\(\int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx\) [255]

Optimal result

Integrand size = 25, antiderivative size = 132 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^3 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2} \] Output:

18/5*e^3*(e*cos(d*x+c))^(5/2)/a^3/d+6*e^6*cos(d*x+c)^(1/2)*InverseJacobiAM 
(1/2*d*x+1/2*c,2^(1/2))/a^3/d/(e*cos(d*x+c))^(1/2)+6*e^5*(e*cos(d*x+c))^(1 
/2)*sin(d*x+c)/a^3/d+4*e*(e*cos(d*x+c))^(9/2)/a/d/(a+a*sin(d*x+c))^2
 

Mathematica [C] (verified)

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

Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.50 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {17}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{13 a^3 d e (1+\sin (c+d x))^{13/4}} \] Input:

Integrate[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

(-2*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[3/4, 13/4, 17/4, (1 
- Sin[c + d*x])/2])/(13*a^3*d*e*(1 + Sin[c + d*x])^(13/4))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3159, 3042, 3161, 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.

\(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {9 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{\sin (c+d x) a+a}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{\sin (c+d x) a+a}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3161

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \int (e \cos (c+d x))^{3/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\)

Input:

Int[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^3,x]
 

Output:

(4*e*(e*Cos[c + d*x])^(9/2))/(a*d*(a + a*Sin[c + d*x])^2) + (9*e^2*((2*e*( 
e*Cos[c + d*x])^(5/2))/(5*a*d) + (e^2*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF 
[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + (2*e*Sqrt[e*Cos[c + d*x]]*S 
in[c + d*x])/(3*d)))/a))/a^2
 

Defintions of rubi rules used

rule 3042

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

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]
 

rule 3159

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

rule 3161

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 113.45 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37

method	result	size
default	\(\frac {2 e^{6} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-34 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+19 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\)	\(181\)

Input:

int((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

2/5/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(8*sin( 
1/2*d*x+1/2*c)^7+20*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x 
+1/2*c)^5-10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-15*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2 
^(1/2))-34*sin(1/2*d*x+1/2*c)^3+19*sin(1/2*d*x+1/2*c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {\frac {1}{2}} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {\frac {1}{2}} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (e^{5} \cos \left (d x + c\right )^{2} + 5 \, e^{5} \sin \left (d x + c\right ) - 20 \, e^{5}\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{5 \, a^{3} d} \] Input:

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-2/5*(15*I*sqrt(1/2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I* 
sin(d*x + c)) - 15*I*sqrt(1/2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x 
 + c) - I*sin(d*x + c)) + (e^5*cos(d*x + c)^2 + 5*e^5*sin(d*x + c) - 20*e^ 
5)*sqrt(e*cos(d*x + c)))/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((e*cos(d*x+c))**(11/2)/(a+a*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^(11/2)/(a*sin(d*x + c) + a)^3, x)
 

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(11/2)/(a*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((e*cos(c + d*x))^(11/2)/(a + a*sin(c + d*x))^3,x)
 

Output:

int((e*cos(c + d*x))^(11/2)/(a + a*sin(c + d*x))^3, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right ) e^{5}}{a^{3}} \] Input:

int((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x)
 

Output:

(sqrt(e)*int((sqrt(cos(c + d*x))*cos(c + d*x)**5)/(sin(c + d*x)**3 + 3*sin 
(c + d*x)**2 + 3*sin(c + d*x) + 1),x)*e**5)/a**3