3.3.1 \(\int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) [201]

3.3.1.1 Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {(b-a \cos (c+d x)) \csc ^2(c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {a \log (1-\cos (c+d x))}{4 (a+b)^2 d}-\frac {a \log (1+\cos (c+d x))}{4 (a-b)^2 d}+\frac {a^2 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^2 d} \]

output

1/2*(b-a*cos(d*x+c))*csc(d*x+c)^2/(a^2-b^2)/d+1/4*a*ln(1-cos(d*x+c))/(a+b) 
^2/d-1/4*a*ln(1+cos(d*x+c))/(a-b)^2/d+a^2*b*ln(b+a*cos(d*x+c))/(a^2-b^2)^2 
/d
 

3.3.1.2 Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-(a-b)^2 (a+b) \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 a \left ((a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (b+a \cos (c+d x))-(a-b)^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(a-b) (a+b)^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 (a-b)^2 (a+b)^2 d} \]

input

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

output

(-((a - b)^2*(a + b)*Csc[(c + d*x)/2]^2) - 4*a*((a + b)^2*Log[Cos[(c + d*x 
)/2]] - 2*a*b*Log[b + a*Cos[c + d*x]] - (a - b)^2*Log[Sin[(c + d*x)/2]]) + 
 (a - b)*(a + b)^2*Sec[(c + d*x)/2]^2)/(8*(a - b)^2*(a + b)^2*d)
 

3.3.1.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.31, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4360, 25, 25, 3042, 25, 3316, 25, 27, 593, 25, 657, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int -\frac {\cot (c+d x) \csc ^2(c+d x)}{-a \cos (c+d x)-b}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\frac {\cot (c+d x) \csc ^2(c+d x)}{b+a \cos (c+d x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a \cos (c+d x)+b}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {a^3 \int -\frac {\cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^3 \int \frac {\cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^2 \int \frac {a \cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 593

\(\displaystyle -\frac {a^2 \left (\frac {\int -\frac {b-a \cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 \left (a^2-b^2\right )}-\frac {b-a \cos (c+d x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^2 \left (-\frac {\int \frac {b-a \cos (c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 \left (a^2-b^2\right )}-\frac {b-a \cos (c+d x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 657

\(\displaystyle -\frac {a^2 \left (-\frac {\int \left (\frac {-a-b}{2 a (a-b) (\cos (c+d x) a+a)}+\frac {b-a}{2 a (a+b) (a-a \cos (c+d x))}+\frac {2 b}{(a-b) (a+b) (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{2 \left (a^2-b^2\right )}-\frac {b-a \cos (c+d x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \left (-\frac {b-a \cos (c+d x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {\frac {2 b \log (a \cos (c+d x)+b)}{a^2-b^2}+\frac {(a-b) \log (a-a \cos (c+d x))}{2 a (a+b)}-\frac {(a+b) \log (a \cos (c+d x)+a)}{2 a (a-b)}}{2 \left (a^2-b^2\right )}\right )}{d}\)

input

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

output

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

3.3.1.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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 4360

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

3.3.1.4 Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (d x +c \right )-1\right )}+\frac {a \ln \left (\cos \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (d x +c \right )+1\right )}-\frac {a \ln \left (\cos \left (d x +c \right )+1\right )}{4 \left (a -b \right )^{2}}+\frac {b \,a^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(110\)
default	\(\frac {\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (d x +c \right )-1\right )}+\frac {a \ln \left (\cos \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (d x +c \right )+1\right )}-\frac {a \ln \left (\cos \left (d x +c \right )+1\right )}{4 \left (a -b \right )^{2}}+\frac {b \,a^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(110\)
parallelrisch	\(\frac {8 \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right ) a^{2} b -\left (a -b \right ) \left (-4 a \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a +b \right ) \left (\left (a -b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a +b \right )\right )\right )}{8 d \left (a -b \right )^{2} \left (a +b \right )^{2}}\)	\(111\)
norman	\(\frac {-\frac {1}{8 d \left (a +b \right )}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d \left (a -b \right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {a^{2} b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}\)	\(137\)
risch	\(-\frac {i a x}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i a c}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {i a x}{2 a^{2}-4 a b +2 b^{2}}+\frac {i a c}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i a^{2} b x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i a^{2} b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} a}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(311\)

input

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

output

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

3.3.1.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.86 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - 4 \, {\left (a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2} - {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d\right )}} \]

input

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

output

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

3.3.1.6 Sympy [F]

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

input

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

output

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

3.3.1.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {4 \, a^{2} b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a \cos \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

input

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

output

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

3.3.1.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.74 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {8 \, a^{2} b \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {{\left (a + b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a - b\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]

input

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

output

1/8*(8*a^2*b*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^4 - 2*a^2*b^2 + b^4) + 2*a*log( 
abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^2 + 2*a*b + b^2) + (a + b 
 - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^2 + 2 
*a*b + b^2)*(cos(d*x + c) - 1)) - (cos(d*x + c) - 1)/((a - b)*(cos(d*x + c 
) + 1)))/d
 

3.3.1.9 Mupad [B] (verification not implemented)

Time = 13.70 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {a\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{4\,d\,{\left (a+b\right )}^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (\frac {a}{4\,{\left (a+b\right )}^2}-\frac {a}{4\,{\left (a-b\right )}^2}\right )}{d}-\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2-1\right )}-\frac {a\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{4\,d\,{\left (a-b\right )}^2} \]

input

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

output

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