\(\int \frac {1}{(a+b \csc ^2(c+d x))^{7/2}} \, dx\) [15]

Optimal result

Integrand size = 16, antiderivative size = 180 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{a^{7/2} d}+\frac {b \cot (c+d x)}{5 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )^{5/2}}+\frac {b (9 a+5 b) \cot (c+d x)}{15 a^2 (a+b)^2 d \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}+\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{15 a^3 (a+b)^3 d \sqrt {a+b+b \cot ^2(c+d x)}} \] Output:

-arctan(a^(1/2)*cot(d*x+c)/(a+b+b*cot(d*x+c)^2)^(1/2))/a^(7/2)/d+1/5*b*cot 
(d*x+c)/a/(a+b)/d/(a+b+b*cot(d*x+c)^2)^(5/2)+1/15*b*(9*a+5*b)*cot(d*x+c)/a 
^2/(a+b)^2/d/(a+b+b*cot(d*x+c)^2)^(3/2)+1/15*b*(33*a^2+40*a*b+15*b^2)*cot( 
d*x+c)/a^3/(a+b)^3/d/(a+b+b*cot(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\frac {\csc ^7(c+d x) \left (\frac {b \cos (c+d x) (a+2 b-a \cos (2 (c+d x))) \left (135 a^4+480 a^3 b+709 a^2 b^2+460 a b^3+120 b^4-4 a \left (45 a^3+135 a^2 b+117 a b^2+35 b^3\right ) \cos (2 (c+d x))+a^2 \left (45 a^2+60 a b+23 b^2\right ) \cos (4 (c+d x))\right )}{15 a^3 (a+b)^3}+\frac {\sqrt {2} (-a-2 b+a \cos (2 (c+d x)))^{7/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )}{a^{7/2}}\right )}{16 d \left (a+b \csc ^2(c+d x)\right )^{7/2}} \] Input:

Integrate[(a + b*Csc[c + d*x]^2)^(-7/2),x]
 

Output:

(Csc[c + d*x]^7*((b*Cos[c + d*x]*(a + 2*b - a*Cos[2*(c + d*x)])*(135*a^4 + 
 480*a^3*b + 709*a^2*b^2 + 460*a*b^3 + 120*b^4 - 4*a*(45*a^3 + 135*a^2*b + 
 117*a*b^2 + 35*b^3)*Cos[2*(c + d*x)] + a^2*(45*a^2 + 60*a*b + 23*b^2)*Cos 
[4*(c + d*x)]))/(15*a^3*(a + b)^3) + (Sqrt[2]*(-a - 2*b + a*Cos[2*(c + d*x 
)])^(7/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + 
d*x)]]])/a^(7/2)))/(16*d*(a + b*Csc[c + d*x]^2)^(7/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 316, 402, 402, 27, 291, 216}

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.

\(\displaystyle \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sec \left (c+d x+\frac {\pi }{2}\right )^2\right )^{7/2}}dx\)

\(\Big \downarrow \) 4616

\(\displaystyle -\frac {\int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{7/2}}d\cot (c+d x)}{d}\)

\(\Big \downarrow \) 316

\(\displaystyle -\frac {\frac {\int \frac {-4 b \cot ^2(c+d x)+5 a+b}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{5/2}}d\cot (c+d x)}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

\(\Big \downarrow \) 402

\(\displaystyle -\frac {\frac {\frac {\int \frac {15 a^2+12 b a+5 b^2-2 b (9 a+5 b) \cot ^2(c+d x)}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a+b\right )^{3/2}}d\cot (c+d x)}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

\(\Big \downarrow \) 402

\(\displaystyle -\frac {\frac {\frac {\frac {\int \frac {15 (a+b)^3}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)}{a (a+b)}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \sqrt {b \cot ^2(c+d x)+a+b}}d\cot (c+d x)}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

\(\Big \downarrow \) 291

\(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\frac {a \cot ^2(c+d x)}{b \cot ^2(c+d x)+a+b}+1}d\frac {\cot (c+d x)}{\sqrt {b \cot ^2(c+d x)+a+b}}}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {\frac {\frac {15 (a+b)^2 \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{a^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \cot (c+d x)}{a (a+b) \sqrt {a+b \cot ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \cot (c+d x)}{3 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \cot (c+d x)}{5 a (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{5/2}}}{d}\)

Input:

Int[(a + b*Csc[c + d*x]^2)^(-7/2),x]
 

Output:

-((-1/5*(b*Cot[c + d*x])/(a*(a + b)*(a + b + b*Cot[c + d*x]^2)^(5/2)) + (- 
1/3*(b*(9*a + 5*b)*Cot[c + d*x])/(a*(a + b)*(a + b + b*Cot[c + d*x]^2)^(3/ 
2)) + ((15*(a + b)^2*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + 
d*x]^2]])/a^(3/2) - (b*(33*a^2 + 40*a*b + 15*b^2)*Cot[c + d*x])/(a*(a + b) 
*Sqrt[a + b + b*Cot[c + d*x]^2]))/(3*a*(a + b)))/(5*a*(a + b)))/d)
 

Defintions of rubi rules used

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 216

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 316

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

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1313\) vs. \(2(162)=324\).

Time = 0.84 (sec) , antiderivative size = 1314, normalized size of antiderivative = 7.30

method	result	size
default	\(\text {Expression too large to display}\)	\(1314\)

Input:

int(1/(a+b*csc(d*x+c)^2)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/30/d*4^(1/2)*b^3/((a*(a+b))^(1/2)-a)^3/((a*(a+b))^(1/2)+a)^3/(-a)^(1/2) 
/(a+b)^3*(-45*cos(d*x+c)*sin(d*x+c)^6*(-a)^(1/2)*a^5*b+60*cos(d*x+c)*(-3+c 
os(d*x+c)^2)*sin(d*x+c)^4*(-a)^(1/2)*a^4*b^2+cos(d*x+c)*(-23*cos(d*x+c)^4+ 
200*cos(d*x+c)^2-285)*sin(d*x+c)^2*(-a)^(1/2)*a^3*b^3+cos(d*x+c)*(-58*cos( 
d*x+c)^4+250*cos(d*x+c)^2-225)*(-a)^(1/2)*a^2*b^4+10*cos(d*x+c)*(-9+5*cos( 
d*x+c)^2)*(-a)^(1/2)*a*b^5-15*(-a)^(1/2)*b^6*cos(d*x+c)+15*(1+cos(d*x+c))* 
sin(d*x+c)^6*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*ln(4*(-a)^(1/2)*( 
(a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)+4*(-a)^(1/2)*((a*sin 
(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)*a^6+45*(2-cos(d*x+c)^ 
3-cos(d*x+c)^2+2*cos(d*x+c))*sin(d*x+c)^4*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c 
))^2)^(1/2)*ln(4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*co 
s(d*x+c)+4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d* 
x+c)*a)*a^5*b+45*(cos(d*x+c)^5+cos(d*x+c)^4-5*cos(d*x+c)^3-5*cos(d*x+c)^2+ 
5*cos(d*x+c)+5)*sin(d*x+c)^2*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*l 
n(4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)+4*(- 
a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos(d*x+c)*a)*a^4*b 
^2+15*(-cos(d*x+c)^7-cos(d*x+c)^6+12*cos(d*x+c)^5+12*cos(d*x+c)^4-30*cos(d 
*x+c)^3-30*cos(d*x+c)^2+20*cos(d*x+c)+20)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c 
))^2)^(1/2)*ln(4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)*co 
s(d*x+c)+4*(-a)^(1/2)*((a*sin(d*x+c)^2+b)/(1+cos(d*x+c))^2)^(1/2)-4*cos...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 667 vs. \(2 (162) = 324\).

Time = 1.73 (sec) , antiderivative size = 1445, normalized size of antiderivative = 8.03 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^(7/2),x, algorithm="fricas")
 

Output:

[-1/120*(15*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(d*x + c)^6 - a^6 - 
6*a^5*b - 15*a^4*b^2 - 20*a^3*b^3 - 15*a^2*b^4 - 6*a*b^5 - b^6 - 3*(a^6 + 
4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*cos(d*x + c)^4 + 3*(a^6 + 5*a^5 
*b + 10*a^4*b^2 + 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)*cos(d*x + c)^2)*sqrt(-a) 
*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*(a^4 
+ 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 
+ b^4 - 32*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2 + 8*(16*a^3* 
cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + a*b 
^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqrt(-a 
)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)) + 8* 
((45*a^5*b + 60*a^4*b^2 + 23*a^3*b^3)*cos(d*x + c)^5 - 5*(18*a^5*b + 39*a^ 
4*b^2 + 28*a^3*b^3 + 7*a^2*b^4)*cos(d*x + c)^3 + 15*(3*a^5*b + 9*a^4*b^2 + 
 10*a^3*b^3 + 5*a^2*b^4 + a*b^5)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a 
- b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))/((a^10 + 3*a^9*b + 3*a^8*b^2 + a^ 
7*b^3)*d*cos(d*x + c)^6 - 3*(a^10 + 4*a^9*b + 6*a^8*b^2 + 4*a^7*b^3 + a^6* 
b^4)*d*cos(d*x + c)^4 + 3*(a^10 + 5*a^9*b + 10*a^8*b^2 + 10*a^7*b^3 + 5*a^ 
6*b^4 + a^5*b^5)*d*cos(d*x + c)^2 - (a^10 + 6*a^9*b + 15*a^8*b^2 + 20*a^7* 
b^3 + 15*a^6*b^4 + 6*a^5*b^5 + a^4*b^6)*d), 1/60*(15*((a^6 + 3*a^5*b + 3*a 
^4*b^2 + a^3*b^3)*cos(d*x + c)^6 - a^6 - 6*a^5*b - 15*a^4*b^2 - 20*a^3*b^3 
 - 15*a^2*b^4 - 6*a*b^5 - b^6 - 3*(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^...
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(a+b*csc(d*x+c)**2)**(7/2),x)
 

Output:

Integral((a + b*csc(c + d*x)**2)**(-7/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^(7/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (162) = 324\).

Time = 0.68 (sec) , antiderivative size = 737, normalized size of antiderivative = 4.09 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*csc(d*x+c)^2)^(7/2),x, algorithm="giac")
 

Output:

-1/15*(((((((33*a^20*b^3*sgn(sin(d*x + c)) + 40*a^19*b^4*sgn(sin(d*x + c)) 
 + 15*a^18*b^5*sgn(sin(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^24 + 3*a^23*b 
+ 3*a^22*b^2 + a^21*b^3) + 5*(60*a^21*b^2*sgn(sin(d*x + c)) + 95*a^20*b^3* 
sgn(sin(d*x + c)) + 52*a^19*b^4*sgn(sin(d*x + c)) + 9*a^18*b^5*sgn(sin(d*x 
 + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/2*c)^2 
+ 10*(72*a^22*b*sgn(sin(d*x + c)) + 126*a^21*b^2*sgn(sin(d*x + c)) + 81*a^ 
20*b^3*sgn(sin(d*x + c)) + 22*a^19*b^4*sgn(sin(d*x + c)) + 3*a^18*b^5*sgn( 
sin(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/ 
2*c)^2 - 10*(72*a^22*b*sgn(sin(d*x + c)) + 126*a^21*b^2*sgn(sin(d*x + c)) 
+ 81*a^20*b^3*sgn(sin(d*x + c)) + 22*a^19*b^4*sgn(sin(d*x + c)) + 3*a^18*b 
^5*sgn(sin(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d 
*x + 1/2*c)^2 - 5*(60*a^21*b^2*sgn(sin(d*x + c)) + 95*a^20*b^3*sgn(sin(d*x 
 + c)) + 52*a^19*b^4*sgn(sin(d*x + c)) + 9*a^18*b^5*sgn(sin(d*x + c)))/(a^ 
24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))*tan(1/2*d*x + 1/2*c)^2 - (33*a^20* 
b^3*sgn(sin(d*x + c)) + 40*a^19*b^4*sgn(sin(d*x + c)) + 15*a^18*b^5*sgn(si 
n(d*x + c)))/(a^24 + 3*a^23*b + 3*a^22*b^2 + a^21*b^3))/(b*tan(1/2*d*x + 1 
/2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c)^2 + b)^(5/ 
2) - 30*arctan(-1/2*(sqrt(b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(b*tan(1/2*d*x + 
 1/2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c)^2 + b) + 
 sqrt(b))/sqrt(a))/(a^(7/2)*sgn(sin(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{7/2}} \,d x \] Input:

int(1/(a + b/sin(c + d*x)^2)^(7/2),x)
 

Output:

int(1/(a + b/sin(c + d*x)^2)^(7/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {\sqrt {\csc \left (d x +c \right )^{2} b +a}}{\csc \left (d x +c \right )^{8} b^{4}+4 \csc \left (d x +c \right )^{6} a \,b^{3}+6 \csc \left (d x +c \right )^{4} a^{2} b^{2}+4 \csc \left (d x +c \right )^{2} a^{3} b +a^{4}}d x \] Input:

int(1/(a+b*csc(d*x+c)^2)^(7/2),x)
 

Output:

int(sqrt(csc(c + d*x)**2*b + a)/(csc(c + d*x)**8*b**4 + 4*csc(c + d*x)**6* 
a*b**3 + 6*csc(c + d*x)**4*a**2*b**2 + 4*csc(c + d*x)**2*a**3*b + a**4),x)