3.3.31 \(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx\) [231]

3.3.31.1 Optimal result
3.3.31.2 Mathematica [A] (verified)
3.3.31.3 Rubi [A] (warning: unable to verify)
3.3.31.4 Maple [B] (verified)
3.3.31.5 Fricas [A] (verification not implemented)
3.3.31.6 Sympy [F(-1)]
3.3.31.7 Maxima [A] (verification not implemented)
3.3.31.8 Giac [F]
3.3.31.9 Mupad [F(-1)]

3.3.31.1 Optimal result

Integrand size = 19, antiderivative size = 103 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{9/2}}+\frac {2}{7 b d (d \cos (a+b x))^{7/2}}+\frac {2}{3 b d^3 (d \cos (a+b x))^{3/2}} \]

output
-arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(9/2)-arctanh((d*cos(b*x+a))^(1/ 
2)/d^(1/2))/b/d^(9/2)+2/7/b/d/(d*cos(b*x+a))^(7/2)+2/3/b/d^3/(d*cos(b*x+a) 
)^(3/2)
 
3.3.31.2 Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\frac {-21 \arctan \left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}-21 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}+14 \sec (a+b x)+6 \sec ^3(a+b x)}{21 b d^4 \sqrt {d \cos (a+b x)}} \]

input
Integrate[Csc[a + b*x]/(d*Cos[a + b*x])^(9/2),x]
 
output
(-21*ArcTan[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]] - 21*ArcTanh[Sqrt[Cos[a 
 + b*x]]]*Sqrt[Cos[a + b*x]] + 14*Sec[a + b*x] + 6*Sec[a + b*x]^3)/(21*b*d 
^4*Sqrt[d*Cos[a + b*x]])
 
3.3.31.3 Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3045, 27, 264, 264, 266, 756, 216, 219}

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.

\(\displaystyle \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x) (d \cos (a+b x))^{9/2}}dx\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {\int \frac {d^2}{(d \cos (a+b x))^{9/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {d \int \frac {1}{(d \cos (a+b x))^{9/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {d \left (\frac {\int \frac {1}{(d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{d^2}-\frac {2}{7 d^2 (d \cos (a+b x))^{7/2}}\right )}{b}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {d \left (\frac {\frac {\int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}}{d^2}-\frac {2}{7 d^2 (d \cos (a+b x))^{7/2}}\right )}{b}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

\(\displaystyle -\frac {d \left (\frac {\frac {2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}}{d^2}-\frac {2}{7 d^2 (d \cos (a+b x))^{7/2}}\right )}{b}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {d \left (\frac {\frac {2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}}{d^2}-\frac {2}{7 d^2 (d \cos (a+b x))^{7/2}}\right )}{b}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {d \left (\frac {\frac {2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{d^2}-\frac {2}{3 d^2 (d \cos (a+b x))^{3/2}}}{d^2}-\frac {2}{7 d^2 (d \cos (a+b x))^{7/2}}\right )}{b}\)

input
Int[Csc[a + b*x]/(d*Cos[a + b*x])^(9/2),x]
 
output
-((d*(-2/(7*d^2*(d*Cos[a + b*x])^(7/2)) + ((2*(ArcTan[Sqrt[d]*Cos[a + b*x] 
]/(2*d^(3/2)) + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2))))/d^2 - 2/(3*d^2 
*(d*Cos[a + b*x])^(3/2)))/d^2))/b)
 

3.3.31.3.1 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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 264
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

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 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

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

rule 3045
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 
3.3.31.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1056\) vs. \(2(83)=166\).

Time = 0.16 (sec) , antiderivative size = 1057, normalized size of antiderivative = 10.26

method result size
default \(\text {Expression too large to display}\) \(1057\)

input
int(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x,method=_RETURNVERBOSE)
 
output
1/42/d^(11/2)/(-d)^(1/2)/(16*sin(1/2*b*x+1/2*a)^8-32*sin(1/2*b*x+1/2*a)^6+ 
24*sin(1/2*b*x+1/2*a)^4-8*sin(1/2*b*x+1/2*a)^2+1)*(42*d^(3/2)*ln(2/cos(1/2 
*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))+40*(-d)^(1 
/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-21*(-d)^(1/2)*ln(2/(cos(1/ 
2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2 
+d)^(1/2)-d))*d-21*(-d)^(1/2)*ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b* 
x+1/2*a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*d-336*(-2*d^(3/2) 
*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d 
))+(-d)^(1/2)*ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2) 
*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*d+(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/ 
2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2 
)-d))*d)*sin(1/2*b*x+1/2*a)^8+672*(-2*d^(3/2)*ln(2/cos(1/2*b*x+1/2*a)*((-d 
)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))+(-d)^(1/2)*ln(-2/(cos(1/2* 
b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d 
)^(1/2)+d))*d+(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2* 
a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))*d)*sin(1/2*b*x+1/2*a)^6 
-56*(6*d^(3/2)*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a 
)^2+d)^(1/2)-d))+2*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)- 
3*(-d)^(1/2)*ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)* 
(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*d-3*(-d)^(1/2)*ln(2/(cos(1/2*b*...
 
3.3.31.5 Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.32 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\left [\frac {42 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{4} - 21 \, \sqrt {-d} \cos \left (b x + a\right )^{4} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} + 3\right )}}{84 \, b d^{5} \cos \left (b x + a\right )^{4}}, -\frac {42 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{4} - 21 \, \sqrt {d} \cos \left (b x + a\right )^{4} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (7 \, \cos \left (b x + a\right )^{2} + 3\right )}}{84 \, b d^{5} \cos \left (b x + a\right )^{4}}\right ] \]

input
integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="fricas")
 
output
[1/84*(42*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) 
+ 1)/(d*cos(b*x + a)))*cos(b*x + a)^4 - 21*sqrt(-d)*cos(b*x + a)^4*log((d* 
cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) - 6*d* 
cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b* 
x + a))*(7*cos(b*x + a)^2 + 3))/(b*d^5*cos(b*x + a)^4), -1/84*(42*sqrt(d)* 
arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a))) 
*cos(b*x + a)^4 - 21*sqrt(d)*cos(b*x + a)^4*log((d*cos(b*x + a)^2 - 4*sqrt 
(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b 
*x + a)^2 - 2*cos(b*x + a) + 1)) - 8*sqrt(d*cos(b*x + a))*(7*cos(b*x + a)^ 
2 + 3))/(b*d^5*cos(b*x + a)^4)]
 
3.3.31.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\text {Timed out} \]

input
integrate(csc(b*x+a)/(d*cos(b*x+a))**(9/2),x)
 
output
Timed out
 
3.3.31.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=-\frac {\frac {42 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {7}{2}}} - \frac {21 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {7}{2}}} - \frac {4 \, {\left (7 \, d^{2} \cos \left (b x + a\right )^{2} + 3 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} d^{2}}}{42 \, b d} \]

input
integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="maxima")
 
output
-1/42*(42*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(7/2) - 21*log((sqrt(d*co 
s(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d)))/d^(7/2) - 4*(7*d^ 
2*cos(b*x + a)^2 + 3*d^2)/((d*cos(b*x + a))^(7/2)*d^2))/(b*d)
 
3.3.31.8 Giac [F]

\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}}} \,d x } \]

input
integrate(csc(b*x+a)/(d*cos(b*x+a))^(9/2),x, algorithm="giac")
 
output
integrate(csc(b*x + a)/(d*cos(b*x + a))^(9/2), x)
 
3.3.31.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}} \,d x \]

input
int(1/(sin(a + b*x)*(d*cos(a + b*x))^(9/2)),x)
 
output
int(1/(sin(a + b*x)*(d*cos(a + b*x))^(9/2)), x)