\(\int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [225]

Optimal result

Integrand size = 19, antiderivative size = 163 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {b^3}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b^2 \left (3 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{2 (a+b)^3 d}-\frac {\log (1+\cos (c+d x))}{2 (a-b)^3 d}+\frac {b \left (3 a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d} \]

[Out]

-1/2*b^3/a^2/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))+1/2*ln(1-cos(d*
x+c))/(a+b)^3/d-1/2*ln(1+cos(d*x+c))/(a-b)^3/d+b*(3*a^2+b^2)*ln(b+a*cos(d*x+c))/(a^2-b^2)^3/d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2916, 12, 1643} \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {b^2 \left (3 a^2-b^2\right )}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {b \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}-\frac {b^3}{2 a^2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)^3}-\frac {\log (\cos (c+d x)+1)}{2 d (a-b)^3} \]

[In]

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

[Out]

-1/2*b^3/(a^2*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + (b^2*(3*a^2 - b^2))/(a^2*(a^2 - b^2)^2*d*(b + a*Cos[c +
d*x])) + Log[1 - Cos[c + d*x]]/(2*(a + b)^3*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)^3*d) + (b*(3*a^2 + b^2)*Log[
b + a*Cos[c + d*x]])/((a^2 - b^2)^3*d)

Rule 12

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

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2916

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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*} \text {integral}& = -\int \frac {\cos ^2(c+d x) \cot (c+d x)}{(-b-a \cos (c+d x))^3} \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {x^3}{a^3 (-b+x)^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{(-b+x)^3 \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2}{2 (a-b)^3 (a-x)}-\frac {b^3}{(a-b) (a+b) (b-x)^3}+\frac {3 a^2 b^2-b^4}{(a-b)^2 (a+b)^2 (b-x)^2}-\frac {a^2 b \left (3 a^2+b^2\right )}{(a-b)^3 (a+b)^3 (b-x)}+\frac {a^2}{2 (a+b)^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = -\frac {b^3}{2 a^2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b^2 \left (3 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{2 (a+b)^3 d}-\frac {\log (1+\cos (c+d x))}{2 (a-b)^3 d}+\frac {b \left (3 a^2+b^2\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.25 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \left (\frac {b^3}{a^2 \left (-a^2+b^2\right )}-\frac {2 b^2 \left (-3 a^2+b^2\right ) (b+a \cos (c+d x))}{a^2 (a-b)^2 (a+b)^2}+\frac {2 (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{(-a+b)^3}+\frac {2 b \left (3 a^2+b^2\right ) (b+a \cos (c+d x))^2 \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3}+\frac {2 (b+a \cos (c+d x))^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{(a+b)^3}\right ) \sec ^3(c+d x)}{2 d (a+b \sec (c+d x))^3} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])*(b^3/(a^2*(-a^2 + b^2)) - (2*b^2*(-3*a^2 + b^2)*(b + a*Cos[c + d*x]))/(a^2*(a - b)^2*(a
+ b)^2) + (2*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2]])/(-a + b)^3 + (2*b*(3*a^2 + b^2)*(b + a*Cos[c + d*x]
)^2*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^3 + (2*(b + a*Cos[c + d*x])^2*Log[Sin[(c + d*x)/2]])/(a + b)^3)*Sec[c
+ d*x]^3)/(2*d*(a + b*Sec[c + d*x])^3)

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {b^{3}}{2 a^{2} \left (a +b \right ) \left (a -b \right ) \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {b^{2} \left (3 a^{2}-b^{2}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2} \left (b +a \cos \left (d x +c \right )\right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{3}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(148\)
default	\(\frac {-\frac {b^{3}}{2 a^{2} \left (a +b \right ) \left (a -b \right ) \left (b +a \cos \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {b^{2} \left (3 a^{2}-b^{2}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2} \left (b +a \cos \left (d x +c \right )\right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{3}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}}{d}\)	\(148\)
parallelrisch	\(\frac {3 \left (a^{2}+\frac {b^{2}}{3}\right ) \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right ) b \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )+\left (a -b \right )^{3} \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (a +b \right ) \left (\frac {\left (a +\frac {b}{2}\right ) \left (a -\frac {b}{3}\right ) \cos \left (2 d x +2 c \right )}{2}+a \left (a +b \right ) \cos \left (d x +c \right )+\frac {a^{2}}{2}+\frac {11 a b}{12}+\frac {b^{2}}{12}\right ) b^{2}}{\left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right )}\)	\(223\)
norman	\(\frac {\frac {6 a \,b^{2}}{d \left (a^{4}-2 a^{3} b +2 a \,b^{3}-b^{4}\right )}-\frac {2 \left (3 a \,b^{2}+b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(226\)
risch	\(\frac {i x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {i c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b \,a^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {6 i b \,a^{2} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 i b^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {2 i b^{3} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 b^{2} \left (-3 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{3} {\mathrm e}^{i \left (d x +c \right )}+b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} \left (-a^{2}+b^{2}\right )^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right ) a^{2}}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(591\)

[In]

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

[Out]

1/d*(-1/2/a^2*b^3/(a+b)/(a-b)/(b+a*cos(d*x+c))^2+b*(3*a^2+b^2)/(a+b)^3/(a-b)^3*ln(b+a*cos(d*x+c))+b^2*(3*a^2-b
^2)/(a+b)^2/(a-b)^2/a^2/(b+a*cos(d*x+c))-1/2/(a-b)^3*ln(cos(d*x+c)+1)+1/2/(a+b)^3*ln(cos(d*x+c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (157) = 314\).

Time = 0.34 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.91 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {5 \, a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7} + 2 \, {\left (3 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) + 2 \, {\left (3 \, a^{4} b^{3} + a^{2} b^{5} + {\left (3 \, a^{6} b + a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5} + {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b - 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} - a^{3} b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{2} - 3 \, a^{6} b^{4} + 3 \, a^{4} b^{6} - a^{2} b^{8}\right )} d\right )}} \]

[In]

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

[Out]

1/2*(5*a^4*b^3 - 6*a^2*b^5 + b^7 + 2*(3*a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c) + 2*(3*a^4*b^3 + a^2*b^5 + (
3*a^6*b + a^4*b^3)*cos(d*x + c)^2 + 2*(3*a^5*b^2 + a^3*b^4)*cos(d*x + c))*log(a*cos(d*x + c) + b) - (a^5*b^2 +
3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5 + (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*cos(d*x + c)^2 + 2*(a^6*b + 3*a^5*b^2
+ 3*a^4*b^3 + a^3*b^4)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5
+ (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*cos(d*x + c)^2 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*cos(d*x
+ c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c)^2 + 2*(a^9*b - 3*
a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c) + (a^8*b^2 - 3*a^6*b^4 + 3*a^4*b^6 - a^2*b^8)*d)

Sympy [F]

\[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\csc {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.48 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{2} b + b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {5 \, a^{2} b^{3} - b^{5} + 2 \, {\left (3 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{a^{6} b^{2} - 2 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \left (d x + c\right )} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(3*a^2*b + b^3)*log(a*cos(d*x + c) + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (5*a^2*b^3 - b^5 + 2*(3*a
^3*b^2 - a*b^4)*cos(d*x + c))/(a^6*b^2 - 2*a^4*b^4 + a^2*b^6 + (a^8 - 2*a^6*b^2 + a^4*b^4)*cos(d*x + c)^2 + 2*
(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)) - log(cos(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + log(cos(
d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (157) = 314\).

Time = 0.39 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.77 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{2} b + b^{3}\right )} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {9 \, a^{3} b + 15 \, a^{2} b^{2} + 3 \, a b^{3} - 3 \, b^{4} + \frac {18 \, a^{3} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {6 \, a^{2} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {10 \, a b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{3} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {9 \, a^{2} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} {\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(3*a^2*b + b^3)*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1)))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^3 + 3
*a^2*b + 3*a*b^2 + b^3) - (9*a^3*b + 15*a^2*b^2 + 3*a*b^3 - 3*b^4 + 18*a^3*b*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 6*a^2*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*a*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b^
4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^3*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 9*a^2*b^2*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*a*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3*b^4*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2)/((a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*(a + b + a*(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2))/d

Mupad [B] (verification not implemented)

Time = 14.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.12 \[ \int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^3}-\frac {\frac {\cos \left (c+d\,x\right )\,\left (b^4-3\,a^2\,b^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\left (b^4-5\,a^2\,b^2\right )}{2\,a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\cos \left (c+d\,x\right )}^2+2\,a\,b\,\cos \left (c+d\,x\right )+b^2\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^3}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (\frac {1}{2\,{\left (a+b\right )}^3}-\frac {1}{2\,{\left (a-b\right )}^3}\right )}{d} \]

[In]

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

[Out]

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