\(\int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx\) [471]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=-\frac {2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)}-\frac {8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac {11}{2}}(e+f x)}-\frac {64 b}{385 f (b \sec (e+f x))^{3/2} \sin ^{\frac {7}{2}}(e+f x)}-\frac {256 b}{1155 f (b \sec (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x)} \] Output:

-2/15*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(15/2)-8/55*b/f/(b*sec(f*x+e))^( 
3/2)/sin(f*x+e)^(11/2)-64/385*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(7/2)-25 
6/1155*b/f/(b*sec(f*x+e))^(3/2)/sin(f*x+e)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\frac {2 b (-195+150 \cos (2 (e+f x))-36 \cos (4 (e+f x))+4 \cos (6 (e+f x)))}{1155 f (b \sec (e+f x))^{3/2} \sin ^{\frac {15}{2}}(e+f x)} \] Input:

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]
 

Output:

(2*b*(-195 + 150*Cos[2*(e + f*x)] - 36*Cos[4*(e + f*x)] + 4*Cos[6*(e + f*x 
)]))/(1155*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(15/2))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3064, 3042, 3064, 3042, 3064, 3042, 3058}

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.

\(\displaystyle \int \frac {1}{\sin ^{\frac {17}{2}}(e+f x) \sqrt {b \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^{17/2} \sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3064

\(\displaystyle \frac {4}{5} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {13}{2}}(e+f x)}dx-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin (e+f x)^{13/2}}dx-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3064

\(\displaystyle \frac {4}{5} \left (\frac {8}{11} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {9}{2}}(e+f x)}dx-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} \left (\frac {8}{11} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin (e+f x)^{9/2}}dx-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3064

\(\displaystyle \frac {4}{5} \left (\frac {8}{11} \left (\frac {4}{7} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {5}{2}}(e+f x)}dx-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} \left (\frac {8}{11} \left (\frac {4}{7} \int \frac {1}{\sqrt {b \sec (e+f x)} \sin (e+f x)^{5/2}}dx-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3058

\(\displaystyle \frac {4}{5} \left (\frac {8}{11} \left (-\frac {8 b}{21 f \sin ^{\frac {3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac {2 b}{7 f \sin ^{\frac {7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{11 f \sin ^{\frac {11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\right )-\frac {2 b}{15 f \sin ^{\frac {15}{2}}(e+f x) (b \sec (e+f x))^{3/2}}\)

Input:

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]
 

Output:

(4*((8*((-2*b)/(7*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (8*b)/(21 
*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(3/2))))/11 - (2*b)/(11*f*(b*Sec[e 
+ f*x])^(3/2)*Sin[e + f*x]^(11/2))))/5 - (2*b)/(15*f*(b*Sec[e + f*x])^(3/2 
)*Sin[e + f*x]^(15/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3058

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

rule 3064

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {\frac {256 \cos \left (f x +e \right )^{7}}{1155}-\frac {64 \cos \left (f x +e \right )^{5}}{77}+\frac {8 \cos \left (f x +e \right )^{3}}{7}-\frac {2 \cos \left (f x +e \right )}{3}}{f \sin \left (f x +e \right )^{\frac {15}{2}} \sqrt {b \sec \left (f x +e \right )}}\)	\(63\)

Input:

int(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(17/2),x,method=_RETURNVERBOSE)
 

Output:

2/1155/f/sin(f*x+e)^(15/2)/(b*sec(f*x+e))^(1/2)*(128*cos(f*x+e)^7-480*cos( 
f*x+e)^5+660*cos(f*x+e)^3-385*cos(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\frac {2 \, {\left (128 \, \cos \left (f x + e\right )^{8} - 480 \, \cos \left (f x + e\right )^{6} + 660 \, \cos \left (f x + e\right )^{4} - 385 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sqrt {\sin \left (f x + e\right )}}{1155 \, {\left (b f \cos \left (f x + e\right )^{8} - 4 \, b f \cos \left (f x + e\right )^{6} + 6 \, b f \cos \left (f x + e\right )^{4} - 4 \, b f \cos \left (f x + e\right )^{2} + b f\right )}} \] Input:

integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(17/2),x, algorithm="fricas")
 

Output:

2/1155*(128*cos(f*x + e)^8 - 480*cos(f*x + e)^6 + 660*cos(f*x + e)^4 - 385 
*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*sqrt(sin(f*x + e))/(b*f*cos(f*x + e) 
^8 - 4*b*f*cos(f*x + e)^6 + 6*b*f*cos(f*x + e)^4 - 4*b*f*cos(f*x + e)^2 + 
b*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\text {Timed out} \] Input:

integrate(1/(b*sec(f*x+e))**(1/2)/sin(f*x+e)**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\int { \frac {1}{\sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac {17}{2}}} \,d x } \] Input:

integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(17/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(17/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\text {Timed out} \] Input:

integrate(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(17/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 32.69 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {\frac {b}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1024{}\mathrm {i}}{77\,b\,f}+\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,384{}\mathrm {i}}{55\,b\,f}-\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,5248{}\mathrm {i}}{1155\,b\,f}+\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,256{}\mathrm {i}}{165\,b\,f}-\frac {{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,256{}\mathrm {i}}{1155\,b\,f}\right )\,1{}\mathrm {i}}{128\,{\sin \left (e+f\,x\right )}^{15/2}} \] Input:

int(1/(sin(e + f*x)^(17/2)*(b/cos(e + f*x))^(1/2)),x)
 

Output:

(exp(- e*8i - f*x*8i)*(b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^ 
(1/2)*((exp(e*8i + f*x*8i)*1024i)/(77*b*f) + (exp(e*8i + f*x*8i)*cos(2*e + 
 2*f*x)*384i)/(55*b*f) - (exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*5248i)/(1155 
*b*f) + (exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*256i)/(165*b*f) - (exp(e*8i + 
 f*x*8i)*cos(8*e + 8*f*x)*256i)/(1155*b*f))*1i)/(128*sin(e + f*x)^(15/2))
 

Reduce [F]

\[ \int \frac {1}{\sqrt {b \sec (e+f x)} \sin ^{\frac {17}{2}}(e+f x)} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right ) \sin \left (f x +e \right )^{9}}d x \right )}{b} \] Input:

int(1/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(17/2),x)
 

Output:

(sqrt(b)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/(sec(e + f*x)*sin(e + 
 f*x)**9),x))/b