\(\int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx\) [268]

Optimal result

Integrand size = 19, antiderivative size = 92 \[ \int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {4 \cos (a+b x)}{7 b \sqrt {\csc (a+b x)}}+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}+\frac {8 \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{7 b} \] Output:

4/7*cos(b*x+a)/b/csc(b*x+a)^(1/2)+2/7*cos(b*x+a)^3/b/csc(b*x+a)^(1/2)+8/7* 
csc(b*x+a)^(1/2)*InverseJacobiAM(1/2*a-1/4*Pi+1/2*b*x,2^(1/2))*sin(b*x+a)^ 
(1/2)/b
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {\sqrt {\csc (a+b x)} \left (-32 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sqrt {\sin (a+b x)}+10 \sin (2 (a+b x))+\sin (4 (a+b x))\right )}{28 b} \] Input:

Integrate[Cos[a + b*x]^4*Sqrt[Csc[a + b*x]],x]
 

Output:

(Sqrt[Csc[a + b*x]]*(-32*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + 
b*x]] + 10*Sin[2*(a + b*x)] + Sin[4*(a + b*x)]))/(28*b)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3108, 3042, 3108, 3042, 4258, 3042, 3120}

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.

\(\displaystyle \int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\csc (a+b x)}}{\sec (a+b x)^4}dx\)

\(\Big \downarrow \) 3108

\(\displaystyle \frac {6}{7} \int \cos ^2(a+b x) \sqrt {\csc (a+b x)}dx+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \int \frac {\sqrt {\csc (a+b x)}}{\sec (a+b x)^2}dx+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 3108

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \sqrt {\csc (a+b x)}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\right )+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \sqrt {\csc (a+b x)}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\right )+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\right )+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}}dx+\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}\right )+\frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \cos ^3(a+b x)}{7 b \sqrt {\csc (a+b x)}}+\frac {6}{7} \left (\frac {2 \cos (a+b x)}{3 b \sqrt {\csc (a+b x)}}+\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b}\right )\)

Input:

Int[Cos[a + b*x]^4*Sqrt[Csc[a + b*x]],x]
 

Output:

(2*Cos[a + b*x]^3)/(7*b*Sqrt[Csc[a + b*x]]) + (6*((2*Cos[a + b*x])/(3*b*Sq 
rt[Csc[a + b*x]]) + (4*Sqrt[Csc[a + b*x]]*EllipticF[(a - Pi/2 + b*x)/2, 2] 
*Sqrt[Sin[a + b*x]])/(3*b)))/7
 

Defintions of rubi rules used

rule 3042

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

rule 3108

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

rule 3120

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

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\frac {4 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{7}+\frac {2 \sin \left (b x +a \right )^{5}}{7}-\frac {8 \sin \left (b x +a \right )^{3}}{7}+\frac {6 \sin \left (b x +a \right )}{7}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\)	\(100\)

Input:

int(cos(b*x+a)^4*csc(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

(4/7*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Elli 
pticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+2/7*sin(b*x+a)^5-8/7*sin(b*x+a)^3+ 
6/7*sin(b*x+a))/cos(b*x+a)/sin(b*x+a)^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx=\frac {2 \, {\left ({\left (\cos \left (b x + a\right )^{3} + 2 \, \cos \left (b x + a\right )\right )} \sqrt {\sin \left (b x + a\right )} - 2 i \, \sqrt {2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, \sqrt {-2 i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}}{7 \, b} \] Input:

integrate(cos(b*x+a)^4*csc(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

2/7*((cos(b*x + a)^3 + 2*cos(b*x + a))*sqrt(sin(b*x + a)) - 2*I*sqrt(2*I)* 
weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) + 2*I*sqrt(-2*I)* 
weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)))/b
 

Sympy [F]

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

integrate(cos(b*x+a)**4*csc(b*x+a)**(1/2),x)
 

Output:

Integral(cos(a + b*x)**4*sqrt(csc(a + b*x)), x)
 

Maxima [F]

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

integrate(cos(b*x+a)^4*csc(b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(cos(b*x + a)^4*sqrt(csc(b*x + a)), x)
 

Giac [F]

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

integrate(cos(b*x+a)^4*csc(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(cos(b*x + a)^4*sqrt(csc(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(a+b x) \sqrt {\csc (a+b x)} \, dx=\int {\cos \left (a+b\,x\right )}^4\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}} \,d x \] Input:

int(cos(a + b*x)^4*(1/sin(a + b*x))^(1/2),x)
 

Output:

int(cos(a + b*x)^4*(1/sin(a + b*x))^(1/2), x)
 

Reduce [F]

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

int(cos(b*x+a)^4*csc(b*x+a)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

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