3.3.77 \(\int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx\) [277]

3.3.77.1 Optimal result
3.3.77.2 Mathematica [A] (verified)
3.3.77.3 Rubi [A] (verified)
3.3.77.4 Maple [A] (warning: unable to verify)
3.3.77.5 Fricas [C] (verification not implemented)
3.3.77.6 Sympy [F(-1)]
3.3.77.7 Maxima [F]
3.3.77.8 Giac [F]
3.3.77.9 Mupad [F(-1)]

3.3.77.1 Optimal result

Integrand size = 25, antiderivative size = 371 \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=-\frac {c}{4 b d \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{7/2}}+\frac {1}{16 b c d \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{3/2}}-\frac {3 \arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {c \sec (a+b x)}}+\frac {3 \arctan \left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}{32 \sqrt {2} b c^2 d^2 \sqrt {c \sec (a+b x)}}-\frac {3 \sqrt {d \csc (a+b x)} \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {c \sec (a+b x)}}+\frac {3 \sqrt {d \csc (a+b x)} \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {\tan (a+b x)}}{64 \sqrt {2} b c^2 d^2 \sqrt {c \sec (a+b x)}} \]

output
-1/4*c/b/d/(c*sec(b*x+a))^(7/2)/(d*csc(b*x+a))^(1/2)+1/16/b/c/d/(c*sec(b*x 
+a))^(3/2)/(d*csc(b*x+a))^(1/2)+3/64*arctan(-1+2^(1/2)*tan(b*x+a)^(1/2))*( 
d*csc(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^2*2^(1/2)/(c*sec(b*x+a))^(1/2 
)+3/64*arctan(1+2^(1/2)*tan(b*x+a)^(1/2))*(d*csc(b*x+a))^(1/2)*tan(b*x+a)^ 
(1/2)/b/c^2/d^2*2^(1/2)/(c*sec(b*x+a))^(1/2)-3/128*ln(1-2^(1/2)*tan(b*x+a) 
^(1/2)+tan(b*x+a))*(d*csc(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^2*2^(1/2) 
/(c*sec(b*x+a))^(1/2)+3/128*ln(1+2^(1/2)*tan(b*x+a)^(1/2)+tan(b*x+a))*(d*c 
sc(b*x+a))^(1/2)*tan(b*x+a)^(1/2)/b/c^2/d^2*2^(1/2)/(c*sec(b*x+a))^(1/2)
 
3.3.77.2 Mathematica [A] (verified)

Time = 3.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=-\frac {\left (4+6 \cos (2 (a+b x))+2 \cos (4 (a+b x))+3 \sqrt {2} \arctan \left (\frac {-1+\sqrt {\cot ^2(a+b x)}}{\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}\right ) \sqrt [4]{\cot ^2(a+b x)}-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}{1+\sqrt {\cot ^2(a+b x)}}\right ) \sqrt [4]{\cot ^2(a+b x)}\right ) \sqrt {c \sec (a+b x)}}{64 b c^3 d \sqrt {d \csc (a+b x)}} \]

input
Integrate[1/((d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(5/2)),x]
 
output
-1/64*((4 + 6*Cos[2*(a + b*x)] + 2*Cos[4*(a + b*x)] + 3*Sqrt[2]*ArcTan[(-1 
 + Sqrt[Cot[a + b*x]^2])/(Sqrt[2]*(Cot[a + b*x]^2)^(1/4))]*(Cot[a + b*x]^2 
)^(1/4) - 3*Sqrt[2]*ArcTanh[(Sqrt[2]*(Cot[a + b*x]^2)^(1/4))/(1 + Sqrt[Cot 
[a + b*x]^2])]*(Cot[a + b*x]^2)^(1/4))*Sqrt[c*Sec[a + b*x]])/(b*c^3*d*Sqrt 
[d*Csc[a + b*x]])
 
3.3.77.3 Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.69, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 3107, 3042, 3108, 3042, 3109, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {1}{(c \sec (a+b x))^{5/2} (d \csc (a+b x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(c \sec (a+b x))^{5/2} (d \csc (a+b x))^{3/2}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {\int \frac {\sqrt {d \csc (a+b x)}}{(c \sec (a+b x))^{5/2}}dx}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {d \csc (a+b x)}}{(c \sec (a+b x))^{5/2}}dx}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3108

\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {d \csc (a+b x)}}{\sqrt {c \sec (a+b x)}}dx}{4 c^2}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {d \csc (a+b x)}}{\sqrt {c \sec (a+b x)}}dx}{4 c^2}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3109

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\tan (a+b x)}}dx}{4 c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\tan (a+b x)}}dx}{4 c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\sqrt {\tan (a+b x)} \left (\tan ^2(a+b x)+1\right )}d\tan (a+b x)}{4 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \int \frac {1}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \int \frac {1-\tan (a+b x)}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}+\frac {1}{2} \int \frac {\tan (a+b x)+1}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \int \frac {1-\tan (a+b x)}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}+\frac {1}{2} \int \frac {1}{\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \int \frac {1-\tan (a+b x)}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}+\frac {1}{2} \left (\frac {\int \frac {1}{-\tan (a+b x)-1}d\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (a+b x)-1}d\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \int \frac {1-\tan (a+b x)}{\tan ^2(a+b x)+1}d\sqrt {\tan (a+b x)}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (a+b x)}}{\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (a+b x)}}{\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (a+b x)}}{\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (a+b x)}+1}{\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1}d\sqrt {\tan (a+b x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {3 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{2 \sqrt {2}}\right )\right )}{2 b c^2 \sqrt {c \sec (a+b x)}}+\frac {d}{2 b c (c \sec (a+b x))^{3/2} \sqrt {d \csc (a+b x)}}}{8 d^2}-\frac {c}{4 b d (c \sec (a+b x))^{7/2} \sqrt {d \csc (a+b x)}}\)

input
Int[1/((d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(5/2)),x]
 
output
-1/4*c/(b*d*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(7/2)) + (d/(2*b*c*Sqrt[ 
d*Csc[a + b*x]]*(c*Sec[a + b*x])^(3/2)) + (3*Sqrt[d*Csc[a + b*x]]*((-(ArcT 
an[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[ 
a + b*x]]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + 
 b*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]/(2*Sqr 
t[2]))/2)*Sqrt[Tan[a + b*x]])/(2*b*c^2*Sqrt[c*Sec[a + b*x]]))/(8*d^2)
 

3.3.77.3.1 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 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3107
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_.), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1) 
/(a*f*(m + n))), x] + Simp[(m + 1)/(a^2*(m + n))   Int[(a*Csc[e + f*x])^(m 
+ 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] 
&& NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3108
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[(-a)*(a*Csc[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n + 
 1)/(b*f*(m + n))), x] + Simp[(n + 1)/(b^2*(m + n))   Int[(a*Csc[e + f*x])^ 
m*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, - 
1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3109
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Csc[e + f*x])^m*((b*Sec[e + f*x])^n/Tan[e + f*x]^n 
)   Int[Tan[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !Integer 
Q[n] && EqQ[m + n, 0]
 

rule 3957
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 
3.3.77.4 Maple [A] (warning: unable to verify)

Time = 49.00 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.39

method result size
default \(-\frac {\sqrt {2}\, \left (16 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )^{4}+16 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )^{3}-4 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \cos \left (b x +a \right )^{2}-4 \cos \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+3 \ln \left (2 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cot \left (b x +a \right )+2 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \csc \left (b x +a \right )-2 \cot \left (b x +a \right )+2\right )-6 \arctan \left (\frac {-\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )+6 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )-3 \ln \left (-2 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cot \left (b x +a \right )-2 \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \csc \left (b x +a \right )-2 \cot \left (b x +a \right )+2\right )\right )}{128 b \left (\cos \left (b x +a \right )+1\right ) \sqrt {c \sec \left (b x +a \right )}\, \sqrt {d \csc \left (b x +a \right )}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, c^{2} d}\) \(517\)

input
int(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 
output
-1/128/b*2^(1/2)*(16*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/ 
2)*cos(b*x+a)^4+16*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2) 
*cos(b*x+a)^3-4*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2)*co 
s(b*x+a)^2-4*cos(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^ 
(1/2)+3*ln(2*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*cot(b 
*x+a)+2*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*csc(b*x+a) 
-2*cot(b*x+a)+2)-6*arctan((-sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(co 
s(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(cos(b*x+a)-1))+6*arctan((sin(b*x+a)*2^ 
(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(cos(b 
*x+a)-1))-3*ln(-2*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)* 
cot(b*x+a)-2*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*csc(b 
*x+a)-2*cot(b*x+a)+2))/(cos(b*x+a)+1)/(c*sec(b*x+a))^(1/2)/(d*csc(b*x+a))^ 
(1/2)/(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)/c^2/d
 
3.3.77.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.30 (sec) , antiderivative size = 1375, normalized size of antiderivative = 3.71 \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=\text {Too large to display} \]

input
integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas 
")
 
output
-1/256*(3*b*c^3*d^2*(-1/(b^4*c^10*d^6))^(1/4)*log(2*(b^3*c^7*d^4*(-1/(b^4* 
c^10*d^6))^(3/4)*cos(b*x + a)^2*sin(b*x + a) + (b*c^2*d*cos(b*x + a)^3 - b 
*c^2*d*cos(b*x + a))*(-1/(b^4*c^10*d^6))^(1/4))*sqrt(c/cos(b*x + a))*sqrt( 
d/sin(b*x + a)) + 1) - 3*b*c^3*d^2*(-1/(b^4*c^10*d^6))^(1/4)*log(-2*(b^3*c 
^7*d^4*(-1/(b^4*c^10*d^6))^(3/4)*cos(b*x + a)^2*sin(b*x + a) + (b*c^2*d*co 
s(b*x + a)^3 - b*c^2*d*cos(b*x + a))*(-1/(b^4*c^10*d^6))^(1/4))*sqrt(c/cos 
(b*x + a))*sqrt(d/sin(b*x + a)) + 1) + 3*I*b*c^3*d^2*(-1/(b^4*c^10*d^6))^( 
1/4)*log(-2*(I*b^3*c^7*d^4*(-1/(b^4*c^10*d^6))^(3/4)*cos(b*x + a)^2*sin(b* 
x + a) + (-I*b*c^2*d*cos(b*x + a)^3 + I*b*c^2*d*cos(b*x + a))*(-1/(b^4*c^1 
0*d^6))^(1/4))*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a)) + 1) - 3*I*b*c^3* 
d^2*(-1/(b^4*c^10*d^6))^(1/4)*log(-2*(-I*b^3*c^7*d^4*(-1/(b^4*c^10*d^6))^( 
3/4)*cos(b*x + a)^2*sin(b*x + a) + (I*b*c^2*d*cos(b*x + a)^3 - I*b*c^2*d*c 
os(b*x + a))*(-1/(b^4*c^10*d^6))^(1/4))*sqrt(c/cos(b*x + a))*sqrt(d/sin(b* 
x + a)) + 1) + 3*b*c^3*d^2*(-1/(b^4*c^10*d^6))^(1/4)*log(1/2*(b*c^2*d*(-1/ 
(b^4*c^10*d^6))^(1/4)*cos(b*x + a)^2*sin(b*x + a) + (b^3*c^7*d^4*cos(b*x + 
 a)^3 - b^3*c^7*d^4*cos(b*x + a))*(-1/(b^4*c^10*d^6))^(3/4))*sqrt(c/cos(b* 
x + a))*sqrt(d/sin(b*x + a)) - 1/2*cos(b*x + a)*sin(b*x + a) - 1/4*(2*b^2* 
c^5*d^3*cos(b*x + a)^2 - b^2*c^5*d^3)*sqrt(-1/(b^4*c^10*d^6))) - 3*b*c^3*d 
^2*(-1/(b^4*c^10*d^6))^(1/4)*log(-1/2*(b*c^2*d*(-1/(b^4*c^10*d^6))^(1/4)*c 
os(b*x + a)^2*sin(b*x + a) + (b^3*c^7*d^4*cos(b*x + a)^3 - b^3*c^7*d^4*...
 
3.3.77.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=\text {Timed out} \]

input
integrate(1/(d*csc(b*x+a))**(3/2)/(c*sec(b*x+a))**(5/2),x)
 
output
Timed out
 
3.3.77.7 Maxima [F]

\[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (d \csc \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima 
")
 
output
integrate(1/((d*csc(b*x + a))^(3/2)*(c*sec(b*x + a))^(5/2)), x)
 
3.3.77.8 Giac [F]

\[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (d \csc \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate(1/(d*csc(b*x+a))^(3/2)/(c*sec(b*x+a))^(5/2),x, algorithm="giac")
 
output
integrate(1/((d*csc(b*x + a))^(3/2)*(c*sec(b*x + a))^(5/2)), x)
 
3.3.77.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2}} \, dx=\int \frac {1}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{3/2}} \,d x \]

input
int(1/((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(3/2)),x)
 
output
int(1/((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(3/2)), x)