3.232 \(\int \frac{\csc ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=515 \[ \frac{3 a^2 b^4 \sin (c+d x)}{2 d \left (a^2-b^2\right )^4 (a \cos (c+d x)+b)}-\frac{a^2 b^3 \sin (c+d x)}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)^2}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^4 (a \cos (c+d x)+b)}-\frac{2 a b^3 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{a b^3 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{2 a b \left (8 a^2 b^2+3 a^4+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{\sin (c+d x)}{12 d (a+b)^3 (1-\cos (c+d x))}-\frac{(a-2 b) \sin (c+d x)}{4 d (a+b)^4 (1-\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 d (a-b)^4 (\cos (c+d x)+1)}+\frac{\sin (c+d x)}{12 d (a-b)^3 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^3 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^3 (\cos (c+d x)+1)^2} \]

[Out]

(-2*a*b^3*(3*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9/2)*d) -
(a*b^3*(a^2 + 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9/2)*d) - (
2*a*b*(3*a^4 + 8*a^2*b^2 + b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9
/2)*d) - Sin[c + d*x]/(12*(a + b)^3*d*(1 - Cos[c + d*x])^2) - ((a - 2*b)*Sin[c + d*x])/(4*(a + b)^4*d*(1 - Cos
[c + d*x])) - Sin[c + d*x]/(12*(a + b)^3*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(12*(a - b)^3*d*(1 + Cos[c + d*x
])^2) + Sin[c + d*x]/(12*(a - b)^3*d*(1 + Cos[c + d*x])) + ((a + 2*b)*Sin[c + d*x])/(4*(a - b)^4*d*(1 + Cos[c
+ d*x])) - (a^2*b^3*Sin[c + d*x])/(2*(a^2 - b^2)^3*d*(b + a*Cos[c + d*x])^2) + (3*a^2*b^4*Sin[c + d*x])/(2*(a^
2 - b^2)^4*d*(b + a*Cos[c + d*x])) + (a^2*b^2*(3*a^2 + b^2)*Sin[c + d*x])/((a^2 - b^2)^4*d*(b + a*Cos[c + d*x]
))

________________________________________________________________________________________

Rubi [A]  time = 0.774694, antiderivative size = 515, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2897, 2650, 2648, 2664, 2754, 12, 2659, 208} \[ \frac{3 a^2 b^4 \sin (c+d x)}{2 d \left (a^2-b^2\right )^4 (a \cos (c+d x)+b)}-\frac{a^2 b^3 \sin (c+d x)}{2 d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)^2}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^4 (a \cos (c+d x)+b)}-\frac{2 a b^3 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{a b^3 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{2 a b \left (8 a^2 b^2+3 a^4+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{9/2} (a+b)^{9/2}}-\frac{\sin (c+d x)}{12 d (a+b)^3 (1-\cos (c+d x))}-\frac{(a-2 b) \sin (c+d x)}{4 d (a+b)^4 (1-\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 d (a-b)^4 (\cos (c+d x)+1)}+\frac{\sin (c+d x)}{12 d (a-b)^3 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^3 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^3 (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Sec[c + d*x])^3,x]

[Out]

(-2*a*b^3*(3*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9/2)*d) -
(a*b^3*(a^2 + 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9/2)*d) - (
2*a*b*(3*a^4 + 8*a^2*b^2 + b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(9/2)*(a + b)^(9
/2)*d) - Sin[c + d*x]/(12*(a + b)^3*d*(1 - Cos[c + d*x])^2) - ((a - 2*b)*Sin[c + d*x])/(4*(a + b)^4*d*(1 - Cos
[c + d*x])) - Sin[c + d*x]/(12*(a + b)^3*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(12*(a - b)^3*d*(1 + Cos[c + d*x
])^2) + Sin[c + d*x]/(12*(a - b)^3*d*(1 + Cos[c + d*x])) + ((a + 2*b)*Sin[c + d*x])/(4*(a - b)^4*d*(1 + Cos[c
+ d*x])) - (a^2*b^3*Sin[c + d*x])/(2*(a^2 - b^2)^3*d*(b + a*Cos[c + d*x])^2) + (3*a^2*b^4*Sin[c + d*x])/(2*(a^
2 - b^2)^4*d*(b + a*Cos[c + d*x])) + (a^2*b^2*(3*a^2 + b^2)*Sin[c + d*x])/((a^2 - b^2)^4*d*(b + a*Cos[c + d*x]
))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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 2650

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

Rule 2648

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

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 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 208

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

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\int \frac{\cot ^3(c+d x) \csc (c+d x)}{(-b-a \cos (c+d x))^3} \, dx\\ &=\int \left (\frac{1}{4 (a-b)^3 (-1-\cos (c+d x))^2}+\frac{-a-2 b}{4 (a-b)^4 (-1-\cos (c+d x))}+\frac{1}{4 (a+b)^3 (1-\cos (c+d x))^2}+\frac{a-2 b}{4 (a+b)^4 (1-\cos (c+d x))}+\frac{a b^3}{\left (a^2-b^2\right )^2 (-b-a \cos (c+d x))^3}+\frac{a b^2 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^3 (-b-a \cos (c+d x))^2}+\frac{a b \left (3 a^4+8 a^2 b^2+b^4\right )}{\left (a^2-b^2\right )^4 (-b-a \cos (c+d x))}\right ) \, dx\\ &=\frac{\int \frac{1}{(-1-\cos (c+d x))^2} \, dx}{4 (a-b)^3}+\frac{(a-2 b) \int \frac{1}{1-\cos (c+d x)} \, dx}{4 (a+b)^4}+\frac{\int \frac{1}{(1-\cos (c+d x))^2} \, dx}{4 (a+b)^3}-\frac{(a+2 b) \int \frac{1}{-1-\cos (c+d x)} \, dx}{4 (a-b)^4}+\frac{\left (a b^3\right ) \int \frac{1}{(-b-a \cos (c+d x))^3} \, dx}{\left (a^2-b^2\right )^2}+\frac{\left (a b^2 \left (3 a^2+b^2\right )\right ) \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{\left (a^2-b^2\right )^3}+\frac{\left (a b \left (3 a^4+8 a^2 b^2+b^4\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^4}\\ &=-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))^2}-\frac{(a-2 b) \sin (c+d x)}{4 (a+b)^4 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))^2}+\frac{(a+2 b) \sin (c+d x)}{4 (a-b)^4 d (1+\cos (c+d x))}-\frac{a^2 b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))^2}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}-\frac{\int \frac{1}{-1-\cos (c+d x)} \, dx}{12 (a-b)^3}+\frac{\int \frac{1}{1-\cos (c+d x)} \, dx}{12 (a+b)^3}+\frac{\left (a b^3\right ) \int \frac{2 b-a \cos (c+d x)}{(-b-a \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )^3}+\frac{\left (a b^2 \left (3 a^2+b^2\right )\right ) \int \frac{b}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^4}+\frac{\left (2 a b \left (3 a^4+8 a^2 b^2+b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=-\frac{2 a b \left (3 a^4+8 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))^2}-\frac{(a-2 b) \sin (c+d x)}{4 (a+b)^4 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 (a-b)^4 d (1+\cos (c+d x))}-\frac{a^2 b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))^2}+\frac{3 a^2 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{\left (a b^3\right ) \int \frac{a^2+2 b^2}{-b-a \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac{\left (a b^3 \left (3 a^2+b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^4}\\ &=-\frac{2 a b \left (3 a^4+8 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))^2}-\frac{(a-2 b) \sin (c+d x)}{4 (a+b)^4 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 (a-b)^4 d (1+\cos (c+d x))}-\frac{a^2 b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))^2}+\frac{3 a^2 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{\left (a b^3 \left (a^2+2 b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^4}+\frac{\left (2 a b^3 \left (3 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=-\frac{2 a b^3 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{2 a b \left (3 a^4+8 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))^2}-\frac{(a-2 b) \sin (c+d x)}{4 (a+b)^4 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 (a-b)^4 d (1+\cos (c+d x))}-\frac{a^2 b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))^2}+\frac{3 a^2 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{\left (a b^3 \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^4 d}\\ &=-\frac{2 a b^3 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{a b^3 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{2 a b \left (3 a^4+8 a^2 b^2+b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{9/2} (a+b)^{9/2} d}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))^2}-\frac{(a-2 b) \sin (c+d x)}{4 (a+b)^4 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 (a-b)^3 d (1+\cos (c+d x))}+\frac{(a+2 b) \sin (c+d x)}{4 (a-b)^4 d (1+\cos (c+d x))}-\frac{a^2 b^3 \sin (c+d x)}{2 \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))^2}+\frac{3 a^2 b^4 \sin (c+d x)}{2 \left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}+\frac{a^2 b^2 \left (3 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^4 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.07206, size = 388, normalized size = 0.75 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b) \left (\csc ^3(c+d x) \left (-154 a^5 b^2 \cos (3 (c+d x))+62 a^5 b^2 \cos (5 (c+d x))+110 a^4 b^3 \cos (4 (c+d x))-205 a^3 b^4 \cos (3 (c+d x))+39 a^3 b^4 \cos (5 (c+d x))+120 a^2 b^5 \cos (4 (c+d x))-2 a \left (-94 a^4 b^2-35 a^2 b^4+16 a^6+8 b^6\right ) \cos (c+d x)+8 \left (-56 a^2 b^5-45 a^4 b^3+2 a^6 b-6 b^7\right ) \cos (2 (c+d x))+154 a^4 b^3+424 a^2 b^5-20 a^6 b \cos (4 (c+d x))+36 a^6 b-4 a^7 \cos (3 (c+d x))+4 a^7 \cos (5 (c+d x))+48 a b^6 \cos (3 (c+d x))+16 b^7\right )+\frac{96 a b \left (23 a^2 b^2+6 a^4+6 b^4\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}\right )}{96 d \left (a^2-b^2\right )^4 (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Sec[c + d*x])^3,x]

[Out]

((b + a*Cos[c + d*x])*((96*a*b*(6*a^4 + 23*a^2*b^2 + 6*b^4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2
]]*(b + a*Cos[c + d*x])^2)/Sqrt[a^2 - b^2] + (36*a^6*b + 154*a^4*b^3 + 424*a^2*b^5 + 16*b^7 - 2*a*(16*a^6 - 94
*a^4*b^2 - 35*a^2*b^4 + 8*b^6)*Cos[c + d*x] + 8*(2*a^6*b - 45*a^4*b^3 - 56*a^2*b^5 - 6*b^7)*Cos[2*(c + d*x)] -
4*a^7*Cos[3*(c + d*x)] - 154*a^5*b^2*Cos[3*(c + d*x)] - 205*a^3*b^4*Cos[3*(c + d*x)] + 48*a*b^6*Cos[3*(c + d*
x)] - 20*a^6*b*Cos[4*(c + d*x)] + 110*a^4*b^3*Cos[4*(c + d*x)] + 120*a^2*b^5*Cos[4*(c + d*x)] + 4*a^7*Cos[5*(c
+ d*x)] + 62*a^5*b^2*Cos[5*(c + d*x)] + 39*a^3*b^4*Cos[5*(c + d*x)])*Csc[c + d*x]^3)*Sec[c + d*x]^3)/(96*(a^2
- b^2)^4*d*(a + b*Sec[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.102, size = 328, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({\frac{1}{ \left ( 8\,{a}^{3}-24\,{a}^{2}b+24\,a{b}^{2}-8\,{b}^{3} \right ) \left ( a-b \right ) } \left ({\frac{a}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+3\,a\tan \left ( 1/2\,dx+c/2 \right ) +3\,b\tan \left ( 1/2\,dx+c/2 \right ) \right ) }+2\,{\frac{ab}{ \left ( a-b \right ) ^{4} \left ( a+b \right ) ^{4}} \left ({\frac{ \left ( 5/2\,{a}^{3}{b}^{2}+3\,a{b}^{4}-3\,{a}^{4}b-5/2\,{a}^{2}{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+ \left ( 5/2\,{a}^{3}{b}^{2}+3\,a{b}^{4}+3\,{a}^{4}b+5/2\,{a}^{2}{b}^{3} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{2}}}-1/2\,{\frac{6\,{a}^{4}+23\,{a}^{2}{b}^{2}+6\,{b}^{4}}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \right ) }-{\frac{1}{24\, \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{3\,a-3\,b}{8\, \left ( a+b \right ) ^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+b*sec(d*x+c))^3,x)

[Out]

1/d*(1/8/(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*(1/3*tan(1/2*d*x+1/2*c)^3*a-1/3*b*tan(1/2*d*x+1/2*c)^3+3*a*tan(1/2*d*
x+1/2*c)+3*b*tan(1/2*d*x+1/2*c))+2*a*b/(a-b)^4/(a+b)^4*(((5/2*a^3*b^2+3*a*b^4-3*a^4*b-5/2*a^2*b^3)*tan(1/2*d*x
+1/2*c)^3+(5/2*a^3*b^2+3*a*b^4+3*a^4*b+5/2*a^2*b^3)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/
2*c)^2*b-a-b)^2-1/2*(6*a^4+23*a^2*b^2+6*b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b)
)^(1/2)))-1/24/(a+b)^3/tan(1/2*d*x+1/2*c)^3-1/8/(a+b)^4*(3*a-3*b)/tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

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(csc(d*x+c)^4/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.72401, size = 3421, normalized size = 6.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/12*(78*a^6*b^3 + 46*a^4*b^5 - 116*a^2*b^7 - 8*b^9 + 2*(4*a^9 + 58*a^7*b^2 - 23*a^5*b^4 - 39*a^3*b^6)*cos(d
*x + c)^5 - 10*(2*a^8*b - 13*a^6*b^3 - a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^4 - 4*(3*a^9 + 55*a^7*b^2 - 8*a^5*b^
4 - 56*a^3*b^6 + 6*a*b^8)*cos(d*x + c)^3 + 3*(6*a^5*b^3 + 23*a^3*b^5 + 6*a*b^7 - (6*a^7*b + 23*a^5*b^3 + 6*a^3
*b^5)*cos(d*x + c)^4 - 2*(6*a^6*b^2 + 23*a^4*b^4 + 6*a^2*b^6)*cos(d*x + c)^3 + (6*a^7*b + 17*a^5*b^3 - 17*a^3*
b^5 - 6*a*b^7)*cos(d*x + c)^2 + 2*(6*a^6*b^2 + 23*a^4*b^4 + 6*a^2*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*
b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 -
b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) + 4*(6*a^8*b - 56*a^6*b^3 - 8*a^4*b^5 + 55*
a^2*b^7 + 3*b^9)*cos(d*x + c)^2 + 10*(12*a^7*b^2 - a^5*b^4 - 13*a^3*b^6 + 2*a*b^8)*cos(d*x + c))/(((a^12 - 5*a
^10*b^2 + 10*a^8*b^4 - 10*a^6*b^6 + 5*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^4 + 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^
5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3 - (a^12 - 6*a^10*b^2 + 15*a^8*b^4 - 20*a^6*b^6 + 15*a^4*
b^8 - 6*a^2*b^10 + b^12)*d*cos(d*x + c)^2 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^
11)*d*cos(d*x + c) - (a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d)*sin(d*x + c)), -1
/6*(39*a^6*b^3 + 23*a^4*b^5 - 58*a^2*b^7 - 4*b^9 + (4*a^9 + 58*a^7*b^2 - 23*a^5*b^4 - 39*a^3*b^6)*cos(d*x + c)
^5 - 5*(2*a^8*b - 13*a^6*b^3 - a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^4 - 2*(3*a^9 + 55*a^7*b^2 - 8*a^5*b^4 - 56*a
^3*b^6 + 6*a*b^8)*cos(d*x + c)^3 - 3*(6*a^5*b^3 + 23*a^3*b^5 + 6*a*b^7 - (6*a^7*b + 23*a^5*b^3 + 6*a^3*b^5)*co
s(d*x + c)^4 - 2*(6*a^6*b^2 + 23*a^4*b^4 + 6*a^2*b^6)*cos(d*x + c)^3 + (6*a^7*b + 17*a^5*b^3 - 17*a^3*b^5 - 6*
a*b^7)*cos(d*x + c)^2 + 2*(6*a^6*b^2 + 23*a^4*b^4 + 6*a^2*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^
2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) + 2*(6*a^8*b - 56*a^6*b^3 - 8*a^4*b^5 +
55*a^2*b^7 + 3*b^9)*cos(d*x + c)^2 + 5*(12*a^7*b^2 - a^5*b^4 - 13*a^3*b^6 + 2*a*b^8)*cos(d*x + c))/(((a^12 -
5*a^10*b^2 + 10*a^8*b^4 - 10*a^6*b^6 + 5*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^4 + 2*(a^11*b - 5*a^9*b^3 + 10*a^7
*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a*b^11)*d*cos(d*x + c)^3 - (a^12 - 6*a^10*b^2 + 15*a^8*b^4 - 20*a^6*b^6 + 15*a
^4*b^8 - 6*a^2*b^10 + b^12)*d*cos(d*x + c)^2 - 2*(a^11*b - 5*a^9*b^3 + 10*a^7*b^5 - 10*a^5*b^7 + 5*a^3*b^9 - a
*b^11)*d*cos(d*x + c) - (a^10*b^2 - 5*a^8*b^4 + 10*a^6*b^6 - 10*a^4*b^8 + 5*a^2*b^10 - b^12)*d)*sin(d*x + c))]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+b*sec(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**4/(a + b*sec(c + d*x))**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.45734, size = 957, normalized size = 1.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(24*(6*a^5*b + 23*a^3*b^3 + 6*a*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2
*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*sq
rt(-a^2 + b^2)) + (a^6*tan(1/2*d*x + 1/2*c)^3 - 6*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 15*a^4*b^2*tan(1/2*d*x + 1/2*
c)^3 - 20*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^5*tan(1/2*d*x + 1/2*c)^3
+ b^6*tan(1/2*d*x + 1/2*c)^3 + 9*a^6*tan(1/2*d*x + 1/2*c) - 36*a^5*b*tan(1/2*d*x + 1/2*c) + 45*a^4*b^2*tan(1/2
*d*x + 1/2*c) - 45*a^2*b^4*tan(1/2*d*x + 1/2*c) + 36*a*b^5*tan(1/2*d*x + 1/2*c) - 9*b^6*tan(1/2*d*x + 1/2*c))/
(a^9 - 9*a^8*b + 36*a^7*b^2 - 84*a^6*b^3 + 126*a^5*b^4 - 126*a^4*b^5 + 84*a^3*b^6 - 36*a^2*b^7 + 9*a*b^8 - b^9
) - 24*(6*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*b^4*tan(1/2*d*x + 1/2*c)^3
- 6*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 - 6*a^5*b^2*tan(1/2*d*x + 1/2*c) - 5*a^4*b^3*tan(1/2*d*x + 1/2*c) - 5*a^3*
b^4*tan(1/2*d*x + 1/2*c) - 6*a^2*b^5*tan(1/2*d*x + 1/2*c))/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a
*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) - (9*a*tan(1/2*d*x + 1/2*c)^2 - 9*b*tan(1/2*d*x
+ 1/2*c)^2 + a + b)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*tan(1/2*d*x + 1/2*c)^3))/d