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

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

[Out]

-2*a*b^3*(3*a^2+b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(9/2)/(a+b)^(9/2)/d-a*b^3*(a^2+
2*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(9/2)/(a+b)^(9/2)/d-2*a*b*(3*a^4+8*a^2*b^2+b^
4)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(9/2)/(a+b)^(9/2)/d-1/12*sin(d*x+c)/(a+b)^3/d/(1-
cos(d*x+c))^2-1/4*(a-2*b)*sin(d*x+c)/(a+b)^4/d/(1-cos(d*x+c))-1/12*sin(d*x+c)/(a+b)^3/d/(1-cos(d*x+c))+1/12*si
n(d*x+c)/(a-b)^3/d/(1+cos(d*x+c))^2+1/12*sin(d*x+c)/(a-b)^3/d/(1+cos(d*x+c))+1/4*(a+2*b)*sin(d*x+c)/(a-b)^4/d/
(1+cos(d*x+c))-1/2*a^2*b^3*sin(d*x+c)/(a^2-b^2)^3/d/(b+a*cos(d*x+c))^2+3/2*a^2*b^4*sin(d*x+c)/(a^2-b^2)^4/d/(b
+a*cos(d*x+c))+a^2*b^2*(3*a^2+b^2)*sin(d*x+c)/(a^2-b^2)^4/d/(b+a*cos(d*x+c))

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Rubi [A]  time = 0.77, antiderivative size = 515, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 1.08, size = 388, normalized size = 0.75 \[ \frac {\sec ^3(c+d x) (a \cos (c+d x)+b) \left (\frac {96 a b \left (6 a^4+23 a^2 b^2+6 b^4\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\csc ^3(c+d x) \left (-4 a^7 \cos (3 (c+d x))+4 a^7 \cos (5 (c+d x))-20 a^6 b \cos (4 (c+d x))+36 a^6 b-154 a^5 b^2 \cos (3 (c+d x))+62 a^5 b^2 \cos (5 (c+d x))+110 a^4 b^3 \cos (4 (c+d x))+154 a^4 b^3-205 a^3 b^4 \cos (3 (c+d x))+39 a^3 b^4 \cos (5 (c+d x))+120 a^2 b^5 \cos (4 (c+d x))+424 a^2 b^5+8 \left (2 a^6 b-45 a^4 b^3-56 a^2 b^5-6 b^7\right ) \cos (2 (c+d x))-2 a \left (16 a^6-94 a^4 b^2-35 a^2 b^4+8 b^6\right ) \cos (c+d x)+48 a b^6 \cos (3 (c+d x))+16 b^7\right )\right )}{96 d \left (a^2-b^2\right )^4 (a+b \sec (c+d x))^3} \]

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)

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fricas [A]  time = 0.83, size = 1550, normalized size = 3.01 \[ \text {result too large to display} \]

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))]

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giac [A]  time = 0.50, size = 709, normalized size = 1.38 \[ \frac {\frac {24 \, {\left (6 \, a^{5} b + 23 \, a^{3} b^{3} + 6 \, a b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 45 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{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}} - \frac {24 \, {\left (6 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}} - \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

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

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maple [A]  time = 0.62, size = 328, normalized size = 0.64 \[ \frac {\frac {\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{3}+3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a -b \right )}+\frac {2 a b \left (\frac {\left (\frac {5}{2} b^{2} a^{3}+3 a \,b^{4}-3 b \,a^{4}-\frac {5}{2} a^{2} b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} b^{2} a^{3}+3 a \,b^{4}+3 b \,a^{4}+\frac {5}{2} a^{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{2}}-\frac {\left (6 a^{4}+23 a^{2} b^{2}+6 b^{4}\right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{4} \left (a +b \right )^{4}}-\frac {1}{24 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a -3 b}{8 \left (a +b \right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

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*a*tan(1/2*d*x+1/2*c)^3-1/3*tan(1/2*d*x+1/2*c)^3*b+3*a*tan(1/2*d*
x+1/2*c)+3*tan(1/2*d*x+1/2*c)*b)+2*a*b/(a-b)^4/(a+b)^4*(((5/2*b^2*a^3+3*a*b^4-3*b*a^4-5/2*a^2*b^3)*tan(1/2*d*x
+1/2*c)^3+(5/2*b^2*a^3+3*a*b^4+3*b*a^4+5/2*a^2*b^3)*tan(1/2*d*x+1/2*c))/(a*tan(1/2*d*x+1/2*c)^2-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(tan(1/2*d*x+1/2*c)*(a-b)/((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))

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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(csc(d*x+c)^4/(a+b*sec(d*x+c))^3,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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.78, size = 588, normalized size = 1.14 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d\,{\left (a-b\right )}^3}+\frac {\frac {a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^7-21\,a^6\,b+111\,a^5\,b^2-145\,a^4\,b^3+145\,a^3\,b^4-111\,a^2\,b^5+21\,a\,b^6-3\,b^7\right )}{{\left (a+b\right )}^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (17\,a^6-102\,a^5\,b+399\,a^4\,b^2-364\,a^3\,b^3+399\,a^2\,b^4-102\,a\,b^5+17\,b^6\right )}{3\,{\left (a+b\right )}^3}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}{3\,{\left (a+b\right )}^2}}{d\,\left (\left (-8\,a^6+48\,a^5\,b-120\,a^4\,b^2+160\,a^3\,b^3-120\,a^2\,b^4+48\,a\,b^5-8\,b^6\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (16\,a^6-64\,a^5\,b+80\,a^4\,b^2-80\,a^2\,b^4+64\,a\,b^5-16\,b^6\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,a^6+16\,a^5\,b+8\,a^4\,b^2-32\,a^3\,b^3+8\,a^2\,b^4+16\,a\,b^5-8\,b^6\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}{8\,d\,{\left (a-b\right )}^4}+\frac {a\,b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^8-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6\,b^2+6{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^4-4{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^6+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^8}{{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{7/2}}\right )\,\left (6\,a^4+23\,a^2\,b^2+6\,b^4\right )\,1{}\mathrm {i}}{d\,{\left (a+b\right )}^{9/2}\,{\left (a-b\right )}^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^4*(a + b/cos(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)^3/(24*d*(a - b)^3) + ((a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)/(3*(a + b)) + (tan(c/2 +
(d*x)/2)^6*(21*a*b^6 - 21*a^6*b + 3*a^7 - 3*b^7 - 111*a^2*b^5 + 145*a^3*b^4 - 145*a^4*b^3 + 111*a^5*b^2))/(a +
b)^4 - (tan(c/2 + (d*x)/2)^4*(17*a^6 - 102*a^5*b - 102*a*b^5 + 17*b^6 + 399*a^2*b^4 - 364*a^3*b^3 + 399*a^4*b
^2))/(3*(a + b)^3) + (7*tan(c/2 + (d*x)/2)^2*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))/(3*(a + b)^2
))/(d*(tan(c/2 + (d*x)/2)^3*(16*a*b^5 + 16*a^5*b - 8*a^6 - 8*b^6 + 8*a^2*b^4 - 32*a^3*b^3 + 8*a^4*b^2) - tan(c
/2 + (d*x)/2)^7*(8*a^6 - 48*a^5*b - 48*a*b^5 + 8*b^6 + 120*a^2*b^4 - 160*a^3*b^3 + 120*a^4*b^2) + tan(c/2 + (d
*x)/2)^5*(64*a*b^5 - 64*a^5*b + 16*a^6 - 16*b^6 - 80*a^2*b^4 + 80*a^4*b^2))) + (3*tan(c/2 + (d*x)/2)*(a + b))/
(8*d*(a - b)^4) + (a*b*atan((a^8*tan(c/2 + (d*x)/2)*1i + b^8*tan(c/2 + (d*x)/2)*1i - a^2*b^6*tan(c/2 + (d*x)/2
)*4i + a^4*b^4*tan(c/2 + (d*x)/2)*6i - a^6*b^2*tan(c/2 + (d*x)/2)*4i)/((a + b)^(9/2)*(a - b)^(7/2)))*(6*a^4 +
6*b^4 + 23*a^2*b^2)*1i)/(d*(a + b)^(9/2)*(a - b)^(9/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

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)

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