\(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx\) [281]

Optimal result

Integrand size = 25, antiderivative size = 168 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2 c (c \sin (a+b x))^{3/2}}{15 b d^3 (d \cos (a+b x))^{5/2}}-\frac {4 c (c \sin (a+b x))^{3/2}}{15 b d^5 \sqrt {d \cos (a+b x)}}+\frac {4 c^2 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{15 b d^6 \sqrt {\sin (2 a+2 b x)}} \] Output:

2/9*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(9/2)-2/15*c*(c*sin(b*x+a))^ 
(3/2)/b/d^3/(d*cos(b*x+a))^(5/2)-4/15*c*(c*sin(b*x+a))^(3/2)/b/d^5/(d*cos( 
b*x+a))^(1/2)-4/15*c^2*(d*cos(b*x+a))^(1/2)*EllipticE(cos(a+1/4*Pi+b*x),2^ 
(1/2))*(c*sin(b*x+a))^(1/2)/b/d^6/sin(2*b*x+2*a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.43 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {2 \cos ^5(a+b x) \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {13}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{7/2}}{7 b c (d \cos (a+b x))^{11/2}} \] Input:

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(11/2),x]
 

Output:

(2*Cos[a + b*x]^5*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[7/4, 13/4, 11/4 
, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(7/2))/(7*b*c*(d*Cos[a + b*x])^(11/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3046, 3042, 3051, 3042, 3051, 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.

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}}dx\)

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{9 b d (d \cos (a+b x))^{9/2}}-\frac {c^2 \left (\frac {2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}}\right )}{5 d^2}+\frac {2 (c \sin (a+b x))^{3/2}}{5 b c d (d \cos (a+b x))^{5/2}}\right )}{3 d^2}\)

Input:

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(11/2),x]
 

Output:

(2*c*(c*Sin[a + b*x])^(3/2))/(9*b*d*(d*Cos[a + b*x])^(9/2)) - (c^2*((2*(c* 
Sin[a + b*x])^(3/2))/(5*b*c*d*(d*Cos[a + b*x])^(5/2)) + (2*((2*(c*Sin[a + 
b*x])^(3/2))/(b*c*d*Sqrt[d*Cos[a + b*x]]) - (2*Sqrt[d*Cos[a + b*x]]*Ellipt 
icE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(b*d^2*Sqrt[Sin[2*a + 2*b*x]] 
)))/(5*d^2)))/(3*d^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3046

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 
1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1)))   Int[(a*Sin[e + f 
*x])^(m - 2)*(b*Cos[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 3051

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

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 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 [A] (verified)

Time = 4.90 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.64

method	result	size
default	\(\frac {2 \sqrt {c \sin \left (b x +a \right )}\, c^{2} \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (3 \cot \left (b x +a \right )+3 \csc \left (b x +a \right )\right )+\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-6 \cot \left (b x +a \right )-6 \csc \left (b x +a \right )\right )+6 \cot \left (b x +a \right )-3 \csc \left (b x +a \right )-8 \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2}+5 \csc \left (b x +a \right ) \sec \left (b x +a \right )^{4}\right )}{45 b \sqrt {d \cos \left (b x +a \right )}\, d^{5}}\)	\(275\)

Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(11/2),x,method=_RETURNVERBOSE)
 

Output:

2/45/b*(c*sin(b*x+a))^(1/2)*c^2/(d*cos(b*x+a))^(1/2)/d^5*((-cot(b*x+a)+csc 
(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a 
))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(3*cot(b* 
x+a)+3*csc(b*x+a))+(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b* 
x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((-cot(b*x+a)+csc(b*x 
+a)+1)^(1/2),1/2*2^(1/2))*(-6*cot(b*x+a)-6*csc(b*x+a))+6*cot(b*x+a)-3*csc( 
b*x+a)-8*csc(b*x+a)*sec(b*x+a)^2+5*csc(b*x+a)*sec(b*x+a)^4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=-\frac {2 \, {\left (-3 i \, \sqrt {i \, c d} c^{2} \cos \left (b x + a\right )^{5} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 3 i \, \sqrt {-i \, c d} c^{2} \cos \left (b x + a\right )^{5} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 3 i \, \sqrt {i \, c d} c^{2} \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 i \, \sqrt {-i \, c d} c^{2} \cos \left (b x + a\right )^{5} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + {\left (6 \, c^{2} \cos \left (b x + a\right )^{4} + 3 \, c^{2} \cos \left (b x + a\right )^{2} - 5 \, c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{45 \, b d^{6} \cos \left (b x + a\right )^{5}} \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(11/2),x, algorithm="fricas" 
)
 

Output:

-2/45*(-3*I*sqrt(I*c*d)*c^2*cos(b*x + a)^5*elliptic_e(arcsin(cos(b*x + a) 
+ I*sin(b*x + a)), -1) + 3*I*sqrt(-I*c*d)*c^2*cos(b*x + a)^5*elliptic_e(ar 
csin(cos(b*x + a) - I*sin(b*x + a)), -1) + 3*I*sqrt(I*c*d)*c^2*cos(b*x + a 
)^5*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) - 3*I*sqrt(-I*c* 
d)*c^2*cos(b*x + a)^5*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1 
) + (6*c^2*cos(b*x + a)^4 + 3*c^2*cos(b*x + a)^2 - 5*c^2)*sqrt(d*cos(b*x + 
 a))*sqrt(c*sin(b*x + a))*sin(b*x + a))/(b*d^6*cos(b*x + a)^5)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(11/2),x, algorithm="maxima" 
)
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(11/2), x)
 

Giac [F]

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

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(11/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(11/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{11/2}} \,d x \] Input:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(11/2),x)
 

Output:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(11/2), x)
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{11/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{6}}d x \right ) c^{2}}{d^{6}} \] Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(11/2),x)
 

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x))*sin(a + b*x)** 
2)/cos(a + b*x)**6,x)*c**2)/d**6