*b^6)*cos(d*x + c)^2 + 2*(2*a^5*b + a^3*b^3 +
12*a*b^5 - (2*a^5*b + a^3*b^3 + 12*a*b^5)*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))) + 4*(a^6*b + 3*a^4*b^3 - 9*a^2*b^5)*cos(d*x + c) - (5*a^4*b^3 - 7*a^2*
b^5 - 12*b^7 + (5*a^2*b^5 - 12*b^7)*cos(d*x + c)^4 - (5*a^4*b^3 - 2*a^2*b^5 - 24*b^7)*cos(d*x + c)^2 + 2*(5*a^
3*b^4 - 12*a*b^6 - (5*a^3*b^4 - 12*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (5*a^4*b
^3 - 7*a^2*b^5 - 12*b^7 + (5*a^2*b^5 - 12*b^7)*cos(d*x + c)^4 - (5*a^4*b^3 - 2*a^2*b^5 - 24*b^7)*cos(d*x + c)^
2 + 2*(5*a^3*b^4 - 12*a*b^6 - (5*a^3*b^4 - 12*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2
) - 2*(4*a^6*b*d*x*cos(d*x + c)^2 - 4*a^6*b*d*x + 3*(a^5*b^2 + a^3*b^4 - 4*a*b^6)*cos(d*x + c)^3 - (3*a^5*b^2
- a^3*b^4 - 12*a*b^6)*cos(d*x + c))*sin(d*x + c))/(a^5*b^5*d*cos(d*x + c)^4 - (a^7*b^3 + 2*a^5*b^5)*d*cos(d*x
+ c)^2 + (a^7*b^3 + a^5*b^5)*d - 2*(a^6*b^4*d*cos(d*x + c)^2 - a^6*b^4*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)**3/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.30663, size = 691, normalized size = 1.75 \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)^3/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(8*(d*x + c)/b^3 + 4*(5*a^2 - 12*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 - (a^3*tan(1/2*d*x + 1/2*c)^2 -
12*a^2*b*tan(1/2*d*x + 1/2*c))/a^6 - 8*(2*a^6 - a^4*b^2 + 11*a^2*b^4 - 12*b^6)*(pi*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^5*b^3) - (10*a^4*b^2*tan(
1/2*d*x + 1/2*c)^6 - 24*a^2*b^4*tan(1/2*d*x + 1/2*c)^6 - 8*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 4*a^3*b^3*tan(1/2*d*
x + 1/2*c)^5 - 32*a*b^5*tan(1/2*d*x + 1/2*c)^5 - 16*a^6*tan(1/2*d*x + 1/2*c)^4 - 53*a^4*b^2*tan(1/2*d*x + 1/2*
c)^4 + 16*a^2*b^4*tan(1/2*d*x + 1/2*c)^4 + 16*b^6*tan(1/2*d*x + 1/2*c)^4 - 56*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 4
4*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 112*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 16*a^6*tan(1/2*d*x + 1/2*c)^2 - 32*a^4*b
^2*tan(1/2*d*x + 1/2*c)^2 + 76*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 8*a^3*b^3*tan(1/2*d*x + 1/2*c) - a^4*b^2)/((a*
tan(1/2*d*x + 1/2*c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))^2*a^5*b^2))/d