\(\int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx\) [232]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=-\frac {d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac {15 d^6 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b \sqrt {d \cos (a+b x)}}-\frac {15 d^5 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b}-\frac {9 d^3 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b} \] Output:

-d*(d*cos(b*x+a))^(9/2)*csc(b*x+a)/b-15/7*d^6*cos(b*x+a)^(1/2)*InverseJaco 
biAM(1/2*a+1/2*b*x,2^(1/2))/b/(d*cos(b*x+a))^(1/2)-15/7*d^5*(d*cos(b*x+a)) 
^(1/2)*sin(b*x+a)/b-9/7*d^3*(d*cos(b*x+a))^(5/2)*sin(b*x+a)/b
 

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.72 \[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\frac {d^5 \sqrt {d \cos (a+b x)} \csc (a+b x) \left (\sqrt {\cos (a+b x)} (-45+16 \cos (2 (a+b x))+\cos (4 (a+b x)))-60 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right ) \sin (a+b x)\right )}{28 b \sqrt {\cos (a+b x)}} \] Input:

Integrate[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^2,x]
 

Output:

(d^5*Sqrt[d*Cos[a + b*x]]*Csc[a + b*x]*(Sqrt[Cos[a + b*x]]*(-45 + 16*Cos[2 
*(a + b*x)] + Cos[4*(a + b*x)]) - 60*EllipticF[(a + b*x)/2, 2]*Sin[a + b*x 
]))/(28*b*Sqrt[Cos[a + b*x]])
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3047, 3042, 3115, 3042, 3115, 3042, 3121, 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.

\(\displaystyle \int \csc ^2(a+b x) (d \cos (a+b x))^{11/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \cos (a+b x))^{11/2}}{\sin (a+b x)^2}dx\)

\(\Big \downarrow \) 3047

\(\displaystyle -\frac {9}{2} d^2 \int (d \cos (a+b x))^{7/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{2} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{7/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \int (d \cos (a+b x))^{3/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {9}{2} d^2 \left (\frac {5}{7} d^2 \left (\frac {2 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b}\)

Input:

Int[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^2,x]
 

Output:

-((d*(d*Cos[a + b*x])^(9/2)*Csc[a + b*x])/b) - (9*d^2*((2*d*(d*Cos[a + b*x 
])^(5/2)*Sin[a + b*x])/(7*b) + (5*d^2*((2*d^2*Sqrt[Cos[a + b*x]]*EllipticF 
[(a + b*x)/2, 2])/(3*b*Sqrt[d*Cos[a + b*x]]) + (2*d*Sqrt[d*Cos[a + b*x]]*S 
in[a + b*x])/(3*b)))/7))/2
 

Defintions of rubi rules used

rule 3042

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

rule 3047

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
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 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(241\) vs. \(2(109)=218\).

Time = 9.92 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.95

method	result	size
default	\(-\frac {\sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{7} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (-128 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{12}+384 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}-576 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}+30 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+512 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}-204 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+12 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+7\right )}{14 \left (-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} d +\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d \right )^{\frac {3}{2}} \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\)	\(242\)

Input:

int((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/14*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^7/(-2*si 
n(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(3/2)/cos(1/2*b*x+1/2*a)*sin( 
1/2*b*x+1/2*a)*(-128*sin(1/2*b*x+1/2*a)^12+384*sin(1/2*b*x+1/2*a)^10-576*s 
in(1/2*b*x+1/2*a)^8+30*cos(1/2*b*x+1/2*a)*(2*sin(1/2*b*x+1/2*a)^2-1)^(3/2) 
*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))+512*si 
n(1/2*b*x+1/2*a)^6-204*sin(1/2*b*x+1/2*a)^4+12*sin(1/2*b*x+1/2*a)^2+7)/(d* 
(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98 \[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\frac {15 i \, \sqrt {\frac {1}{2}} d^{\frac {11}{2}} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 15 i \, \sqrt {\frac {1}{2}} d^{\frac {11}{2}} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (2 \, d^{5} \cos \left (b x + a\right )^{4} + 6 \, d^{5} \cos \left (b x + a\right )^{2} - 15 \, d^{5}\right )} \sqrt {d \cos \left (b x + a\right )}}{7 \, b \sin \left (b x + a\right )} \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="fricas")
 

Output:

1/7*(15*I*sqrt(1/2)*d^(11/2)*sin(b*x + a)*weierstrassPInverse(-4, 0, cos(b 
*x + a) + I*sin(b*x + a)) - 15*I*sqrt(1/2)*d^(11/2)*sin(b*x + a)*weierstra 
ssPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + (2*d^5*cos(b*x + a)^4 + 
 6*d^5*cos(b*x + a)^2 - 15*d^5)*sqrt(d*cos(b*x + a)))/(b*sin(b*x + a))
 

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\text {Timed out} \] Input:

integrate((d*cos(b*x+a))**(11/2)*csc(b*x+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}} \csc \left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="maxima")
 

Output:

integrate((d*cos(b*x + a))^(11/2)*csc(b*x + a)^2, x)
 

Giac [F]

\[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}} \csc \left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((d*cos(b*x + a))^(11/2)*csc(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{11/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \] Input:

int((d*cos(a + b*x))^(11/2)/sin(a + b*x)^2,x)
 

Output:

int((d*cos(a + b*x))^(11/2)/sin(a + b*x)^2, x)
 

Reduce [F]

\[ \int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx=\sqrt {d}\, \left (\int \sqrt {\cos \left (b x +a \right )}\, \cos \left (b x +a \right )^{5} \csc \left (b x +a \right )^{2}d x \right ) d^{5} \] Input:

int((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x)
 

Output:

sqrt(d)*int(sqrt(cos(a + b*x))*cos(a + b*x)**5*csc(a + b*x)**2,x)*d**5