\(\int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [461]

Optimal result

Integrand size = 23, antiderivative size = 115 \[ \int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {7 b \sin ^{\frac {3}{2}}(e+f x)}{30 f (b \sec (e+f x))^{3/2}}-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {7 E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (e+f x)}}{20 f \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}} \] Output:

-7/30*b*sin(f*x+e)^(3/2)/f/(b*sec(f*x+e))^(3/2)-1/5*b*sin(f*x+e)^(7/2)/f/( 
b*sec(f*x+e))^(3/2)-7/20*EllipticE(cos(e+1/4*Pi+f*x),2^(1/2))*sin(f*x+e)^( 
1/2)/f/(b*sec(f*x+e))^(1/2)/sin(2*f*x+2*e)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 11.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {b \left (23-26 \cos (2 (e+f x))+3 \cos (4 (e+f x))+42 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{120 f (b \sec (e+f x))^{3/2} \sqrt {\sin (e+f x)}} \] Input:

Integrate[Sin[e + f*x]^(9/2)/Sqrt[b*Sec[e + f*x]],x]
 

Output:

-1/120*(b*(23 - 26*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)] + 42*Hypergeometr 
ic2F1[-1/2, 1/4, 1/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(1/4)))/(f*(b*Sec[ 
e + f*x])^(3/2)*Sqrt[Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3063, 3042, 3063, 3042, 3065, 3042, 3052, 3042, 3119}

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.

\(\displaystyle \int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^{9/2}}{\sqrt {b \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {7}{10} \int \frac {\sin ^{\frac {5}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \int \frac {\sin (e+f x)^{5/2}}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {7}{10} \left (\frac {1}{2} \int \frac {\sqrt {\sin (e+f x)}}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {1}{2} \int \frac {\sqrt {\sin (e+f x)}}{\sqrt {b \sec (e+f x)}}dx-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3065

\(\displaystyle \frac {7}{10} \left (\frac {\int \sqrt {b \cos (e+f x)} \sqrt {\sin (e+f x)}dx}{2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {\int \sqrt {b \cos (e+f x)} \sqrt {\sin (e+f x)}dx}{2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {7}{10} \left (\frac {\sqrt {\sin (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} \left (\frac {\sqrt {\sin (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {7}{10} \left (\frac {\sqrt {\sin (e+f x)} E\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{2 f \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)}}-\frac {b \sin ^{\frac {3}{2}}(e+f x)}{3 f (b \sec (e+f x))^{3/2}}\right )-\frac {b \sin ^{\frac {7}{2}}(e+f x)}{5 f (b \sec (e+f x))^{3/2}}\)

Input:

Int[Sin[e + f*x]^(9/2)/Sqrt[b*Sec[e + f*x]],x]
 

Output:

-1/5*(b*Sin[e + f*x]^(7/2))/(f*(b*Sec[e + f*x])^(3/2)) + (7*(-1/3*(b*Sin[e 
 + f*x]^(3/2))/(f*(b*Sec[e + f*x])^(3/2)) + (EllipticE[e - Pi/4 + f*x, 2]* 
Sqrt[Sin[e + f*x]])/(2*f*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])))/10
 

Defintions of rubi rules used

rule 3042

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

rule 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
 

rule 3063

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

rule 3065

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(96)=192\).

Time = 0.56 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.03

method	result	size
default	\(-\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-21-21 \sec \left (f x +e \right )\right )+\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (42+42 \sec \left (f x +e \right )\right )+24 \cos \left (f x +e \right )^{5}-76 \cos \left (f x +e \right )^{3}+94 \cos \left (f x +e \right )-42}{120 f \sqrt {b \sec \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right )}}\)	\(234\)

Input:

int(sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/120/f/(b*sec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)*((csc(f*x+e)-cot(f*x+e)+1)^ 
(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)* 
EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*(-21-21*sec(f*x+e)) 
+(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-cs 
c(f*x+e)+cot(f*x+e))^(1/2)*EllipticE((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2 
^(1/2))*(42+42*sec(f*x+e))+24*cos(f*x+e)^5-76*cos(f*x+e)^3+94*cos(f*x+e)-4 
2)
 

Fricas [F]

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

integrate(sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b*sec(f*x + e))*sqrt 
(sin(f*x + e))/(b*sec(f*x + e)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**(9/2)/(b*sec(f*x+e))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sin(f*x + e)^(9/2)/sqrt(b*sec(f*x + e)), x)
 

Giac [F]

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

integrate(sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

integrate(sin(f*x + e)^(9/2)/sqrt(b*sec(f*x + e)), x)
 

Mupad [F(-1)]

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

int(sin(e + f*x)^(9/2)/(b/cos(e + f*x))^(1/2),x)
 

Output:

int(sin(e + f*x)^(9/2)/(b/cos(e + f*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sin ^{\frac {9}{2}}(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )^{4}}{\sec \left (f x +e \right )}d x \right )}{b} \] Input:

int(sin(f*x+e)^(9/2)/(b*sec(f*x+e))^(1/2),x)
 

Output:

(sqrt(b)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x))*sin(e + f*x)**4)/sec(e 
 + f*x),x))/b