\(\int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [97]

Optimal result

Integrand size = 19, antiderivative size = 82 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {1}{6 d (a+a \cos (c+d x))^3}+\frac {5}{8 a d (a+a \cos (c+d x))^2}-\frac {7}{8 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3957, 2915, 12, 90, 212} \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {7}{8 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {5}{8 a d (a \cos (c+d x)+a)^2}-\frac {1}{6 d (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

-1/8*ArcTanh[Cos[c + d*x]]/(a^3*d) - 1/(6*d*(a + a*Cos[c + d*x])^3) + 5/(8*a*d*(a + a*Cos[c + d*x])^2) - 7/(8*
d*(a^3 + a^3*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 212

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

Rule 2915

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[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)}{(-a-a \cos (c+d x))^3} \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {x^3}{a^3 (-a-x) (-a+x)^4} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{(-a-x) (-a+x)^4} \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a^2}{2 (a-x)^4}+\frac {5 a}{4 (a-x)^3}-\frac {7}{8 (a-x)^2}+\frac {1}{8 \left (a^2-x^2\right )}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = -\frac {1}{6 d (a+a \cos (c+d x))^3}+\frac {5}{8 a d (a+a \cos (c+d x))^2}-\frac {7}{8 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,-a \cos (c+d x)\right )}{8 a^2 d} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{8 a^3 d}-\frac {1}{6 d (a+a \cos (c+d x))^3}+\frac {5}{8 a d (a+a \cos (c+d x))^2}-\frac {7}{8 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\left (2-15 \cos ^2\left (\frac {1}{2} (c+d x)\right )+42 \cos ^4\left (\frac {1}{2} (c+d x)\right )+12 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sec ^3(c+d x)}{12 a^3 d (1+\sec (c+d x))^3} \]

[In]

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

[Out]

-1/12*((2 - 15*Cos[(c + d*x)/2]^2 + 42*Cos[(c + d*x)/2]^4 + 12*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2]] - Log
[Sin[(c + d*x)/2]]))*Sec[c + d*x]^3)/(a^3*d*(1 + Sec[c + d*x])^3)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(\frac {-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 a^{3} d}\)	\(61\)
derivativedivides	\(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}-\frac {1}{6 \left (\cos \left (d x +c \right )+1\right )^{3}}+\frac {5}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {7}{8 \left (\cos \left (d x +c \right )+1\right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}}{d \,a^{3}}\)	\(67\)
default	\(\frac {\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}-\frac {1}{6 \left (\cos \left (d x +c \right )+1\right )^{3}}+\frac {5}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {7}{8 \left (\cos \left (d x +c \right )+1\right )}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}}{d \,a^{3}}\)	\(67\)
norman	\(\frac {-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{32 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{48 d a}}{a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{3} d}\)	\(82\)
risch	\(-\frac {21 \,{\mathrm e}^{5 i \left (d x +c \right )}+54 \,{\mathrm e}^{4 i \left (d x +c \right )}+82 \,{\mathrm e}^{3 i \left (d x +c \right )}+54 \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{i \left (d x +c \right )}}{12 a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 a^{3} d}\)	\(119\)

[In]

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

[Out]

1/96*(-2*tan(1/2*d*x+1/2*c)^6+9*tan(1/2*d*x+1/2*c)^4-18*tan(1/2*d*x+1/2*c)^2+12*ln(tan(1/2*d*x+1/2*c)))/a^3/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (74) = 148\).

Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {42 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 54 \, \cos \left (d x + c\right ) + 20}{48 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

-1/48*(42*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1
/2) - 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 54*cos(d*x + c
) + 20)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F]

\[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\csc {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

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

[Out]

Integral(csc(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.20 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (21 \, \cos \left (d x + c\right )^{2} + 27 \, \cos \left (d x + c\right ) + 10\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \]

[In]

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

[Out]

-1/48*(2*(21*cos(d*x + c)^2 + 27*cos(d*x + c) + 10)/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x
+ c) + a^3) + 3*log(cos(d*x + c) + 1)/a^3 - 3*log(cos(d*x + c) - 1)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.38 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {6 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac {\frac {18 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{9}}}{96 \, d} \]

[In]

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

[Out]

1/96*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (18*a^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
+ 9*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 2*a^6*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^9)/d

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {7\,{\cos \left (c+d\,x\right )}^2}{8}+\frac {9\,\cos \left (c+d\,x\right )}{8}+\frac {5}{12}}{d\,\left (a^3\,{\cos \left (c+d\,x\right )}^3+3\,a^3\,{\cos \left (c+d\,x\right )}^2+3\,a^3\,\cos \left (c+d\,x\right )+a^3\right )}-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a^3\,d} \]

[In]

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

[Out]

- ((9*cos(c + d*x))/8 + (7*cos(c + d*x)^2)/8 + 5/12)/(d*(3*a^3*cos(c + d*x) + a^3 + 3*a^3*cos(c + d*x)^2 + a^3
*cos(c + d*x)^3)) - atanh(cos(c + d*x))/(8*a^3*d)