\(\int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx\) [63]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\csc ^4(c+d x)}{4 a d} \]

[Out]

-1/8*arctanh(cos(d*x+c))/a/d-1/8*cot(d*x+c)*csc(d*x+c)/a/d+1/4*cot(d*x+c)*csc(d*x+c)^3/a/d-1/4*csc(d*x+c)^4/a/
d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2914, 2686, 30, 2691, 3853, 3855} \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d} \]

[In]

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

[Out]

-1/8*ArcTanh[Cos[c + d*x]]/(a*d) - (Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d)
- Csc[c + d*x]^4/(4*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3853

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 3855

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

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 {\cot (c+d x) \csc ^2(c+d x)}{-a-a \cos (c+d x)} \, dx \\ & = -\frac {\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a}+\frac {\int \cot (c+d x) \csc ^4(c+d x) \, dx}{a} \\ & = \frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \csc ^3(c+d x) \, dx}{4 a}-\frac {\text {Subst}\left (\int x^3 \, dx,x,\csc (c+d x)\right )}{a d} \\ & = -\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\int \csc (c+d x) \, dx}{8 a} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\csc ^4(c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\left (2 \cot ^2\left (\frac {1}{2} (c+d x)\right )+4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{16 a d (1+\sec (c+d x))} \]

[In]

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

[Out]

-1/16*((2*Cot[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^2*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) + Sec[(c +
d*x)/2]^2)*Sec[c + d*x])/(a*d*(1 + Sec[c + d*x]))

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {-\frac {1}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}+\frac {1}{8 \cos \left (d x +c \right )-8}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}}{d a}\)	\(55\)
default	\(\frac {-\frac {1}{8 \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{16}+\frac {1}{8 \cos \left (d x +c \right )-8}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{16}}{d a}\)	\(55\)
parallelrisch	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}\)	\(61\)
norman	\(\frac {-\frac {1}{16 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{32 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\)	\(79\)
risch	\(\frac {{\mathrm e}^{5 i \left (d x +c \right )}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{4 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\)	\(128\)

[In]

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

[Out]

1/d/a*(-1/8/(cos(d*x+c)+1)^2-1/16*ln(cos(d*x+c)+1)+1/8/(cos(d*x+c)-1)+1/16*ln(cos(d*x+c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, \cos \left (d x + c\right ) + 4}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}} \]

[In]

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

[Out]

1/16*(2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + (c
os(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*cos(d*x + c) + 4)/(a*d*cos
(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 2\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]

[In]

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

[Out]

1/16*(2*(cos(d*x + c)^2 + cos(d*x + c) + 2)/(a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) - a) - log(c
os(d*x + c) + 1)/a + log(cos(d*x + c) - 1)/a)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.57 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {2 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac {\frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}}}{32 \, d} \]

[In]

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

[Out]

-1/32*(2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)/(a*(cos(d*x + c) - 1)) - 2*log(abs(-co
s(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a - (2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2)/a^2)/d

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {\csc ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a\,d}-\frac {\frac {{\cos \left (c+d\,x\right )}^2}{8}+\frac {\cos \left (c+d\,x\right )}{8}+\frac {1}{4}}{d\,\left (-a\,{\cos \left (c+d\,x\right )}^3-a\,{\cos \left (c+d\,x\right )}^2+a\,\cos \left (c+d\,x\right )+a\right )} \]

[In]

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

[Out]

- atanh(cos(c + d*x))/(8*a*d) - (cos(c + d*x)/8 + cos(c + d*x)^2/8 + 1/4)/(d*(a + a*cos(c + d*x) - a*cos(c + d
*x)^2 - a*cos(c + d*x)^3))