\(\int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\) [189]

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {a^2 \left (b \left (3+\frac {b^2}{a^2}\right )+a \left (1+\frac {3 b^2}{a^2}\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 d}+\frac {(a+b)^2 (a+4 b) \log (1-\cos (c+d x))}{4 d}-\frac {b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))}{d}-\frac {(a-4 b) (a-b)^2 \log (1+\cos (c+d x))}{4 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec ^2(c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 6.46 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.60 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {24 a b^2-(a+b)^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-4 (a-4 b) (a-b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b \left (3 a^2+2 b^2\right ) \log (\cos (c+d x))+4 (a+b)^2 (a+4 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+(a-b)^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {2 b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {24 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {24 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{8 d} \] Input:

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

Output:

(24*a*b^2 - (a + b)^3*Csc[(c + d*x)/2]^2 - 4*(a - 4*b)*(a - b)^2*Log[Cos[( 
c + d*x)/2]] - 8*b*(3*a^2 + 2*b^2)*Log[Cos[c + d*x]] + 4*(a + b)^2*(a + 4* 
b)*Log[Sin[(c + d*x)/2]] + (a - b)^3*Sec[(c + d*x)/2]^2 + (2*b^3)/(Cos[(c 
+ d*x)/2] - Sin[(c + d*x)/2])^2 + (24*a*b^2*Sin[(c + d*x)/2])/(Cos[(c + d* 
x)/2] - Sin[(c + d*x)/2]) + (2*b^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^ 
2 - (24*a*b^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(8* 
d)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.16, 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, 532, 25, 2333, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int \csc ^3(c+d x) \sec ^3(c+d x) \left (-(-a \cos (c+d x)-b)^3\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -(b+a \cos (c+d x))^3 \csc ^3(c+d x) \sec ^3(c+d x)dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \csc ^3(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^3dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^6 \int \frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x)}{a^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 532

\(\displaystyle -\frac {a^6 \left (\frac {a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )}{2 a^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {\int -\frac {\left (2 b^3+6 a \cos (c+d x) b^2+2 a^2 \left (\frac {b^2}{a^2}+3\right ) \cos ^2(c+d x) b+a^3 \left (\frac {3 b^2}{a^2}+1\right ) \cos ^3(c+d x)\right ) \sec ^3(c+d x)}{a^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^6 \left (\frac {\int \frac {\left (2 b^3+6 a \cos (c+d x) b^2+2 a^2 \left (\frac {b^2}{a^2}+3\right ) \cos ^2(c+d x) b+a^3 \left (\frac {3 b^2}{a^2}+1\right ) \cos ^3(c+d x)\right ) \sec ^3(c+d x)}{a^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}+\frac {a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )}{2 a^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 2333

\(\displaystyle -\frac {a^6 \left (\frac {\int \left (\frac {2 b^3 \sec ^3(c+d x)}{a^5}+\frac {6 b^2 \sec ^2(c+d x)}{a^4}+\frac {2 \left (2 b^3+3 a^2 b\right ) \sec (c+d x)}{a^5}+\frac {(a+b)^2 (a+4 b)}{2 a^4 (a-a \cos (c+d x))}+\frac {(a-4 b) (a-b)^2}{2 a^4 (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))}{2 a^2}+\frac {a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )}{2 a^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^6 \left (\frac {a \left (\frac {3 b^2}{a^2}+1\right ) \cos (c+d x)+b \left (\frac {b^2}{a^2}+3\right )}{2 a^2 \left (a^2-a^2 \cos ^2(c+d x)\right )}+\frac {-\frac {b^3 \sec ^2(c+d x)}{a^4}-\frac {(a+b)^2 (a+4 b) \log (a-a \cos (c+d x))}{2 a^4}+\frac {(a-4 b) (a-b)^2 \log (a \cos (c+d x)+a)}{2 a^4}-\frac {6 b^2 \sec (c+d x)}{a^3}+\frac {2 b \left (3 a^2+2 b^2\right ) \log (a \cos (c+d x))}{a^4}}{2 a^2}\right )}{d}\)

Input:

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

Output:

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

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 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 2009

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

rule 2333

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

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]
 

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(162\)
default	\(\frac {a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(162\)
norman	\(\frac {-\frac {a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}{8 d}+\frac {\left (a^{3}+15 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d}+\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 d}-\frac {\left (2 a^{3}-3 a^{2} b +30 a \,b^{2}-9 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {\left (a^{3}+6 a^{2} b +9 a \,b^{2}+4 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(250\)
parallelrisch	\(\frac {-192 b \left (a^{2}+\frac {2 b^{2}}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-192 b \left (a^{2}+\frac {2 b^{2}}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (a +4 b \right ) \left (a +b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (-16 \cos \left (2 d x +2 c \right ) b^{3}+\left (-a^{3}-15 a \,b^{2}+2 b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (-4 a^{3}-36 a \,b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (-12 a^{3}-12 a \,b^{2}\right ) \cos \left (d x +c \right )+a^{3}-48 a^{2} b +15 a \,b^{2}-2 b^{3}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+192 a^{2} b \right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-192 a^{2} b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(274\)
risch	\(\frac {a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+9 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+6 \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2} b +4 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 \,{\mathrm e}^{4 i \left (d x +c \right )} a^{2} b +3 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{3}+3 \,{\mathrm e}^{3 i \left (d x +c \right )} a \,b^{2}+6 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +4 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{3}+a^{3} {\mathrm e}^{i \left (d x +c \right )}+9 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a^{2} b}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a \,b^{2}}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a^{2} b}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a \,b^{2}}{2 d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{3}}{d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(425\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.79 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {12 \, a b^{2} \cos \left (d x + c\right ) - 2 \, {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + {\left ({\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

-1/4*(12*a*b^2*cos(d*x + c) - 2*(a^3 + 9*a*b^2)*cos(d*x + c)^3 + 2*b^3 - 2 
*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 + 4*((3*a^2*b + 2*b^3)*cos(d*x + c)^4 - 
(3*a^2*b + 2*b^3)*cos(d*x + c)^2)*log(-cos(d*x + c)) + ((a^3 - 6*a^2*b + 9 
*a*b^2 - 4*b^3)*cos(d*x + c)^4 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(d*x 
 + c)^2)*log(1/2*cos(d*x + c) + 1/2) - ((a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)* 
cos(d*x + c)^4 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*cos(d*x + c)^2)*log(-1/ 
2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^4 - d*cos(d*x + c)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, {\left (6 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )}}{\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2}}}{4 \, d} \] Input:

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

Output:

-1/4*((a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*log(cos(d*x + c) + 1) - (a^3 + 6*a 
^2*b + 9*a*b^2 + 4*b^3)*log(cos(d*x + c) - 1) + 4*(3*a^2*b + 2*b^3)*log(co 
s(d*x + c)) + 2*(6*a*b^2*cos(d*x + c) - (a^3 + 9*a*b^2)*cos(d*x + c)^3 + b 
^3 - (3*a^2*b + 2*b^3)*cos(d*x + c)^2)/(cos(d*x + c)^4 - cos(d*x + c)^2))/ 
d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.17 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {{\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, d} + \frac {{\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} - \frac {{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{d} - \frac {6 \, a b^{2} \cos \left (d x + c\right ) - {\left (a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} - {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{2 \, d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )} \cos \left (d x + c\right )^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 10.74 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{4\,d}-\frac {\ln \left (\cos \left (c+d\,x\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3}{2}+\frac {9\,a\,b^2}{2}\right )-\frac {b^3}{2}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}+b^3\right )-3\,a\,b^2\,\cos \left (c+d\,x\right )}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^4\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2\,\left (a-4\,b\right )}{4\,d} \] Input:

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

Output:

(log(cos(c + d*x) - 1)*(a + b)^2*(a + 4*b))/(4*d) - (log(cos(c + d*x))*(3* 
a^2*b + 2*b^3))/d - (cos(c + d*x)^3*((9*a*b^2)/2 + a^3/2) - b^3/2 + cos(c 
+ d*x)^2*((3*a^2*b)/2 + b^3) - 3*a*b^2*cos(c + d*x))/(d*(cos(c + d*x)^2 - 
cos(c + d*x)^4)) - (log(cos(c + d*x) + 1)*(a - b)^2*(a - 4*b))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 584, normalized size of antiderivative = 3.60 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^3 \, dx =\text {Too large to display} \] Input:

int(csc(d*x+c)^3*(a+b*sec(d*x+c))^3,x)
 

Output:

( - 2*cos(c + d*x)*sin(c + d*x)**2*a**3 - 18*cos(c + d*x)*sin(c + d*x)**2* 
a*b**2 + 2*cos(c + d*x)*a**3 + 6*cos(c + d*x)*a*b**2 - 12*log(tan((c + d*x 
)/2) - 1)*sin(c + d*x)**4*a**2*b - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**4*b**3 + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + 8*log(ta 
n((c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 - 12*log(tan((c + d*x)/2) + 1)*si 
n(c + d*x)**4*a**2*b - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**3 + 
12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b + 8*log(tan((c + d*x)/ 
2) + 1)*sin(c + d*x)**2*b**3 + 2*log(tan((c + d*x)/2))*sin(c + d*x)**4*a** 
3 + 12*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**2*b + 18*log(tan((c + d*x) 
/2))*sin(c + d*x)**4*a*b**2 + 8*log(tan((c + d*x)/2))*sin(c + d*x)**4*b**3 
 - 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**3 - 12*log(tan((c + d*x)/2)) 
*sin(c + d*x)**2*a**2*b - 18*log(tan((c + d*x)/2))*sin(c + d*x)**2*a*b**2 
- 8*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**3 - sin(c + d*x)**4*a**3 - 15 
*sin(c + d*x)**4*a*b**2 - 2*sin(c + d*x)**4*b**3 + sin(c + d*x)**2*a**3 - 
6*sin(c + d*x)**2*a**2*b + 15*sin(c + d*x)**2*a*b**2 - 2*sin(c + d*x)**2*b 
**3 + 6*a**2*b + 2*b**3)/(4*sin(c + d*x)**2*d*(sin(c + d*x)**2 - 1))