\(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [230]

Optimal result

Integrand size = 19, antiderivative size = 100 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{7/2}}+\frac {2}{5 b d (d \cos (a+b x))^{5/2}}+\frac {2}{b d^3 \sqrt {d \cos (a+b x)}} \] Output:

arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)-arctanh((d*cos(b*x+a))^(1/2 
)/d^(1/2))/b/d^(7/2)+2/5/b/d/(d*cos(b*x+a))^(5/2)+2/b/d^3/(d*cos(b*x+a))^( 
1/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {5 \arctan \left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}-5 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}+2 \left (5+\sec ^2(a+b x)\right )}{5 b d^3 \sqrt {d \cos (a+b x)}} \] Input:

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

Output:

(5*ArcTan[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]] - 5*ArcTanh[Sqrt[Cos[a + 
b*x]]]*Sqrt[Cos[a + b*x]] + 2*(5 + Sec[a + b*x]^2))/(5*b*d^3*Sqrt[d*Cos[a 
+ b*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98, 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, 827, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x) (d \cos (a+b x))^{7/2}}dx\)

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {d \left (\frac {\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{b}\)

Input:

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

Output:

-((d*(-2/(5*d^2*(d*Cos[a + b*x])^(5/2)) + ((2*(-1/2*ArcTan[Sqrt[d]*Cos[a + 
 b*x]]/Sqrt[d] + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*Sqrt[d])))/d^2 - 2/(d^2* 
Sqrt[d*Cos[a + b*x]]))/d^2))/b)
 

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 827

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(861\) vs. \(2(82)=164\).

Time = 2.78 (sec) , antiderivative size = 862, normalized size of antiderivative = 8.62

method	result	size
default	\(\text {Expression too large to display}\)	\(862\)

Input:

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

Output:

1/10/d^(9/2)/(-d)^(1/2)/(8*sin(1/2*b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6* 
sin(1/2*b*x+1/2*a)^2-1)*(10*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/ 
2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(3/2)-24*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/ 
2)*(-d)^(1/2)*d^(1/2)+5*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)*d+5*ln(2/(co 
s(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*sin(1/2*b*x+1/2*a) 
^2*d+d)^(1/2)-d))*(-d)^(1/2)*d-40*(2*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*( 
-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(3/2)+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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*( 
-d)^(1/2)*d+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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)*d)*sin(1/2*b*x+1/2*a)^6- 
20*(-6*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^( 
1/2)-d))*d^(3/2)+4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d)^(1/2)*d^(1/2)- 
3*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*sin(1/2* 
b*x+1/2*a)^2*d+d)^(1/2)-d))*(-d)^(1/2)*d-3*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d) 
^(1/2)*d)*sin(1/2*b*x+1/2*a)^4+10*(-6*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)* 
(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(3/2)+8*(-2*sin(1/2*b*x+1/2*a)^2 
*d+d)^(1/2)*(-d)^(1/2)*d^(1/2)-3*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*(-d)^(1/2)*d...
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.33 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\left [\frac {10 \, \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) \cos \left (b x + a\right )^{3} - 5 \, \sqrt {-d} \cos \left (b x + a\right )^{3} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} + 1\right )}}{20 \, b d^{4} \cos \left (b x + a\right )^{3}}, -\frac {10 \, \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) \cos \left (b x + a\right )^{3} - 5 \, \sqrt {d} \cos \left (b x + a\right )^{3} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} + 1\right )}}{20 \, b d^{4} \cos \left (b x + a\right )^{3}}\right ] \] Input:

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

Output:

[1/20*(10*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) 
+ d))*cos(b*x + a)^3 - 5*sqrt(-d)*cos(b*x + a)^3*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))*(5*cos(b*x 
 + a)^2 + 1))/(b*d^4*cos(b*x + a)^3), -1/20*(10*sqrt(d)*arctan(2*sqrt(d*co 
s(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d))*cos(b*x + a)^3 - 5*sqrt(d)*cos(b 
*x + a)^3*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))*(5*cos(b*x + a)^2 + 1))/(b*d^4*cos(b*x + a)^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\frac {10 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {5 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {4 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} + d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{2}}}{10 \, b d} \] Input:

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

Output:

1/10*(10*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 5*log((sqrt(d*cos( 
b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d)))/d^(5/2) + 4*(5*d^2* 
cos(b*x + a)^2 + d^2)/((d*cos(b*x + a))^(5/2)*d^2))/(b*d)
 

Giac [F]

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

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

Output:

integrate(csc(b*x + a)/(d*cos(b*x + a))^(7/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*int((sqrt(cos(a + b*x))*csc(a + b*x))/cos(a + b*x)**4,x))/d**4